BIOCHEMISTRY
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A
A 55-year -oia farmer is brought to the emergency department after his daughter found
him confused and disoriented in the tool shed at home He has been otherwise healthy
and does not take any medications . On physical examination , biood pressure is 110/70
mm Hg and pulse is 50, min The patient's pupils are symmetric , 2 mm and reactive to
light His eyes are tearing considerably There are scattered wheezes bilaterally on lung
auscultation The patient' s skin is clammy and he is sweating profusely , impairment of
which of the following steps at the neuromuscular junction is most likely responsible for
his presentation?
Choline
A
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* Choline
B
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Depolarization
c
Acetyl
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Acetylcholine
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Acetate
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Inhibitors of acetylcholine
neurotransmission
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Notes
(
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Choline
n
Hemkholimum
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^ Choline
! Acetyl
gh fl ' CoA
* '
Vesamitol
Acetylcholine
Brom oacet y Itholme
H
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SNARES
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toxin
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Choline
Acetylcholine
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receptor
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ti
antagonists
Acetate
Acetylcholinesterase
I
inhibitor (orqanophosphates)
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This patient' s confusion miosis and lacrimation bradycardia diaphoresis , and
bronchospasm ( bilateral wheezes ) indicate a state of global muscarinic cholinergic
overstimulation This can be the result of a pharmacologic cholinomimetic ( eg.
carbachol methacholine ) or cholinesterase inhibitor ( blocks breakdown of
endogenous acetylcholine ).
,
,
Organophosphates are cholinesterase inhibitors that are widely used as pesticides in
agriculture They are lipid soluble ; rapidly absorbed via oral , cutaneous and inhalationai
routes of exposure ; and readily penetrate the blood brain barrier Organophosphates are
irreversible cholinesterase inhibitors they elicit cholinergic stimulation that lasts until
new cholinesterase enzymes are synthesized Organophosphate poisoning is treated
with muscarinic antagonists ( eg . atropine ) and pralidoxime (PAM), a drug that
reactivates acetylcholinesterase by binding organophosphates and decoupling them from
the enzyme .
.
.
{ Choices A B . and C ) Blockade of neuronal choline uptake and inhibition of choline
acetyltransferase would dimmish acetylcholine synthesis Blockade of acetylcholine
uptake into axoplasmic vesicles would reduce acetyicholine release from presynaptic
neurons . A decrease In acetylcholine synthesis or release would have effects opposite
those observed in this patient .
{ Choice D ) Botulinum toxin inhibits acetylcholine release from presynaptic neurons .
Botulinum toxicity { botulism ) causes descending skeletal muscle paralysis typically
beginning with the cranial nerves.
{ Choice F) Atropine is an anticholinergic agent used pharmacologically to block
post- junctional acetylcholine receptors and is the antidote for organophosphate
poisoning ,
i
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A 29-year -oia man comes to the physician after discovering a painless scrotal mass on
self-examination He also has increased sweating and heat intolerance . He has no
significant past medical history . Physical examination shows an enlarged nontender right
testicle Laboratory evaluation shows increased serum T4 and T 3 concentrations
Scrotal ultrasound shows a hypoechoic mass within the right testicle The constellation
of findings seen in this patient most likely suggests an elevation of which of the following
serum markers?
13
O A . Alpha-fetoprotein
O B. Follicle - stimulating hormone
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C
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C. Human chorionic gonadotropin
D. Lactate dehydrogenase
O E. Placenta -tike alkaline phosphatase
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A 29- year-oia man comes to the physician after discovering a painless scrotal mass on
self-examination He also has increased sweating and heat intolerance . He has no
significant past medical history . Physical examination shows an enlarged nontender right
testicle Laboratory evaluation shows increased serum T4 and T 3 concentrations
Scrotal ultrasound shows a hypoechoic mass within the right testicle The constellation
of findings seen in this patient most likely suggests an elevation of which of the following
serum markers?
O A . Alpha-fetoprotein [22%)
O B . Follicle-stimulating hormone [16% ]
* <# C . Human chorionic gonadotropin [51%]
O D . Lactate dehydrogenase [5%]
i
-
E . Placenta - tike alkaline phosphatase [5 %J
Explanation:
This patient has a testicular malignancy that Is most likely secreting human chorionic
gonadotropin ( hCG ) . This peptide hormone is normally produced by the placenta
However , it is also produced by malignant testicular tumors , particularly
nonseminomatous germ cell tumors , which can secrete very high levels of hCG . The
alpha subunits of hCG . TSH, LH , and FSH are identical, and the beta subunits of hCG
and TSH share significant sequence homology . Because of this structural similarity .
hCG can bind to the TSH receptor ( although with much lower affinity than TSH ) . As a
resuit , very high circulating levels of hCG can over- stimulatethe thyroid gland, causing
paraneoplastic hyperthyroidism
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{ Choice A ) Alpha- fetoprotein ( AFP ) can be used as a tumor marker for a number of
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I iilculdlor
O E . Placenta -like alkaline phosphatase [ 5 % ]
/s
Explanation:
This patient has a testicular malignancy that is most likely secreting human chorionic
gonadotropin ( hCG ). This peptide hormone is normally produced by the placenta
However It is also produced by malignant testicular tumors particularly
nonseminomatous germ cell tumors , which can secrete very high levels of hCG , The
alpha subunits of hCG . TSH , LH. and FSH are identical, and the beta subunits of hCG
and TSH share significant sequence homology . Because of this structural similarity .
hCG can bind to the TSH receptor ( although with much lower affinity than TSH ) . As a
result , very high circulating levels of hCG can over - stimuiate the thyroid gland , causing
paraneoplastic hyperthyroidism .
b
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(Choice A ) Alpha- fetoprotein ( AFP ) can be used as a tumor marker for a number of
malignancies including hepatocellular carcinoma and nonseminomatous germ cell
tumors of the testes . However elevated AFP levels would not result in hyperthyroid
symptoms .
,
,
( Choice B ) FSH is produced by gonadotrophs in the anterior pituitary cells FSH is not
a marker for testicular tumors and has no affinity for thyroid receptors .
(Choice D ) Lactate dehydrogenase (LDH ) Is an enzyme involved in anaerobic
glycolysis Although increased LDH levels can occur with both seminomatous and
nonseminomatous tumors of the testes . LOH does not interact with TSH receptors
(Choice E ) The majority of circulating alkaline phosphatase comes from the bone , liver ,
gastrointestinal tract , and placenta Placenta-like alkaline phosphatase (PLAP ) is a
nonspecific tumor marker that can be increased in testicular seminoma and other
malignancies .
_ __
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Notes
t alcufdtor
can secrete very mgn ieveis <
. ini
alpha subunits of hCG. TSH , LH, and FSH are identical, and the beta subunits of hCG
and TSH share significant sequence homology . Because of this structural similarity
hCG can bind to the TSH receptor ( although with much lower affinity than TSH). As a
result , very high circulating levels of hCG can over - stimulate the thyroid gland , causing
paraneoplastic hyperthyroidism.
nonseminomatous >T-
i ci
tumors , wnicn
nuij
A
.
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IS
IS
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(Choice A ) Alpha -fetoprotein ( AFP ) can be used as a tumor marker for a number of
malignancies including hepatocellular carcinoma and nonseminomatous germ cell
tumors of the testes However , elevated AFP levels would not result in hyperthyroid
symptoms
(Choice B ) FSH is produced by gonadotrophs in the anterior pituitary cells FSH is not
a marker for testicular tumors and has no affinity for thyroid receptors .
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(Choice D ) Lactate dehydrogenase (LDH) is an enzyme involved in anaerobic
glycolysis Although increased LDH levels can occur with both seminomatous and
nonseminomatous tumors of the testes . LDH does not interact with TSH receptors
(Choice E ) The majority of circulating alkaline phosphatase comes from the bone , liver,
gastrointestinal tract , and placenta Placenta -like alkaline phosphatase (PLAP ) is a
nonspecific tumor marker that can be increased in testicular seminoma and other
malignancies.
Educational objective:
Human chorionic gonadotropin (hCG ) has a structure similar to TSH. Patients with
testicular germ cell tumors or gestational trophoblastic disease may develop very high
serum hCG concentrations , which can stimulate TSH receptors and cause
hyperthyroidism.
References:
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A diagram illustrating the DMA replication process is shown below. Which of the points
marked by letters represents the site of action of helicase in the following diagram'?
y
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U
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c
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t 4ilc uldlor
A diagram illustrating the DMA replication process is shown below. Which of the points
marked by letters represents the site of action of helicase in the following diagram'?
i
r
n
11
12
13
U
IB
1S
17
5
C
U
3
B
H
D
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O A . A [2%)
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O B . S [2%J
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O G. C [4%]
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*
O . D [2%1
«: E.i[90%]
3S
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37
Explanation:
38
DMA polymerase 1 ~ 5' cxcmuclease activity
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Explanation:
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DNA pclymcfase I S' ex <yiunease aclivily
n
Ppfruisc
= forms RNA primer
( DNA dep RNA polymerase )
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3‘
It
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Hoi ' Crise
Okazaki fragment
RNA pnmer
DNA replication occurs during the S-phase (synthesis phase ) of the cell cycle
Replication of the genome is performed by the coordinated effects of more than thirty
proteins to ensure that the daughter strands are the exact replica of the parent strands
DNA polymerases are the main enzymes responsible for the synthesis of new strands of
0NA In E. coti , there are three major types of DNA polymerases : I II and III In
prokaryotes DNA polymerase III performs the bulk of the DNA replication ( Choice C)
.
DNA polymerases form new daughter strands in the 5' to 3 ' direction using the parent
3 sinale stranded DNA : therefore
t
strand as aWUM
•il occ
*JR«te5 . DNAn=WLL« mml can onlv
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I altuldior
A
DNA replication occurs during the S- phase { synthesis phase ) of the cell cycle.
Replication of the genome is performed by the coordinated effects of more than thirty
proteins to ensure that the daughter strands are the exact replica of the parent strands.
DNA polymerases are the main enzymes responsible for the synthesis of new strands of
ONA . In E. coli . there are three major types of DNA polymerases : I. II and III In
prokaryotes DNA polymerase III performs the bulk of the DNA replication ( Choice C)
DNA polymerases form new daughter strands in the 5 ' to 3’ direction using the parent
strand as a template DNA replication can only occur on single stranded DNA ; therefore
unwinding and dissociation of the parent DNA strands is necessary before replication can
proceed . First the origin of replication is identified and bound by several monomers of
the DnaA protein which serves to locally dissociate double stranded DNA ( dsDNA ) into
single stranded DNA ( ssONA ) at the origin of replication . Single strand binding proteins
( SSBs , Choice D ) then bind to the ssDNA and stabilize it preventing premature
reannealing of the ssDNA to dsDNA . Helicase (Choice E ) then binds the ssDNA at the
origin of replication , moves into the replication fork , and then proceeds to separate and
unwind the dsDNA
,
The topoisomerase enzymes ( not pictured ) include topoisomerase I and II , which relieve
supercoiling tension of the dsDNA strand caused by the unwinding action of helicase
Topoisomerase IE is also known as gyrase in prokaryotes . If pictured on the above
diagram these proteins would be ahead of helicase on the dsDNA segment
DNA polymerases synthesize new DNA strands in the 5 ' to 3 ' direction and require a free
3 -hydroxyl group upon which to begin polymenzation This 3‘- hydroxyl group is provided
by the enzyme primase . a DNA-dependent RNA polymerase which forms RNA primers
(Choice A ) . Both daughter strands are not produced continuously; the strand that forms
continuously from the 5 r to the 3' direction is called the leading strand (upper strand in the
above diagram ) and the second strand , the lagging strand , is generated by the formation
of short fragments called Okazaki fragments (lower strand on the above diagram ) In the 5 '
f
,
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1 4ilc uldlor
canon Cc oniy occu
as a
si
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sirai
*
unwinding and dissociation of the parent DNA strands is necessary before replication can
proceed . First , the origin of replication is Identified and bound by several monomers of
the DnaA protein which serves to locally dissociate double stranded DNA ( dsDNA ) into
single stranded DNA ( ssDNA ) at the origin of replication Single strand binding proteins
( SSBs . Choice D ) then bind to the ssDNA and stabilize it . preventing premature
reannealing of the ssDNA to dsDNA . Helicase (Choice E ) then binds the ssDNA at the
origin of replication , moves into the replication fork and then proceeds to separate and
lempiaie . UJVM H
'
le
l
ft
,
unwind the dsDNA .
The topoisomerase enzymes ( not pictured ) include topoisomerase I and II , which relieve
supercoiling tension of the dsDNA strand caused by the unwinding action of helicase
Topoisomerase (i is aiso known as gyrase in prokaryotes . If pictured on the above
diagram these proteins would be ahead of helicase on the dsDNA segment
,
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DNA polymerases synthesize new DNA strands in the 5 ' to 3' direction and require a free
3 ' -hydroxyl group upon which to begin polymerization This 3'-hydroxyl group is provided
by the enzyme primase. a DNA-dependent RNA polymerase which forms RNA primers
(Choice A ) . Both daughter strands are not produced continuously ; the strand that forms
continuously from the 5‘to the 3 ' direction is called the leading strand (upper strand in the
above diagram ), and the second strand , the lagging strand , is generated by the formation
of short fragments called Okazaki fragments (lower strand on the above diagram ) in the 5 '
to 3 direction . The fragments of the lagging strand are bound together by the enzyme
ligase ( Choice B )
Educational Objective:
Helicase unwinds DNA at the replication fork and separates dsDNA into ssDNA dunng
the replication process . Initial separation of dsDNA at the origin of replication is
facilitated by DnaA protein and strand binding proteins ( SS 3 ) proteins .
Time Spent 15 seconds
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Notes
I akulator
Elastin fibers in the alveolar walls of the lungs can be stretched easily during inspiration
and recoil to their onginal shape once the force is released This process facilitates
expiration . The property described can be best explained by:
10
ii
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A . Heavy posttranslationai hydroxylation
B . High content of polar
amino
acids
C . Chain assembly to form a triple helix
D . Interchain crosslinks involving lysine
E . Abundant interchain disulfide bridges
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t alculator
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Elastin fibers in the alveolar walls of the lungs can be stretched easily during inspiration
and recoil to their onginal shape once the force is released This process facilitates
expiration The property described can be best explained by :
n
A . Heavy posttranslational hydroxylatlon [9%]
13
i
is
15
B . High content of polar amino acids [ 11%]
*
17
ia
O C. Chain assembly to form a triple helix [ 18%)
*
[39%)
G E . Abundant interchain disulfide bridges [ 23%]
D . ntercham crosslinks involving
.
H
20
21
Explanation:
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24
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Elastin is a fibrous protein in the connective tissue that gets its name because of its
elastic properties Elastin fibers can be stretched to several times their length but recoil
back when stretching forces are withdrawn Elastin assembly appears to be closely
n
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related to that of collagen Similar to collagen , elastin is synthesized as a large
polypeptide precursor composed of about 700 amino acids called tropoelastin Elastin is
primarily composed of the non-polar amino acids glycine , alanine and valine , The elastin
molecule also contains proline and lysine ; however , in contrast to collagen , few of these
amino acids are hydroxylated . Tropoelastin is secreted into the extracellular space
where it interacts with microfibrils called fibrillin Once in the extracellular space , side
chains of some of the lysine residues are covalently bound to form a desmosine
crosslink , Extensive desmosine crosslinking accounts for eiastin' s resilient properties ,
allowing it to stretch and bend in any direction on applying force only to recoil to its
original size when the stretching force Is withdrawn
,
,
Elastin gives elastic properties to the skin , blood vessels and lung alveolae . A number of
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I dlcufdtor
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7
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1?
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15
*
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13
H
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Explanation:
Elastin is a fibrous protein in the connective tissue that gets its name because of its
elastic properties Elastin fibers can be stretched to several times their length but recoil
back when stretching forces are withdrawn Elastin assembly appears to be closely
related to that of collagen Similar to collagen elastin is synthesized as a large
polypeptide precursor composed of about 700 amino acids called tropoelastin Elastin is
,
primarily composed of the non-poiar amino acids glycine , alanine and valine . The elastin
molecule also contains proline and lysine ; however , in contrast to collagen few of these
amino acids are hydroxylated . Tropoelastin is secreted into the extracellular space
where It interacts with microfibrils called fibrillin . Once in the extracellular space , side
chains of some of the lysine residues are covalently bound to form a desmosine
crosslink . Extensive desmosine crosslinking accounts for elastin s resilient properties
allowing it to stretch and bend in any direction on applying force only to recoil to its
original size when the stretching force is withdrawn
,
,
Elastin gives elastic properties to the skin , blood vessels and lung alveolae . A number of
endogenous enzymes called proteinases hydrolyze and destroy such proteins . For
elastin. the most important proteinase is neutrophil- secreted etastase . a 1 - antitrypsin
inhibits the action ofthese endogenous proteolytic enzymes , thereby preventing damage
to essential structures within organs, A congenital deficiency of a 1- antitrypsin results in
excessive degradation of elastin in the lungs and liver , causing panacinar emphysema
and cirrhosis, respectively .
( Choices A , C and E ) Elastin differs from collagen in a number of ways :
1. Very few proiine and lysine residues are hydroxylated In elastin .
2 . Whereas triple helix formation is the basis of the collagen molecule elastin does
not form triple helices.
3. Triple helix formation in collagen is initiated by hydroxylation . glycosylation and
J1JL
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crosslink . Extensive aesmosine crosslinking accounts for eiastin ' s resilient properties ,
allowing It to stretch and bend in any direction on applying force only to recoil to its
original size when the stretching force is withdrawn
A
Elastin gives elastic properties to the skin blood vessels and lung alveolae A number of
endogenous enzymes called proteinases hydrolyze and destroy such proteins . For
elastin . the most important proteinase is neutrophil- secreted elastase a 1-antitrypsin
inhibits the action of these endogenous proteolytic enzymes thereby preventing damage
to essential structures within organs, A congenital deficiency of a 1 - antitrypsin results in
excessive degradation of elastin in the lungs and liver , causing panacinar emphysema
and cirrhosis, respectively .
,
17
ia
(Choices A , C and E) Elastin differs from collagen in a number of ways :
H
20
21
n
23
24
25
25
1 Very few prolme and lysine residues are hydroxylated In elastin.
2 . Whereas tnpie helix formation is the basis of the collagen molecule , elastin does
not form triple helices .
3. Triple helix formation in collagen is initiated by hydroxylation , glycosylation and
interchain disulfide bridges at the C- terminus of procollagen molecule These
modifications do not occur in the formation of elastin molecules
T
27
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40
u
( Choice B ) Collagen and elastin are both composed of a large number of nonpolar
amino acids .
Educational Objective:
Eiastin' s plasticity and ability to recoil upon release of tension is attributable to a unique
form of desmosme crosslinking between four different lysine residues on four different
elastin chains . This crosslinking is accomplished by the action of extracellular lysyl
hydroxylase.
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a
A 2- year -old Caucasian rrale is being evaluated for progressive neurological
deterioration Laboratory evaluation including leukocyte enzyme activity analysis ,
suggests Niemann-Pick disease. This patient most likely has a deficiency of which of the
following enzymes?
,
9
to
it
12
13
A . Arylsuifatase A
U
IS
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17
ia
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20
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p-Hexosaminidase A
C. a-Galaclosidase
B.
;
O 0 . p-Glucosidase
O E . Neuraminidase
F Sphingomyelinase
C G. Ceramioase
zs
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A
A 2-year -old Caucasian male is being evaluated for progressive neurological
deterioration Laboratory evaluation, including leukocyte enzyme activity analysis ,
suggests Niemann-Pick disease This patient most likely has a deficiency of which of the
following enzymes?
O A. Aryisulfatase A [2%]
J B. p-Hexosaminidase A [5%]
O C a-Galactosidase [3%]
0. p -Giucosidase [3%]
O E. Neuraminidase [3%J
v ® F, Sphingomyelinase [83%J
O G. Ceramidase [1% J
Explanation:
Niemann-Pick disease Type A is an autosomal recessive disorder most common in
individuals of Ashkenazi Jewish descent Affected infants present in the first year of life
with hepatospienomegaly and progressive hypotonia and mental retardation following a
period of normal early development . The cause is deficiency of the sphingomyelinase
enzyme, which in normal Individuals is responsible for cleaving sphingomyelin into
phosphorylcholine and ceramide In infants with Niemann -Pick disease , sphingomyelin
accumulates within phagocytes, producing characteristic "foamy histiocytes " These
foamy- appearing sphingomyelln- laden histiocytes accumulate in the liver and spleen
causing massive hepatospienomegaly. Progressive sphingomyelin accumulation in the
central nervous system is responsible for the neurologic degeneration that
occurs. Sphingomyelin deposition in the retina causes blindness as well. A cherry -red
V
it
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Explanation:
6
7
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1?
13
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35
Niemann-PicK disease Type A is an autosomal recessive disorder most common in
individuals of Ashkenazi Jewish descent Affected infants present in the first year of life
with hepatosplenomegaly and progressive hypotonia and mental retardation following a
period of normal early development The cause is deficiency of the sphingomyelinase
enzyme , which in normal individuals is responsible for cleaving sphingomyelin into
phosphorylcholine and ceramide In infants with Niemann -Pick disease , sphingomyelin
accumulates within phagocytes , producing characteristic "foamy histiocytes , " These
foamy - appearing. sphingomyelin- laden histiocytes accumulate in the liver and spleen
causing massive hepatosplenomegaly Progressive sphingomyelin accumulation In the
centra! nervous system is responsible for the neurologic degeneration that
occurs. Sphingomyelin deposition in the retina causes blindness as well. A cherry -red
macular spot similar to that seen in Tay-Sachs disease is also often found . Death
usually occurs before age three .
b
(Choice A ) In patients with the autosomal recessive condition metachromatic
leukodystrophy , a deficiency of the enzyme arylsulfatase A causes sulfatides to
accumulate within tissues
(Choice B ) In the autosomal recessive condition Tay -Sachs disease , a deficiency of the
enzyme p -hexosaminidase A causes G , . ganglioside to accumulate within neurons.
(Choice C ) In Fabry disease, an X -!inked recessive condition, deficiency of the enzyme o galactosidase A causes ceramide trihexoside to accumulate in tissues
(Choice D )
p -Glucosidase is a plant enzyme used for the breakdown of starch.
3b
37
33
(Choice E ) Neuraminidase ( sialidase ) deficiency is the cause of human
sialidosis Neuraminidase is also a surface enzyme found on the influenza virus
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(Choice B ) In the autosomal recessive condition Tay - Sachs disease , a deficiency of the
enzyme (3-hexosaminidase A causes G,, ganglioside to accumulate within neurons .
CJ In Fabry disease an X -linked recessive condition , deficiency of the enzyme a-
21
galactosidase A causes ceramide trihexoside to accumulate in tissues .
23
24
(Choice D ) p-Glucosidase is a plant enzyme used for the breakdown of starch.
26
(Choice E ) Neuraminidase ( sialidase ) deficiency is the cause of human
sialidosls Neuraminidase is also a surface enzyme found on the influenza virus.
37
I aiculalor
(Choice A ) In patients with the autosomal recessive condition metachromatic
leukodystrophy , a deficiency of the enzyme arylsulfatase A causes sulfatides to
accumulate within tissues
(Choice
26
21
23
29
30
31
32
33
34
3$
36
iNotrs
foamy - appearing sphingomyelin-laden histiocytes accumulate in the liver and spleen
causing massive hepatosplenomegaiy . Progressive sphingomyelin accumulation in the
central nervous system is responsible for the neurologic degeneration that
occurs. Sphingomyelin deposition in the retina causes blindness as well A cherry-red
macular spot similar to that seen in Tay-Sachs disease, is also often found . Death
usually occurs before age three .
20
n
Lab Valuer
Next
,
(Choice G ) Ceramldase deficiency causes Farber disease , an autosomal recessive
condition character zed by ceramide accumulation within neurons and within granulomas
in the skin .
-
Educational Objective :
In Niemann-Pick disease , deficiency of sphingomyelinase causes abnormal
accumulations of the ceramide phospholipid sphingomyelin and neurologic deterioration
wrthln the first year of life.
33
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A pharmaceutical corporation is investigating new therapeutic agents for treatment of
Burkitt lymphoma A double - stranded RNA molecule consisting of 21 base pairs is
created that is complementary to a region of mRNA encoding c -Myc . Introduction of this
molecule into tumor cells results in a significant reduction in cell growth Western blot
analysis of equivalent numbers of treated and untreated cells is shown below .
13
Untreated
cells
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Which of the following processes was most likely interrupted in the treated cells ?
31
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O A . DNA replication
34
B . DNA transcription
35
C C mRNA translation
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:
30
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O E. Splicing
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D. Proteasome activity
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© UWorid
Which of the following processes was most likely interrupted in the treated cells ?
27
O A. DNA replication [ 12%]
I B. DNA transcription [27%J
23
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^
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C mRNA translator [52%]
O D Proteasome activity [4%]
O E. Splicing [4%]
35
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Explanation:
33
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7
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RNA interference
Exogenous <3 RNA
(
es. lrRNA* ) /
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Translation repression
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RNA interference is an important mechanism by which short ( 20- 30 base pair )
non-coding RNA sequences induce posttranscnptional gene silencing Types of
silencing RNA include small interfering RNA ( siRNA ) and microRNA ( miRNAl The
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mRNA cleavage
*i« ton
TrBFiliation repr
UiVWl
RNA interference is an important mechanism by which short ( 20-30 base pair )
non-coding RNA sequences induce posttranscrjptional gene silencing Types of
silencing RNA include small interfering RNA ( siRNA ) and microRNA ( miRNA ) The
human genome encodes over 1000 miRNA genes each one capable of repressing
hundreds of target genes Altered expression of even a few miRNA genes can lead to
cellular dysregulation and has been implicated in the development of many diseases ,
including hematologic and solid malignancies In addition synthetic siRNA sequences
can be introduced into cells to silence specific pathogenic genes ( eg c- Myc oncogene ;
and are being explored as possible therapeutic agents .
After being transcribed . miRNA undergoes processing in the nucleus to form a
double -stranded precursor that is then exported into the cytoplasm . There , the
precursor is cleaved into a short RNA helix by a ribonuclease protein called dicer . The
individual strands are then separated and incorporated into RNA -induced silencing
complex (RISC ) . This multiprotein complex uses Its associated miRNA as a template to
bind to complementary sequences found on target mRNAs An exact match generally
results in mRNA degradation , but a partial match also causes translational repression by
preventing ribosome and transcription factor binding
(Choice A ) DNA polymerase requires a short nucleic acid sequence primer for initiation
of DNA synthesis During DNA replication , these primers are formed from RNA bases by
the enzyme DNA primase ,
"
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( Choice
B ) DNA
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Notes
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results in mRNA degradation, but a partial match also causes translational repression by
preventing ribosome and transcription factor binding .
A
{ Choice A ) DNA polymerase requires a short nucleic acid sequence primer for initiation
of DNA synthesis During DNA replication , these primers are formed from RNA bases by
the enzyme DNA primase.
{ Choice B ) DNA transcription is the process in which RNA is transcribed from a DNA
template by an RNA polymerase enzyme Although certain miRNA sequences can cause
transcriptional inhibition posttranscriptional silencing is the predominant means of RNA
interference
{ Choice D ) Degradation of proteins and polypeptides occurs in proteasomes and
lysosomes Proteasomes mainly degrade nuclear and cytoplasmic proteins ; lysosomes
degrade cellular organelles and extracellular proteins.
{ Choice E ) Small nuclear RNA ( snRNA ) molecules bind to specific proteins to form small
nuclear nbonucieoproteins isnRNPs ) , These snRNPs associate with pre-mRNA to form
spliceosomes which function to remove inirons from pre -mRNA during processing within
the nucleus .
,
I
27
28
29
30
31
32
33
34
3S
36
Educational objective:
Short non- coding RNA sequences ( eg microRNA and small interfering RNA } induce
posttranscnptionai gene silencing by base-pairing with complementary sequences within
target mRNA molecules
References:
.
1. Origins and mechanisms of miRNAs and siRNAs
37
38
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Cultured fibroblasts taken from an infant suffering from hypotonia and seizures show an
impaired ability to oxidize very long chain fatty acids (VLCFA ) and phytanic acid The
defect is most likely localized to:
O A . Mitochondria
b
B . Rough endoplasmic reticuium
O Proteasomes
O D. Lysosomes
O E. Peroxisomes
O F. Golgi apparatus
n
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t ale ulalor
4
&
6
7
a
9
10
it
/%
Cultured fibroblasts taken from an infant suffering from hypotonia and seizures show an
impaired ability to oxidize very long chain fatty acids (VLCFA ) and phytanic acid The
defect is most likely localized to:
n
O A . Mitochondria [38%1
13
11
is
is
17
C B . Rough endoplasmic reticulum [5%J
n
H
20
:
O C. Proteasomes [4%]
O D. Lysosomes [12%]
* <9 E. Peroxisomes [ 38%J
O F. Golgi apparatus [4%]
21
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21
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27
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33
31
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41
Explanation:
The patient described in the question stem is most likely suffenng from a peroxisomal
disease Peroxisomal diseases are rare inborn errors of metabolism where peroxisomes
are either absent or nonfunctional . Very long chain and some branched chain fatty acids
cannot undergo mitochondnai beta -oxidation These fatty acids are metabolized by a
special form of beta oxidation ( very long chain fatty acids ) or by alpha oxidation
{branched chain fatty acids such as phytanic acid ) within peroxisomes When
peroxisomes are absent or nonfunctional these fatty acids accumulate within the
tissues One example of a peroxisomal disease is Zellweger syndrome . In this condition,
infants are unable to properly form myelin in the CNS . Symptoms of this disease include
hypotonia and seizures as mentioned in the question stem as well as hepatomegaly ,
mental retardation and early death within months of initial presentation. Refsum disease
results from a defect in peroxisomal alpha oxidation and leads to neurologic disturbances
in response to accumulation of phytanic acid within the body . Treatment of this disease
.
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t tikufdior
/%
{ Choice A ) The mitochondria is the primary site of beta oxidation , ketogenesis . the TCA
cycle , the electron transport chain, the Initial and final steps of the urea cycle , and of
decarboxylation reactions among others it does not metabolize very long chain fatty
acids or fatty acids with branch points at odd-numbered carbons.
(Choice B ) The rough endoplasmic reticulum is the site of synthesis of proteins destined
for organelles cell membrane proteins and extracellular proteins . Proteins synthesized in
the RER are sent to the Golgi apparatus for sorting .
( Choice C ) Proteasomes function to degrade unneeded or improperly formed
intracellular proteins to small polypeptides or to amino acids The proteasome is
essential for regulation of cellular processes because it degrades proteins that express a
function that is no longer needed by the cell Proteasomes also function to degrade viral
proteins for expression on MHC Class I molecules for recognition by T lymphocytes .
(Choice D ) Lysosomes are organelles containing an acidic fluid with various proteins for
degrading fatty acids , carbohydrates , proteins and nucleic acids Diseases resulting from
lysosomal dysfunction include the mucopolysaccharidoses most classically among others
k
25
26
27
28
29
30
31
32
33
3
*
35
35
37
(Choice F) The Golgi apparatus serves to sort proteins from the rough ER and route
them to their ultimate location within membrane- bound vesicles
Educational Objective:
Peroxisomal diseases are rare inborn errors of metabolism where peroxisomes are either
absent or nonfunctional. Very long chain fatty acids or fatty acids with branch points at
odd-numbered carbons can not undergo mitochondrial beta-oxidation: these fatty acids
are metabolized by a special form of beta oxidation (very long chain fatty acids ) or by
alpha oxidation (branched chain fatty acids such as phytanic acid ) within peroxisomes .
These diseases commonly lead to neurologic defects from improper CNS myelination.
38
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b
6
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ii
12
A 35-year -old female is hospitalized with headaches and vomiting She has a long
history of psychiatric illness and is known to practice eccentric dietary habits Physical
findings include papilledema dry skin and hepatospienomegaly. Head CT scan is
ordered immediately but is negative for intracranial mass Which of the following is a
likely cause of this patient's condition?
b
13
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is
115
17
I it
19
20
21
22
23
24
2b
26
C A . Thiamine deficiency
O B Niacin deficiency
,
C. Vitamin 8,, deficiency
D Riboflavin deficiency
O E. Vitamin C overuse
C F Vitamin E overuse
C G. Vitamin A overuse
.
27
28
29
30
31
32
33
34
3S
38
37
38
"
i
1i
41
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TLX Of
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2
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O. Id
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o
Lab Value
Sent
*
Notes
Iulcufdtor
4
&
6
7
9
10
11
12
13
14
is
17
13
H
20
21
22
21
24
25
26
27
28
29
30
31
32
33
34
35
3S
37
38
39
40
41
A
A 35-year -old female is hospitalized with headaches and vomiting She has a long
history of psychiatric illness and is known to practice eccentric dietary habits Physical
findings include papilledema , dry skin and hepatospienomegaly. Head CT scan is
ordered immediately but is negative for intracranial mass Which of the following ES a
likely cause of this patient's condition?
k
()
A . Thiamine deficiency [9%]
O B. Niacin deficiency [12%}
C C Vitamin B : deficiency [ 6%J
,
D . Riboflavin deficiency [ 4%]
O E. Vitamin C overuse [3%]
G F. Vitamin E overuse [5%}
* # 0. Vitamin A overuse [61% ]
Explanation:
Individuals who consume more than 10 times the Daily Value (Recommended Dietary
Allowance) of vitamin A are prone to developing toxicity and may suffer hepatic injury so
severe as to cause cirrhosis
Vitamin A toxicity has been subdivided into three syndromes acute , chronic and
teratogenic . The signs and symptoms of acute toxicity occur after the ingestion of a
single high dose of vitamin A and include nausea , vomiting , vertigo, and blurred vision
The signs and symptoms of chronic toxicity occur after the long-term ingestion of high
doses of vitamin A , and Include alopecia dry skin , hyperlipidemia , hepatotoxicity
hepatospienomegaly , and visual difficulties Papilledema when present, Js suggestive of
_
. - .
.
- .
—
.1
M
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>
[
Lab Value
Sent
*
Notes
t alcufdtor
4
6
r\
Explanation:
7
9
10
11
12
13
n
is
15
17
13
19
20
21
n
23
n
2S
26
27
26
29
30
31
32
33
3
*
3S
35
37
38
39
40
41
Individuals who consume more than 10 times the Daily Value (Recommended Dietary
Allowance ) of vitamin A are prone to developing toxicity and may suffer hepatic injury so
severe as to cause cirrhosis
Vitamin A toxicity has been subdivided into three syndromes acute chronic , and
teratogenic . The signs and symptoms of acute toxicity occur after the ingestion of a
single high dose of vitamin A and include nausea vomiting , vertigo, and blurred vision .
The signs and symptoms of chronic toxicity occur after the long-term ingestion of high
doses of vitamin A , and include alopecia dry skin , hyperlipidemia , hepatotoxicity
hepatosplenomegaly. and visual difficulties . Papilledema when present, is suggestive of
cerebral edema in the setting of benign intracranial hypertension ( pseudotumor cerebri ).
Teratogenic effects of excessive vitamin A ingestion include microcephaly , cardiac
anomalies , and fetal death ( especially in the first trimester of pregnancy ).
*
(Choice A ) Thiamine deficiency is associated with infantile and adult benberi as well as
Wemicke-Korsakoff syndrome in alcoholics .
(Choice B ) Niacin deficiency is characterized by the 3 D' s of pellagra (dementia ,
dermatitis, and diarrhea ).
(Choice C ) Vitamin B . ( cobalamin ) deficiency is frequently associated with pernicious
anemia The classic presentation of pernicious anemia is an older , mentally slow woman
of northern European descent who is 'lemon colored * ( anemic and icteric ) , has a smooth
shiny tongue indicative of atrophic glossitis , and demonstrates a shuffling broad- based
gait.
1
(Choice D ) Vitamin 82 ( riboflavin ) deficiency is characterized by cheilosis stomatitis ,
glossitis , dermatitis, corneal vascularization, and ariboflavinosis.
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o
7
cerebral edema m the setting of benign intracranial hypertension ( pseudotumor cerebri )
Teratogenic effects of excessive vitamin A ingestion include microcephaly , cardiac
anomalies, and fetal death ( especially in the first trimester of pregnancy ).
9
10
(Choice A ) Thiamine deficiency is associated with infantile and adult beriberi as well as
Wernieke-KorsakofT syndrome in alcoholics .
n
(Choice B ) Niacin deficiency is characterized by the 3 D s of pellagra (dementia ,
A
&
6
13
u
IS
IS
17
1j
19
20
Ldb Value
Newt
Notes
*
Idkufdlor
/v
b
dermatitis, and diarrhea ).
(Choice C ) Vitamin B,; (cobalamin ) deficiency is frequently associated with pernicious
anemia The classic presentation of pernicious anemia is an older , mentally slow woman
of northern European descent who is " lemon colored" ( anemic and icteric ) , has a smooth ,
shiny tongue indicative of atrophic glossitis , and demonstrates a shuffling broad- based
gait
21
n
21
n
25
25
27
28
29
(Choice D ) Vitamin B2 ( riboflavin) deficiency is characterized by cheilosis stomatitis,
glossitis, dermatitis, corneal vascularization, and ahboflavinosls.
(Choice E ) Large doses of vitamin C can give false negative stool guaiac results and are
associated with diarrhea and abdominal bloating Some studies suggest an association
between high doses of vitamin C and calcium oxalate nephrolithiasis , though this remains
controversial
30
31
32
'5
3*S
3
35
37
38
39
40
n
(Choice F) Large doses of vitamin E have been associated with higher mortality rates
due to hemorrhagic stroke in adults and higher rates of necrotizing enterocolitis in infants.
Educational Objective:
Vitamin A overuse can result in intracranial hypertension , skin changes and
hepatospIenomegaly.
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2
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*
Notes
t dlcufdtor
4
&
6
7
e
An infant bom to a 23-year -okt female is diagnosed with an inherited condition that
results in impaired transport of ornithine from the cytosol to the mitochondria Restriction
of which of the following substances in the diet can improve this patients condition?
10
11
n
13
u
IS
115
17
13
H
20
21
t
O A. Fatty acids
O B. Protein
O C. Cholesterol
*
D Monosaccharides
OE
,
Disaccharides
O F. Purines
n
23
24
2&
25
27
28
29
30
31
32
33
34
3S
35
37
38
39
li
41
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Notes
(
ulcufalor
4
6
7
£
to
11
12
13
H
IS
15
17
13
H
20
ft
An infant born to a 23- year -old female is diagnosed with an inherited condition that
results in impaired transport of ornithine from the cytosol to the mitochondria Restriction
of which of the following substances in the diet can improve this patient s condition?
O A . Fatty acids [15%]
* •> B. Protein [59%]
(
O C. Cholesterol [2%]
C D. Monosaccharides [2%]
O E. Disaccharides [ 2%\
O F . Purines [20%J
21
n
2\
24
2&
26
Explanation:
27
28
29
30
31
32
33
34
3S
N- acaiytglutnmaie
CO. * NH '
*
Cartfsmayl
*
+
phosphate
synthetase t
T
Arq- nmasMOcmate
T
Cartumoyf phosphalo
QmUNfte
transeartutmoylose
VBM>
»
35
Fumarale
*
Arginine
37
33
39
Argionsv
Oforthlne
11
41
Aiportate
Cil 'iHine
2 ATP
Block Time Remaining :
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*
Notes
t dlculdlor
4
6
/%
Explanation:
7
a
10
11
n
13
U
IS
16
17
13
N- ce(ytgliJtamfiT(3
^
+
*
CO: * NH/ + 2 ATP
C&rbamoy!
phospnara
synthotaso /
.Crtrullifw
C
I
I
Aspartate
<.
fynjh
#
*
Argin masticori te
QmMMte
ranspsrttarrtoy/astf
T
Carbamoyl phosphate
*
^
If***
H
t
20
Fumaroto
* Ine
Argin
21
Arpjnaae
22
23
24
2&
25
G
Ornithine
OnrihUiioe '
Urea
MITOCHONDRIA
27
23
29
30
31
32
33
34
3S
35
Ammonia Is generated from the metabolism of alpha amino acids and Is converted into
urea through the hepatic urea cycle Defects in any of the urea cycle steps , including the
transport of ornithine from the cytosol into the mitochondria result in disorders of the
urea cycle . The common problem resulting from disorders of the urea cycle is increased
blood concentration of ammonia , leading to central nervous system dysfunction . Severe
defects manifest during early infancy and childhood , while milder defects may not
manifest until adulthood
37
33
39
40
ti
The treatment of urea cycle disorders consist of balancing dietary protein intake and
protein output , such that the body receives the essential amino acids needed for growth
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Lab Value
*
Note
t ole ufdtor
*
4
&
6
7
a
a
10
n
12
13
H
IS
is
17
U
14
20
21
n
2\
24
25
26
27
23
29
30
31
32
33
34
3$
36
37
33
39
40
41
A
Ammonia is generated from the metabolism of alpha ammo acids and is converted into
urea through the hepatic urea cycle . Defects in any of the urea cycle steps , including the
transport of ornithine from the cytosol into the mitochondria resuit in disorders of the
urea cycle. The common problem resulting from disorders of the urea cycle is increased
blood concentration of ammonia , leading to central nervous system dysfunction . Severe
defects manifest during early infancy and childhood , while milder defects may not
manifest until adulthood
,
The treatment of urea cycle disorders consist of balancing dietary protein intake and
protein output , such that the body receives the essential amino acids needed for growth
and development but not in excess such that excessive ammonia is formed Thus ,
protein restriction is the main form of therapy for urea cycle disorders Medications that
provide alternative pathways for the removal of ammonia from the blood can be combined
with protein restriction .
b
(Choices A , C , Dh E and F) Restriction of these substances in the diet does not result in
a decrease in blood ammonia concentration Restriction of fats and carbohydrates is
essential for weight loss in obesity and is recommended for patients with hyperlipidemia
and diabetes mellitus A low cholesterol diet is used for the treatment of
hypercholesterolemia . Restriction of purines Is required for treating hyperuricemia
Educational Objective:
Ornithine transport into mitochondria is essentia! for urea formation , as ornithine is
needed to combine with carbamoyl phosphate within the mitochondria to form citrulline in
the second step of the urea cycle . Urea cycle defects cause neurological damage
primarily due to the accumulation of ammonia Protein restriction would improve this
condition .
Time Spent : 14 seconds
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'
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*
Notes
(
dlculdior
4
&
6
7
8
n
12
13
14
IS
15
17
ia
A 4-day -olti girl is brought to the office for a routine newborn visit She was born at 39
weeks gestation via normal spontaneous vaginal delivery to a gravida 5 para 4 woman
The patient has been breastfed exclusively but has had increasing difficulty feeding over
the past 24 hours Her parents say that she Is ’loo sleepy to feed" and has been
vomiting The infant has 3 healthy living siblings and a brother who died in infancy from
"low sugar and acid in his blood Physical examination shows tachypnea and signs of
dehydration The patient is responsive to painful stimuli only . After acute treatment and
stabilization , urine testing reveals significantly elevated levels of methylmalonic acid
Which of the following sets of laboratory values would most likely result from this patient's
condition?
p
Urine propionic
Serum glucose Urine ketones Serum ammonia
acid
20
21
22
23
24
2&
25
21
23
29
30
31
32
33
3
*
35
O A.
t
Normal
Negative
Normal
O 8.
T
1
T
T
Oc
Normal
i
Negative
f
C D.
Normal
Normal
Negative
t
O E.
Normal
1
t
i
35
37
33
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&
JT
Lab Values
NCKt
Notes
I alculaior
A
&
6
7
a
9
i
11
12
13
1
IS
115
17
13
19
*
A
A 4-day -ola girl is brought to the office for a routine newborn visit She was born at 39
weeks gestation via normal spontaneous vaginal delivery to a gravida 5 para 4 woman
The patient has been breastfed exclusively but has had increasing difficulty feeding over
the past 24 hours Her parents say that she is 'loo sleepy to feed" and has been
vomiting The infant has 3 healthy living siblings and a brother who died in infancy from
"low sugar and acid in his blood Physical examination shows tachypnea and signs of
dehydration The patient is responsive to painful stimuli only . After acute treatment and
stabilization urine testing reveals significantly elevated levels of methylmalonic acid
Which of the following sets of laboratory values would most likely result from this patient’s
condition?
p
1»
Urine propionic
Serum glucose Urine ketones Serum ammonia
acid
20
21
22
23
24
2&
26
27
28
29
O A.
t
Normal
Negative
Normal
* :• B
T
4
T
I
[ 57%]
.
[6%]
30
O c.
Normal
i
Negative
f
[21%]
31
32
33
34
O D.
O E.
Norma:
Normal
Negative
t
[ 5%]
Normal
4
t
4
[ 11
3S
36
37
*1
Explanation:
38
39
40
4’
V
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M „f t
Ldto Values
Notes
I olcuhilor
4
A
Methylmalonic acidemia
&
e
7
Phenylalanine
a
11
12
13
1
19
16
17
1B
19
Leucine
Acetyl CoA
9
Oxidative decarboxylation
( Branched chaw o- keto
acid dehydrogenase }
Tyrosine
*
TCA
Cycle
Fumarate
20
21
22
Succinyl CoA
23
24
2S
26
Methyimatonyt
CoA Mutase
77
Vitamin Bi 2
23
29
30
Methylmalonyl CoA
31
32
33
34
Pn>p <onyl CoA
carboxylase
3S
36
37
33
39
40
n
Threonine
Methionine
Biotin
Propionyl CoA
-
4
Oxidative decarboxylation
( Branched chain o- keto
acid dehydrogenase )
Valine
Isoteucine
© UWbfW
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O Id
4
6
G
7
£
9
11
12
13
14
IS
16
17
ia
1 -1
20
21
n
23
24
2&
26
27
23
29
30
31
32
33
34
35
36
37
33
"
i
l
i
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tyu
tf
'
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<1
Previous
o
i
Lab Values
Sent
Notes
t akulator
»vonu
A
This patients presentation is consistent with methylmalonic acidemia , an autosomal
recessive organic acidemia resulting from complete or partial deficiency of the enzyme
methylmalonyl CoA mutase.
-
-
Catabolism of isoleucine , valine, threonine , methionine , and odd chain fatty acids
normally leads to formation of propionyl CoA , which is then converted to methylmaionyl
CoA by biotin-dependent carboxyiation . Isomerization of methylmaionyl CoA through a
vitamir 812 -dependent reaction forms succinyi CoA. which subsequently enters the TCA
cycle. Mutations in methylmalonyl- CoA mutase result in buildup of methylmalonic acid
and propionic acid , leading to a metabolic acidosis . Hypoglycemia results from
overall increased metabolic rate leading to increased glucose utilization and direct toxic
inhibition of giucoreogenesis by the organic acids. The presence of hypoglycemia leads
to increased free fatty acid metabolism that produces ketones , resulting in a further
anion gap metabolic acidosis Finally , organic acids also directly inhibit the urea cycle ,
leading to hyperammonemia
.
Complete deficiency of methylmalonyl-CoA mutase results in an anion gap metabolic
acidosis hypoglycemia ketosis and hyperammonemia (Choice A ) , These metabolic
derangements manifests as hypotonia lethargy , vomiting , and respiratory distress
{ tachypnea due to acidosis i in the neonatal period Diagnosis is confirmed by the
presence of elevated urine methylmalonic acid and propionic acid .
L;
Propionic acidemia a deficiency in propionyi-CoA carboxylase also results m
hyperammonemia , hypoglycemia , and metabolic acidosis , although it would not have
elevated levels of urine methylmalonic acid .
(Choice C ) Fatty acid oxidation disorders , such as medium-chain acyi- CoA
dehydrogenase (MCAD ) deficiency , can present with hypoglycemia hyperammonemia
and metabolic acidosis, but these disorders lack an appropriate ketosis .
1
F jfc |
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i
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IS
1&
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*
(Notes
I alculaior
derangements manifests as hypotonia lethargy , vomiting , and respiratory distress
( tachypnea due to acidosis ) in the neonatal period Diagnosis is confirmed by the
presence of elevated urine methylmalonic acid and propionic acid
A
Propionic acidemia , a deficiency in propionyl-CoA carboxylase, also results in
hyperammonemia , hypoglycemia and metabolic acidosis , although it would not have
elevated levels of urine methylmalonic acid,
(Choice C ) Fatty acid oxidation disorders , such as medium-chain acyl-CoA
dehydrogenase (MCAD) deficiency , can present with hypoglycemia hyperammonemia
and metabolic acidosis , but these disorders lack an appropriate ketosis
17
13
(Choice D ) Urea cycle defects typically present with hyperammonemia without
H
hypoglycemia
2Q
2\
n
23
24
2&
26
27
28
29
30
31
32
33
34
3S
3S
or ketosis.
(Choice E ) Propionic acid levels are elevated in patients with methylmalonic acidemia
due to downstream enzymatic deficiency of methylmalonyl - CoA mutase.
Educational objective:
Methylmalonic acidemia is an organic acidemia due to complete or partial deficiency of
methylmalonyl-CoA mutase. Complete deficiency classically presents with lethargy,
vomiting , and tachypnea in a newborn Laboratory testing shows hyperammonemia
ketotic hypoglycemia and metabolic acidosis The diagnosis is confirmed by elevated
urine methylmalonic acid and propionic acid .
.
References:
.
1 Proposed guidelines for the diagnosis and management of
.
methylmalonic and propionic acidemia
37
38
39
40
n
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*
Notes
(
u l r u f -i t o r
4
&
6
7
a
9
10
A 64 -year -old male hospitalized with severe abdominal pain and hypotension begins to
hyperventilate Laboratory testing reveals metabolic acidosis , an increased anion gap
and a high plasma lactate level . This patients findings are best explained by a low
activity of:
1?
13
A . Pyruvate kinase
u
B. Lactate dehydrogenase
is
115
&
C Pyruvate dehydrogenase
17
ia
D Pyruvate carboxylase
H
2Q
C
E. Enolase
21
n
23
24
2&
26
71
23
29
30
31
32
33
34
3S
36
37
33
39
li
41
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Notes
£ dlculdtor
4
&
6
7
a
9
n
12
13
H
IS
13
17
13
H
20
<*S
A 64 -year -oia male hospitalized with severe abdominal pain and hypotension begins to
hyperventilate Laboratory testing reveals metabolic acidosis , an increased anion gap
and a high plasma lactate level. This patients findings are best explained by a low
activity of
O A . Pyruvate kinase [9%]
O B. Lactate dehydrogenase [21%]
* <§' C. Pyruvate dehydrogenase [61%]
b
C D. Pyruvate carboxylase [8%|
O E. Enolase [1%]
21
22
Explanation:
23
Glucose
2&
25
27
28
29
Initial Steps of Glycolysis
30
31
32
33
34
I
Phosphoenolpyruvatc
3S
35
37
Pyruvat Kinase
*
33
"
i
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*3
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Notes
t ole uKii or
Explanation:
&
6
Glucose
7
a
9
10
initial Steps of Glycolysis
«
1?
13
H
15
1C
Phosphocnol pyruvate
17
13
19
Pyruvate Kinase
20
21
22
23
2i
Pyruvate
25
25
2T
23
.
{ derate Dehydrogenase
Pyruvate Dehydrogenase
( Anaerobic )
29
30
( Aerobic )
Pyruvate
Carboxylase
31
32
33
34
ACETYL*Co A
Lactic Acid
35
Oxaloacetate
36
TQA cycle
37
38
"
i
40
41
i
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6
6
7
£
9
TO
12
13
U
IS
115
17
n
19
2Q
21
22
23
21
25
25
27
23
29
30
31
32
33
34
M
'
<1
M „f t
l>
Previous
i
Lab Values
Newt
Notes
I diculdlor
*v
<
The patient described in the vignette has lactic acidosis secondary to hypoperfusion of
his tissues and subsequent metabolism of glucose via anaerobic glycolysis In the
process of glycolysis glucose is ultimately converted to pyruvate . Glycolysis takes place
in the cytosol , and oxidative phosphorylation which requires the presence of oxygen ,
occurs in the mitochondna The fate of pyruvate generated during glycolysis is
dependent on the presence of oxygen . When inadequate amounts of oxygen are present
In the tissues , pyruvate is converted to lactate by the enzyme lactate dehydrogenase in
,
+
to regenerate the NAD " from NADH H . Thus under anaerobic conditions
increased amounts of lactate are generated because pyruvate is preferentially converted
to lactate by the enzyme lactate dehydrogenase . Increased lactate levels result in
metabolic acidosis, and patients with metabolic acidosis will attempt to compensate by
causing a respiratory alkalosis. This is accomplished by hyperventilation and loss of
COj ( an acid when dissolved in the blood ) in the presence of oxygen , pyruvate is
order
,
,
preferentially converted to acetyl coenzyme A by the enzyme pyruvate dehydrogenase
( choice C ). Acetyl coenzyme A enters the mitochondria and undergoes oxidative
phosphorylation in the citric acid cycle .
{ Choice A ) The final step of glycolysis is conversion of phosphoenolpyruvate to pyruvate
by the enzyme pyruvate kinase .
{ Choice B ) Lactate dehydrogenase is a bidirectional enzyme serves to interconvert
pyruvate and lactate . This enzyme converts pyruvate to lactate in anaerobic conditions,
most classically In exercising skeletal muscle . It also plays a key role in the liver where
lactate usually generated by working skeletal muscles is taken up from the blood and
converted to pyruvate for giuconeogenesls .
,
,
35
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{ Choice D ) Pyruvate can also be converted to oxaloacetate by the enzyme pyruvate
carboxylase This enzyme is involved in regeneration of glucose from pyruvate by the
process of gluconeogenesis.
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Notes
(
dlculdlor
order to regenerate the NAD from NADH H , Thus under anaerobic conditions
increased amounts of lactate are generated because pyruvate is preferentially converted
to lactate by the enzyme lactate dehydrogenase Increased lactate levels result in
metabolic acidosis , and patients with metabolic acidosis will attempt to compensate by
causing a respiratory alkalosis. This is accomplished by hyperventilation and loss of
C
tan acid when dissolved in the blood ) . In the presence of oxygen, pyruvate Is
A
02
preferentially converted to acetyl coenzyme A by the enzyme pyruvate dehydrogenase
( choice C ). Acetyl coenzyme A enters the mitochondria and undergoes oxidative
phosphorylation in the citric acid cycle.
b
(Choice A ) The final step of glycolysis is conversion of phosphoenoipyruvate to pyruvate
by the enzyme pyruvate kinase
(Choice B ) Lactate dehydrogenase is a bidirectional enzyme serves to interconvert
pyruvate and lactate , This enzyme converts pyruvate to lactate in anaerobic conditions ,
most classically in exercising skeletal muscle . It also plays a key role in the liver where
lactate , usually generated by working skeletal muscles , is taken up from the blood and
converted to pyruvate for gluconeogenesis
(Choice D ) Pyruvate can also be converted to oxaloacetate by the enzyme pyruvate
carboxylase . This enzyme is involved in regeneration of glucose from pyruvate by the
process of gluconeogenesis.
30
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(Choice E ) Conversion of 2- phosphoglycerate to phosphoenoipyruvate Is accomplished
by the enzyme enolase .
Educational Objective:
Hypoxia- induced lactic acidosis is caused by a low activity of pyruvate dehydrogenase
( oxidative phosphorylation pathway ) and a high activity of lactate dehydrogenase .
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t alcufator
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A 42- year-old woman comes to the neurologist for enrollment rn a research study She
has a 15- year history of resting tremor bradyhinesia and cogwheel rigidity consistent
with Parkinson ' s disease . One of her siblings recently started having similar
symptoms Genetic analysis is performed on the patient and her affected sibling The
results show a loss-of-funclion mutation in a gene that leads to an accumulation of
misfolded proteins Which of the following biochemical processes is most likely defective
In this patient?
C
I
A . Acetylation
B . Gamma-carboxylation
C. Glucuronidation
O Phosphorylation
Q E . Ubiqurtination
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(Notes
I dlculdior
A
A
5
6
7
8
9
10
ii
13
14
IS
15
17
13
A 42 -year -old woman comes to the neurologist for enrollment in a research study . She
has a 15-year history of resting tremor , bradykinesia and cogwheel rigidity consistent
with Parkinson's disease . One of her siblings recently started having similar
symptoms Genetic analysis is performed on the patient and her affected sibling The
results show a loss-of-function mutation in a gene that leads to an accumulation of
misfolded proteins Which of the following biochemical processes is most likely defective
in this patient?
O A . Acetylation [8%]
B . Gamma -carboxylation J 6% j
O C . Glucuronidation [3%]
O D . Phosphorylation [6%]
20
21
n
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2i
2&
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# E . Ubiqultmation [77% j
Explanation:
27
Ubiquilln dependent protein catabolism
23
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Ubiquitw> craved
from tafgel prate o
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3
& recycled
*
3S
_
—
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*
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Neat
Notes
I akulator
A
* # E. Ubiquitination [77%]
Explanation:
9
Ubiquitin dependent protein catabolism
10
It
llbiquitm ek> Jvrd
from urget protein
4 recycled
13
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Proteavome
HydfolyzHl
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© UWorld
'
Ubiqurtin is a protein found in all eukaryotic cells that undergoes ATP -dependent
attachment to other proteins , labeling them for degradation . The proteasome then
recognizes these ubiquinated proteins and uses ATP energy to drive them through its
tubular structure , degrading them into small peptides in the process Attachment of 4 or
, *
,
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-
I <ilc ul stor
1/ '
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© UWcrld
A
Ubiqurtin is a protein found in all eukaryotic cells that undergoes ATP - dependent
attachment to other proteins , labeling them for degradation . The proteasome then
recognizes these ubiquinated proteins and uses ATP energy to drive them through its
tubuiar structure, degrading them into small peptides in the process Attachment of 4 or
more ubiquitin monomers is required before most proteins are allowed entry into the
proteasome Ubiquitination plays an important role in many cell functions, including
antigen processing muscle wasting, cell cycle regulation, QNA repair, and disposal of
misfolded proteins and regulatory enzymes .
Impairment of the ubiquitin- proteasome system can contribute to the development of
neurodegenerative disorders such as Parkinson s and Alzheimer s diseases Failure of
the system to properly degrade abnormal proteins causes protein misfolding,
aggregation , and eventual obstruction of intracellular molecular traffic , leading to cell
death Together , the Parkin. PINK1, and DJ- 1 genes code for a protein complex that
promotes the degradation of misfolded proteins via the ubiquitin-proteasome
system . Mutations in Parkin PINK 1, and DJ- 1 are each associated with autosomal
recessive forms of Parkinson's disease that have an early age of onset ( < 50 years )
*
(Choice A ) Heterochromalin is condensed and methylated DNA that has a low level of
transcriptional activity. In contrast to heterochromatin, euchromatin (loosely arranged
chromatin } has very high levels of transcriptional activity . Histone acetylation promotes
the formation of euchromatin.
{ Choice B ) Vitamin K -dependent gamma - carboxylation is critical for the functioning of
clotting factors II. VII IX. and X and of the anticoagulative proteins C and S. Warfarin
inhibits carboxylatlon of these proteins.
.
(Choice C ) The hepatic processing of bilirubin is accomplished in three key steps :
nacekro
iin
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Newt
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*
(Notes
I alcufaior
A
(Choice B ) Vitamin K -dependent gamma - carboxyiation is critical for the functioning of
clotting factors H VII IX and X and of the anticoagulative proteins C and S Warfarin
inhibits carboxylation of these proteins .
.
.
.
{ Choice C ) The hepatic processing of bilirubin is accomplished in three Key steps:
carrier -mediated passive uptake of bilirubin at the sinusoidal membrane ; conjugation of
bilirubin with glucuronic acid in the endoplasmic reticulum and active biliary excretion of
the water -soluble nontoxic bilirubin-giucuronides Disruption of this process occurs in
Crigler -Najjar syndrome a condition in which patients lack the enzyme needed to
catalyze bilirubin glucuronidation .
{ Choice D ) Phosphorylation is the addition of a phosphate group ( PO ) to a protein or
other organic molecule Phosphorylation is commonly Involved in the regulation of
enzymatic activity.
^
.
I
Educational objective:
Ubiqurtin is a protein that undergoes ATP -dependent attachment to other proteins ,
labeling them for degradation These modified proteins enter the proteasome and are
degraded into small peptides Impairment of the ubiquitin- proteasome system can
contribute to the development of reurodegenerative disorders including Parkinson s and
Alzheimer's diseases
References:
1. Parkin and defective ubiquitination in Parkinson' s disease .
2 . Parkin . PINK 1, and DJ-1 form a ubiqurtin E3 ligase complex promoting
unfolded protein degradation.
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I alculdior
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Concerned parents bnng their previously healthy 8 -month-oid son for evaluation because
he has been losing previously acquired motor skills Hepatosplenomegaly is noted on
physical examination Liver biopsy reveals the presence of leukocytes with prominent
intracellular sphingomyelin accumulations The most likely diagnosis is :
11
12
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A . Tay-Sachs disease
O B. Hurler syndrome
I C. Niemann-PIck disease
O D. Gaucher disease
C E. Von Gierke disease
O F Pompe disease
.
O G Fabry disease
H. Lesch-Nyhan syndrome
,
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I olc uhnor
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&
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a
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n
ft
Concerned parents bring their previously healthy 8 -month-old son for evaluation because
he has been losing previously acquired motor shills Hepatosplenomegaly fs noted on
physical examination Liver biopsy reveals the presence of leukocytes with prominent
intracellular sphingomyelin accumulations. The most likely diagnosis is :
11
12
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n
C A. Tay-Sachs disease [7%]
O B. Hurler syndrome [2%]
v (§) C. Niemann-Pick disease [80%]
L,
C D. Gaucher disease [ 4%]
O E. Von Gierke disease [2% ]
.
O F Pompe disease [1%J
C G. Fabry disease [2%]
H. Lesch-Nyhan syndrome [2%]
Explanation:
Niemann-Pick disease Type A is an autosomal recessive disorder that typically presents
in infants of Ashkenazi Jewish descent . The disease is characterized by a deficiency of
sphingomyelinase that causes sphingomyelin to accumulate within phagocytes. The
resultant hToamy histiocytes* accumulate in the liver , spleen and skin. Additionally , there
is gradual sphingomyelin deposition in the CNS which causes neurologic degeneration.
Affected infants classically present with loss of previously acquired motor capabilities
following a penod of normal development . The neurologic detenoration typically
progresses to hypotonia and blindness by age 1. A cherry-red macular spot (as in
Tay-Sachs disease ) and hepatosplenomegaly are common findings on physical
examination . Death usually occurs before age 3 .
.
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Notes
I dlculaior
Explanation:
A
Niemann- Pick disease Type A is an autosomal recessive disorder that typically presents
In infants of Ashkenazi Jewish descent . The disease Is characterized by a deficiency of
sphingomyelinase that causes sphingomyelin to accumulate within phagocytes The
resultant ' foamy histiocytes" accumulate in the liver , spleen and skin. Additionally , there
is gradual sphingomyelin deposition in the CNS , which causes neurologic degeneration
Affected infants classically present with loss of previously acquired motor capabilities
following a penod of normal development. The neurologic deterioration typically
progresses to hypotonia and blindness by age 1 A cherry-red macular spot as in
Tay - Sachs disease } and hepatospienomegaly are common findings on physical
examination. Death usually occurs before age 3.
(Choice A ) In Tay-Sachs disease deficiency of the enzyme hexosaminidase A causes
ganglioside accumulation .
:
{ Choice B ) Hurler syndrome is a mucopolysaccharidosis where heparan sulfate and
dermatan sulfate accumulate due to a deficiency of alpha -L- iduronidase
(Choice D ) In Gaucher disease , glucocerebroside accumulates within phagocytes due to
a deficiency of glucocerebrosidase .
(Choices E and F) Von Gierke disease and Pompe disease are glycogen storage
diseases In von Gierke disease there is a deficiency of glucose-6-phosphatase . In
Pompe disease there is a deficiency of lysosomal alpha -1 4-glucosidase
(Choice G) In Fabry disease a deficiency of alpha-galactosidase A causes the ceramide
trihexoside to accumulate Disease manifestations include angiokeratomas ,
acroparesthesia , hypohidrosis and renal failure.
,
(Choice H) Lesch-Nyhan syndrome is a disorder of uric acid metabolism caused by a
deficiency of the enzyme hypoxanthine guanine phosphoribosyl transferase ( HGPRT )
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progresses to hypotonia and blindness by age 1 A cherry-red macular spot ( as in
Tay -Sachs disease ) and hepatospienomegaly are common findings on physical
examination. Death usually occurs before age 3.
3S
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41
£ dlculdtor
A
{ Choices E and F) Von Gierke disease and Pompe disease are glycogen storage
diseases in von Gierke disease there is a deficiency of giucose 6-phosphatase . In
Pompe disease there is a deficiency of lysosomal alpha-1 , 4 -glucosidase .
-
,
(Choice H) Lesch-Nyhan syndrome is a disorder of uric acid metabolism caused by a
deficiency of the enzyme hypoxanthine guanine phosphoribosyl transferase (HGPRT),
3S
*
{ Choice D ) In Gaucher disease , glucocerebroside accumulates within phagocytes due to
a deficiency of glucocerebrosidase.
27
30
Note
(Choice B ) Hurler syndrome is a mucopolysacchandosis where heparan sulfate and
dermatan sutfate accumulate due to a deficiency of alpha -L- iduronidase .
{ Choice G ) In Fabry disease a deficiency of alpha-gaiactosidase A causes the ceramide
trihexoside to accumulate Disease manifestations include angiokeratomas ,
acroparesthesia hypohidrosis and renal failure .
31
32
33
34
*
{ Choice A ) In Tay-Sachs disease deficiency of the enzyme hexosaminidase A causes
Gu. ganglioside accumulation .
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Lab Value
Next
,
Educational Objective:
Niemann- Pick disease is an autosomal recessive disorder characterized by a deficiency
of the sphingomyelinase enzyme and resultant accumulation of sphingomyelin. Patients
present In Infancy with loss of motor skills , hepatospienomegaly. hypotonia and a
cherry-red macular spot . Foamy histiocytes are the classic finding on tissue histology .
Death occurs before age 3.
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13 : 36
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(
alculator
A schematic structure of tRNA is shown on the slide below Which of the following sites
is responsible for amino acid binding?
7
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c
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OA . A
O B. B
O c. c
O D. O
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I olc uhnor
(Notes
4
5
6
7
a
*v
<
A schematic structure of tRNA is shown on the slide below . Which of the following sites
is responsible for amino acid binding?
9
to
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c
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O
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H
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A, A [62% ]
O B. B [2%J
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O C.
O D.
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O E. E [11%J
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^
C [23%]
D [2%]
Explanation:
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(Notes
I olcuhHor
O C . C [23% ]
C D D [2%]
O E , E [11%I
5
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e
/%
9
10
Explanation:
12
Codon
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51
GCC
3’
CGG
16
Anticodon
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T C arm
^
D arm
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tRNA is the smallest of the RNA subtypes The image above shows the tRNA secondary
structure. The 3' and 5 ' ends of the molecule form the so - called " acceptor stem '* The 3'
end is the amino acid attachment site The anticodon is located on the opposite end of
the tRNA molecule ( site C). It recognizes and binds the mRNA codon and assures
placement of the proper amino acid In the growing polypeptide chain . The TijjC arm ,
which contains a sequence of three nucleotides oresent in sll tRNA molecules
,
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t akulator
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5
G
7
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O
TqC arm
D arm
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tRNA Is the smallest of the RNA subtypes The image above shows the tRNA secondary
structure The 3' and 5 ' ends of the molecule form the so-cafled acceptor stem The 3'
end is the amino acid attachment site. The anticodon is located on the opposite end of
the tRNA molecule ( site C ). It recognizes and binds the mRNA codon and assures
placement of the proper amino acid in the growing polypeptide chain The TIJJC arm ,
which contains a sequence of three nucleotides present in all 1RNA molecules
(ribothymidine . pseudouridine and cytosine ) and the 0 loop have unpaired bases The
TqjC arm. D loop and anticodon are responsible for the ’cloverlear secondary structure
of the molecule The D and TipC loops help to determine the tRNA tertiary structure as
well. The 5 ’ end of tRNA generally is composed of a terminal guanosine and does not
participate in amino acid or mRNA binding .
.
,
Educational Objective:
tRNA is the smallest subtype of cellular RNA It is responsible for transporting amino
acids to the site of protein synthesis and introducing them into the growing polypeptide
chain at the correct locations The 3 -end of the tRNA molecule is the site of amino acid
binding . The opposite side of the molecule contains the anticodon loop , which
recognizes a specific codon on the mRNA molecule .
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( <ilc
ufdtor
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A 5- year -oid boy with developmental delay is brought to the office due to difficulty 'seeing
the board" at school Examination shows a boy with a tall , thin habitus with elongated
Jimbs Funduscopy shows bilateral lens subluxation Four years later , the patient dies
suddenly of a massive cerebrovascular accident . Autopsy shows middle cerebral artery
thrombosis and old renal infarcts His parents wish to know if anything could have been
done to have prevented his death Which of the following would have been the most
appropriate supplementation for this patient?
A . Ascorbic acid
O 8 - Carnitine
C C. Pyhdoxine
O D. Thiamine
O E. Tyrosine
O F. Vitamin K
27
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ulruldtor
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U
115
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H
2Q
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A
A 5- year -otd boy with developmental delay is brought to the office due to difficulty " seeing
the board* at school . Examination shows a boy with a tall , thin habitus with elongated
imnbs Funduscopy snows bilateral lens subluxation . Four years later , the patient dies
suddenly of a massive cerebrovascular accident . Autopsy shows middle cerebral artery
thrombosis and old renal infarcts. His parents wish to know if anything could have been
done to have prevented his death . Which of the following would have been the most
appropriate supplementation for this patient?
: A. Ascorbic acid [30%]
O B. Carnitine [ 8%J
* # C. ^ yndoxine [39%]
O D. Thiamine [6%]
O E. Tyrosine [7%J
O F. Vitamin K [10%J
Explanation:
28
29
Methionine cycle
30
31
32
33
34
THF
Methionine
35
35
V
Methionine synthase
Vitamin B 12
38
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Notes
t akufdlor
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4
As
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6
7
THF
e
Methionine
9
1.1
Methionine synthase
it
Vitamin Bi ?
12
13
14
15
17
ia
19
20
5 - Methyl- THF
Homocysteine
Metbylcobaiamin
S -adenosyl
methionine
Senre
21
22
23
24
2b
26
27
23
29
Cystathionine
synthase
X
Vitamin B&
Methy (transferase
S-adenosyl
homocysteine
Cystathionine
Methyl- X
30
31
32
33
34
Cystathtonase
Vitamin Be
35
3S
37
38
"
i
Cysteine
© VWortd
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*
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A
This patient s presentation Is most consistent with homocystinuria . the most common
Inborn error of methionine metabolism . Most patients present at age 3- 10 with ectopia
lentis ( dislocated lens ) . About half of patients have intellectual disability . Other
clinical manifestations include a Marfanoid habitus ( eg . elongated limbs
arachnodactyly scoliosis ) . Patients are at high risk for thromboembolic occlusion of
both large and small vessels , especially those of the brain heart and kidneys .
Thromboembolic complications are the major cause of morbidity and mortality in these
patients .
,
15
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Homocystinuria is most frequently caused by an autosomal recessive deficiency of
cystathionine beta-synthase an enzyme that requires pyhdoxine ( vitamin B6 ) as a
cofactor . Approximately 50 % of affected patients respond to high doses of pyridoxine ,
which improves residual enzymatic activity and reduces plasma homocysteine levels .
Additional treatment includes dietary restriction of methionine .
(Choice A ) Vitamin C is a necessary cofactor for the hydroxylation of proline and lysine
residues in collagen Vitamin C deficiency results in decreased strength of collagen
fibers and causes scurvy .
( Choice B ) Carnitine assists with long-chain fatty acid transport into mitochondria .
Deficiencies in carnitine ( eg . primary systemic carnitine deficiency ) lead to fatty acid
oxidation defects in cardiac and skeletal muscle .
(Choice D ) Thiamine ( vitamin 81 ) deficiency can cause dry and wet benben and
Wernicke-Korsakoff syndrome.
35
3S
37
33
39
11
41
( Choice E ) Tyrosine is a nonessential amino acid and the precursor for catecholamines
such as dopamine , epinephrine , and norepinephrine .
(Choice F ) Vitamin K is involved in the post-transiational conversion of glutamate to
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{ Choice A ) Vitamin C is a necessary cofactor for the hydroxylation of proline and lysine
residues in collagen Vitamin C deficiency results in decreased strength of collagen
fibers and causes scurvy .
{ Choice D ) Thiamine ( vitamin B 1 } deficiency can cause dry and wet beriberi and
is
17
13
(Choice E ) Tyrosine is a nonessential amino acid and the
H
{ Choice F ) Vitamin K Is involved in the post-translational conversion of glutamate to
gamma - carboxyglutamic acid . This modification is necessary for the function of many
clotting factors and regulatory proteins involved in the coagulation cascade .
21
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*
(Notes
t alculator
{ Choice B ) Carnitine assists with long-chain fatty acid transport into mitochondria .,
Deficiencies in carnitine ( eg , primary systemic carnitine deficiency ) lead to fatty acid
oxidation defects in cardiac and skeletal muscle .
u
u
20
Lab Value
Sent
Wernicke-KorsakofT syndrome .
precursor for catecholamines
such as dopamine , epinephrine , and norepinephrine .
k
Educational objective:
Homocystinuria is most commonly caused by cystathionine synthase deficiency Affected
individuals have marfanoid habitus , ectopia lentis and developmental delay Significant
morbidity ana mortality are due primarily to thromboembolism Many patients with
homocystinuria respond dramatically to pyridoxine ( vitamin B6 ) supplementation
30
31
32
33
31
35
38
References:
-
t . Hypermethionlnemias of genetic and non genetic origin: A review.
.
.
2 Overview of homocysteine and folate metabolism With special
references to cardiovascular disease and neural tube defects.
37
33
39
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As part of an experiment healthy volunteers undergo a 12- hour fast and then drink a
solution containing radiolabeled alanine Consecutive blood samples are drawn every 15
minutes for the next 3 hours initial blood samples detect the radiolabeled alanine , but
analysis of later samples shows that the radiotracer is present in blood primarily in the
form of glucose Before alanine can be converted to glucose its amino group is
transferred to which of the following?
,
H
IS
17
13
19
20
21
n
21
24
A . a-Ketoglutarate
&
O B . L-cItrulline
O C. Malate
O D Citrate
I E Oxaloacetate
,
,
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7
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15
17
13
19
20
21
22
21
24
25
A
As part of an experiment, healthy volunteers undergo a 12- hour fast and then drink a
solution containing radiolabeled alanine , Consecutive blood samples are drawn every 15
minutes for the next 3 hours . Initial blood samples detect the radiolabeled alanine , but
analysis of later samples shows that the radiotracer is present in blood primarily in the
form of glucose Before alanine can be converted to glucose its amino group is
transferred to which of the following?
v
-
u
# A, a Ketoglutarate [61%]
O B. L-crtrulline [6%]
O C. Malate [?% ]
O D . Citrate [6%)
O E. Oxaloacetate [20%J
Explanation:
25
27
G U«K6
23
29
*
pucrfcm
30
31
32
33
34
3S
36
37
33
39
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4
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A
Explanation:
r
8
9
10
11
GFUGOM
T
12
13
H
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Alanrn
Tf
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^
Pyruvate
Afejkri
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13
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dehydrogenase
22
23
24
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o-KeKifliularalQ
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21
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urea
27
28
29
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30
31
32
33
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35
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Notes
£ alculalor
4
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7
£
9
10
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A
Alanine and glutamine play an important role in transporting nitrogen throughout the
body. Glutamine is produced by most body tissues and is catabollzed primarily by the gut
and kidney for maintenance of cellular metabolism and acid- base regulation
respectively . A significant portion of the glutamine used by these tissues is converted to
alanine and released into the circulation . Alanine is also released by skeletal muscle
tissue during protein catabolism as part of the giucosc aianine cycle that helps remove
excess nitrogen . Alanine is then transported to the liver , where it serves as a vehicle for
nitrogen disposal and as a source of carbon skeletons for gluconeogenesis .
,
-
IS
17
13
19
20
21
22
23
24
25
2&
28
29
30
31
32
33
34
35
3S
37
33
"
i
40
41
in the liver , alanine istransammated by alanine aminotransferase to pyruvate with the
amino group being transferred to a-ketoglutarate to form glutamate. Almost all
aminotransferase enzymes use a-ketoglutarate as the amino group acceptor Thus
amino groups are funneied into glutamate duhng protein catabolism Glutamate is further
metabolized by the enzyme glutamate dehydrogenase , which liberates free ammonia and
regenerates a-ketoglutarate . Ammonia then enters the urea cycle to form urea , the
primary disposal form of nitrogen In humans . Urea subsequently enters the blood and is
excreted in the urine .
,
&
(Choices C , D , and E ) Maiate citrate , and oxaloacetate are all intermediates of the
tricarboxylic acid cycle .
(Choice B ) L-citrullme is an amino acid produced as an intermediate in the conversion
of ornithine to arglninosucclnate during the hepatic urea cycle .
Educational objective:
Alanine Is the major amino acid responsible for transferring nitrogen to the liver for
disposal . During the catabolism of proteins , amino groups are transferred to aketogiutarate to form glutamate . Glutamate is then processed in the liver to form urea
the primary disposal form of nitrogen jn humans . Free ammonia is also excreted into the
urine by the kidney for regulation of acid-base status .
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Nc Kt
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Notes
£ dlculdtor
4
&
6
7
a
9
n
A 3-year -old male is hospitalized with progressive spastic paresis of his lower extremities
and choreoathetoid movements Comprehensive laboratory testing reveals very high
arginine levels in both the child' s plasma and cerebrospinal fluid The enzyme deficient
in this patient is normally involved in the production of which of the following?
11
n
13
14
IS
15
13
19
20
21
22
A . Serotonin
O B. Glutamine
O C. Orotic acid
C D. Homocysteine
I E . y-aminobutyric acid
o F. Urea
&
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24
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*
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*
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£ dlculdtor
4
5
6
7
a
9
n
A
A 3-year -old male is hospitalized with progressive spastic paresis of his lower extremities
and choreoathetoid movements Comprehensive laboratory testing reveals very high
arginine levels in both the child' s plasma and cerebrospinal fluid The enzyme deficient
in this patient is normally involved In the production of which of the following?
11
n
13
U
15
IS
13
19
20
21
n
21
24
25
25
O A . Serotonin [3%]
O B. Glutamine [10%]
O C. Orotic acid [11%]
C D. Homocysteine [7%J
O E. y-amlnobutync acid |13% ]
* # F, Urea [55%]
Explanation:
27
28
29
30
31
32
33
34
35
3S
N-AOetyfglutarnate
+
*
.
CrfJ'UllHVJ
CO, * NH»* + 2 ATP
Ovbarnoyf
pftcspnate
syvihetase I
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As pan file
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yjitr
> Fufrwraie
37
*
Argimno
33
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Ldb Values
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Explanation:
(Notes
£ dlculdtor
&
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a
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n
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n
15
15
N- rKver /tglutamate
+
*
CO! + NH ' -* 2 ATP
*
CartHirftoy )
phosphate
synthetase /
.
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Cftnjll no
jyniMs
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iranscartamoyUise
forbarney! phosphate
tyvie
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19
Anjjuaso
20
OmrUune
21
Onmiftino *
Ur
22
21
24
*
*
*
Argininowiccinrttt
^
MITOCHONDRIA
25
26
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35
The patient described in the question stem exhibits features of arginase
deficiency . Arginase is an enzyme of the urea cycle that produces urea and ornithine
from arginine Arginase deficiency is likely underdiagnosed because the spasticity seen
commonly in this disorder may simply be attributed to cerebral palsy . Treatment of
arginase deficiency consists of a low-protein diet that is devoid of
arginine Administration of a synthetic protein made of essential amino acids usually
results in a dramatic decrease in plasma arginine concentration and an improvement in
neurologic abnormalities
36
37
33
"
i
40
41
(Choice A ) Serotonin ( 5 -hydroxytryptamlne . 5HT ) is formed by the hydroxylation and
decarboxylation of tryptophan by tryptophan hydroxylase It is degraded by monoamine
oxidase and also undergoes neuronal reuptake .
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7
8
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13
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41
A
The patient described in the question stem exhibits features of arginase
deficiency . Arginase is an enzyme of the urea cycle that produces urea and ornithine
from arginine. Arginase deficiency is likely underdiagnosed because the spasticity seen
commonly in this disorder may simply be attributed to cerebral palsy . Treatment of
arginase deficiency consists of a low-protein diet that is devoid of
arginine. Administration of a synthetic protein made of essential amino acids usually
results in a dramatic decrease in plasma arginine concentration and an improvement in
neurologic abnormalities
(Choice A ) Serotonin ( 5-hydroxytryptamine , 5HT ) Is formed by the hydroxylation and
decarboxylation of tryptophan by tryptophan hydroxylase. It is degraded by monoamine
oxidase and also undergoes neuronal reuptake .
(Choice B ) Glutamine is one of the 20 major amino acids Glutamine is the major amino
acid found in the blood because it transports ammonia from peripheral tissues to the
kidney. In the nephron, the amide nitrogen is hydrolyzed by the enzyme glutaminase to
regenerate glutamate and a free ammonium ion , which can then be excreted in the urine
(Choice C ) Orotic acid is a chemical overproduced from carbamoyl phosphate by
carbamoyl phosphate synthetase II ( CPS II ) in an alternative pathway when there is a
block in the urea cycle Excessive amounts of orotic acid are usually found in OTC
( ornithine transcarbamyiase ) deficiency , citmlJinemia , and often in argininosuccinic
aciduria
(Choice D ) Deficiencies of vitamins
B 12. and folate are associated with high levels of
plasma homocysteine which in turn is associated with atherosclerosis and thrombotic
events.
(Choice E ) The amino acid derivative y-aminobutyrate . also called GABA , Is a wellknown inhibitor of presynaptic transmission in the CNS and in the retina The formation
i
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Lab Value
(Choice A ) Serotonin ( 5 -hydroxytryptamine 5HT ) Is formed by the hydroxylation and
decarboxylation of tryptophan by tryptophan hydroxylase. It is degraded by monoamine
10
oxidase and also undergoes neuronal reuptake.
12
13
14
15
IS
13
19
20
21
22
21
24
2S
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V
33
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41
Notes
Calculator
A
results in a dramatic decrease in plasma arginine concentration and an improvement in
neurologic abnormalities,
8
9
11
*
(Choice B ) Glutamine is one of the 20 major amino acids . Glutamine is the major amino
acid found in the blood because it transports ammonia from peripheral tissues to the
kidney . In the nephron, the amide nitrogen is hydrolyzed by the enzyme glutaminase to
regenerate glutamate and a free ammonium ion which can then be excreted in the urine .
.
(Choice C ) Orotic acid is a chemical overproduced from carbamoyl phosphate by
carbamoyl phosphate synthetase II ( CPS 11 } in an alternative pathway when there is a
block in the urea cycle Excessive amounts of orotic acid are usually found in OTC
( ornithine transcarbamyiase ) deficiency , citrullinemia , and often in argininosuccinic
aciduria
. B 12, and folate are associated with high levels of
^
plasma homocysteine , which in turn is associated with atherosclerosis and thrombotic
(Choice D ) Deficiencies of vitamins B
events.
-
(Choice E ) The amino acid derivative y-aminobutyrate also called GABA , is a well
known inhibitor of presynaptic transmission In the CNS and In the retina The formation
of GABA occurs by the decarboxylation of glutamate catalyzed by glutamate
decarboxylase ( GAD ).
Educational Objective:
Argmase is an enzyme of the urea cycle that produces urea and ornithine from arginine.
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*
Motes
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A
&
6
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a
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11
12
13
U
IS
IS
A 3-year -old male is seen by his pediatrician with complaints of recurrent respiratory
infections . The patient ' s mother has been exasperated with the bo/ s medical problems
since coming to the country soon after he was bom. She asks ‘why does my baby keep
getting sick when all of his siblings are fine?' The pediatrician orders laboratory tests
including genetic testing It is determined that he has a mutation in an exon of a gene
coding for a transmembrane chloride channel. The mRNA is isolated from cultured
fibroblasts and cDNA is synthesized and amplified The results are compared to the
product obtained from a healthy sibling and are shown below
1
17
Sibling
19
20
21
T 26 bp
22
Which of the following is most likely responsible for this patient ' s condition?
23
2
2&
25
*
27
28
29
30
31
32
33
3
Patient
lM t
f
O A . Codon deletion
B. Frameshift mutation
C C . Missense mutation
O 0. Silent mutation
C E . Trinucleotide expansion
*
35
3S
37
38
39
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11
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Newt
(Note
*
*
(
olc uhHor
4
A
5
6
7
a
9
10
11
n
u
U
15
15
A 3- year-old male is seen by his pediatrician with complaints of recurrent respiratory
infections . The patient ' s mother has been exasperated with the boy' s medical problems
since coming to the country soon after he was bom . She asks "why does my baby keep
getting sick when all of his siblings are fine?" The pediatrician orders laboratory tests
including genetic testing It is determined that he has a mutation in an exon of a gene
coding for a transmembrane chloride channel The mRMA is isolated from cultured
fibroblasts and cDNA is synthesized and amplified The results are compared to the
product obtained from a healthy sibling and are shown below
17
Siting
19
i
&
Palienl
bp
20
124 bp
21
n
23
24
2S
26
27
23
29
30
31
32
33
34
35
35
37
33
"
I
40
tl
Which of the following is most likely responsible for this patient' s condition?
C A . Codon deletion [23%]
#
B.
"
;
.i
-
ft
tation [57% ]
C C . Missense mutation [18%]
O D . Silent mutation [1%)
’
E . Trinucleotide expansion [1 % ]
Explanation:
DNA and RNA are composed of sequences of four bases arranged into codons
composed of three sequential bases Each codon calls for a particular amino acid except
for one codon that signals the initiation of protein synthesis (AUG ) and three that stop
protein synthesis fUAA UAG and UGA 1 Changes in the DNA sequences are called
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(Notes
I alculator
Explanation:
A
DNA and RNA are composed of sequences of four bases arranged into codons
composed of three sequential bases Each codon calls for a particular amino acid except
for one codon that signals the initiation of protein synthesis (AUG ) and three that stop
protein synthesis ( UAA , UAG , and UGA ). Changes in the DNA sequences are caEled
mutations Some changes in the genetic code can result in the formation of altered
proteins such as enzymes , channels and structural proteins , which may lead to serious
clinical manifestations .
i
,
There are several types of mutations :
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Base substitutions ( point mutations ), where one base is substituted with another base
are the most common type of mutation These point mutations can be of three types:
,
1. Silent mutations ( Choice D) : These mutations result from codon base
substitutions which code for the same amino acid. For example , a single base
substitution in UCA to UCU will not result in any change in protein structure
because both code for the same amino acid serine . Silent mutations do not cause
amino acid changes within proteins
2 . Missense mutations (Choice C) : These mutations are characterized by base
substitutions that result in the placement of an incorrect amino acid in a protein
sequence . For instance , a change in the code from UUU to UCU changes the
translated amino acid from phenylalanine (UUU ) to serine (UCU ).
3 . Nonsense mutations: These mutations introduce a stop codon within gene
sequences resulting In the formation of shorter, truncated proteins . An example of
a nonsense mutation is a mutational change in the codon UCA ( serine ) to UAA
( stop codon ).
Frameshift mutations result from deletion or insertion of bases that are not a multiple of
three . As their name implies , frameshift mutations alter the reading frame of the genetic
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A
a nonsense mutation is a mutational change in the codon UCA ( serine ) to UM
( stop codon ).
&
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IS
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Frameshift mutations result from deletion or insertion of bases that are not a multiple of
three As their name implies , frameshift mutations alter the reading frame of the genetic
code resulting in the formation of non- functional proteins . The patient in the vignette
appears to have cystic fibrosis , a genetically inherited condition with hundreds of
implicated mutations. The most common mutation is a codon deletion of the
phenylalanine at position 508 in the CFTR protein. However , it appears that this
patient's mutation is a frameshift mutation caused by a 2 base -pair deletion in the CFTR
protein.
,
(Choice E ) Trinucleotide expansions increase the number of trinucleotide repeats In the
coding region of a gene , resulting in large , often unstable , proteins .
(Choice A ) Codon deletions are deletions of one or more entire codons (genetic triplet
codes). These deletions usually result in the formation of shorter , potentially
nonfunctional proteins.
Educational Objective:
Deletions or the additions of a number of base pairs which are not a multiple of three
indicate that a frameshift mutation has occurred. Frameshift mutations alter the reading
frame of the genetic code, resulting in the formation of non-functional proteins .
References:
t . The CFTR frameshift mutation 3905insT and its effect at transcript and
protein level.
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t dlculdlor
A 54 - year -old woman is evaluated in the clinic for exertional dyspnea and easy
fatigability The patient has no chest pain , cough , or wheezing She does not use
tobacco , alcohol, or illicit drugs On physical examination, her gait is unstable when her
eyes are closed and there is impaired vibratory sensation in the lower extremities .
Marked pallor of the conjunctivae . nail beds and palms is present Which of the
following laboratory tests would help confirm the most likely diagnosis in this patient?
C A . Erythrocyte glucose-6-phosphate dehydrogenase activity
O B. Erythrocyte glutathione reductase activity
O C. Erythrocytic pyruvate kinase activity
O D. Erythrocyte transketolase activity
O E , Serum methylmalonic acid level
O F Serum protoporphyrin level
.
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I dlculaior
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ia
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A 54 -year -old woman is evaluated in the clinic for exertional dyspnea and easy
fatigability The patient has no chest pain, cough , or wheezing She does not use
tobacco , alcohol, or iilicit drugs On physical examination, her gait is unstable when her
eyes are closed and there is impaired vibratory sensation in the lower extremities .
Marked pallor of the conjunctivae nail beds and palms is present Which of the
following laboratory tests would help confirm the most likely diagnosis in this patient?
-
O A . Erythrocyte glucose 6-phosphate dehydrogenase activity [5%]
O B, Erythrocyte glutathione reductase activity [5%]
O C. Erythrocytic pyruvate kinase activity [5% ]
O P, Erythrocyte transketoiase activity [8%]
* % E. Serum methylmalonic acid level [70%]
O F. Serum protoporphyrin level [8%|
25
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Explanation:
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I dlculdior
/%
Explanation:
6
7
Vitamin B 12 deficiency
8
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It
12
Methylmalonic acid
13
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115
Buildup
Impaired
myelin synthesis
Melhylmafonyl - CoA
17
ia
Methyimalonyf - CoA rrmiase
20
Vitamin B 12
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t
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Suconyl- CoA
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77
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Impaired
DNA synthesis
Homocysteine
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Metbyl tetrahydrofolate
(methyl- THF )
-
Methionine synthase
Vitamin B12
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*
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Vi
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Tetrahydrofolate (THF )
!
Methionine
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© LMtertd
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l
This patient has features of anemia { eg . exertional dyspnea , fatigue , pallor ) with
associated neurologic deficits that are highly suggestive of vitamin B„deficiency The
hematologic manifestations of B deficiency teg megaloblastic anemia pancytopenia )
are related to Impaired ONA synthesis .
k
Vitamin Bi: (hydroxocobalamin ) also serves as a cofactor for methyimalonyFCoA mutase
{ converts methylmalonyl-CoA to succinyl-CoA ) and methionine synthase ( converts
homocysteine and foiic acid to methionine ). B , ; deficiency consequently results in
elevated levels of serum methylmalonic acid and homocysteine . Increased
methylmalonic acid levels can disrupt myelin synthesis and result in subacute
combined degeneration of the dorsal columns ( eg , loss of proprioception/vibration .
Romberg sign ) and lateral corticospinal tract ( eg spastic muscle weakness ,
hyperreflexia ) Axonal degeneration of peripheral nerves can also be seen
(Choice A ) Giucose-6-phosphate dehydrogenase deficiency ( G6P0 ) leads to acute
hemolytic anemia in response to oxidative stress Neurologic manifestations are not
present in G6PD-associated anemia .
(Choice B ) Erythrocyte glutathione reductase activity may be decreased in patients with
vitamin B (riboflavin ) deficiency because glutathione reductase uses FAD (derived from
vitamin B. ) as a cofactor to reduce disulfide bonds . Vitamin B . deficiency typically
presents with normocytic anemia and inflammation of the lips ( cheilosis ), mouth
( stomatitis ), and /or tongue (glossitis ).
(Choice C ) Pyruvate kinase generates ATP through the conversion of
phosphoenofpyruvate to pyruvate Pyruvate kinase deficiency is an autosomal recessive
condition that typically presents with congenital hemolytic anemia due to impaired
glycolytic ATP generation.
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( stomatitis), and/or tongue (glossitis).
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Notes
I die uldtor
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A
(Choice C ) Pyruvate kinase generates ATP through the conversion of
phosphoenolpyruvate to pyruvate. Pyruvate kinase deficiency is an autosomal recessive
condition that typically presents with congenital hemolytic anemia due to impaired
glycolytic ATP generation.
(Choice D ) Transketolase is an enzyme of the hexose monophosphate pathway that
utilizes thiamine ( vitamin B . j as a coenzyme Erythrocyte transketolase activity is
decreased in thiamine deficiency which causes Wernicke-KorsakofT syndrome and
beriberi
(Choice F) Serum protoporphyrin levels are increased in iron deficiency anemia , lead
poisoning , and erythropoietic protoporphyria . Lead poisoning can cause sideroblastic
anemia and peripheral neuropathy, but other characteristic features ( eg ,
abdominaL/musculoskeietal pain cognitive impairment , nephropathy ) are not evident in
this patient .
Educational objective:
Vitamin Bi; deficiency often presents with megaloblastic anemia ( impaired DMA synthesis )
and neurologic deficits (impaired myelin synthesis ) . Charactenstic neurologic findings
include subacute combined degeneration of the dorsal columns and lateral corticospinal
tract Elevations in methylmalonic acid and homocysteine levels occur due to decreased
metabolism of these molecules
References:
.
1. Clinical practice Vitamin 812 deficiency
.
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NPKt
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N o t e*
(
akutdtor
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A
An infant born to a 32-year -old female demonstrates lethargy , vomiting , and seizures
after her first few feedings Initial laboratory evaluation demonstrates a markedly
increased blood ammonium level Liver biopsy suggests impaired formation of
N acetylglutamate as the cause of the child' s problems Which of the following reactions
is most likely impaired in this patient?
-
COJ + NH .
L- Arginine
p
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n
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Carbamoyl
Phosphate
L-Ornithine
Arginmosuccinaie
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L -Cilrulline
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O A. A
O B. B
O c. c
O D. D
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ft
An infant born to a 32- year -oid female demonstrates lethargy , vomiting , and seizures
after her first few feedings Initial laboratory evaluation demonstrates a markedly
increased blood ammonium level Liver biopsy suggests impaired formation of
N-acetylglutamate as the cause of the child's problems . Which of the following reactions
is most likely impaired in this patient^
,
CO
t
NH .
L- Argimne
17
13
19
*
20
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24
2S
25
Carbamoyl
Phosphate
L - Ornithine
Arginnosuccmaie
27
26
L -Ciirulline
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35
3&
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* m A, A [54%]
O B. B [21%I
O C. C [7%]
O D. D [8%]
O E. E [11%J
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A
Explanation:
G
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a
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19
*
N-acotytgkJtarnate
+
CO. * NH/ * 2 ATP
Csrbamoyf
* pftosjffldta
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synthetase I
Ornithine
f
Carbanx>yt phosphate
fnwiscartwmo> /aw
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.
Argifime
/
Ornithine
*
r\
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Ar&wxmrr
» Funwai*
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25
Aspartate
vOtrulline
Ornithine *
L
Urea
MITOCHONDRIA
Ammonia is generated from the metabolism of alpha amino acids and is normally
converted to urea by the urea cycle . The urea cycle involves five enzymatic steps : two in
the mitochondrial matrix and three in the cytosol . One more enzyme that indirectly
participates in the urea cycle is N- acetylglutamate synthetase (NAGS ) The first step of
urea cycle combines CO ammonia and ATP to form carbamoyl phosphate in a reaction
catalyzed by the enzyme carbamoyl phosphate synthetase I , the rate-limiting step in the
urea cycie. Carbamoyl phosphate synthetase i ( CPS ) requires the presence of
N- acetylglutamate (NAG ), a molecule formed by NAGS as this molecule acts as an
allosteric activator of CPS . None of the other steps in the urea cycle require NAG as an
activator .
.,
,
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Arginine
5
6
Argmase
7
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Lab Valuer
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4
9
&
Orritn jie '
Urea
MITOCHONDRIA
it
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u
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Ammonia is generated from the metabolism of alpha amino acids and is normally
converted to urea by the urea cycle . The urea cycle involves five enzymatic steps : two in
the mitochondhal matrix and three in the cytosol One more enzyme that indirectly
participates in the urea cycle is N- acetylglutamate synthetase ( NAGS ). The first step of
urea cycle combines CO;. ammonia , and ATP to form carbamoyl phosphate in a reaction
catalyzed by the enzyme carbamoyl phosphate synthetase I, the rate- limiting step in the
urea cycle . Carbamoyl phosphate synthetase I (CPS ) requires the presence of
N-acetylglutamate ( NAG ), a molecule formed by NAGS, as this molecule ads as an
allosteric activator of CPS . None of the other steps in the urea cycle require NAG as an
activator .
l,
The symptoms seen in the patient in this question stem are the result of toxic effects of
ammonia accumulation within the child' s blood and tissues The first few feedings
provide a protein load to the infant that results in amino acids being available for
metabolism , but the defect in the urea cycle prevents the disposal of toxic ammonia from
the child' s body , leading to lethargy, vomiting , and seizures .
Educational Objective:
N- acetylglutamate is an essential activator of carbamoyl phosphate synthase I and is
formed by the enzyme N- acetylglutamate synthetase from the precursors acetyl- CoA and
glutamate ,
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I alculdtor
Notes
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&
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i
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n
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U
IS
IS
1?
lj
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2Q
n
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2&
25
A 34 -year -olo woman wrth a history of recurrent unnary tract infections comes to the
physician with dysuria and increased urinary frequency Her urine culture grows colonies
of Gram-negative bacteria The bacteria are isolated and placed in a growth-enhancing
nutrient solution , where they undergo rapid cellular division As they are actively dividing
the bacterial cells are lysed and their DNA is extracted and purified. Analysis of the
partially replicated DMA fragments shows the presence of uracil This finding is most
likely mediated by which of the following enzymes?
it
O A . DNA ligase
B. DNA polymerase I
C. DNA polymerase III
O 0 Gyrase
O E. Helicase
C F. Primase
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er*
*
Note
I olc uhilor
*
4
&
6
7
8
9
n
11
12
13
U
15
I?
ft
A 34 -year -old woman with a history of recurrent urinary tract infections comes to the
physician with dysuria and increased urinary frequency Her urine culture grows colonies
of Gram-negative bacteria The bacteria are isolated and placed in a growth- enhancing
nutrient solution , where they undergo rapid cellular division As they are actively dividing,
the bacterial cells are lysed and their DNA is extracted and purified Analysis of the
partially replicated DMA fragments shows the presence of uracil This finding is most
likely mediated by which of the following enzymes?
17
O A ONA Eigase [3%J
13
19
OB. DNA polymerase I [16%J
20
n
21
24
25
25
27
23
29
,
(
C DNA polymerase 111 ( 19%]
O D. Gyrase [3%]
O E. Heiicase [2%]
* # F . Primase [57%]
Explanation:
DNA replication fork
30
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3*
r
3’
Leading
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(
olc uhilor
4
A
&
6
Explanation:
7
DNA replication fork
8
9
10
11
3'
n
5*
13
14
IB
15
trifling
} tr nd
b
*
17
13
polymerdw III
N
DMA
20
polymerase i
n
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24
2S
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"
i
Primase
© UWorld
DNA ligaw
Lagging
«rand
S'
1'
5'
3'
5'
Single stranded
DNA binding protein
This question describes a scenario in which uracil is found in association with bacterial
DNA during prokaryotic DNA replication . In general , uracil is found only in RNA so the
question essentially asks which enzyme involved in DNA synthesis catalyzes the
formation of RNA strands in prokaryotic DNA replication , primase ( an RNA polymerase )
is responsible for synthesizing a short RNA primer using the separated strands of DNA at
the replication fork as templates DNA replication then proceeds , with DNA polymerase
using the 3' hydroxyl group of the RNA primer as a starting point for synthesis Primase
is a crucial enzyme for bacterial replication as DNA polymerase cannot initiate DNA
synthesis without this short nucleic acid sequence primer .
.
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*
I alc uhilor
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&
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a
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IS
1S
A
This question describes a scenario in which uracil is found in association with bacterial
DNA during prokaryotic DNA replication . In general, uracil Is found only in RNA so the
question essentially asks which enzyme involved in DNA synthesis catalyzes the
formation of RNA strands In prokaryotic DNA replication , primase ( an RNA polymerase )
is responsible for synthesizJng a short RNA primer using the separated strands of DNA at
the replication fork as templates DNA replication then proceeds , with DNA polymerase
using the 3 ' hydroxyl group of the RNA primer as a starting point for synthesis. Primase
is a crucial enzyme for bacterial replication as DNA polymerase cannot initiate DNA
synthesis without this short nucleic acid sequence primer .
.
17
{ Choice A ) DNA Ugase is the enzyme that repairs single - strand breaks in duplex DNA
13
during DNA replication and repair
H
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?&
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3S
3S
37
.
(Choices 8 and C ) During replication , DNA polymerase III is the primary enzyme
responsible for synthesis of daughter DNA strands; DNA polymerase I functions chiefly to
replace the RNA primers with DNA segments , Unlike DNA polymerase III DNA
polymerase I has 5 ’ - 3 exonuclease activity that can remove RNA primers and damaged
DNA segments The 3' — 5' exonuclease activity of DNA polymerase I and III provides a
proofreading function that fixes mismatched nucleotides in the newly formed daughter
strands
f
(Choices D and E ) Heiicase unwinds DNA at the replication fork However this
process resuits in supercoiling of the DNA DNA gyrase is a type II topoisomerase that
helps to relieve the resultant strain .
Educational objective:
Primase Is a DNA- dependent RNA polymerase that incorporates short RNA primers into
replicating DNA.
38
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ulalor
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It
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IS
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ia
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29
An 18-month-old boy is brought to the office due to language regression. He said several
words at hts 1-year appointment but now no longer speaks any words at all . His moods
have also become more unpredictable over the past 4 months with frequent tantrums
The parents tried to bring him in sooner for evaluation but they live in an impoverished
part of the city and experienced financial difficulties with transportation to the office On
physical examination the boy is quiet arid maintains appropriate eye contact throughout
the visit . Hemoglobin is 9 g/dL. Which of the following enzymes is most likely inhibited in
this patient?
C
[,
A , C-Amlnolevulinate dehydratase
B. Bilirubin giucuronyl transferase
‘
C Porphobilinogen deaminase
O 0 Pyruvate kinase
E. Uroporphyrinogen decarboxylase
30
31
32
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A
An 18 -month-old boy is brought to the office due to language regression. He said several
words at his 1-year appointment but now no longer speaks any words at all . His moods
have also become more unpredictable over the past 4 months with frequent tantrums
The parents tried to bring him in sooner for evaluation , but they live in an impoverished
part of the city and experienced financial difficulties with transportation to the office On
physical examination the boy is quiet and maintains appropriate eye contact throughout
the visit . Hemoglobin is 9 g/dL. Which of the following enzymes is most likely inhibited in
this patient?
b
* # A. 5-Amlnolevulinate dehydratase [46%]
B. Bilirubin glucuronyl transferase [21%]
C C Porphobilinogen deaminase [14%]
O D. Pyruvate kinase [13%]
E. Uroporphyrinogen decarboxylase [ 5%]
25
25
27
Explanation:
28
29
Lead toxicity
30
31
32
33
34
X
1
Cytoplasm
5 - Aminolevulinic acid { ALA)
ALA synthase
Succinyl
CoA
-+
I
+
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35
33
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\
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Mitochondria
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Lead toxicity
*5
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7
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23
24
Hydroxymethylbilane (HMB )
25
Fe:
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25
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Ferrochelatase
Uroporphyrinogen ill
23
29
synthase
Protoporphyrin IX
30
31
32
33
34
Uroporphyrinogen III
Uroporphynnogen
decarboxylase
35
3S
37
38
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41
Coproporphynnogen III
Protoporphynnogen IX
Copt oporphvnnopen
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Coproporphynnogen
oxidase
&
6
© UWorid
7
8
9
10
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13
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35
37
38
This child ' s language regression and anemia are most likely due to lead poisoning . Lead
toxicity is most prevalent among impoverished children residing in deteriorating urban
housing built before 1978 . Young children are particularly susceptible to lead
poisoning via inhalation and ingestion of lead- based paint dust or chips due to normal
crawling and mouthing behaviors The incomplete blood- brain- bamer in children is
vulnerable to the neurotoxic effects of lead , which include long- standing behavioral
problems and developmental delay or regression .
Anemia in lead poisoning results from inhibition of 6-aminolevulinic acid ( ALA )
synthase, ALA dehydratase , and ferrocheiatase in the heme biosynthesis
pathway , Because protoporphyrin IX cannot combine with iron (Fe; ) to form heme due to
ferrocheiatase inhibition , it instead incorporates a zinc ion , leading to elevated zinc
protoporphyrin levels In addition , ALA levels are increased . Lead poisoning also
commonly coexists with iron deficiency anemia and severe lead poisoning can also
induce hemolytic anemia .
t
( Choice B ) Giucuronyl transferase (uridine 5 -diphospho-glucuronosyltransferase ) is
necessary for hepatic bilirubin conjugation. Gilbert syndrome , a condition marked by
jaundice elevated unconjugated bilirubin levels ) during times of stress results from
^
mutations In the gene encoding giucuronyl transferase .
,
( Choice C ) Defects in porphobilinogen deaminase result in acute intermittent porphyria
(AIP ), a disorder characterized by acute attacks of abdominal pain neuropsychiatric
symptoms, and red or brown urine . The chronicrty of symptoms, lack of abdominal
pain and anemia make AIP unlikely in this patient .
,
,
"
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i
41
( Choice D ) Pyruvate kinase deficiency is typically inherited in an autosomal recessive
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Notes
(
dkuldtsr
a disorder characterized by acute attacks of abdominal pain , neuropsychiatric
symptoms , and red or brown urine The chronicity of symptoms, Jack of abdominal
pain, and anemia make AIP unlikely In this patient.
( AIP )
/\
(Choice D ) Pyruvate kinase deficiency is typically Inherited in an autosomal recessive
pattern and leads to hemolytic anemia Pyruvate kinase deficiency can present with
pallor , scleral Icterus and splenomegaly , but it does not present with behavioral
regression or language difficulties
,
17
(Choice E ) Defects in uroporphyrinogen decarboxylase cause porphyria cutanea tarda
{PCT }, the most common form of porphyria Patients with PCT exhibit chronic
photosensitivity with blistering in areas of sun exposure and elevated levels of
13
uroporphyrinogen in the urine.
H
20
21
’f
.
23
24
25
Educational objective:
Young children who reside in homes built before 1978 are at significant nsk for lead
toxicity Lead directly inhibits 6-aminolevulinic acid ( ALA ) synthase , ALA dehydratase,
and ferrocheiatase . resulting in anemia. ALA accumulation , and elevated zinc
protoporphyrin levels Neurotoxicity is also a significant long-term complication ,
25
27
28
29
30
31
32
33
34
35
35
References:
1. Lead toxicity , a review of the literature . Part 1: Exposure, evaluation,
and treatment,
2 . A study on the ALAD gene polymorphisms associated with lead
exposure .
3. The important health impact of where a child lives: neighborhood
characteristics and the burden of lead poisoning
.
37
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40
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A
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6
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9
n
11
12
13
U
15
15
17
13
H
20
21
22
21
25
25
A 3- year-oid boy who recently immigrated to the United States is brought to the physician
by his parents because he has not yet begun to walk or speak . Assessment of his
developmental milestones shows severe intellectual disability He dies 6 months later
from refractory seizures resulting in respiratory failure Autopsy shows pallor of the
substantia nigra locus ceruleus and vagal nucleus dorsalis The underlying condition
responsible for this patient s death is most likely caused by a deficiency of which of the
following enzymes?
I
A . Branched-chain ketoacid dehydrogenase
O B . Dopamine hydroxylase
O C. Homogentlsic acid oxidase
C D Phenylalanine hydroxylase
O E . Tyrosinase
27
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Notes
4
&
6
7
6
9
n
11
12
13
U
15
15
ft
A 3-year -oid boy who recently Immigrated to the United States is brought to the physician
by his parents because he has not yet begun to walk or speak . Assessment of his
developmental milestones shows severe intellectual disability He dies 6 months later
from refractory seizures resulting in respiratory failure . Autopsy shows pallor of the
substantia nigra locus ceruleus and vagal nucleus dorsalis The underlying condition
responsible for this patient' s death is most likely caused by a deficiency of which of the
following enzymes?
A . Branched-chain ketoacid dehydrogenase [6%]
17
O B. Dopamine hydroxylase [24%]
13
H
O C. Homogentisic acid oxidase [4%]
20
21
22
24
25
26
b
*
* D. Phenylalanine hydroxylase [51%[
O E. Tyrosinase [ 14%]
Explanation:
27
23
29
Phenylketonuria
30
31
32
33
34
Dthydropteridine
reductase
35
BH4
35
BHt
Melanin
37
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*
(
olcuhilor
4
A
6
G
Phenylketonuria
7
a
Dihyaroptendine
reductase
9
10
11
n
Melanin
13
U
19
16
Phenylalanine
Tyrosine
17
Phenyfa&ntne
13
19
hydroxylase
20
Phenylpyruvate
21
DOPA
Catecholamines
Homogentisate
22
24
2S
26
HOfTtcgentmc ac < d
dioxygenasc
Phenylacetate
Phenyllactate
27
MaIeylaceto acetate
28
29
30
31
32
33
34
Fumarylacetoacetate
3S
TCA
Cycle
Fumarate
36
37
38
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© UWortd
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*
I ale ulalor
4
&
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7
6
9
n
11
12
13
14
15
1b
17
U
H
20
2%
22
24
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26
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41
This patient s severe intellectual disability history of seizures and abnormal pallor of
catecholaminergic brain nuclei on autopsy are suggestive of phenylketonuria
( PKU } _ PKU results from the inability to convert phenylalanine into tyrosine , a reaction
which is normally catalyzed by phenylalanine hydroxylase . This enzyme requires the
cofactor telrahydro bio pterin ( BHj , which Is regenerated from dlhydroblopterln (BH,) by
the enzyme dihydropteridine reductase Although neonatal hyperphenylalaninemia can
be caused by deficiency of either enzyme , most cases are attributable to abnormalities in
phenylalanine hydroxylase.
I
It is believed that excess phenylalanine and the presence of large concentrations of
phenylalanine metabolites ( eg phenyilaclate & phenylacetate } contribute to the brain
damage seen in PKU. Hypopigmentation involving the skin, hair , eyes , and
catecholaminergic brain nuclei ( which produce a dark pigment known as neuromelanin )
results from the inhibitory effect of excess phenylalanine on melanin synthesis . The
classic musty or mousy body odor is due to the accumulation of abnormal phenylalanine
metabolites .
(Choice A ) Branched-chain ketoacid dehydrogenase catalyzes decarboxylation of the o ketoacid derivatives of all 3 branched chain amino acids leucine , isoleucine , and
valine . Deficiency causes maple syrup urine disease , which is characterized by a burntsugar smell in the urine of affected patients . Hypopigmentation is not seen in the
condition .
(Choice B ) Dopamine hydroxylase is the enzyme that catalyzes the biosynthesis of
norepinephrine from dopamine Deficiency of this enzyme causes a rare form of
dysautonomia characterized by ptosis orthostatic hypotension hypoglycemia and
hypothermia .
,
,
( Choice C ) Alkaptonuria is an autosomal recessive disorder of tyrosine degradation
caused by a deficiency of homogentisic acid oxidase . This condition results in the
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I ole uf Jtor
4
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6
7
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10
11
12
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U
IS
15
17
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20
21
A
(Choice A ) Branched-chain ketoacid dehydrogenase catalyzes decarboxylation of the a ketoacid derivatives of all 3 branched chain amino acids: leuclner Isoleucine , and
valine Deficiency causes maple syrup urine disease , which is charactenzed by a burntsugar smell in the urine of affected patients. Hypopigmentation is not seen in the
condition .
(Choice C ) Alkaptonuria is an autosomal recessive disorder of tyrosine degradation
caused by a deficiency of homogentisic acid oxidase. This condition results in the
accumulation of large amounts of homogentisic acid , leading to connective tissue
22
hyperpigmentation and degenerative joint disease ,
2-t
2&
26
(Choice E ) Albinism is an autosomal recessive disorder caused by defects in the
biosynthesis and distribution of melanin ft is most commonly due to depressed or absent
tyrosinase activity Melanin synthesis begins with the conversion of tyrosine to DOPA
and DOPA to dopaquinone by the enzyme tyrosinase in melanocytes . Albinism does not
present with neurologic dysfunction.
27
28
29
30
31
32
33
3
*
35
35
37
I
(Choice B ) Dopamine hydroxylase is the enzyme that catalyzes the biosynthesis of
norepinephrine from dopamine Deficiency of this enzyme causes a rare form of
dysautonomia characterized by ptosis orthostatic hypotension hypoglycemia and
hypothermia.
Educational objective:
Deficiency of the enzyme phenylalanine hydroxylase or its cofactor tetrahydrobiopterin
causes accumulation of phenylalanine in body fluids and the central nervous
system . Homozygous infants are normal at birth but gradually develop severe intellectual
disability and seizures if left untreated . Hypopigmentation of the skin , hair, eyes , and
catecholaminerglc brain nuclei is also frequently seen .
33
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Last updated [ 9 / 26 2015 ]
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*
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Note
A
5
6
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a
9
n
12
13
14
15
IS
17
A newborn is diagnosed with hyperphenyialaninemia during newborn screening After
extensive workup the child is placed on a special phenylalanine-restricted diet with
tyrosine supplementation and is instructed to return to his physician for regular follow-up
visits The parents are extensively counseled on the child' s condition , including the
necessary dietary restrictions that the child must adhere to. Several months later
laboratory tests indicate that the infant has a normal serum phenylalanine level ; however ,
after careful examination the physician observes some neurological abnormalities
Further workup includes a serum prolactin level , which is elevated Which of the
following enzymes is most likely deficient in this patient?
13
H
20
21
22
21
2S
25
21
23
29
A . Phenylalanine hydroxylase
O B Olhydrobioptedn reductase
O C. Dopamine hydroxylase
O D. Homogentisate oxidase
E . Phenylethanoiamine N-methyltransferase
O F Tyrosinase
.
30
31
32
33
34
3S
35
37
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4
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6
7
a
9
10
11
n
13
14
15
15
17
ft
A newborn is diagnosed with hyperphenyialaninemia during newborn screening After
extensive workup the child is placed on a special phenylalanine- restricted diet with
tyrosine supplementation and is instructed to return to his physician for regular follow- up
visits The parents are extensively counseled on the child's condition , including the
necessary dietary restrictions that the child must adhere to . Several months later
laboratory tests indicate that the infant has a normal serum phenylalanine level; however ,
after careful examination the physician observes some neurological abnormalities
Further workup includes a serum prolactin level , which is elevated Which of the
following enzymes is most likely deficient In this patient?
13
H
20
21
22
23
25
25
27
23
29
30
A . Phenylalanine hydroxylase |19%]
v @ B. Dihydrobiopterin reductase [33%]
O C . Dopamine hydroxylase [28%)
O D . Homogentisate oxidase [2%]
E . Phenyiethanolamine N- methyltransferase [5 %]
O F. Tyrosinase [13 % ]
Explanation:
31
32
33
34
Qihytfrvtifeotvrin
redact 3 se
Dthydfubioptenn
BH ,
35
BH ;
35
37
30
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Phenylalanine c
Ptienyialofirw
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^
E Phenylethanolamine N-methyltransferase [5%]
4
&
6
/%
O F , Tyrosinase [13% ]
7
a
9
10
11
Explanation:
DhytfiQbtwtQrin
OrhydrotMoptenn
n
reductase
u
14
15
IS
17
13
&
Phenyialarnne c
Pnmyiaiafiw
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dtcnrt .»iyi6s0
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22
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Dopamine
25
25
Dupannn
27
hydroxyuisa
*0
23
29
Norepinephrine
30
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32
33
34
S faJenosytmetbumine iSAM >
-
35
-
Pnanyieihanoi&n nine
N rn thyUrjtnshmrsa
^
( PNMT )
S Aden01 , ihomocy&tein
*
35
Epinephrine
37
38
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The above image illustrates the biochemical reaction whereby phenylalanine is converted
4
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(
dlculdior
( PNMT)
4
5
S Artsrios>1homocyjti;iiw *
G
£
9
to
it
n
13
14
15
IS
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13
H
20
2%
22
23
24
2S
25
27
23
29
30
31
32
33
34
35
36
37
33
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A
-
Epinephrine
7
*
Lab Values
Sent
The above image illustrates the biochemical reaction whereby phenylalanine is converted
to tyrosine by phenylalanine hydroxylase using tetrahydrobiopterin ( BH, ) as a cofactor .
Tyrosine is a non- essential amino acid that becomes essential in the setting of
phenylketonuria (PKU ). The next step in this pathway is the conversion of tyrosine to
DOPA catalyzed by the enzyme tyrosine hydroxylase which also uses BH, as a
cofactor . When there is deficiency of dihydrobiopterin reductase this reaction is
compromised (the phenylalanine hydroxylase reaction is also compromised but the
effects are unseen in this case because the patient was receiving tyrosine
supplementation , thus bypassing the impaired phenylalanine hydroxylase reaction ) .
Once DOPA is synthesized , it is decarboxylated to dopamine by the enzyme DOPA
decarboxylase Dopamine ultimately serves as the precursor molecule to the
catecholamines epinephrine and norepinephrine .
&
Deficiency of dihydrobiopterin reductase , the enzyme responsible for reduction of
dihydrobiopterin (BH. ) to BH, . is the most common cause for a deficiency of BHr This
results in what is known as atypical or malignant phenylketonuria Tetrahydrobioptenn is
a cofactor for enzymes that participate in the synthesis of tyrosine , DOPA. serotonin and
nitric oxide Under normal conditions dopamine from the tuberoinfundibular system
tonicatly inhibits the release of prolactin . Decreased 8H, causes decreased levels of
dopamine which therefore cause Increased levels of prolactin .
,
,
{ Choice A ) Phenylalanine hydroxylase is an enzyme that converts ingested
phenylalanine to tyrosine The above patient's phenylalanine levels have normalized
with diet therapy , but he has low dopamine levels . Patients with classic phenylketonuria
or phenylalanine hydroxylase deficiency do not have any metabolic errors producing
dopamine when their diet contains adequate amounts of tyrosine. Therefore this patient
,
V
11
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*
N o t e*
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(Choice A ) Phenylalanine hydroxylase is an enzyme that converts ingested
phenylalanine to tyrosine The above patient's phenylalanine levels have normalized
with diet therapy , but he has low dopamine levels. Patients with classic phenylketonuria
or phenylalanine hydroxylase deficiency , do not have any metabolic errors producing
dopamine when their diet contains adequate amounts of tyrosine . Therefore this patient
must have dihydrobioptenn reductase deficiency which even with phenylalanine
restriction and tyrosine supplementation, causes low dopamine levels ( see diagram) .
A
,
,
(Choice C ) Dopamine hydroxylase is the enzyme that catalyzes the biosynthesis of
norepinephrine from dopamine it does not use BH, as a cofactor .
( Choice D ) Alkaptonuria is an autosomal- recessive disorder caused by a deficiency of
the enzyme homogentisic acid oxidase , which normally breaks down homogentisic acid
( also called alkapton ) , a toxic tyrosine byproduct that is harmful to bones and cartilage .
b
( Choice E ) Phenyiethanolamine N- methyltransferase (PNMT ) converts norepinephnne
into epinephrine it requires S-adenosyi-methionine ( SAM ) as a cofactor
(Choice F ) Albinism is caused by defects in the biosynthesis and distribution of
melanin Melanin is synthesized in melanocytes from tyrosine by the enzyme tyrosinase .
27
28
29
30
31
32
33
34
3$
3&
Educational objective:
Tetrahydrobiopterin (BHJ is a cofactor used in the synthesis of tyrosine , DOPA ,
serotonin , and nitric oxide Initially , tyrosine is converted to DOPA by the enzyme
tyrosine hydroxylase, with BHt used as a cofactor . Next , DOPA is decarboxylated to
dopamine by the enzyme DOPA decarboxylase . In atypical phenylketonuria ( PKU ) with
tyrosine supplementation only the catecholamine synthesis reactions downstream of
tyrosine are compromised .
,
37
33
References:
"
i
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41
1.
Tyrosine ohenyjalanine .
,
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synthesis
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Notes
I <i l t uldlor
A
5
6
7
a
9
n
Autopsy of a 61- year - old homeless male who died in the ER shows foci of hemorrhage
and necrosis in the mamillary bodies and the gray matter surrounding the third and fourth
ventricles . This patient’s condition could have been diagnosed early by measuring which
of the following?
11
n
13
14
15
IS
17
ia
H
20
21
22
O A . Erythrocyte transketolase activity
O B. Erythrocyte glutathione reductase activity
C. Erythrocyte giucose-6-phosphate dehydrogenase activity
O D Serum NAD level
O E. Serum methylmalonic acid level
O F Blood protoporphyrin level
.
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ft
Autopsy of a 61' year- old homeless male who died in the ER shows foci of hemorrhage
and necrosis in the mamillary bodies and the gray matter surrounding the third and fourth
ventricles . This patient's condition could have been diagnosed early by measuring which
of the following?
11
12
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17
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20
21
*
•A . Erythrocyte transketolase activity [36%]
(
O B. Erythrocyte glutathione reductase activity [8%)
C Erythrocyte giucose-6-phosphate dehydrogenase activity [7%]
O D . Serum NAD level [24 %]
O E . Serum methylmalonic acid level [19% ]
O F . Blood protoporphyrin level [ 5% J
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Explanation:
Wernicke syndrome manifests with the triad of ophthalmoplegia ataxia , and confusion . It
is lethal in 10-20% of patients . Foci of hemorrhage and necrosis in the mamillary
bodies and penaqueductal gray matter are round on autopsy . This condition occurs due
to chronic thiamine deficiency , a condition common in patients with alcoholism .
,
Thiamine ( vitamin B 1 ) participates In a number of reactions of glucose metabolism It is a
cofactor for the following enzymes :
1. Pyruvate dehydrogenase converts pyruvate (the end- product of glycolysis ; into
acetyl CoA (which enters the citric acid cycle ).
2 . o -ketoglutarate dehydrogenase Is an enzyme of the citric acid cycle .
3. Transketolase is an enzyme of the hexose monophosphate pathway . It converts
pentoses ( derived from glucose ) to glyceraldehyde 3P ( an intermediary of
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Wernicke syndrome manifests with the triad of ophthalmoplegia , ataxia and confusion n
is lethal in 10^20% of patients Foci of hemorrhage and necrosis in the mamillary
bodies and periaqueductal gray matter are found on autopsy . This condition occurs due
to chronic thiamine deficiency , a condition common in patients with alcoholism ,
,
&
Thiamine ( vitamin B 1 } participates In a number of reactions of glucose metabolism . It is a
cofactor for the following enzymes :
1. Pyruvate dehydrogenase converts pyruvate (the end- product of glycolysis ) into
acetyl CoA (which enters the citric acid cycle ).
2 . o -ketoglutarate dehydrogenase is an enzyme of the citric acid cycle .
3 . TransketoJase is an enzyme of the hexose monophosphate pathway . It converts
pentoses (derived from glucose ) to glyceraidehyde 3P ( an intermediary of
glycolysis ).
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Thiamine deficiency , therefore , results In decreased glucose utilization , which is
especially pronounced in the CNS . If a patient with chronic thiamine deficiency Is given a
glucose infusion without thiamine supplementation , acute cerebral damage occurs An
increase in erythrocyte transketoJase levels after thiamine infusion is diagnostic for
thiamine deficiency. ( In actual practice , if a patient might be an alcoholic or appears to
be very malnourished , presume that the patient is thiamine deficient and give thiamine
supplementation with glucose infusion.)
{ Choices B and D ) Neither erythrocyte glutathione reductase nor NAD are used for the
diagnosis of thiamine deficiency.
(Choice C ) Erythrocyte giucose-6- phosphate dehydrogenase (G6PD ) catalyzes a ratelimiting step in the pentose phosphate pathway This pathway is necessary for NADPH
production and the function of the erythrocyte antioxidant system . Decreased glucose-6phosphale dehydrogenase levels cause hemolytic anemia
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glucose infusion without thiamine supplementation, acute cerebral damage occurs An
Increase in erythrocyte transketoJase levels after thiamine infusion is diagnostic for
thiamine deficiency , (in actual practice if a patient might be an alcoholic or appears to
be very malnourished , presume that the patient is thiamine deficient and give thiamine
supplementation with glucose infusion.)
&
(Choices B and D ) Neither erythrocyte glutathione reductase nor NAD are used for the
diagnosis of thiamine deficiency.
(Choice C ) Erythrocyte g!ucose -6 -phosphate dehydrogenase ( G6PD ) catalyzes a rate limiting step in the pentose phosphate pathway . This pathway is necessary for NADPH
production and the function of the erythrocyte antioxidant system Decreased glucose-6 -
phosphate dehydrogenase levels cause hemolytic anemia.
(Choice E ) Methylmalonic acid is a product of fatty acid oxidation. It is converted to
succinyl CoA by methylmalonyl CoA mutase This enzyme uses B12 as a
coenzyme Methylmalonic acid levels are increased in vitamin B12 deficiency
(Choice F) Protoporphyrin is one of the precursors of heme An increased erythrocyte
protoporphyrin concentration is the hallmark of erythropoietic protoporphyria (EPP );
however , this elevation is nonspecific and can be seen in other conditions such as iron deficient anemia and lead poisoning.
Educational Objective:
Chronic thiamine (B1 ) deficiency leads to the diminished ability of cerebral cells to utilize
glucose . The mechanism is decreased function of the enzymes that use vitamin B1 as a
cofactor (pyruvate dehydrogenase o -hetoglutarate dehydrogenase , and
transketolase ). Thiamine deficiency car be diagnosed by measuring erythrocyte
transketolase activity.
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DNA exonucleases hydrolytically remove one nucleotide at a time from the end of a DNA
chain Which of the following enzymes has 5' to 3' exonuclease activity?
( A . Helicase
O B Prlmase
,
O
L,
C Gyrase
C 0 DNA polymerase 111
(
E . DNA polymerase I
.
O F Ligase
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NA exonucleases hydrolytically remove one nucleotide at a time from the end of a DNA
chain Which of the following enzymes has 5' to 3* exonuclease activity?
9
10
O A . Heltcase [3%]
11
12
13
14
15
115
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O B . Primase [4%]
O C. Gyrase [3%J
I D . DNA polymerase 111126% ]
* It E . DNA polymerase J [56%]
O F . Llgase [5%|
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Explanation:
23
Topoisometases I and (I relieve coiling
tension by introducing txoak & to alto
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DNA polymerases are the primary enzymes responsible for DNA replication . In E . coif ,
there are three major DNA polymerases : DNA polymerase I II , and ML DNA replication
requires a high degree of fidelity in order to preserve the genetic code in daughter cells
and prevent potentially lethal mutations . This high fidelity replication is accomplished by
the 3’ to 5 ’ 'proofreading" exonuclease activity of DNA polymerase . ( Only DNA
polymerase I has 5 ' to 3 ' exonuclease activity ). This 5' to 3' exonuclease activity of DNA
polymerase I functions to remove the RNA primer ( 3 -hydroxyl group ) which is used by
NA polymerase III for the initiation of DNA replication . The 5 ' to 3 ' exonuclease activity
of DNA polymerase I also performs exonuclease excision and repair of damage to parent
DNA
( Choice AJ Before the process of DNA replication begins the parent strand of DNA
unwinds and dissociates secondary to the action of the enzyme helicase
(Choice C ) The enzymes topoisomerase I and II release tension caused during DNA
strand unwinding by relieving both negative and positive supercoils in eukaryotic
ceils The enzyme topoisomerase II , also known as DNA gyrase in prokaryotic cells , has
a sliohtlv different function than the eukaryotic toooisomerase Hand does not have 5 ' to 3'
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and prevent potentially lethal mutations This high fidelity replication is accomplished by
the 3' to 5‘ "proofreading * exonuclease activity of DNA polymerase . ( Only DNA
polymerase I has 5 ' to 3 ' exonuclease activity ). This 5 ' to 3' exonuclease activity of DNA
polymerase f functions to remove the RNA primer ( 3 -hydroxyl group ) which is used by
DNA polymerase HI for the initiation of DNA replication . The 5 ' to 3 ' exonuclease activity
of DNA polymerase I also performs exonuclease excision and repair of damage to parent
12
DNA .
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( Choice A ) Before the process of DNA replication begins the parent strand of DNA
unwinds and dissociates secondary to the action of the enzyme helicase .
L.v
( Choice C ) The enzymes topoisomerase I and It release tension caused during DNA
strand unwinding by relieving both negative and positive supercoils in eukaryotic
cells . The enzyme topoisomerase II. also known as DNAgyrase in prokaryotic cells , has
a slightly different function than the eukaryotic topoisomerase II and does not have 5 h to 31
exonuclease activity .
(Choice B ) Once a strand is unwound , DNA polymerases begin to synthesize
complementary strands in the 5 ' to 3 ' direction but they require a free 3’-hydroxyi group
primer (RNA primer } This free 3 -hydroxyl group is placed on the strand to be duplicated
by an RNA polymerase called primase .
,
(Choice F) The Okazaki fragments of the lagging strand are bound together by the
enzyme llgase
Educational Objective:
DNA polymerase I has 5 ' to 3' exonuclease activity in addition to Its 5 ' to 3' polymerase
and 3 ' to 5 ' exonuclease activities . This 5 ' to 3 exonuclease activity is used to remove
the RNA primer ( which initiates DNA polymerization ) and to remove damaged DNA .
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1 4i11 u l a t o r
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A 6- year -old Caucasian male experiences recurrent skin lesions on his face and upper
extremities that rapidly progress to cancer You suspect xeroderma pigmentosum as a
diagnosis . Which of the following enzymes is most likely nonfunctional if this is the
correct diagnosis?
12
A . 3 ' — S ' exonuclease
13
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O B. Endonuclease
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C. Topoisomerase
O D. DNA ligase
O E. Helicase
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5
6
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10
A
A 6- year -oid Caucasian male experiences recurrent skin lesions on his face and upper
extremities that rapidiy progress to cancer You suspect xeroderma pigmentosum as a
diagnosis . Which of the following enzymes is most likely nonfunctional if this is the
correct diagnosis?
11
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17
-
A . 3 ' — 5 ' exonuclease [31%]
v @ B. Endonuclease [58%]
O C . Topoisomerase [5%]
O D. DNA iigase [5% J
O E. Helicase [1%J
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Explanation:
5‘
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Ultraviolet light causes
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'
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Photolyase directly corrects 1iie
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DNA can be damaged by ultraviolet rays , which leads to the formation of thymine dimers
from two adjacent thymine residues These thymine dimers are repaired by several
mechanisms The most common defect that causes xeroderma pigmentosum is the
absence of UV -specific endonuclease This UV-specific endonuclease recognizes
distortions in the structure of DMA caused by thymine dimers, and subsequently excises
*
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/%
NA can be damaged by ultraviolet rays , which leads to the formation of thymine dimers
from two adjacent thymine residues These thymine dimers are repaired by several
mechanisms The most common defect that causes xeroderma pigmentosum Is the
absence of UV-specific endonuclease This UV- specIfic endonuclease recognizes
distortions in the structure of DNA caused by thymine dimers , and subsequently excises
stretches of single stranded DNA which contain these defects . The gap created
following this excision is then filled in by DNA polymerase which uses the opposite DNA
strand as a template . The new strand of DNA is then Joined on both ends to the existing
NA by the enzyme ligase
.
,
Remember that xeroderma pigmentosum ( XP ). like most enzymatic disorders is an
autosomal recessive disease Patients suffering from xeroderma pigmentosum ( XP )
exhibit photosensitivity poikiloderma , and hyperpigmentation in sun- exposed areas and
also possess a markedly increased nsk of developing skin cancers . There are at least
eight different forms of XPr and all involve defects in nucleotide excision repair .
,
,
(Choice A ) 3 ' to 5 exonuclease activity describes the "proofreading" ability of DNA
polymerase . This proofreading ability allows for the recognition and repair of mismatched
bases during DNA replication : defective repair of mismatched bases is associated with
hereditary nonpolyposis colon cancer .
(Choice C ) Topoisomerase enzymes relieve DNA supercoiling produced during
unwinding and separation by helicase Topoisomerase also known as gyrase in
prokaryotes is the target of the anticancer drug etoposide and the fluoroquinolone group
of antibiotics
(Choice D ) DNA ligase is responsible for creating a phosphodiester linkage between the
phosphate group of the 5 ' end of a DNA fragment and the hydroxyl group of the 3 ' end .
DNA ligase is particularly active in joining the numerous fragments of DNA that result
from discontinuous replication of the lagging strand .
V
11
41
*
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ONA by the enzyme ligase
7
Remember that xeroderma pigmentosum ( XP ) , like most enzymatic disorders is an
autosomal recessive disease Patients suffering from xeroderma pigmentosum (XP )
exhibit photosensitivity poikiloderma , and hyperpigmentation In sun-exposed areas and
also possess a markedly increased risk of developing skin cancers . There are at least
eight different forms of XPr and alJ involve defects In nucleotide excision repair.
£
9
10
11
12
13
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IS
IS
17
13
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20
21
22
23
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29
JUIUCU
*
Notes
(
alcuFator
tfAiMIMy
*
(Choice A ) 3 h to 5 exonuclease activity descnbes the "proofreading" ability of DNA
polymerase . This proofreading ability allows for the recognition and repair of mismatched
tr
bases dunng DNA replication ; defective repair of mismatched bases is associated with
hereditary nonpolyposis coion cancer .
(Choice C ) Topoisomerase enzymes relieve DNA supercoiling produced during
unwinding and separation by helicase Topoisomerase . also known as gyrase in
prokaryotes is the target of the anticancer drug etoposide and the fluoroquinolone group
of antibiotics
(Choice D ) DNA ligase is responsible for creating a phosphodiester linkage between the
phosphate group of the 5’ end of a ONA fragment and the hydroxyl group of the 3' end
DNA ligase is particularly active In joining the numerous fragments of DNA that result
from discontinuous replication of the lagging strand .
30
31
32
33
34
3S
35
37
33
(Choice E ) Helicases are responsible for unwinding and separating the double stranded
DNA into single stranded DNA in preparation for DNA replication.
Educational Objective:
UV-specrfic endonuclease deficiency is the most common cause of the autosomal
recessive disorder xeroderma pigmentosum
.
"
i
40
41
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Notes
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(
ulcuhitor
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a
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11
12
A research group is studying the pathogenesis of Alzheimer disease A protein isolated
from brain tissue of an affected patient has a conformation consisting of mostly
beta -pleated sheets A sample of a new medication is applied to the protein , and the
prevailing structure changes to an alpha -helical structure . This conformational change is
the result of reorganization of which of the following:
-
I
13
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17
13
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2Q
21
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O A . Ionic interactions
O B. Hydrophobic interactions
O C Peptide bonds
.
O 0.
Hydrogen bonds
E . Disulfide bonds
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24
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A
A research group is studying the pathogenesis of Alzheimer disease A protein isolated
from brain tissue of an affected patient has a conformation consisting of mostly
beta -pleated sheets A sample of a new medication is applied to the protein , and the
prevailing structure changes to an alpha -helical structure . This conformational change is
the result of reorganization of which of the following:
&
O A. Ionic interactions [3%]
B. Hydrophobic interactions [ 12% ]
O C. Peptide bonds [1Q%]
v <§) D . Hydrogen bonas [56 %]
O E. Disulfide bonds [19%]
22
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Explanation:
Primary
S
Slructure
Amino
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Explanation:
a
-
9
p^> 0-K>0-C)-CM^
^
^
10
Amino
acids
1?
13
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15
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Primary
Structure
linked by
peptide bonds
fe
17
a -Helix
n
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H
20
2\
r
22
23
24
25
25
21
29
30
Hydrogen bond
between every
:
4 amino add
l
Secondary
Structure
rr^TJ
-
1
-
L
31
32
33
34
35
Hydrogen bond
between all residues
of antiparallef strands
35
3?
38
"
i
40
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A protein' s primary structure is the
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of amino acids linked by covalent peptide
A
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7
a
9
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IS
17
13
H
20
21
22
23
24
25
25
27
29
30
31
32
33
34
3S
35
3?
30
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Newt
Lab Value
yroiiHporonerairongs
A
5
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Previous
Mdrk
*
Note
*
t atcufaior
/%
A protein's primary structure is the sequence or amino acids linked by covalent peptide
bonds ( Choice C ) . Proteins may also assume a secondary structure , such as the
alpha - helix or beta-sheet , due to subsequent hydrogen bonding . In patients with
Alzheimer disease beta- amyloid protein loses its alpha - helical configuration and forms
beta - sheets which are less soluble and therefore prone to aggregating. Aggregations of
beta - sheets are the primary component of the extracellular senile ( neuriticj plaques found
in Alzheimer patients .
I
The conversion of alpha - helices to beta - sheets involves the breaking and reforming of
hydrogen bonds ( Choice D) Tertiary structure is the overall shape that a single
polypeptide chain assumes following compact folding of the secondary structure Many
forces combine to stabilize the tertiary structure , including ionic bonds ( Choice A ) ,
hydrophobic interactions ( Choice 8) , hydrogen bonds ( Choice D) . and disulfide bonds
(Choice E ) . Remember that disulfide bonds are very strong covalent bonds between two
cysteine residues within the same polypeptide chain that enhance a protein s ability to
withstand denaturation
-
Educational objective:
Hydrogen bonds are the principal stabilizing force for the secondary structure of proteins .
References:
1. Membrane - mediated peptide conformation change from
alpha-monomers to beta-aggregates,
2 . Transformation of amyloid B ( 1 *40 ) oligomers into fibrils
by a major change in secondary structure.
is characterized
3 . Alpha -helix structure in Alzheimer ' s disease aggregates of tau-protein.
"
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Sent
Notes
(
alcufator
A
&
6
7
8
9
10
11
12
13
14
15
15
A 1-month-old full-term African -American boy is brought to the physician for a routine
check-up He is exclusively breastfed His mother consumes a well balanced diet , takes
no medications , and does not drink alcohol The boy is at the 56th percentile for weight
and the 60 th percentile for height His physical examination is normal . This infant
requires supplementation with which of the following vitamins ?
-
C A . Folic acid
O B Iron
17
C. Riboflavin
13
H
20
21
22
23
i)
,
(
0. Thiamine
O E. Vitamin A
C F . Vitamin D
24
2S
25
27
23
30
31
32
33
34
35
35
37
38
33
li
11
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Lab Value
*
Notes
I akulaior
4
5
6
7
a
9
n
it
12
ft
A 1-month-old full-term African -American boy is brought to the physician for a routine
check -up He is exclusively breastfed His mother consumes a well-balanced diet , takes
no medications , and does not drink alcohol The boy is at the 56th percentile for weight
and the 60 th percentile for height . His physical examination is normal . This infant
requires supplementation with which of the following vitamins ?
n
u
O A . Folic acid [5%J
O 3. Iron [15%1
O C. Riboflavin [3%J
15
15
17
13
H
O D. Thiamine [3%]
20
O E. Vitamin A [ 5%)
21
22
23
24
25
E>
*
* F.
itamin C [ 69%l
Explanation:
25
27
23
30
31
32
33
34
35
35
37
33
Breast milk is the gold standard of infant nutrition: it contains proteins , carbohydrates,
fats vitamins trace minerals, immunoglobulins amino acids and enzymes All vitamins
and trace minerals are present in adequate amounts except vitamins D and K Vitamin
K is generally supplemented by an intramuscular injection at delivery to prevent
hemorrhagic disease in the newborn Regular sunlight exposure typically provides
sufficient amounts of vitamin . However , infants are generally shielded from direct
sunlight due to sunburn risk . In addition infants with dark skin pigmentation require more
sunlight exposure to produce adequate vitamin D Melanin is a natural sunblock and
prevents ultraviolet rays from reaching the shin for vitamin D synthesis Therefore all
exclusively breastfed infants should receive vitamin D supplementation to prevent rickets .
,
"
i
40
n
^Choices A
.
V
R C and D 1 Human breast milk normally nrovides adequate amounts of
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Explanation:
Lab Value
W
*
Note
*
£ alculdtor
User U
A
6
e
7
s
9
10
11
12
13
14
IS
IS
17
Breast milk is the gold standard of infant nutrition: it contains proteins , carbohydrates
fats , vitamins trace minerals , immunoglobulins , amino acids and enzymes All vitamins
and trace minerals are present in adequate amounts except vitamins D and K . Vitamin
K Is generally supplemented by an intramuscular injection at delivery to prevent
hemorrhagic disease in the newborn. Regular sunlight exposure typically provides
sufficient amounts of vitamin D . However , infants are generally shielded from direct
sunlight due to sunburn risk . In addition , infants with dark skin pigmentation require more
sunlight exposure to produce adequate vitamin 0. Melanin is a natural sunblock and
prevents ultraviolet rays from reaching the skin for vitamin D synthesis . Therefore all
exclusively breastfed infants should receive vitamin supplementation to prevent rickets.
ti
13
14
20
21
22
23
24
2S
26
27
23
30
31
32
33
34
35
36
37
38
39
11
41
{ Choices A, B. C , and D) Human breast milk normally provides adequate amounts of
folic acid, iron , riboflavin , and thiamine for full-term infants .
(Choice E ) Vitamin A stores in the liver are low at birth but rapidly increase due to the
large amount in colostrum and breast milk .
Educational objective:
The breast milk content of vitamins D and K is typically insufficient for the nutritional
needs of the newborn Vitamin K is given parenterally at birth to prevent hemorrhagic
disease in the newborn Exclusively breastfed infants may develop vitamin D deficiency
due to lack of sunlight exposure ,
References:
t . Prevalence and risk factors for vitamin D deficiency among healthy
infants and young children in Sacramento California
.
.
.
2. Adherence to vitamin D recommendations among US infants
3. Use of supplemental vitamin d among infants breastfed for prolonged
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Nent
*
Notes
(
dkuldior
A
5
6
7
a
9
10
11
12
13
14
15
IS
17
U
A 4 - year -old boy is brought to the physician for fatigue and persistent bone pain .
Physical examination shows diffuse lymphadenopathy and multiple purpura over his arms
and legs . Laboratory analysis reveals anemia and thrombocytopenia , and a peripheral
blood smear shows lymphoblasts After further workup , he is diagnosed with acute
lymphoblastic leukemia and started on a chemotherapy regimen that includes
doxorubicin. This agent intercalates between DNA base pairs and inhibits DNA
replication, a process that normally occurs at sites known as replication forks As the
replication forks move across the DNA molecule 2 distinct daughter strands are formed
Which of the following is unique to the daughter strand that is synthesized in the opposite
direction of the growing replication fork ?
k
14
20
21
C
A . Synthesis of multiple , short DNA fragments
22
B. 5' -+3’ exonuclease activity of DNA polymerase
21
C. 3' +5' exonuclease activity of DNA polymerase
n
-
25
25
D. 3’^5’ polymerase activity of DNA polymerase
27
E. RNA primer synthesis before DNA strand synthesis
28
29
31
32
33
34
3$
3&
37
38
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Ldb Value
Newt
*
Notes
I alculdtor
A
6
7
8
9
to
11
n
13
14
IS
115
1?
U
r\
A 4-year -old boy is brought to the physician for fatigue and persistent bone pain.
Physical examination shows diffuse lymphadenopathy and multiple purpura over his arms
and legs. Laboratory analysis reveals anemia and thrombocytopenia , and a peripheral
blood smear shows lymphoblasts After further workup , he is diagnosed with acute
lymphoblastic leukemia and started on a chemotherapy regimen that includes
doxorubicin. This agent intercalates between DNA base pairs and inhibits DNA
replication, a process that normally occurs at sites known as replication forks As the
replication forks move across the DNA molecule 2 distinct daughter strands are formed
Which of the following Is unique to the daughter strand that is synthesized in the opposite
direction of the growing replication fork?
14
20
21
22
23
n
25
25
27
23
29
I
* @ A. Synthesis of multiple, short DNA fragments [79%]
C B. 5' ^3’ exonuclease activity of DNA polymerase ( 5%)
C . 3‘— 5' exonuclease activity of DNA polymerase [ 8%]
D , 3’-* 5 polymerase activity of DNA polymerase [3%]
C E. RNA primer synthesis before DNA strand synthesis [ 5%]
-
'
Explanation:
31
32
33
34
DNA replication fork
35
3'
5'
35
37
38
39
40
n
DNA
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Leading strand
DNA polymerase
A
A
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8
9
10
Explanation:
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6
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Lab values
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(Notes
t dlcufdior
(JUiyiMtJI die [ OVoj
/%
D 3’ 5 ' polymerase activity of DNA polymerase [3%]
.
primer synthesis before DNA strand synthesis [5%]
n
12
DNA replication fork
13
U
U
IS
IS
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y
17
13
H
DNA
helicase jf
20
21
m-
S’
Leading strand
.fir
ONA polymerase
22
23
24
2&
25
Movement of replication fork
%
27
RNA primer
23
29
Prima. se
i
31
32
33
34
35
36
37
33
39
40
41
Lagging strand
Single stranded
DNA binding protein
© UWorld
Okazaki fragments
DNA replication begins at multiple sites within eukaryotic chromosomes called origins of
replication . At these sites the parent DNA double helix is separated and unwound In a
process facilitated by the heilcase enzyme and single-stranded ONA- binding
,
nrnleinc
Th &
Inralinnq where
Block Time Remaining:
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unwound HNA moete the non- ^oneraterl Honhle hnliir aro
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i Mdrk
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Notes
I alculdior
A
5
G
7
a
9
10
11
12
13
14
IS
15
17
13
H
20
21
22
21
24
2&
25
27
28
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31
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34
35
35
37
38
A
DMA replication begins at multiple sites within eukaryotic chromosomes called origins of
replication. At these sites the parent DNA double helix is separated and unwound in a
process facilitated by the heiicase enzyme and single -stranded ONA -bindlng
proteins The locations where unwound DNA meets the non- separated double helix are
known as replication forks . Replication forks travel bidirectionally away from the origin of
replication as DNA polymerase synthesizes complementary daughter DNA strands .
Synthesis of the daughter strands occurs simultaneously from both parent
strands . Because DNA synthesis can occur only in the 5r * 3' direction one
daughter strand is synthesized continuously toward the replication fork (leading
strand ) . However, the other strand must be synthesized discontinuousJy in a
direction away from the replication fork ( lagging strand) with more and more segments
being added as the replication forK moves across the DNA double helix . This results in
the formation of Okazaki fragments , short stretches of newly synthesized DNA that are
separated by RNA pnmers These primers are removed and replaced with DNA and the
Okazaki fragments are subsequently joined together by DNA ligase.
-
>
b
(Choices B and C ) 5 — >3' and 3 — ^ 5 ' exonuclease activity is needed during the synthesis
of both daughter strands The 5 »3 ' exonuclease activity removes RNA primers as well
as damaged DNA segments. The 3^5' exonuclease activity performs a proofreading
function that removes and replaces mismatched nucleotides on the newly formed
daughter strands
r
-
f
{ Choice D ) DNA polymerases do not have 3’— > 5’ polymerase activity .
(Choice E ) Before DNA polymerase can initiate DNA synthesis RNA primers must first
be synthesized by the enzyme primase (DNA- dependent RNA polymerase ) Only one
primer is needed for synthesis of each leading strand , but the lagging strand requires
synthesis of many RNA primers.
"
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13
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&
Lab Valuer
strand I, However , the other strand must be synthesized discontinuous in a
direction away from the replication fork (lagging strand ) with more and more segments
being added as the replication fork moves across the DNA double helix . This results in
the formation of Okazaki fragments , short stretches of newly synthesized DNA that are
separated by RNA primers These primers are removed and replaced with DNA . and the
Okazaki fragments are subsequently joined together by DNA ligase .
^
( Choices B and C ) 5H-> 3 ' and 3 ' — > 5' exonuclease activity is needed during the synthesis
of both daughter strands The 5 f 3 ' exonuclease activity removes RNA primers as well
as damaged DNA segments The 3‘ * 5 exonuclease activity performs a proofreading
function that removes and replaces mismatched nucleotides on the newly formed
-
f
daughter strands .
23
24
25
synthesis of many RNA primers.
25
27
23
29
31
32
33
34
Educational objective:
DNA synthesis can occur only in the 5' — 3’ direction. Okazaki fragments are short
stretches of newly synthesized DNA that are separated by RNA primers . They are
formed by the discontinuous synthesis of DNA on the lagging strand during replication .
-
References:
1. Reconstitution of eukaryotic lagging strand DNA replication.
35
3S
37
33
39
40
41
k
( Choice D ) DNA polymerases do not have 3’-* 5 ' polymerase activity .
( Choice E ) Before DNA polymerase can initiate DNA synthesis . RNA primers must first
be synthesized by the enzyme pnmase (DNA-dependent RNA polymerase ) . Only one
primer is needed for synthesis of each leading strand , but the lagging strand requires
22
I alculdtor
Notes
2 . Timing , coordination, and rhythm: acrobatics at the DNA replication fork.
Tlmim
O ** ** ** * ** ~fl **
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A
b
6
7
a
9
n
11
12
13
U
15
IS
1?
1B
19
20
21
22
23
2i
2b
25
27
23
29
*
Notes
I <alcufdtor
A 6 -month-old boy is brought to the office by his mother out of concern that he is not
developing normally He has been feeding regularly and has had no medical problems
other than a mild respiratory infection a month earlier However, the mother says , "he
doesn't seem to be as interactive as my other children were at his age * Physical
examination reveals delayed developmental milestones and hypotonia Two years later
the child is found to have involuntary movements and demonstrates a tendency to
aggressively bite his own lips and fingers. Laboratory analysis shows an elevated blood
uric acid level Activity of which of the following enzymes is most likely increased as a
result of this patient's condition?
A. Aspartate carbamoyttransferase
O B. Dlhydroorotase
C. Hypoxanthine-guanine phosphoribosyltransferase
O. Phosphoribosyl pyrophosphate amidotransferase
O E Ribonucleotide reductase
O F Thymidylate synthase
,
.
30
32
33
34
35
35
37
33
39
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Notes
t alculdtor
4
5
6
7
8
9
10
11
12
13
U
15
«
1
17
ft
A 6-month-old boy is brought to the office by his mother out of concern that he is not
developing normally He has been feeding regularly and has had no medical problems
other than a mild respiratory infection a month earlier However, the mother says , "he
doesn't seem to be as interactive as my other children were at his age .' Physical
examination reveals delayed developmental milestones and hypotonia Two years later
the child is found to have involuntary movements and demonstrates a tendency to
aggressively bite his own lips and fingers Laboratory analysis shows an elevated blood
uric acid level Activity of which of the following enzymes is most likely increased as a
result of this patient's condition?
1
U
A . Aspartate carbamoyltransferase [4%]
H
20
21
n
23
24
25
25
27
23
29
30
O B. Dihydroorotase [3%]
C Hypoxanthine-guanine phosphoribosyttransferase [45% ]
D. ^ hosphonbosyl pyrophosphate amiootransfenise [39%]
O E. Ribonucleotide reductase [4%]
O F. Thymidylate synthase [4%]
* *
Explanation:
32
33
34
De novo purine synthesis
35
Ribose 5 ‘phosphate
35
I
37
33
"
i
40
41
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PRPP
synthetase
V
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Lob Values
SCKt
Notes
t ole ultilor
4
A
&
6
Ribose 5 phosphate
-
I
7
a
9
10
11
PRPP
tynthetase
PRPP
12
13
H
15
15
AMP
IMP
GMP
17
*
PRPP
amidotransferase
13
5 Phosphoribosylanrtine
H
-
20
21
Nitrogen sou fee
22
21
2
*
25
Carbon donor
Glycine
Aspartate
Tetrahydrofolate
Glutamine
25
27
28
29
Inosine monophosphate
(IMP)
30
32
33
34
"
ADP
35
35
37
38
39
l
© UWorld
v
Y
i
V
GMP
GDP
AMP
i
41
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I M V J H I L
4
&
6
a
A DP
9
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11
13
H
15
15
17
13
H
20
21
22
21
24
25
25
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30
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35
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1
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Lab Valur
SCKt
' V i l «* .
W
V
t
*
Notes
I dkuldlor
A
GDP
AMP
GMP
OUWarld
K• <
<
Mdrk
( IMP)
"
7
12
H I W I I W
'
-
This patient has Lesch Nyhan syndrome , an X -ImKed recessive disorder characterized
by the development of dystonia , choreoathetosis , self mutilation and hyperuricemia
within the first few years of life . The condition is caused by deficiency of
hypoxanthine guanine phosphoribosyltransferase (HGPRT ) , an enzyme that
normally functions in the purine salvage pathway to convert hypoxanthine back to
inosine monophosphate and guanine back into guanosine monophosphate. The
absence of HGPRT results in increased degradation of guanine and hypoxanthine bases
into uric acid, which increases the demand for de novo purine synthesis .
-
-
The first step of purine synthesis is the formation of phosphoribosyl pyrophosphate
( PRPP ) by PRPP synthetase PRPP can be used by adenine phosphoribosyltransferase
and HGPRT for purine salvage , or it can be converted to phosphoribosylamine by PRPP
amidotransferase in the first committed step of de novo purine synthesis Because
purine salvage is impaired in Lesch -Nyhan syndrome , the activity of PRPP
amidotransferase must increase to supply a sufficient quantity of purine nucleotides
.
(Choices A B , and F) These enzymes are involved in pyrimidine synthesis Their
activity would not be significantly altered in Lesch-Nyhan syndrome , which impairs purine
salvage.
(Choice C ) The activity of HGPRT is decreased (not increased ) in Lesch-Nyhan
syndrome.
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Notes
l alculator
/T
J
/ I
A
normally functions in the purine salvage pathway to convert hypoxanthine back to
inosine monophosphate and guanine back into guanosine monophosphate. The
absence of HGPRT results in increased degradation of guanine and hypoxanthine bases
Into uric acid , which increases the demand for de novo purine synthesis .
The first step of purine synthesis is the formation of phosphorlbosyl pyrophosphate
( PRPP ) by PRPP synthetase PRPP can be used by adenine phosphonbosyitransferase
and HGPRT for purine salvage , or rt can be converted to phosphoribosyiamine by PRPP
amidotransferase in the first committed step of de novo purine synthesis Because
purine salvage is impaired in Lesch-Nyhan syndrome , the activity of PRPP
amidotransferase must increase to supply a sufficient quantity of purine nucleotides .
.
( Choices A B . and F) These enzymes are involved in pyrimidine synthesis Their
activity would not be significantly altered in Lesch-Nyhan syndrome which impairs purine
U
salvage .
(Choice C ) The activity of HGPRT is decreased (not increased ) in Lesch-Nyhan
syndrome ,
25
25
27
28
29
30
32
33
34
35
35
37
33
39
40
41
(Choice E ) Purine and pyrimidine nucleotides are initially synthesized as bases attached
to a ribose sugar . The enzyme ribonucleotide reductase converts ribose sugars to their
deoxyribose forms for use in DNA synthesis This enzyme is negatively regulated by
Increased levels of deoxyribose nucleotides
.
Educational objective:
Lesch-Nyhan syndrome is an X-linkeo recessive disorder caused by a defect in
hypoxanthlne-guanme phosphoribosyltransferase ( HGPRT ). This resuits in failure of the
purine salvage pathway , leading to increased degradation of hypoxanthine and guanine
to uric acid . De novo purine synthesis must increase to replace the lost bases .
Time Spent 4 seconds
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2 A : $4
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Last updated [ 12 /3 / 2015 ]
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Lab Value
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dlcufdtor
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n
A 22-year -old Caucasian male is being evaluated for skin lesions on his lower abdomen .
A sample of his fibroblasts is obtained and cultured as part of the investigative work -up
The cultured fibroblasts fail to metabolize ceramide trihexoside . This patient is at
greatest risk for developing which of the following?
11
12
13
U
15
IS
17
13
19
20
O A. Pulmonary emphysema
C B. Hepatic cirrhosis
C. Spastic tetraparesis
b
O D Acute leukemia
E. Renal failure
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I alculdtor
4
5
6
7
a
9
n
ft
A 22- year-old Caucasian male is being evaluated for shin lesions on his lower abdomen .
A sample of his fibroblasts is obtained and cultured as part of the Investigative work-up .
The cultured fibroblasts fail to metabolize ceramide trihexoside This patient is at
greatest risk for developing which of the following?
it
12
13
U
15
IS
17
13
19
20
A . Pulmonary emphysema [ 11% ]
O B . Hepatic cirrhosis [31%J
C C. Spastic tetraparesis [16%]
C D . Acute leukemia [6% J
* m E. Reral failure [35%]
b
zt
22
Explanation:
24
Z&
25
Z7
23
29
This patient has Fabry disease iangiokeratoma corporis diffusum ) , which results from an
Inherited deficiency of alpha -galactosidase A . In patients with Fabry disease the
globoside ceramide trihexoside accumulates in tissues . The earliest disease
manifestations are hypohidrosis acroparesthesia and angiokeratomas Acroparesthesia
is episodic , often debilitating burning neuropathic pain in the extremities
Angiokeratomas are punctuate , dark red , non-blanching macules and papules that
classically occur between the umbilicus and the knees Without enzyme replacement
therapy , progressive renal insufficiency leading to renal failure and death may occur
30
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33
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40
n
(Choices A and B ) Alpha - 1 - antitrypsin is a serine protease inhibitor that inactivates the
enzymes elastase and trypsin . Complete deficiency of alpha-1- antitrypsin causes
panactnar pulmonary emphysema and liver cirrhosis
( Choice C ) Spasticity is a sign of upper motor neuron pathology . Spastic tetraparesis
t
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i
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*
Note
*
I alculdior
r\
Explanation:
This patient has Fabry disease ( angiokeratoma corporis dlffusum ) , which results from an
inherited deficiency of alpha -galactosidase A In patients with Fabry disease the
globoside ceramide trihexoside accumulates in tissues . The earliest disease
manifestations are hypohidrosis acroparesthesia and angiokeratomas . Acroparesthesia
is episodic , often debilitating , burning neuropathic pain in the extremities.
Angiokeratomas are punctuate dark red non- blanching macules and papules that
classically occur between the umbilicus and the knees . Without enzyme replacement
therapy , progressive renal Insufficiency leading to renal failure and death may occur .
,
Alpha - 1 - antitrypsin is a serine protease inhibitor that inactivates the
enzymes elastase and trypsin Complete deficiency of aipha- 1- antitrypsin causes
panacinar pulmonary emphysema and liver cintiosis .
( Choices A and B )
L-
(Choice C ) Spasticity is a sign of upper motor neuron pathology Spastic tetraparesis
may be seen in cerebral palsy and fucosldosis Spasticity is also a manifestation of
Niemann-Pick disease In Lesch- Nyhan syndrome , patients exhibit spasticity ,
choreoathetoid movements and self-mutilation
(Choice D ) Patients with Down syndrome Patau syndrome , ataxia- telangiectasia , Bloom
syndrome and Fanconi anemia are at increased risk for developing acute leukemia
Educational Objective:
Fabry disease is an inherited deficiency of alpha-galactosidase A that causes
accumulation of the globoside ceramide trihexoside in tissues . The earliest
manifestations of Fabry disease are angiokeratomas , hypohidrosis and acroparesthesia .
Without enzyme replacement , patients typically develop progressive renal failure .
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Lab Values
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Notes
1 olcubnor
4
5
6
7
a
9
n
11
12
13
U
15
is
17
18
14
2Q
21
22
23
n
A newborn develops vomiting, irritability, and lethargy several days after birth . His
mother states that his diapers smell like "burned sugar ' Laboratory studies confirm the
diagnosis , and the patient is started on the appropriate dietary restrictions with
subsequent improvement in his symptoms The defective enzyme responsible for this
patient's condition normally catalyzes a reaction involving which of the following
substances?
k
O A . Galactocerebroside
O B. Pyridoxlne
O C. Thiamine
O D. Folic acid
O E. Tyrosine
F . Tetrahydrobiopterin
25
25
21
2H
29
30
31
32
34
35
35
37
38
"
i
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11
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t alculdtor
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i
8
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n
11
12
13
A newborn develops vomiting, irritability, and lethargy several days after birth . His
mother states that his diapers smell like 'burned sugar ' Laboratory studies confirm the
diagnosis , and the patient is started on the appropriate dietary restrictions with
subsequent improvement in his symptoms The defective enzyme responsible for this
patient's condition normally catalyzes a reaction involving which of the following
substances?
n
15
15
17
U
14
20
21
22
21
24
25
25
A. Galactocerebroside [10%]
I.
O B. Pyridoxine [16%]
C. Thiamine [36%]
O 0. Folic acid [1%]
O E. Tyrosine [21%J
F . Tetrahydroblopterin [16%]
**
Explanation:
27
21
29
30
31
32
Pftenyial^irw
Acetyl CoA
1
35
35
Leoono
Oipda /'vi?
cfovarboxyidii jn
'
iBwTChvd c /Mun a- Xeto
BCttf dehydrogenase )
Tyrosine
I
14
—
-
i
* Fumarate
TCA cycle
37
38
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Lab Values
Notes
Calculator
fWJ
A
h
5
F . Tetrahydrobiopterin [16%]
I
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NCKt
A
6
7
8
9
Explanation:
10
Phenylalanine
11
n
1J
14
15
16
17
1j
Acetyl CoA
1
\
i
Levcrne
Ot «ial >ve dacartoxyiahon
( Branched chain a- keto
acid fchydrugenas ( ‘ •
T yrosme
E
* F-
uTiarait}
TCA Cycle
:
14
20
2\
n
21
24
25
25
Succmyi CoA
T
T
27
23
29
act d tiGhydrogenasa )
Proponyl CoA
30
31
32
34
35
36
37
38
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40
41
Oxrdafn^ dec&rtoxyiahon
( B h iched chain a-ktfo
' '
Mctnyloidionyi Co A
+-
1
Valine
Isoleuone
This infant is most likely suffering from maple syrup urine disease (MSUD), a disorder
characterized by the defective breakdown of branched chain amino acids (leucine .
Isoleucine and valine Degradation of these amino acids first involves transamination to
^
their respective Q-ketoacids . which are subsequently metabolized by an enzyme complex
referred to as branched- cham a -ketoacld dehydrogenase . MSUD can result from
mutations in any of the 4 genes coding for the 3 catalytic subunits of this
complex. Neurotoxicity results primarily from the accumulation of leucine in the serum
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i
Lab value
Newt
*
Notes
t a l l ufdlor
4
5
e
7
8
9
TO
11
12
13
H
15
IS
17
13
H
20
21
22
23
2t
25
2S
27
28
29
30
31
32
34
3S
36
/%
This infant is most likely suffering from maple syrup urine disease (MSUD ) . a disorder
characterized by the defective breakdown of branched chain amino acids (leucine ,
isoleucine , and valine ). Degradation of these amino acids first involves transamination to
their respective a-ketoacids . which are subsequently metabolized by an enzyme complex
referred to as branched-chain o -ketoacid dehydrogenase . MSUD can result from
mutations in any of the 4 genes coding for the 3 catalytic subunits of this
complex. Neurotoxicity results primaniy from the accumulation of leucine in the serum
and tissues . A metabolite of isoleucine gives the urine of affected infants a distinctive
sweet odor much like burned caramel. MSUD can be life -threatening if untreated, but
dietary restriction of branched chain amino acids can lessen the seventy of symptoms
Branched-chain a -ketoacid dehydrogenase pyruvate dehydrogenase , anda ketoglutarate dehydrogenase all require five cofactors Thiamine pyrophosphate
Lipoate , Coenzyme A , FAD NAD (mnemonic : Tender Loving Care For Nancy ) , Some
patients with MSUD improve with high-dose thiamine treatment (thiamine-responsive ) ,
but most still require lifelong dietary restrictions.
(Choice A ) Galactocerebrosidase catalyzes the liposomal hydrolysis of
galactocerebroside a galactolipid that is found in abundance in myelin Krabbe disease
( globoid cell leukodystrophy ) is a rare autosomal recessive disorder caused by deficiency
of this enzyme The infantile form of this disease typically manifests between 2-5 months
of age with irritability, developmental delay or regression and muscle tone abnormalities
(Choice B ) Pyricoxine ( vitamin B6 ) as pyridoxal phosphate is involved in the
transamination and decarboxylation steps in amino acid metabolism as well as heme and
neurotransmitter synthesis . Pyridoxine supplementation is used in the treatment of
sideroblastic anemia and hyperhomocysteinemla .
37
33
39
40
41
(Choice D ) Folic acid, frequently deficient in alcoholics , is responsible for the transfer of
single carbon moieties during nucleic acid synthesis . Folic acid supplementation is used
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NPKt
Lab Value
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*
I alculdtor
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but most still require lifelong dietary restrictions
{ Choice A ) Galactocerebrosidase catalyzes the liposomal hydrolysis of
galactocerebroside a gaiactohpid that is found in abundance in myelin Krabbe disease
( globoid cell leukodystrophy ) is a rare autosomal recessive disorder caused by deficiency
of this enzyme The infantile form of this disease typically manifests between 2- 5 months
of age with irritability , developmental delay or regression and muscle tone abnormalities
{ Choice B ) Pyridoxine ( vitamin B6 ) as pyridoxal phosphate is involved in the
transamination and decarboxylation steps in amino acid metabolism as well as heme and
neurotransmrtter synthesis . Pyndoxine supplementation is used in the treatment of
sideroblastic anemia and hyperhomocysteinemia .
{ Choice D ) Folic acid , frequently deficient in alcoholics is responsible for the transfer of
single carbon moieties during nucleic acid synthesis Folic acid supplementation is used
for the treatment of hyperhomocysteinemla and in the prevention of neural tube defects in
newborns
,
n
25
25
27
23
29
30
31
32
34
35
3S
37
33
39
11
41
{ Choices E and F) A variant of phenylketonuria is due to deficiency of
tetrabydrobiopterin a cofactor for phenylalanine hydroxyfase , the enzyme that converts
phenylalanine to tyrosine Tetrahydrobiopterin supplementation can reduce
phenylalanine levels In these particular patients.
Educational objective:
Branched-chain o -ketoacid dehydrogenase similar to pyruvate and a -ketoglutarate
dehydrogenase requires several coenzymes : Thiamine pyrophosphate, Lipoate,
Coenzyme A FAD . NAD (mnemonic Tender Loving Care For Nancyi Some patients
with maple syrup urine disease improve with high- dose thiamine treatment ( thiamineresponslve ), but most still require lifelong dietary restnctions.
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(
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&
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10
11
A 2- year-old male demonstrates periodic lethargy , vomiting , and confusion Laboratory
testing reveals increased blood ammonium levels during these episodes as well as
markedly increased urine orotic acid excretion Which of the following enzymes is most
likely to be deficient In this patient?
,
,
12
A , Carbamoyl phosphate synthase I
13
U
IS
IS
17
ia
H
20
O B . Ornithine transcarbamoylase
O C . N- acetyiglutamate synthase
v.
D . Hypoxanthine-guanine phosphoribosyltransferase
C E Adenosine deaminase
,
21
22
23
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2&
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*
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33
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*
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4
5
6
7
a
9
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11
12
13
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15
IS
17
ia
14
20
A
A 2- year-oid male demonstrates periodic lethargy vomiting , and confusion Laboratory
testing reveals increased blood ammonium levels during these episodes as well as
markedly increased urine orotic acid excretion . Which of the following enzymes is most
likely to be deficient In this patient?
,
L
C A . Carbamoyl phosphate synthase I [19%]
v (@ B. Ornithine transcarbamoylase [67%]
O C . N- acetylglutamate synthase [4 %]
D Hypoxanthine -guamne phosphonbosyltransferase [6% ]
O E . Adenosine deaminase [2%]
21
22
Explanation:
21
24
25
25
27
2H
29
30
31
32
33
*
-
35
N- aceiytgiuiamate
+
*
,CilruiNFW
CO . * ISIH,‘* 2 ATP
Caftamoyt
phosphate
synthetase f
f
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trsnscsrti& rnoyiasQ
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*
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37
*
-
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38
39
‘ 40
- 41
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Ornithine
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Notes
I <ilc uLftior
4
5
6
/%
Explanation:
7
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it
12
13
N‘ACo( yigiLrtamato
U
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17
18
+
*
.
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Carbamoyl
Curull ne
JrnffWM ,
Arr ninosucctnal®
phosphate
synthetase I
Aapoitdte
^
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*
Carbamoyl phosphate
han jea/
19
hr« s#
*
*
Arginine
20
21
23
24
25
25
noso
^
Ornithine
A
22
Ornithine
*
Urea
VITOCHONDRIA
27
23
29
30
31
32
33
*
*
-
*
-
35
35
37
38
39
40
41
Ammonia generated from the metabolism of alpha amino acids is converted into urea by
the urea cycle . The urea cycle involves five enzymatic steps , two in mitochondrial matrix ,
and three in the cytosol The combination of CO . , ammonia, and ATP , catalyzed by
carbamoyl phosphate synthase (the rate -limiting enzyme in the urea cycle), forms
carbamoyl phosphate as the first step of the urea cycle . Carbamoyl phosphate then
combines with ornithine to form cltrulline in a reaction catalyzed by ornithine
transcarbamoylase in the mitochondrial matrix . Citrulline then enters the cytosol and is
converted to argininosuccinate. which is then converted to arginine. The conversion of
arginine to ornithine by the cytosolic enzyme arginase completes the urea cycle by
releasing a urea molecule .
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4
5
6
7
8
9
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17
13
H
20
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22
23
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Ldb Value
Sent
*
Notes
l dlculdlor
carbamoyl phosphate synthase (the rate- limiting enzyme in the urea cycle ), forms
carbamoyl phosphate as the first step of the urea cycle . Carbamoyl phosphate then
combines with ornithine to form citrulline in a reaction catalyzed by ornithine
transcarbamoylase in the mitochondrial matrix Citrulline then enters the cytosol and is
converted to argininosuccinate which is then converted to arginine . The conversion of
arginine to ornithine by the cytosolic enzyme arginase completes the urea cycle by
releasing a urea molecule .
Remember that urea synthesis is a cyclic process and that while ammonium ion. CO
ATP . and aspartate are consumed in this process , there is no net loss or gain of
ornithine citrulline , argininosuccinate , or arginine . The molecule N- acetylglutamate
serves as a regulator of the urea cycle through allosteric activation of carbamoyl
phosphate synthetase I.
,
t
Disorders of the urea cycle can result from defects in any of the following six enzymes
1. Carbamoyl phosphate synthetase ( CPS )
2. Ornithine transcarbamoylase (OTC )
3 . Argininosuccinic acid synthetase (AS )
4 . Argininosuccinic acid lyase (AL )
5. Arginase (AG )
6 . N-Acetylglutamate synthetase (NAGS)
30
31
32
33
*
-
35
35
37
<
-
.
-
38
39
40
41
The first five enzymes are directly involved in the urea cycle whereas the sixth enzyme is
involved in the production of N- acetylglutamate , the allosteric activator of carbamoyl
phosphate synthase I . Patients with urea cyde disorders display clinical symptoms oF
neurological damage secondary to increased serum ammonia levels Typically , patients
present early in childhood , however milder defects can present for the first time during
adulthood OTC deficiency is the most common urea cycle disorder , resulting in
increased levels of carbamoyl phosphate and impaired disposal of ammonia ,
V
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Lab Valuer
Notes
I olc uhnor
5 . Argirase ( AG )
6 . N-Acetylglutamate synthetase ( NAGS )
A
The first five enzymes are directly involved in the urea cycle whereas the sixth enzyme is
involved in the production of N- acetylglutamate . the allosteric activator of carbamoyl
phosphate synthase I . Patients with urea cycle disorders display clinical symptoms of
neurological damage secondary to Increased serum ammonia levels Typically , patients
present early in childhood, however , milder defects can present for the first time during
adulthood OTC deficiency is the most common urea cycle disorder, resulting in
increased levels of carbamoyl phosphate and impaired disposal of ammonia .
(Choices A and C ) CPS and NAGS defects result in increased blood levels of ammonia
and neurological disorders , but low levels of carbamoyl phosphate and no elevation in
urinary orotic acid (Remember that in OTC deficiency , accumulated carbamoyl
phosphate is converted into orotic acid ).
(Choice D ) Deficiency of HPRT (Lesch- Nyhan syndrome ) results in excessive uric acid
production because purines cannot be salvaged from degraded DNA . The clinical
manifestations of HPRT deficiency include hyperuricemia urate kidney stones
self-mutilation , and involuntary movements
( Choice £ ) Adenosine deaminase (ADA ) is an enzyme involved in purine metabolism .
Decreased expression of ADA causes severe combined immune deficiency ( SCID ) as the
accumulation of adenosine is toxic to lymphocytes .
Educational Objective:
Ornithine transcarbamoyiase deficiency is the most common disorder of the urea cycle
resulting in severe neurological abnormalities due to high blood and tissue ammonia
levels. Increased urine orotic acid excretion Is typical.
,
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&
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a
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11
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13
u
is
is
An infant born to a 23-year -old female develops vomiting and lethargy several days after
birth Physical examination reveals hypertonicity and muscle rigidity The mother also
notices an odor of burnt sugar in her diapers Which of the following amino acids should
most likely be restricted In this infant s diet?
A . Phenylalanine
O B . Tyrosine
17
O C. Leucine
13
O 0. Methionine
H
20
21
22
O E. Histidine
O F. Lysine
21
24
2S
25
27
23
29
30
31
32
33
34
*
36
37
*
-
.
*
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Lab Values
Notes
(
alculaior
4
5
6
7
8
9
n
A
An infant born to a 23- year-old female develops vomiting and lethargy several days after
birth Physical examination reveals hypertonicity and muscle rigidity The mother also
notices an odor of burnt sugar in her diapers Which of the following amino acids should
most likely be restricted in this infant 's diet?
k
11
n
A . Phenylalanine [21 %]
13
U
15
IS
17
13
H
20
21
O B. Tyrosine [5%J
* # C . Leucine [64 %I
O D Methionine ( 4 %)
O B . Histidine [1% J
C F . Lysine [3% ]
22
23
24
25
25
27
23
29
Explanation:
Phenylalanine
ACfMyl CoA
i
Oxidative dtfCurtoxyiQtion
( Branched chatn a-keto
aud dehydrttgenaw )
Tyros- ne
30
31
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33
34
»
-
*
* Fumarate
TCA cycle
35
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‘ 40
41
-
Leucine
Succmyi CoA
I
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(
dkuldtor
A
Phenyiaiarirne
Acetyl CoA
i
Tytosme
12
I
U
Lab Valuer
Sent
Explanation:
9
10
11
13
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1 Mift
tf
Lflucuw
Qxtdahrt d&cdrtoxytahon
Ifranked cftflm o
'rf dehydrogenase;
dt
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i
- + F-
jnnarate
TCA cycle
19
19
b
17
U
Succnyl CoA
14
T
20
21
n
r
23
24
29
acid dahydrogaroie \
Propionyl CoA
25
27
2a
29
30
31
32
33
34
»
*
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37
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39
40
41
Oxidative dec&rboxytet /on
tBrw&ied diem ato
Melnylrriafunyl CdA
*
\
Valmu
Isoltjcne
The infant described in the question is most likely suffering from maple syrup urine
disease a disorder characterized by defective breakdown of the branched chain amino
acids leucine isoleucine and valine . The specific defect of maple syrup urine disease
occurs in the enzyme branched chain a -keto acid dehydrogenase Because their
degradation is inhibited at the a-keto acid stage, tissue and serum levels of these
branched chain d-ketd acids increase , which leads to neurotoxicity . Maple syrup urine
disease usually manifests within the first few days of life , and classically, the urine of
affected infants has a distinctive sweet odor much like burned caramel. Maple syrup
urine disease can be life threatening if left untreated , but dietary restriction of branched
chain amino acids , such as leucine, can lessen the severity of symptoms .
,
,
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i
Lab Value
Scut
*
Note
*
(
ukuldtor
The infant described in the question is most likely suffering from maple syrup urine
disease , a disorder characterized by defective breakdown of the branched chain amino
acids leucine , isoleucine , and valine . The specific defect of maple syrup urine disease
occurs in the enzyme branched chain cr-keto acid dehydrogenase Because their
degradation is inhibited at the a - keto acid stage , tissue and serum levels of these
branched chain a -keto acids increase , which leads to neurotoxicity . Maple syrup urine
disease usually manifests within the first few days of life , and classically the urine of
affected infants has a distinctive sweet odor, much like burned caramel. Maple syrup
urine disease can be life threatening if left untreated but dietary restriction of branched
chain amino acids , such as leucine, can lessen the severity of symptoms
A
,
{ Choice A) The ammo acid phenylalanine is restricted in patients with phenylketonuria .
U
H
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2&
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35
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38
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*
39
40
41
(Choice B ) Defective breakdown of tyrosine results in hypertyrosinemia or
alkaptonuria A diet low in tyrosine and phenylalanine can be of benefit for the treatment
of alkaptonuria and a certain forms of hypertyrosinemia.
(Choices D, E and F) Hypermethioninemia typically a benign disorder , results from
defective metabolism of methionine by the enzyme methionine
adenosyitransferase. Similarly , most enzymatic defects resulting in hyperiysinemia and
histidinemia are also benign disorders . Methionine restriction and cysteine
supplementation is required for treatment of classic homocystlnurla .
Educational Objective:
Maple syrup urine disease ( MSUD ) is caused by a defect in a-keto acid dehydrogenase,
leading to an inability to degrade branched chain amino acids beyond their deaminated a ketc acid state This illness classically results in dystonia and poor feeding as well as
the maple syrup scent" of the patient ' s urine within the first few days of life Treatment
rests on dietary restriction of branched-chain amino acids.
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Notes
(
ulcufdtor
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9
to
11
12
13
H
IS
IS
17
U
H
Experiments have shown that the tRNA molecule with the UCU anticodon can effectively
bind to both AGA and AGG mRNA codons The finding described above is best referred
to as:
C A . Transition
O B. Ambiguity
I
C C. Universality
D No punctuation
O E. Wobble
20
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*
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O . Id
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Lab Values
(Notes
(
alculator
4
&
6
7
a
9
Experiments have shown that the tRNA molecule with the UCU anticodon can effectively
bind to both AGA and AGG mRNA codons The finding described above is best referred
to as:
n
11
12
13
U
15
15
17
ia
19
20
21
22
21
24
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34
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-
.
-
37
38
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40
41
O A . Transition [3%]
O B. Ambiguity [12%]
O C. Universality [7%l
C D. No punctuation [0%]
® E.
Aobbie [77% )
Explanation:
There are 61 codons that code for amino acids , but only 20 amino acids used in protein
synthesis . The genetic code is thus considered "degenerate* because more than one
codon can code for a particular amino acid For instance , the codons GGU , GGC , GGA
and GGG all correspond to the amino acid glycine
Individual tRNA molecules are specific for certain amino acids and recognize the mRNA
codons corresponding to that amino acid Because of the degeneracy of the code ,
certain tRNA molecules can recognize multiple different codons coding for the same
amino acid a phenomenon explained by the ’wobble * hypothesis . According to this
hypothesis the 5 ' base in the tRNA anticodon often has a different spatial orientation
than the other two bases The 5' nucleotide of the anticodon may be inosine , a
nucleotide not found in mRNA Inosine can form hydrogen bonds with three bases
uracil, adenine , and cytosine. In the case of glycine , for example , one tRNA molecule
recognizes 3 codons ( GGU . GGC and GGG ) because only two base pairs
( corresponding to the GG of the codon ) form "traditional* bonds .
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O C. Universality [7%[
D. No punctuation [0%]
# E . Vobble [77%]
(
H I
M!
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Lab Values
Newt
Notes
(
alculaior
f c,
10
11
Explanation:
u
There are 61 codons that code for ammo acids , but only 20 amino acids used in protein
synthesis . The genetic code is thus considered "degenerate" because more than one
codon can code for a particular amino acid For instance , the codons GGU , GGC . GGA
and GGG all correspond to the amino acid glycine.
n
14
15
15
17
13
H
20
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22
21
2
25
25
*
27
23
29
30
31
32
33
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35
37
•
.-
-
38
39
40
41
Individual tRNA molecules are specific for certain amino acids and recognize the mRNA
codons corresponding to that amino acid Because of the degeneracy of the code ,
certain tRNA molecules can recognize multiple different codons coding for the same
amino acid, a phenomenon explained by the "wobble " hypothesis . According to this
hypothesis the 5 base in the tRNA anticodon often has a different spatial orientation
than the other two bases The 5’ nucleotide of the anticodon may be inosine , a
nucleotide not found in mRNA . Inosine can form hydrogen bonds with three bases
uracil , adenine , and cytosine In the case of glycine, for example one tRNA molecule
recognizes 3 codons ( GGU , GGC and GGG ) because only two base pairs
( corresponding to the GG of the codon) form "traditional" bonds .
f
Educational Objective:
The genetic code is "degenerate." meaning that there are more codons < 61) than amino
acids ( 20). Each tRNA molecule is specific for a given amino acid. Many tRNA
anticodons can bind to a few different codons coding for the same amino acid . This is
called the "wobble" phenomenon.
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Newt
*
Notes
(
dlculdtor
4
5
6
7
a
9
Some proteins that participate in bacterial DNA synthesis have specific exonuclease
activity Which of the following is the best statement about the 3' to 5 ' exonuclease
activity of DNA polymerase III?
n
11
12
13
Q A . It cuts DNA at specific DNA sequences
17
13
H
&
B. It nicks the DNA strands that have formed thymidine dimers
H
15
15
C. It removes an improper base - pair nucleotide during replication
OD
It cleaves DNA strands to relax positive supercoils
; ) E. It can remove groups of nucleotides ( up to ten ) at a time
'
20
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33
34
3S
35
*
38
.
-
39
40
41
-
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Sent
Notes
I akulator
4
r\
5
6
7
a
9
Some proteins that participate in bacterial DNA synthesis have specific exonuclease
activity Which of the following is the best statement about the 3' to 5 ' exonuclease
activity of DNA polymerase III?
n
11
12
13
H
15
15
17
13
*
O A . It cuts DNA at specific DNA sequences [5%]
C B . It nicks the DNA strands that have formed thymidine dimers [7%]
(ft C . t removes an improper base -pair rucleot de during replication [77%]
O
0 . It cleaves DNA strands to relax positive supercoils [4 %]
O E . It can remove groups of nucleotides ( up to ten ) at a time [7%]
H
20
21
Explanation:
22
23
24
Prokaryotic cells
Removal of suporcoils
Toporsomeras® II
Unwinding of double helix
He' ic se
30
Stabilization of unwound template strands
( bind only lo single- stranded ONA )
31
32
33
34
proteins ( SSB )
Synthesis of RNA primer
Pnmase
25
25
27
28
29
3S
35
^
( RNA polymerase )
ONA synihesis
Leading strand
ONA porymerase III
ONA polymerase 111
ng strand
*
.-
-
Single strand binding
38
39
40
41
Removal of RNA pnmer A
of RNA with DNA
ONA polymerase I
replacement
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Notes
t die ultftlor
4
5
r%
Explanation:
6
7
Prokaryotic cells
a
9
Removal of
Topoisofncrase tl
sujwcoiis
10
11
12
13
n
15
IS
17
Unwinding of double fieri *
Heiicase
Slaoiiizaliofl of unwound temple strands
slrorxtcd ONA )
proteins ( SSB )
(bind only to smgte
Pnmase
Synthesis of RNA primer
{RNA polymerase )
13
H
20
21
Single strand landing
ON A synihesis
Leading strand
Lagging strand
ONA polymerase 111
DNA polymerase III
22
23
2
*
25
Removal ol RNA pnm©r &
replacement of RNA wilh DNA
( Proof reading )
ONA polymerase I
( 5' - e onucJeaae]
*
25
27
23
29
Joining of Okazaki fragments
t agging strand )
ONA ligaso
30
31
32
33
34
3S
35
•
38
-
39
40
41
.-
During the process of cell division. DNA replication occurs secondary to the coordinated
effects of multiple enzymes and proteins. DNA polymerases are the primary enzymes
responsible for DNA replication , but they can not function without the assistance of other
enzymes such as primase , helicase, ligase . and topoisomerase I and If. In E. coli, there
are three primary types of DNA polymerases : DNA polymerase I. II, and III. Pnmase forms
the 3’ OH group primer to initiate replication of daughter strands while helicase promotes
unwinding and dissociation of the parent strands. On the other hand , topoisomerases
reduce positive and negative supercoilinq in order to relieve the strain produced by DNA
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Next
responsible for DNA replication , but they can not function without the assistance of other
enzymes such as primase , helicase ligase . and topoisomerase I and II In E. coli , there
are three primary types of DNA polymerases DNA polymerase I II . and III . Primase forms
the 3’ OH group primer to initiate replication of daughter strands while helicase promotes
unwinding and dissociation of the parent strands . On the other hand , topoisomerases
reduce positive and negative supercoiling in order to relieve the strain produced by DNA
unwinding .
*
Calculator
Notes
A
,
I
DNA replication requires a high degree of fidelity ; therefore as synthesis of the daughter
strands proceeds . DNA polymerases proof read to ensure that the daughter DNA Is the
exact complement of the parent DNA . All three prokaryotic DNA polymerases have proof
reading activity and remove mismatched nucleotides via a 3hto 5 exonuclease activity
( Choice C ) . Only DNA polymerase I has 5 ' to 3 ' exonuclease activity which is used to
excise and replace RNA primers and damaged DNA sequences , which are identified by
endonucleases ( Choice E ).
-
,
( Choice A ) In contrast to exonucleases, which remove nucleotides from the end of a
DNA molecule , endonucleases cut ONA at very specific DNA sequences within the
molecule . Restriction endonucleases digest DNA into smaller fragments in a
sequence- specific manner
(Choice B ) One of the major methods of DNA damage by ultraviolet light is the
dimenzation of adjacent pyrimidine bases to form thymidine dimers These dimers are
routinely formed after exposure to sunlight , but are usually removed by protective
enzymatic mechanisms.
Educational Objective:
All three prokaryotic DNA polymerases have proof reading activity and remove
mismatched nucleotides via 3 to 5 ' exonuclease activity Only DNA polymerase I has 5'
to 3 ' exonuclease activity which is used to excise and replace RNA primers and damaged
DNA sequences
r
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Newt
Notes
t dkuldtsr
4
5
6
7
8
9
10
11
12
A newborn experiences lethargy , vomiting , and hypotonia during the first few days of life .
Laboratory examination reveals a metabolic acidosis with a large anion gap , ketosis and
hypoglycemia The concentration of propionic acid is markedly increased in the plasma
and urine . Metabolism of which of the following amino acids contributes to this patient ' s
condition?
u
14
15
16
17
13
H
20
21
22
23
24
2S
25
C A . Phenylalanine
O B Valine
,
C C Asparagine
C
0 Histidine
O E . Proline
O F. lysine
27
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33
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*
Notes
(
dlculdtor
4
5
6
i
8
9
n
11
n
ft
A newborn experiences lethargy , vomiting , and hypotonia during the first few days of life .
Laboratory examination reveals a metabolic acidosis with a large anion gap , ketosis and
hypoglycemia The concentration of propionic acid is markedly increased in the plasma
and urine . Metabolism of which of the following amino acids contributes to this patient s
condition?
13
u
15
15
1?
n
H
20
2\
22
21
24
25
25
27
A . Phenylalanine [26%]
V
•B
. anne [36%]
O C . Asparagine [13%]
O D. Histidine [5%]
O E. Proline [5%]
O F , Lysine [14%J
Explanation:
Phenylalanine
2H
Acetyl CoA
29
30
31
32
33
34
3S
-
39
40
41
Oxidative d&cdrboxytdtton
lBranched chain a heto
aad dehydrogenase )
Tyrosine
t
l
i
> Fumarate
35
37
-
t
Leudne
TCA cycle
V
J
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Ldb Values
Notes
t die uftilor
A
Explanation:
Phenylalanine
a
9
10
it
i Mdrk
Acetyl CoA
Leucine
Oxidates decarboxylation
( Branched chain o - keto
acid dehydrogenase )
Tyrosine
12
13
i
u
f Fumarate
15
15
17
13
19
TCA cycle
20
Succirryl CoA
21
22
23
Methyimakmyr Vftfito B 12
CoA iscmofaso
n
25
25
Methylmalonyi CoA
27
23
29
30
Propiony }
CoA
carboxylase
31
32
33
34
Threonine
Methionine
35
35
37
-
39
40
41
Oxidative decarboxylation
( Branched chain p - ketp
Bl 0(tn
add dehydrogenase )
Valine
Isoleudne
PropionyI CoA
Catabolism of isoleucine . valine , threonine , methionine , cholesterol , and odd- chain fatty
acids leads to the formation of propionic acid , which is then converted to methylmalonic
acid by biotin- dependent carboxylation . Isomerization of methylmaionyi CoA forms
#
1
4* « »
i * hjrK h
i r^A
*
*
'
*K
*ln
*
if
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Propionyi CoA
carboxylase
n
11
n
13
H
IS
IS
17
13
Threonire
Biotin
i
Lab Valor
*
Notes
l iilculdtor
add dehydrogenase )
I
Propionyi CoA
Methionine
o
Sent
Oxidative decarboxylation
( Branched chain o- keto
7
£
9
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Previous
Valine
Isoleudne
Catabolism of isoleucine . valine , threonine , methionine cholesterol , and odd- chain fatty
acids leads to the formation of propionic acid , which is then converted to methylmalonic
acid by biotin- depenaent carboxylation Isomerization of methylmalonyl CoA forms
succinyl CoA . which then enters the TCA cycle , A congenital deficiency of propionyi CoA
carboxylase , the enzyme responsible for the conversion of propionyi CoA to
methylmalonyl CoA leads to the development of propionic acidemia , as propionyi CoA
accumulates Propionic acidemia is clinically characterized by poor feeding vomiting
hypotonia lethargy dehydration , and an anion gap acidosis Propionic acid is the
intermediate in the catabolism of branched chain amino acids , such as valine , and is not
produced during the catabolism of the other amino acids listed .
,
,
20
2\
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23
n
2&
25
?7
23
29
30
31
32
33
34
35
35
37
-
39
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41
,
(Choice A ) Phenylalanine is converted to tyrosine by the enzyme phenylalanine
hydroxylase , which is defective in phenylketonuria (PKU ).
(Choice C ) Asparagine Is a nonessential amino acid that Is catabolized initially to
aspartate by the enzyme asparaginase In rapidly dividing leukemic cells the synthesis
of asparagine is impaired so these cells survive by collecting asparagine from circulating
plasma . L-asparaginase works as an antineoplastic agent by lowering circulating
asparagine levels.
( Choice D ) Histidine is an essential amino acid in children . Histidine is deaminated to
urocanlc acid , which is then converted to N- formlnlno-glutamate ( FIGIu ) . The formlnlno
group of FIGIu Is donated to tetrahydrofolate to form glutamate and
form inino- tetrahydrofolate . Oxidative decarboxylation of histidine forms histamine, which
is released by mast cells in Type I Hypersensitivity reactions.
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A
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6
6
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a
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n
it
12
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15
17
1B
H
20
21
22
21
24
2S
25
77
23
29
30
31
32
33
34
35
*
-
*
36
37
33
39
40
41
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produced during the catabolism of the other amino acids listed .
IML IINEUIDLTI in
me LdLduuirsm
Lab Value
NCKt
.
LM uiemuieu UICJMI dimnu duu:» , buf ii di> vdiine. diiu i:> IILH
*
Note
*
I dkuldior
A
(Choice A ) Phenylalanine is converted to tyrosine by the enzyme phenylalanine
hydroxylase , which is defective in phenylketonuria (PKU).
(Choice C ) Asparagine Is a nonessential amino acid that is catabolized initially to
aspartate by the enzyme asparaginase In rapidly dividing leukemic cells the synthesis
of asparagine is impaired, so these cells survive by collecting asparagine from circulating
plasma L-asparaginase works as an antineoplastic agent by lowering circulating
asparagine levels.
(Choice D ) Histidine is an essential amino acid in children . Histidine is deaminated to
urocanic acid , which is then converted to N-forminino -glutamate ( FiGlu . The fomninino
group of FiGlu is donated to tetrahydrofolate to form glutamate and
form inino-tetrahydrofolate . Oxidative decarboxylation of histidine forms histamine which
is released by mast cells in Type I Hypersensitivity reactions.
>
(Choice E ) Proline is a nonessential amino acid that Is oxidized to glutamate , which in
turn is transaminated to alpha -ketoglutarate,
(Choice F) Lysine is an essential amino acid that is strictly ketogenic . The metabolism of
lysine is unique compared with the other amino acids , as it is not transaminated as an
Initial step
.
Educational Objective:
Propionyl CoA is derived from amino acids (Vai, He Met. and Thr ) , odd-numbered fatty
acids and cholesterol side chams Congenital deficiency of propionyl CoA carboxylase
the enzyme responsible for the conversion of propionyl CoA to methylmalonyl CoA . leads
to the development of propionic acidemia.
.
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Notes
(
ole ufdtor
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it
12
A research scientist is studying biochemical reactions that take place in the liver . He
cultures hepatocyles in a growth media enriched with glutamate labeled with nitrogen
isotopes . After sometime , he finds that the nitrogen isotopes are transferred to
oxaloacetate . forming aspartate in the process Which of the following substances Is
most likely involved in this reaction?
u
13
u
IS
15
17
18
H
20
21
22
23
24
25
25
77
23
29
O A . Biotin
O B Folic acid
O C Niacin
.
0 Pyridoxine
E Riboflavin
OF
Thiamine
30
31
32
33
34
35
3&
37
38
.
-
40
41
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Notes
I dlculdior
A
5
6
7
a
9
10
11
12
A
A research scientist is studying biochemical reactions that take pface in the liver. He
cultures hepatocyles in a growth media enriched with glutamate labeled with nitrogen
isotopes . After sometime , he finds that the nitrogen isotopes are transferred to
oxaloacetate, forming aspartate in the process Which of the following substances is
most likely involved in this reaction?
13
u
O A. Biotin [21%]
15
16
O B. Folic acid [3%]
17
O C. Niacin [11%]
13
20
21
22
21
24
25
* ® D. Pyridoxine [47%]
O E. Riboflavin [ 6%]
O F. Thiamine [13%]
Explanation:
26
27
GkJcote
2H
29
Pnoldin
30
31
32
33
34
35
36
37
Pyruvate
TCA
Trarttamlnatloo
( Vitamin
CfCl
o-KelooKitarate
Gh/ t&mntv
40
*
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38
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Acciyl CoA
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Notes
(
alculdtor
4
&
6
Explanation:
7
a
Glucose
9
Protwn
10
»
11
12
13
14
15
15
17
f
Alanine
Pyruvate
Acetyt CoA
4
*
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TrinsanmniHon
( Vitamin M)
-
a - Kologkjtarate
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19
Glutamate
20
dehydrogenase
21
TrlitUrnlrllllArt
( Vilimkn B &|
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22
23
24
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25
25
27
2H
29
30
31
32
33
34
3S
35
37
30
.
*
40
41
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I olcuhilor
A
&
6
7
a
9
n
11
12
13
14
IS
15
17
13
19
20
21
22
23
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2&
25
27
28
29
30
31
32
33
34
3S
35
37
33
.
-
40
41
A
Pyridoxine ( vitamin B. ) is necessary for the transamination and decarboxylation of
amino acids , for gluconeogenesis . and for other essential biochemical
processes . Transamination reactions typically occur between an amino acid and an
a keto acid . The amino group is transferred to the a -keto acid from the amino acid , and
the a -keto acid thereby becomes an amino acid. For example, glutamate ( amino acid)
reacts with oxaloacetate (a-keto acid) to form aspartate (the resulting amino acid ) and a ketoglutarate (the resulting a-keto acid ).
-
Transaminases ( aminotransferases) are the enzymes that catalyze transamination
reactions and pyridoxal phosphate ( active vitamin BJ serves as an essential cofactor for
the transaminase
(Choice A ) Biotin ( vitamin B.) is a cofactor for all 4 carboxylase enzymes : pyruvate
carboxylase , acetyl-CoA carboxylase , propionyl- CoA carboxylase , and 3 -methylcrotonyl-
CoA-carboxylase.
(Choice B ) Folic acid (vitamin Bj is an essential cofactor in nucleic acid synthesis and
a deficiency of either folate or vitamin B results in megaloblastic anemia
.
(Choice C ) Many dehydrogenases use NAD+ and NADP+ which are formed from
niacin. Niacin (vitamin 8 .. or nicotinic acid) deficiency is known as pellagra and is
classically associated with the 4 Ds: dermatitis dementia , diarrhea and , if untreated ,
death .
(Choice E ) Riboflavin (vitamin B . ) is used in dehydrogenase reactions involving FMN
and FAD.
(Choice F) Thiamine ( vitamin B . ) serves as a coenzyme for a number of important
dehydrogenase enzymes, including transketoiase, a -ketogiutarate dehydrogenase , and
pyruvate dehydrogenase .
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*
Notes
I olcuhilor
4
&
6
7
8
9
10
11
12
1J
U
is
is
1?
13
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20
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22
23
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2&
25
27
23
29
30
31
32
33
34
3$
35
rs
( Choice B ) Folic acid (vitamin B . ) is an essential cofactor in nucleic acid synthesis and
a deficiency of either folate or vitamin EL results in megaloblastic anemia .
,
(Choice C ) Many dehydrogenases use NAD+ and NADP + . which are formed from
niacin Niacin ( vitamin B . , or nicotinic acid } deficiency is known as pellagra and is
classically associated with the 4 Ds: dermatitis dementia , diarrhea and , If untreated ,
death
,
( Choice E ) Riboflavin (vitamin B:) is used in dehydrogenase reactions involving FMN
and FAD
.
(Choice F) Thiamine ( vitamin B , ) serves as a coenzyme for a number of important
dehydrogenase enzymes, including transkelolase , o-ketoglutarate dehydrogenase , and
pyruvate dehydrogenase
.
Educational objective:
Transamination reactions typically occur between an amino acid and an a- keto acid . The
ammo group from the amino acid is transferred to the a -keto acid and the a - keto acid in
turn becomes an amino acid . Pyridoxal phosphate ( active vitamin B. ) serves as a
cofactor in amino acid transamination and decarboxylation reactions
References:
1. Review: the role of coenzymes in clinical enzymology .
2. An appreciation of Professor Alexander E . Braunstein . The discovery
and scope of enzymatic transamination.
3 . Aspartate aminotransferase: an old dog teaches new tricks.
37
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41
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4
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6
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9
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11
n
13
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15
IS
17
13
19
20
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23
n
25
25
27
A 5-month-old boy is brought to the office by his parents who are concerned that he has
developmental delay He has 3 older siblings and the parents report that his siblings
were much more active and interactive at the boy' s age. The family has just moved to
the United Slates from South America and did not have consistent primary care
previously Physical examination shows that the boy is unable to roll from front to bach or
back to front and does not seem to recognize his parents. Comprehensive laboratory
evaluation reveals impaired tetrahydrobiopterin synthesis Which of the following is most
likely deficient in this patient?
O A . Acetylcholine
B. Gamma - aminobutyric
'
i
acid
O C. Glutamate
O 0 . Glycine
O E. Phenylalanine
C F Serotonin
.
28
29
30
31
32
33
34
3S
36
37
33
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40
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4
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6
7
a
9
10
11
12
13
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is
115
A
A 5-month-old boy is brought to the office by his parents who are concerned that he has
developmental delay He has 3 older siblings , and the parents report that his siblings
were much more active and interactive at the boy ' s age . The family has just moved to
the United States from South America and did not have consistent pnmary care
previously . Physical examination shows that the boy is unable to roll from front to back or
back to front and does not seem to recognize his parents . Comprehensive laboratory
evaluation reveals impaired tetrahydrobiopterin synthesis Which of the following is most
likely deficient in this patient?
17
OA
13
Acetylcholine [6%]
B . Gamma - aminobutyric acid [ t 0%]
20
O C. Glutamate [14%J
O D. Glycine [5%]
O E. Phenylalanine [28%]
21
22
21
24
2&
25
,
b
V
m F. Serotonin [36%]
27
23
29
Explanation:
30
31
32
33
34
3S
Dtbydfopien&ne
Oinyftoeton&ne
reductase
reductase
BH4
35
BH4
BHI
BH2
Melanin
37
33
39
i
-
41
Tyrosine
Phenylalanine
DOPA
Phenytaian\ne
Tyrosine
V
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E . Phenylalanine [ 28 % ]
<§ F . Serotonin [36% ]
*
Note
*
t dlculdlor
7
a
9
Explanation:
10
11
12
13
Otny&fOQten&n?
Dtnydroptertftme
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reductase
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115
BH*
BH4
BH ?
Bhh
Melanin
17
13
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Phenylalanine
20
« DOPA
+ Tyrosine
Phenyt$ter )\ ne
Tyrosme
hydroxylase
hydroxylase
21
Catecholamines
22
23
24
Dthydroplendina
25
reductase
25
27
BH4
23
29
30
31
32
33
34
35
35
37
38
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41
BHi
Tryptophan
5 hydroxy tryptophan
-
Typioohan
hydroxylase
5 hydroxytryptamine
(Serotonin)
-
©EWorld
Tetrah ydrob topterin (8HJ is a cofactor in the synthesis of serotonin ( a major
neurotransmitter ) , tyrosine ( a precursor of DOPA ) , and DOPA ( the antecedent of the
neurotransmitters dooamine . norebineDhrine and epinebhrine ) . Dihydrobiobterin
Block Time Remaininq :
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(Notes
(
aiculalor
4
&
6
7
a
9
10
11
12
13
U
IS
15
17
U
14
20
21
22
23
24
2&
25
27
23
29
30
31
32
33
34
35
35
37
33
39
-
ii
41
A
©UWoHd
Tetrahydrobiopterin ( SHJ is a cofactor in the synthesis of serotonin ( a major
neurotransmitter ) , tyrosine { a precursor of DOPA ) , and DOPA (the antecedent of the
neurotransmitters dopamine , norepinephrine and epinephrine ). Dihydrobiopterin
reductase enzymatically reduces dihydrobiopterin (BH.) to tetrahydrobiopterin (BHJ.
Serotonin ( 5-hydroxytryptamine , or 5HT ) is formed through hydroxylation and
decarboxylation of the amino acid tryptophan .
b
Most cases of phenylketonuria are due to phenylalanine hydroxylase deficiency . Less
commonly , the etiology is due to BHt deficiency secondary to dihydropterjdine
reductase deficiency The consequences of defective phenylalanine and tryptophan
metabolism are phenylalanine accumulation ( Choice E ) and low levels of serotonin
and other neurotransmitters respectively.
,
The combination of high phenylalanine levels which may disrupt neuronal and glial
development and low serotonin and other neurotransmitters results in progressive
neurologic deterioration in untreated patients. Manifestations include developmental
delay hypotonia , dystonia and seizures . Treatment Includes both a low phenylalanine
diet and BH, supplementation .
,
{ Choice A ) Acetylcholine is a neuromuscular junction neurotransmitter synthesized from
choline and acetyl-CoA by choline acetyltransferase In myasthenia gravis production of
antibodies against acetylcholine receptors leads to muscle weakness
{ Choice B ) Gamma- aminobutyric acid ( GABA ) is an inhibitor of presynaptic transmission
in the central nervous system that is formed by glutamate decarboxylation a reaction
catalyzed by glutamate decarboxylase Phenobarbital an antiepileptic medication
potentiates GABA activity to decrease or cease seizure activity .
,
,
/ fihnirp Cl Glytamat« K
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{ Choice B ) Gamma - aminobutyric acid ( GABA ) is an inhibitor of pre synaptic transmission
in the central nervous system that is formed by glutamate decarboxylation a reaction
A
catalyzed by glutamate decarboxylase Phenobarbital an antiepileptic medication ,
potentiates GABA activity to decrease or cease seizure activity .
9
10
11
12
13
14
IS
IS
17
13
H
20
21
22
23
24
26
25
27
23
29
30
31
32
33
34
3$
36
(Choice C ) Glutamate is an excitatory neurotransmitter that is synthesized from
glutamine by the enzyme glutaminase Ketamine is an N-methyl-D- aspartate ( NMDA )
receptor noncompetitive antagonist that blocks glutamate , which can result in anesthesia ,
sedation , and memory loss.
(Choice D ) Glycine is an amino acid and inhibitory neurotransmitter synthesized from
serine , with particular influence in the spinal cord Deficiency or mutation in glycine
receptors leads to hyperekplexia a disorder characterized by hypertonia and an
exaggerated startle response ,
Educational objective:
Tetrahydrobiopterln (BH, ) Is a cofactor used by hydroxylase enzymes In the synthesis of
tyrosine dopamine , and serotonin . Phenylketonuria can result from BHt deficiency due
to dihydropteridine reductase deficiency Intellectual disability is the hallmark of this
condition and results in neurotransmitter ( eg , serotonin) deficiency and
hyperphenylalanemia . Treatment involves a low phenylalanine diet and BHt
supplementation .
References:
1. Diagnosis , classification , and genetics of phenylketonuria and
tetrahydrobiopterin ( BH4 ) deficiencies.
2 . Disorders of Tetrahydrobiopterin Metabolism and their Treatment
37
38
39
-
41
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A
b
6
r
a
9
1C
11
12
13
U
15
15
17
Ij
H
20
21
22
21
24
A 57-year -oio man comes to the office for a follow -up appointment . He has a history of
systolic heart failure which has been managed with appropnate medical therapy The
patient experiences significant functional impairment at his baseline and is able to walk
only short distances His most recent echocardiogram showed a left ventricular ejection
fraction of 30% (normal >55% ). The physician decides to start him on
spironolactone The addition of this medication to the patient s current regimen is most
likely to cause a decrease in which of the following renal functions ?
A . Hydrogen ion generation by the proximal convoluted tubule
B. Hydrogen Ion secretion from the collecting tubules
C . Na7K-/2Ch cotransport in the thick ascending limb
D Proximal convoluted tubule brush border transport capacity
C E. Urea reabsorption in the collecting tubules
25
25
27
23
29
30
31
32
33
34
3S
35
37
33
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I calculator
A
5
6
7
a
9
1C
11
n
13
U
ft
A 57-year -old man comes to the office for a follow -up appointment . He has a history of
systolic heart failure which has been managed with appropriate medical therapy . The
patient experiences significant functional impairment at his baseline and is able to walk
only short distances His most recent echocardiogram showed a left ventricular ejection
fraction of 30% (normal >55% ). The physician decides to start him on
spironolactone The addition of this medication to the patients current regimen is most
likely to cause a decrease in which of the following renal functions ?
15
15
A . Hydrogen ion generation by the proximal convoluted tubule [3%]
17
13
H
20
21
22
23
24
25
25
*»
(
B . Hydrogen Ion secretion from the collecting tubules [77%]
C . Na 7K * /2CI cotransport in the thick ascending limb [9%]
C D . Proximal convoluted tubule brush border transport capacity [ 2%]
O E. Urea reabsorption In the collecting tubules [9%]
Explanation:
27
Site of action for various diuretics
23
29
30
31
32
33
34
35
35
37
33
39
40
1: carbonic anhydrase
proximal tubule
( aceiajrolamtde )
2: osmotic diuretics - primarily affects
the descending limb of He ole's loop
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I dkuldtor
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4
5
6
A
Explanation:
7
Site of action for various diuretics
a
9
10
b
HCOi
11
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15
15
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H
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33
34
1 carbonic anbydrase proximal Tubule
( acetazolam «JrM
2 : osmotic diuretics
-
primarily affects
the descending limb of Henle's loop
and proximal tubule
(mannitol )
3; loop diuretics - thick ascending
limb of Henle's loop
( furosemide )
thiazide diuretics distal convoluted tubule
( hydrochlorothiazide)
5 potassium sparing diuretics collecting duct
•
Sodium channel blockers
35
(amiloride)
35
37
38
39
40
*
Aldosterone receptor antagonists
( spironolactone)
V
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Notes
4
&
6
7
a
9
10
It
12
13
ft
©UWwId
Aldosterone is a mineralocorticoid hormone synthesized and released by the zona
glomerulosa ceils of the adrenal cortex . It functions as a component of the
renin aldosterone system, which is normally activated by low blood pressure and
reduced renal blood flow . Under these conditions aldosterone release is stimulated by
angiotensin II. High serum potassium ion concentrations and increased ACTH levels
(transient effect ) can also cause aldosterone secretion.
-
k
u
IS
IS
17
1B
14
20
21
22
23
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2&
25
27
23
29
30
31
32
Aldosterone increases the number of basolateral NaVK * -ATPase pumps and apical
sodium channels found on pnncipal cells in the cortical collecting ducts, Increasing
sodium and water reabsorption. It also promotes potassium and hydrogen ion
secretion from the pnncipal and intercalated cells of the collecting tubules , respectively;
Aldosterone receptor antagonists ( eg spironolactone, eplerenone ) inhibit the effects
of aldosterone and reduce secretion of K * and H* by the collecting tubule
(Choice A ) Carbonic anhyorase within proximal tubule cells synthesizes H\ which is
then secreted into tubular fluid and used by brush border carbonic anhydrase to help
resorb filtered HCO . . Carbonic anhydrase inhibitors such as aceiazoiamide inhibit both
membrane-bound and cytoplasmic forms of this enzyme .
(Choice C ) The Na" / K /2CI- cotransporter in the thick ascending limb is the target of loop
diuretics ieg. furosemide, ethacrynic acid). These potent diuretics cause brisk diuresis
by inhibiting solute reabsorption , which prevents the formation of a concentrated
*
33
34
medullary gradient.
35
(Choice D ) The brush border of the proximal convoluted tubule is responsible for
reabsorbing two -thirds of the sodium and water filtered by the glomerulus. Transport
proteins found in the brush border reabsorb filtered glucose, amino acids phosphate
and lactate via cotransport with sodium.
36
37
33
39
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20
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25
*
21
23
29
30
.
(Choice A ) Carbonic anhydrase within proximal tubule cells synthesizes H* which is
then secreted into tubular fluid and used by brush border carbonic anhydrase to help
resorb filtered HCO , . Carbonic anhydrase inhibitors such as acetazoiamide inhibit both
membrane -bound and cytoplasmic forms of this enzyme .
33
34
3S
35
37
33
39
40
(
dlculdior
b
.'
(Choice C ) The Na / K 2CI cotransporter in the thick ascending limb is the target of loop
diuretics eg . furosemide , ethacrynic acid). These potent diuretics cause brisk diuresis
by inhibiting solute reabsorption, which prevents the formation of a concentrated
medullary gradient .
(Choice D ) The brush border of the proximal convoluted tubule is responsible for
reabsorbing two -thirds of the sodium and water filtered by the glomerulus Transport
proteins found in the brush border reabsorb filtered glucose, amino acids phosphate ,
and lactate via cotransport with sodium .
(Choice E ) Vasopressin ( antidiuretic hormone ) increases urea reabsorption in the
medullary collecting tubules by increasing the number of cell surface urea transporters
This helps to strengthen the corticomedullary interstitial osmotic gradient and is
necessary to produce maximally concentrated urine
.
31
32
Notes
*
U
IS
15
*
secretion from the principal and intercalated cells of the collecting tubules , respectively .
Aldosterone receptor antagonists ( eg spironolactone , eplerenone ) inhibit the effects
of aldosterone and reduce secretion of K * and H by the collecting tubule
a
9
10
It
Lab Value
Newt
Educational objective:
Aldosterone is a component of the renin- angiotensin- aldosterone system that acts on the
principal and intercalated cells of the renal collecting tubules to cause resorption of
sodium and water and loss of potassium and hydrogen ions. Aldosterone receptor
antagonists ( eg . spironolactone , eplerenone ) Inhibit these effects .
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Notes
I dlculdtor
4
5
6
7
8
9
n
n
A 17-year -old female is being evaluated for amenorrhea and short stature
Karyotype analysis reveals 46 chromosomes that contain DNA material tightly
packed with additional proteins . Which of the following proteins outside the
nucleosome core facilitates nucleosome packing into a more compact structure?
12
13
u
15
115
17
ia
H
20
21
22
23
24
25
26
O A * Topoisomerase II
O B . snRNP
O C Ubiquitin
O D . Histone H1
O E Histone H3
b
O F. Histone H4
27
28
29
30
31
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Notes
tdkutdtor
packed with additional proteins Which of the following proteins outside the
nucleosome core facilitates nucleosome packing into a more compact structure?
A
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11
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13
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IS
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17
A. Topoisomerase il [4%J
O B . snRNP [4%]
O C . Ubiquitin [4%J
* > D . Histone H1 [77%J
O E, Histone H3 [6%]
O F. Histone H4 [6%]
.
13
H
20
Explanation:
Hitime HI
(oulikfe coral
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( H2A. H2B H3 H4 L
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Nudeosomes are structural subunits present inside the nucleus composed of
nuclear proteins called histones There are five major subtypes of histones Ht ,
H2A H2B H3 and H4 The nucleosome core is composed of two molecules each
of H2A , H2 B , H3, and H4 . making eight total histone proteins in each nucleosome
core . During the initial steps of DNA packaging into chromatin , the DNA double helix
wraps around the nucleosome core twice , but in contrast to the other histone
proteins . H1 histones are not part of the nucleosome H1 histones participate in
03 : 4 4
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Explanation:
A
Hi&lone Hi
fouliidfl coro }
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Nucwosom core
( H2A. H2B H3 H4 >,
*
h
Nucleosomes are structural subunits present inside the nucleus composed of
nuclear proteins called histones . There are five major subtypes of histones H1
H2A . H2B. H3, and H4 The nucleosome core is composed of two molecules each
of H2A . H2 B , H3 , and H4 , making eight total histone proteins in each nucleosome
core During the initial steps of DNA packaging into chromatin , the DNA double helix
wraps around the nucleosome core twice , but in contrast to the other histone
proteins . H1 histones are not part of the nucleosome. H1 histones participate in
DNA packaging by binding the segment of DNA that lies between nucleosomes and
facilitating the packaging of nucleosomes into more compact structures . The
association of DNA with histones gives the appearance of a "beaded chain, ' as this
structure undergoes further rounds of coiling and association with other structural
proteins , such as nuclear scaffold proteins , before ultimately forming chromosomes .
t
1
(Choice A ) Chromatin also contains non-histone proteins, such as enzymes that are
required for DNA replication and transcription. Topoisomerase II, also called DNA
gyrase , reduces DNA strand tension during DNA replication.
37
(Choice B) snRNPs , also known as "snurps," are small nuclear ribonucleoproteins
that bind pre -mRNA and participate in the formation of spliceosomes . which
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nuclear proteins called histones T here are frve major subtypes ot histones H1
H2A , H2B. H3 and H4 The nucleosome core is composed of two molecules each
of H2A H2 B : H3, and H4 . making eight total histone proteins in each nuf leosome
core During the initial steps of DNA packaging into chromatin , the DNA double helix
wraps around the nucleosome core twice , but in contrast to the other histone
proteins . H1 histones are not part of the nucleosome H1 histones participate in
DNA packaging by binding the segment of DNA that lies between nucleosomes and
facilitating the packaging of nucleosomes into more compact structures . The
association of DNA with histones gives the appearance of a "beaded chain," as this
structure undergoes further rounds of coiling and association with other structural
proteins , such as nuclear scaffold proteins , before ultimately forming chromosomes
*
Notes
(
dlculdior
.
{ Choice A) Chromatin also contains non-histone proteins such as enzymes that are
required for DNA replication and transcription . Topoisomerase II also called DNA
gyrase . reduces DNA strand tension during DNA replication.
(Choice B) snRNPs , also known as “ snurps . " are small nuclear ribonucleoproteins
that bind pre - mRNA and participate in the formation of spliceosomes , which
participate in the processing of pre-mRNA into mature RNA .
(Choice C) Ubiquitm is a small protein present in the cytoplasm and nucleus of all
eukaryotes that attaches covalently to various proteins and provides intracellular
signals for the programmed degradation of 'tagged' proteins by the proteasome.
Educational Objective:
Histone H1 is located outside of the nucleosome core and helps to package
nucleosomes into more compact structures by binding and linking DNA between
adjacent nucleosomes .
Time Spent 219 seconds
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Motes
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G
7
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9
10
A 12-year -old male is evaluated for ataxia accompanied by episodic erythematous
and pruritic skin lesions and loose stools Laboratory evaluation reveals loss of
neutral aromatic amino acids in the urine . This patient' s symptoms would most likely
respond to which of the following supplements?
b
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is
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ia
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O A. Thiamine
B Riboflavin
C C . Folic acid
OD
Niacin
E Pyndoxine
F , Tocopherol
G . Ascorbate
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Newt
*
Note
*
(
alculdtor
4
&
6
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&
9
10
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*v
<
A 12-year-old male is evaluated for ataxia accompanied by episodic erythematous
and pruritic skin lesions and loose stools Laboratory evaluation reveals loss of
neutral aromatic amino acids in the urine . This patient' s symptoms would most likely
respond to which of the following supplements?
n
13
u
IB
16
ir
13
H
20
21
n
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2
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37
C A Thiamine [11%J
O B . Riboflavin [5%]
O C . Folic acid [3%]
V :
o . Niacin [52%]
O E. Pyridoxine [19%]
•
F . Tocopherol [6%]
G Ascorbate [3%J
Explanation:
At least three specific small bowel enterocyte apical transport proteins appear to be
involved in the absorption of amino acids from the diet . In Hartnup disease , the
intestinal and renal absorption of tryptophan is defective . Tryptophan is an essential
amino acid and a precursor for nicotinic acid serotonin , and melatonin The clinical
manifestations of Hartnup disease are primarily due to the malabsorption of
tryptophan, resulting in niacin ( Vitamin BJ deficiency , because niacin is synthesized
from tryptophan .
Most children with Hartnup disease are asymptomatic , but some children experience
photosensitivity and pellagra-like skin rashes as in the case described
i
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l tiiculdtor
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*
3S
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37
ft
Explanation:
At least three specific small bowel enterocyte apical transport proteins appear to be
involved in the absorption of amino acids from the diet In Hartnup disease , the
intestinal and renal absorption of tryptophan is defective Tryptophan is an essential
amino acid and a precursor for nicotinic acid, serotonin , and melatonin. The clinical
manifestations of Hartnup disease are pnmanly due to the malabsorption of
tryptophan , resulting in niacin ( Vitamin B, } deficiency, because niacin is synthesized
from tryptophan .
-
L
Most children with Hartnup disease are asymptomatic , but some children expenence
photosensitivity and pellagra -like skin rashes as in the case described
above . Neurologic involvement can occur most commonly leading to
ataxia Neurologic and skin symptoms typically wax and wane during the course of
this disease The main laboratory finding in Hartnup disease is aminoaciduria,
restricted to the neutral amino acids ( alanine, serine , threonine , valine , leucine ,
isoleucine phenylalanine tyrosine , tryptophan , and histidine ) The urinary excretion
of proline , hydroxyprolme , and arginine remains unchanged, and this important
finding differentiates Hartnup disease from other causes of generalized
aminoaciduria such as Fanconi syndrome Treatment with nicotinic acid or
nicotinamide and a high-protein diet generally results in significant improvement of
symptoms.
(Choice A) Thiamine use by the body is maximal in states of accelerated
carbohydrate metabolism because it acts as a cofactor for the enzyme transketolase
in the pentose phosphate pathway as well as the enzymes a -ketoglutarate
dehydrogenase and pyruvate dehydrogenase , both of which require thiamine as a
c of actor.
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SPKt
Note
(Choice A) Thiamine use by the body is maximal in states of accelerated
carbohydrate metabolism because it acts as a cofactor for the enzyme transketolase
in the pentose phosphate pathway as well as the enzymes a-ketoglutarate
dehydrogenase and pyruvate dehydrogenase , both of which require thiamine as a
*
t <i l t ufdlor
A
cofactor.
(Choice B) The coenzymes flavin mononucleotide (FMN ) and flavin adenine
dinucleotide (FAD) form the prosthetic groups of several enzymes important in
electron transport . Both flavin mononucleotide (FMN) and flavin adenine dinucleotide
( FAD ) are synthesized from riboflavin (Vitamin B.). Clinical features of riboflavin
deficiency include cheilosis (perleche ), glossitis , keratitis conjunctivitis
photophobia , lacrimation , marked corneal vascularization , and seborrheic dermatitis
(Choice C) Folic acid deficiency may result in megaloblastic anemia , but does not
result in the neurological manifestations of subacute combined degeneration of the
posterior and lateral columns seen specifically with vitamin B , deficiency,
(Choice E) Pyridoxin © ( Vitamin B,) is converted to pyridoxal-5-phosphate, which acts
as a coenzyme in the decarboxylation and transamination of amino acids Deficiency
of pyridoxine leads to anemia , peripheral neuropathy , and dermatitis .
(Choice F) Tocopherol ( Vitamin E ) is a fat -soluble vitamin that functions as a
scavenger of free radicals ( antioxidant ). Deficiency of Vitamin E is very uncommon,
but when it occurs , it can result in myelopathy or neurologic dysfunction .
Educational Objective:
Hartnup disease can result in niacin deficiency due to an excess loss of dietary
tryptophan , resulting from defective intestinal and renal tubular absorption of that
amino acid . Remember that niacin ( nicotinamide / Vitamin B; ) is synthesized from
tryptophan and that tryptophan is an essential amino acid.
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A laboratory is performing experiments to determine the structure and function of
several different types of RNA molecules . An RNA molecule is isolated from a
culture of gram- positive bacteria that consists of 90 nucleotides . It is found to
contain dihydrouracil , thymidine , and acetylcytosine residues . Which of the following
is the most likely composition of the 3 '-end of this molecule?
O A. TATA
O B. CCA
O C Poly-A
O D Methylguanosine triphosphate
O E. AUG
O F. UAG
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ii
n
A laboratory is performing experiments to determine the structure and function of
several different types of RNA molecules An RNA molecule is isolated from a
culture of gram -positive bacteria that consists of 90 nucleotides. It is found to
contain dihydrouracil , thymidine , and acetylcytosine residues Which of the following
is the most likely composition of the 3’-end of this molecule?
13
14
IB
16
17
13
H
20
21
n
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24
2&
26
O A . TATA [ 4%]
^ IB CCA [29%]
O C. Poly-A [39%]
O D . Methylguanosine triphosphate [5%]
O E. AUG [7%]
,
O F. UAG [16%]
Explanation:
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3$
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ANTICODON
'ITF
VARIAFU F
LOOP
D LOOP
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3'
T K LOOP
'
5*
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&
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A
Explanation:
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iHNft
ANTICODON
SITE
VARIABLE
LlXlP
LU MJP
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ACCEPTOR
/
AMINO ACID
BINDING SITE
STEM
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3$
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37
Transfer RNA (tRNA ) is one form of non-coding RNA composed of between 74-93
nucleotides . Specific molecules of tRNA transfer certain amino acid residues to the
growing polypeptide chain during translation The tRNA molecule functions by
recognizing the three base codon on the mRNA molecule being translated through
its anticodon region , which contains complementary bases The secondary structure
of tRNA resembles a cloverleaf and contains the following regions:
The acceptor stem is created through the base pairing of the 5‘-terminai
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M „f t
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SI LM
*
Notes
t dlculdtor
ft
Transfer RNA ( tRNA ) is one form of non-coding RNA composed of between 74 - 93
nucleotides . Specific molecules of tRNA transfer certain amino acid residues to the
growing polypeptide chain during translation The tRNA molecule functions by
recognizing the three base codon on the mRNA molecule being translated through
its anticodon region which contains complementary bases . The secondary structure
of tRNA resembles a cloverleaf and contains the following regions:
•The acceptor stem is created through the base pairing of the 5 ' -terminal
nucleotides with the 3'- terminal nucleotides The CCA tail hangs off the 3 ' end,
with the amino acid bound to the 3' terminal hydroxyl group . tRNA is "loaded"
with the appropnate amino acid through the process of aminoacylation, which
is catalyzed by aminoacyl tRNA synthetase. The acceptor stem helps to
mediate correct tRNA recognition by the proper aminoacyl tRNA synthetase
•A 3' CCA tail is added to the 3' end of tRNA as a posttranscriptiona!
modification in eukaryotes and in most prokaryotes . In some prokaryotic
tRNAs , the tail region is directly transcribed from the genome . Several
enzymes utilize this ta ;l to help recognize tRNA molecules , and its presence is
necessary for protein translation.
The D arm contains numerous dihydrouracil residues , which are modified
bases that are often present in tRNA . The D arm ( along with the acceptor
stem and anticodon arm ) facilitates correct tRNA recognition by the proper
aminoacyl tRNA synthetase .
The anticodon arm contains sequences which are complementary to the
mRNA codon and is read in the 3' to 5 ' direction. During translation, the
ribosome complex selects the proper tRNA molecule based solely upon its
anticodon sequence .
•The T arm contains the TYC sequence that is necessary for binding of tRNA
tft riKftCft
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T h a TlJLJf"* c o n n o n r a roforc
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>1 hfl 1 %#
Si 1
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Scut
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N o t e*
t alculator
A
The anticodon arm contains sequences which are complementary to the
mRNA codon and is read in the 3’ to 5 ' direction Dunng translation , the
ribosome complex selects the proper tRNA molecule based solely upon its
anticodon sequence,
•The T arm contains the TYC sequence that is necessary for binding of tRNA
to ribosomes. The TYC sequence refers to the presence of thymidine ,
pseudouridine , and cytidine residues in this arm of tRNA . tRNA is the only
RNA species that contains the nucleoside thymidine ,
A 5' terminal phosphate
U
(Choice A) A TATA box is an upstream promoter region associated with some
genes in eukaryotic organisms . TATA binding protein binds to this promoter during
transcription , unwinding the DNA and initiating separation of the strands
(Choices C and D) After transcription , eukaryotic pre-mRNA undergoes
posttranscriptional modification The maturation process of precursor mRNA
includes the addition of a poly A tail at the 3’ end and a methylguanosme cap at the 5’
end as well as the removal of rntrons (non coding RNA ).
-
27
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29
!>
<3
Mdrk
(Choices E and F ) AUG and UAG are mRNA start and stop codons that initiate and
terminate translation, respectively.
Educational objective:
tRNA is a small , noncoding form of RNA that contains unusual nucleosides such as
pseudouridine and thymidine . Remember that tRNA has a CCA sequence at its 3"
end that is used as a recognition sequence by proteins , and that the 3’ terminal
hydroxyl group of the CCA tail is used as the binding site for the amino acid
-
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Lab Valuer
A 34-year -old woman comes to the physician with abdominal pain and melena She
also complains of progressive fatigue and a 5 kg ( 11 lb ) weight loss over the last 2
months . She has a strong family history of colon , endometrial , and ovarian cancer.
Colonoscopy shows a protuberant , friable mass in the ascending colon, and biopsy
is diagnostic for colon adenocarcinoma Genetic analysis confirms a mutation
consistent with Lynch syndrome (hereditary nonpolyposis colon cancer ) . Which of
the following is most likely responsible for the development of colon cancer in this
patient ?
Notes
I alculdtor
b
17
U
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29
C A . Nucleotide mismatches that escape repair
B Covalent bonds between adjacent pyrimidines
C Insertion of abnormal bases ( eg, uracil) into DNA
D Empty sugar -phosphate residues in the DNA molecule
C E . Double - strand breaks in DNA
30
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3$
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37
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NPKt
*
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*
I alculdtor
A
&
6
7
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10
11
12
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14
IS
1&
A 34-year -old woman comes to the physician with abdominal pain and melena She
also complains of progressive fatigue and a 5 kg ( 11 1b ) weight loss over the last 2
months . She has a strong family history of colon , endometrial , and ovarian cancer.
Colonoscopy shows a protuberant , friable mass m the ascending colon, and biopsy
is diagnostic for colon adenocarcinoma Genetic analysis confirms a mutation
consistent with Lynch syndrome (hereditary nonpolyposis colon cancer ) Which of
the following is most likely responsible for the development of colon cancer in this
patient ?
b
17
18
19
20
mismatches that escape
[ 89% ]
* @O AB.. Nucleotide
Covalent bonds between adjacent pyrimidines [1%]
repair
21
n
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26
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3S
37
-
C . Insertion of abnormal bases ( eg, uracil) into DNA [3%]
D. Empty sugar -phosphate residues in the DNA molecule [!%]
E. Double - strand breaks in DNA [5%|
Explanation:
Lynch syndrome (hereditary nonpolyposis colon cancer ) is an autosomal dominant
disease caused by defective DNA mismatch repair . DNA replication occurs with a
high degree of fidelity because mismatched nucleotides are repaired through the
proofreading activity of DNA polymerases delta and epsilon . However, this
proofreading functionality is not infallible; base substitutions and small
insertion-deletion mismatches occur due to errors in base pairing every 10 bases on
average . It is the function of the DNA mismatch repair system to fix these errors
shortly after the daughter strands are synthesized . The mismatch repair system
munkrae
e
a
i
m
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Newt
*
Notes
I dlcufdtor
A
E . Double - strand breaks in DNA [5%]
Explanation:
9
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u
u
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13
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2&
25
Lynch syndrome (hereditary nonpolyposis colon cancer ) is an autosomal dominant
disease caused by defective DNA mismatch repair DNA replication occurs with a
high degree of fidelity because mismatched nucleotides are repaired through the
proofreading activity of DNA polymerases delta and epsilon However this
proofreading functionality is not infallible base substitutions and small
insertion-deletion mismatches occur due to errors in base pairing every 10 bases on
average . It is the function of the DNA mismatch repair system to fix these errors
shortly after the daughter strands are synthesized The mismatch repair system
involves several genes , including MSH2 and MLH1 which code for components of
the human MutS and MutL homologs . Mutations in these 2 genes account for around
90% of cases of Lynch syndrome .,
Mismatch repair system
Mismatched ba &e
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Nick
Vs
5’
3' Template strand
S' Daughter strand
3*
Newty synthesized daughter strand with mismatch
MutS
3S
MutL
/
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y
S'
S'
y
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t dlculdtor
Mismatch repair system
&
6
Mismatched base
7
Nick
s
9
10
/
5'
3*
i?
13
H
A
3' Template strand
& Daughter strand
J
Newly synthesi / ed daughter strand with mismatch
MutS
IS
IS
17
13
5'
H
3'
20
MutL
-
y
5'
21
Mismatch detection by MutS homotog and recruitment of MutL homolog
22
21
24
2&
26
Exonudease 1
27
2a
29
30
Complex slides along DNA until nick In daughter strand detected
Exonuclease l binds to complex
31
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ssDNA binding protein
35
3S
37
Degradation of daughter strand past the mismatch point
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Mismatch detection by MutS homofog and recruitment of Mutl homotog
7
s
9
10
Exonuclease 1
&
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Complex slides along DNA until nick In daughter strand detected
Exonuclease l binds to complex
ssONA binding protein
H
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5'
3'
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1'
S'
Degradation of daughter strand past the mismatch point
Dissociation of repair complex
27
,
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29
DNA polymerase delta
S'
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3
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31
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Synthesis of new daughter strand segment
35
51
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*
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Llg I
S'
DNA ligase seals fin-il remaining nick
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lil
Mismatch repair begins with MutS homolog detecting a mismatch on the newly
created daughter strand which is distinguished from the parent strand by occasional
nicks in the phosphodiester bonds MutL homolog is then recruited and the
resulting complex slides along the DNA molecule until 1 of the daughter strand nicks
is encountered . At this point exonuclease 1 is loaded onto and activated by the
repair complex. The daughter strand is then degraded backward past the initial
mismatch point , leaving a variable gap of single - stranded DNA that is stabilized by
ssDNA -bindmg protein The complex then dissociates while DNA polymerase delta
loads at the 3' end of the discontinuity and begins synthesizing a new daughter
strand segment . Finally. DNA ligase I seals the remaining nick to complete the
repair process.
H
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23
24
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27
23
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33
34
(Choice B) Exposure to ultraviolet light can cause pyrimidine (usually thymine )
dimers to form due to covalent joining of adjacent pyrimidines Pyrimidine dimers
interfere with DNA replication and are removed by nucleotide excision repair .
(Choices C and D) Several types of insults can alter the DNA bases . For example ,
nitrous acid can deaminate C , A . and G. There are also spontaneous changes , such
as deamination of C to U and the constant low-level loss of punnes via thermal
disruption. Glycosyiases are enzymes that detect and remove abnormal bases from
DNA , creating an empty sugar -phosphate residue that is subsequently removed and
replaced by the correct nucleotide (base excision repair).
35
(Choice E) Exposure to ionizing radiation causes double - stranded DNA breaks that
are repaired by end- joining repair mechanisms Non-homologous end joining , the
main mechanism in primates , is more prone to cause mutations than homologous
37
recombination
35
Educational objective:
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(Choices C and D) Several types of insults can alter the DNA bases. For example
nitrous acid can deammate C , A . and G, There are also spontaneous changes , such
as deamination of C to U and the constant low-level loss of punnes via thermal
disruption. Glycosylases are enzymes that detect and remove abnormal bases from
DNA , creating an empty sugar phosphate residue that is subsequently removed and
replaced by the correct nucleotide (base excision repair ).
(Choice E) Exposure to ionizing radiation causes double - stranded DNA breaks that
are repaired by end- joining repair mechanisms Non-homologous end joining, the
main mechanism in primates , is more prone to cause mutations than homologous
Educational objective:
Lynch syndrome is an autosomal dominant disease caused by abnormal nucleotide
mismatch repair The mismatch repair system involves several genes , including
MSH2 and MLH1, which code for components of the human MutS and MutL
homologs Mutations in these 2 genes account for around 90% of cases of Lynch
syndrome .
37
b
-
30
35
A
(Choice B) Exposure to ultraviolet light can cause pyrimidine (usually thymine )
dimers to form due to covalent joining of adjacent pyrimidines Pyrimidine dimers
interfere with DNA replication and are removed by nucleotide excision repair .
recombination
35
t alculator
Notes
mismatch point , leaving a variable gap of single -stranded DNA that is stabilized by
ssDNA -binding protein The complex then dissociates while DNA polymerase delta
loads at the 3' end of the discontinuity and begins synthesizing a new daughter
strand segment Finally. DNA ligase I seals the remaining nick to complete the
repair process.
23
29
31
32
33
34
Lab Values
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6
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s
9
to
11
A 79-year-old female presents to your office with leg pain and fatigue She lives
alone and has little money to spend on food. The patient has tibial subperiosteal
hematomas and painful gums . Which of the following nutrient deficiencies is most
likely responsible for this patient' s symptoms?
12
13
u
IS
16
£ B . Vitamin B?
17
13
C
H
C D. Fotic acid
20
21
n
21
24
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25
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O A . Vitamin S 1
C . Pyndoxine
£ Ascorbic acid
O F. Vitamin K
O G . Zinc
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37
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7
a
9
io
11
A
A 79-year-old female presents to your office with leg pain and fatigue . She lives
alone and has little money to spend on food. The patient has tibial subperiosteal
hematomas and painful gums . Which of the following nutrient deficiencies is most
likely responsible for this patient' s symptoms?
12
13
H
15
15
c A . Vitamin 81 [1%]
17
O C Pyndoxine [2%}
O D . Folic acid [2%]
O B . Vitamin B? [1%]
1B
19
20
21
n
23
24
25
25
27
23
29
30
31
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35
3S
37
*
E . Ascorbic acid [86%]
O F . Vitamin K [6%]
O G . Zinc [2%]
Explanation:
Vitamin C cannot be synthesized endogenously and therefore must be consumed in
the human diet This is typically not a problem , as ascorbic acid is abundantly found
in fruits and vegetables (while also being present to a lesser extent in milk liver and
fish). Deficiencies of vitamin C are therefore rare in developed countries , but
continue to be a concern in those with inconsistent eating patterns - including the
elderly, alcoholics , and persons who live alone .
,
Vitamin C ( ascorbic acid) deficiency eventually results in scurvy, a disease
characterized by hemorrhages , subperiosteal hematomas bleeding into joint
snaces oinoivai swellinn secondary neriodontal infection anemia hvnerkeratotic
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^Linc |Z 7o j
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Notes
t dlcufdtor
A
Explanation:
Vitamin C cannot be synthesized endogenously and therefore must be consumed in
the human diet This is typically not a problem, as ascorbic acid is abundantly found
in fruits and vegetables (while also being present to a lesser extent in milk , liver, and
fish). Deficiencies of vitamin C are therefore rare in developed countries, but
continue to be a concern in those with inconsistent eating patterns - including the
elderly, alcoholics, and persons who five alone .
Vitamin C ( ascorbic acid) deficiency eventually results in scurvy , a disease
characterized by hemorrhages, subperiosteal hematomas, bleeding into joint
spaces , gingival swelling, secondary periodontal infection , anemia , hyperkeratotic
papular rashes , impaired wound healing and weakened immune response to local
infections.
Because ascorbic acid accelerates hydroxylation and amidation reactions it plays a
crucial role in numerous biosynthetic pathways. One of the most important functions
of ascorbic acid is its activation of prolyl and lysyl hydroxylase precursors , both of
which are necessary for the hydroxylation of procollagen. As collagen contains
considerable hydroxyproline , the quantity and quality of the collagen produced is
dramatically impaired by any reduction in available ascorbic acid
(Choice A) Vitamin B1 (thiamine ) deficiency is characterized by beriberi and
Wernicke syndrome .
35
37
(Choice B) Vitamin B (riboflavin) deficiency is characterized by cheilosis stomatitis,
^
glossitis , dermatitis , corneal vascularization and ariboflavinosis
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Notes
t dkuldior
of ascorbic acid is rts activation of prolyl and lysyl hydroxylase precursors , both of
which are necessary for the hydroxylation of procollagen As collagen contains
considerable hydroxyproline , the quantity and quality of the collagen produced is
dramatically impaired by any reduction in available ascorbic acid
,
(Choice A) Vitamin B (thiamine ) deficiency is characterized by benben and
Wernicke syndrome.
(Choice B) Vitamin
( riboflavin ) deficiency
is characterized by cheilosis stomatitis
glossitis, dermatitis, corneal vascularization , and ariboflavinosis .
(Choice C) Pyridoxine (vitamin B } deficiency is charactenzed by cheilosis glossitis
^
dermatitis, and peripheral neuropathy .
(Choice D) Folic acid deficiency is characterized by megaloblastic anemia and
neural tube defects in the fetus .
(Choice F) Vitamin K deficiency is characterized by a bleeding diathesis ( but not
painful gums).
27
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31
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33
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35
3S
37
(Choice G) Ztnc deficiency is characterized by acrodermatitis enteropathica , growth
retardation, and infertility .
Educational Objective;
The symptoms of scurvy are primanly caused by impaired collagen formation, and
include hemorrhages , subperiosteal hematomas , bleeding into joint spaces gingival
swelling , secondary periodontal infection , anemia hyperkeratotic papular rashes
impaired wound healing, and weakened immune response to local infections
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Notes
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l
5
7
8
9
10
11
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13
H
15
15
17
li
19
20
21
n
An infant who fails to gain weight has no enteropeptidase activity on the surface of
her duodenal epithelium . Formation of which of the following substance is most
likely impaired by this patient's condition?
C A . Lipase
OB. Pepsin
C. Amylase
I
O D. Trypsin
E. Lactase
F . Secretin
23
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I alculalor
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A
&
7
s
9
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11
An infant who fails to gain weight has no enteropeptidase activity on the surface of
her duodenal epithelium Formation of which of the following substance is most
likely impaired by this patient's condition?
n
O A . Lipase [4%)
13
U
IS
115
1?
13
O B. Pepsin [11%]
O C. Amylase [3%]
H
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v
D. Trypsin [65%]
O E . Lactase [10%]
O F. Secretin [ 7%J
Explanation:
Digestion and absorption of nutrients primarily occurs in the small intestine Small
intestinal epithelial cells produce several enzymes responsible for nutrient
absorption. Proteins in ingested food exist primarily as polypeptides and require
hydrolysis to dipeptides , tnpeptides and amino acids for absorption Hydrolysis of
these polypeptides is accomplished by proteolytic enzymes such as pepsin and
trypsin. These enzymes are secreted as the inactive proenzymes pepsinogen and
trypsinogen from the stomach and pancreas respectively. Trypsin , in turn , activates
other proteolytic enzymes including chymotrypsin , carboxypeptidase and elastase .
Actuation of trypsinogen to trypsin is achieved by enteropeptidase ( or enterokinase ),
an enzyme produced in the duodenum . Enteropeptidase deficiency results in
defective conversion of the proenzyme trypsinogen to the active enzyme trypsin ,
Typical clinical manifestations of enteropeptidase deficiency are diarrhea, growth
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A
Digestion and absorption of nutrients pnmanly occurs in the small intestine Small
intestinal epithelial cells produce several enzymes responsible for nutrient
absorption Proteins in ingested food exist primarily as polypeptides and require
hydrolysis to dipeptides tripeptides and amino acids for absorption. Hydrolysis of
these polypeptides is accomplished by proteolytic enzymes such as pepsin and
trypsin . These enzymes are secreted as the inactive proenzymes pepsinogen and
trypsmogen from the stomach and pancreas , respectively. Trypsin, in turn , activates
other proteolytic enzymes including chymotrypsin carboxypeptidase and elastase
Activation of trypsinogen to trypsin is achieved by enteropeptidase ( or enterokinase ) ,
an enzyme produced in the duodenum . Enteropeptidase deficiency results in
defective conversion of the proenzyme trypsinogen to the active enzyme trypsin .
Typical clinical manifestations of enteropeptidase deficiency are diarrhea, growth
retardation and hypoproteinemia .
(Choice A ) Lipase secreted from the exocrine pancreas is the most important
enzyme for the digestion of triglycerides. Chronic pancreatitis is a painful condition
that causes iipase deficiency. This leads to poor fat absorption and steatorrhea i. foul
smelling bulky stools containing undigested fat ).
(Choice B) Activation of pepsinogen to pepsin requires an acidic pH and a small
amount of preexisting pepsin Pepsin initiates the digestive process of proteins in
the stomach that is completed by trypsin and other enzymes in the proximal small
intestine.
(Choices C and E) Hydrolysis of complex carbohydrates to oligo - di- and
monosaccharides is carried out by pancreatic amylase Lactase produced in the
intestinal brush border is responsible for the conversion of lactose to glucose and
galactose . Deficiency of lactase is common and causes lactose intolerance
V
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*
Motes
I iilculdlor
4
&
7
a
9
10
11
13
14
19
115
17
ia
19
2Q
21
n
23
2
29
25
21
23
29
*
30
31
32
33
34
39
35
37
A
(Choice A) Lipase secreted from the exocrine pancreas is the most important
enzyme for the digestion of triglycerides. Chronic pancreatitis is a painful condition
that causes lipase deficiency. This leads to poor fat absorption and steatorrhea ( foul
smelling bulky stools containing undigested fat ).
.
t
(Choice B) Activation of pepsinogen to pepsin requires an acidic pH and a small
amount of preexisting pepsin Pepsin initiates the digestive process of proteins in
the stomach that is completed by trypsin and other enzymes in the proximal small
intestine.
(Choices C and E) Hydrolysis of complex carbohydrates to ohgo -, di- and
monosacchandes is earned out by pancreatic amylase Lactase produced in the
intestinal brush border is responsible for the conversion of lactose to glucose and
galactose Deficiency of lactase is common and causes lactose intolerance .
(Choice F) Secretin is a peptide hormone secreted by the S-cells of the duodenum
In response to low duodenal pH. Secretin stimulates the secretion of bicarbonate
from the pancreas and gall bladder and reduces acid secretion in the stomach by
reducing the production of gastrin Neutralizing the acidic pH of food entenng the
duodenum from the stomach is necessary for the proper function of pancreatic
enzymes ( amylase , lipase ).
Educational Objective:
Trypsinogen is activated to trypsin by duodenal enteropephdase . Trypsin is
essential for protein digestion and absorption in two ways It degrades complex
peptides to dipeptides and amino acids , and it activates other proteases such as
carboxypeptidase . elastase and chymotrypsin.
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dlculdtor
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&
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9
10
ii
Cells obtained from a 73 -year-oid male demonstrate a high activity of an enzyme that
has reverse transcriptase activity . It adds TTAGGG repeats to the 3'-ends of
chromosomes Which of the following cells were most likely obtained from the
patient?
n
13
14
18
115
17
18
19
20
21
A. Epidermal basal cells
C B Pancreatic islet (5-cells
O C Neurons
O 0 . Erythrocytes
O E . Myocardial cells
22
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A
5
6
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11
n
13
u
is
15
17
13
14
20
A
Cells obtained from a 73 -year-oid male demonstrate a high activity of an enzyme that
has reverse transcriptase activity It adds TTAGGG repeats to the 3'-ends of
chromosomes Which of the following cells were most likely obtained from the
patient?
&
* # A. Epidermal basal cells [69%]
B. Pancreatic isiet p- cells [10%]
O C . Neurons [12%1
O 0 . Erythrocytes [6%)
O E. Myocardial cells [2%]
21
n
Explanation:
21
25
25
Embryonic
stem cell
27
23
29
Adult
stem cell
f1
30
31
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35
Telomerase
35
37
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/s
O E . Myocardial cells [2%]
&
6
8
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10
11
M
Explanation:
Embryonic
stem cell
n
b
Adult
stem cell
13
\
u
IS
IS
17
13
Telomerase
14
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21
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21
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spa
A A
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-^Cy\VAN»
/
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3$
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Long telomere
Short telomere
© UWorld
Telomerase is a reverse transcriptase enzyme (RNA - dependent DNA polymerase )
that adds TTAGGG repeats to the 3' end of DNA strands at the terminal end of
chromosomes the telomere region Telomerase is similar to other reverse
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t dkuldlor
A
© UWorld
Telomerase is a reverse transcriptase enzyme (RNA - dependent DNA polymerase )
that adds TTAGGG repeats to the 3' end of DNA strands at the terminal end of
chromosomes the telomere region . Telomerase is similar to other reverse
transcriptase enzymes in that it synthesizes single - stranded DNA using
single - stranded RNA as a template Telomerase is composed of two main subunits
the reverse transcriptase TERT and the RNA template TERC The TERC template
is a ''built-in" part of the telomerase enzyme The TERC RNA template is repeatedly
read by the reverse transcriptase to add TTAGGG DNA sequence repeats to the
ends of chromosomes (telomeres ).
&
Stem cells have very long telomeres and active telomerase , but with every cell
division the length of telomeres progressively shortens Terminally differentiated
adult somatic cells have very short telomeres. Critical shortening in telomere length
is thought to be one signal for programmed cell death. On the other hand, cancer
cells up -regulate their telomerase activity , preventing cellular death by maintaining
the length of their telomeres . Cancer cells are considered immortal because these
cells continue to divide without aging or shortening of their telomeres, thus ,
telomerase is a potential target for the treatment of cancers . Syndromes of
premature aging such as Bloom syndrome are associated with shortened telomeres.
Stem cells are undifferentiated cells that have the potential to differentiate into other
cell types . Embryonic and adult stem cells are two major types of stem cells . While
embryonic stem cells are present in the very early stages of embryogenesis and
give rise to every cell type in adult humans , adult stem cells are thought to be
present in most tissues where they are responsible for replacement of dead cells
For instance , the epidermis is continuously replaced from stem cells present in the
basal cell layers . Similarly , bone marrow stem cells replace peripheral red and white
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Notes
(
iikufdtor
thought to t>e one signal for programmed cell death On the other hand cancer
cells up -regulate their telomerase activity , preventing cellular death by maintaining
the length of their telomeres . Cancer cells are considered immortal because these
cells continue to divide without aging or shortening of their telomeres , thus ,
telomerase is a potential target for the treatment of cancers. Syndromes of
premature aging such as Bloom syndrome are associated with shortened telomeres
is
A
Stem cells are undifferentiated cells that have the potential to differentiate into other
cell types. Embryonic and adult stem cells are two major types of stem cells While
embryonic stem cells are present in the very early stages of embryogenesis and
give rise to every cell type in adult humans, adult stem cells are thought to be
present in most tissues where they are responsible for replacement of dead cells
For instance , the epidermis is continuously replaced from stem cells present in the
basal cell layers Similarly , bone marrow stem cells replace peripheral red and white
blood cells Stem cells have very long telomeres they are able to proliferate
indefinitely m a controlled manner giving rise to replacement stem cells and
daughter cells that differentiate into the desired tissue
.
(Choices B , C and E ) Pancreatic beta *cells, neurons, and myocardial cells have a
very low potential to replicate and have low telomerase activity
(Choice D) Erythrocytes are devoid of a nucleus and have no potential to divide .
Educational Objective:
Telomerase is an enzyme that possesses reverse transcriptase (RNA - dependent
DNA polymerase ) activity and is normally expressed in stem cells as well as cancer
cells. However , cancer cells are immortal because these cells continue to divide
without aging and shortening of their telomeres .
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Notes
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A
6
6
7
6
9
10
11
12
A 63-year-old female presents to clinic for a routine examination Her diet consists
mainly of fruit and vegetables and she takes a daily multivitamin Her last menstrual
period was five years ago She expresses concern about wrinkles around her eyes
that make her "look old /' A decrease in which of the following is most likely
responsible for this patient's complaint?
13
IB
115
17
lj
14
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n
A . Collagen fibril production
O B. Proline hydroxylation
C Collagen cross - linking
:
’
D Collagenase synthesis
O E. Fibrillin synthesis
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Notes
I alculalor
A
&
6
7
9
10
ii
n
/%
A 63-year-old female presents to clinic for a routine examination Her diet consists
mainly of fruit and vegetables and she takes a daily multivitamin Her last menstrual
period was five years ago She expresses concern about wrinkles around her eyes
that make her "look old.” A decrease in which of the following is most likely
responsible for this patient's complaint?
13
b
fibnl production [35%]
- O* BA. Proline-hydroxylation
[11 ]
U
tj
is
15
%
17
(.
13
H
C. Collagen cross - linking [35%]
O D . Coilagenase synthesis [10%]
< E. Fibrillin synthesis [9%]
20
21
n
21
Explanation:
2S
25
27
28
29
30
31
32
33
34
35
35
37
Human skin exhibits evidence of aging by 30 to 35 years. Gradual thinning of the
epidermis is seen , with an associated reduction in subcutaneous fat , blood vessels
hair follicles sweat ducts and sebaceous glands This loss of subcutaneous tissue
causes the skin to become atrophic and more vulnerable to damage Of greatest
aesthetic impact is the decrease in the amount of dermal collagen and elastic fibers
present . Without this intrinsic reticular support, the inelastic skin sags and
demonstrates fine , shallow wrinkling .
(Choice B) Proline hydroxyiation would indirectly decrease during aging because the
amount of end product (collagen ) created decreases , but the more direct cause of
wrinkling is the reduced synthesis of collagen fibrils.
V
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Explanation:
Lab Value
I
*
Notes
I iilculalor
A
Human skin exhibits evidence of aging by 30 to 35 years . Gradual thinning of the
epidermis is seen with an associated reduction in subcutaneous fat , blood vessels
hair follicles sweat ducts and sebaceous glands This loss of subcutaneous tissue
causes the skin to become atrophic and more vulnerable to damage Of greatest
aesthetic impact is the decrease in the amount of dermal collagen and elastic fibers
present Without this intrinsic reticular support , the inelastic skin sags and
demonstrates fine , shallow wrinkling .
(Choice B) Proline hydroxylation would indirectly decrease during aging because the
amount of end product ( collagen ) created decreases , but the more direct cause of
wrinkling is the reduced synthesis of collagen fibrils.
(Choice C) Although aging reduces the total amount of collagen present in the
dermis it does not necessarily affect the quality of interstrand collagen crosslinking
,
(Choice D) The collagenases ( eg . MMP -1, MMP-8 . MMP - 13 ) cleave collagen into
fragments . Decreased collagenase synthesis would therefore reduce matrix
collagen turnover, promoting a net increase in dermal collagen
.
(Choice E) A protein found in a multitude of connective tissues , fibrillin- 1 aids in the
development of extracellular matrix microfibrils These microfibrils create
scaffolding for the deposition of elastic fibers An inherited defect in the gene that
codes for fibrillin-1 is responsible for Marfan syndrome (MRS ).
Educational Objective:
As a consequence of aging fine skin wnnkles appear secondary to the decreased
synthesis and net loss of dermal collagen and elastin .
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*
Notes
I alculalor
4
b
6
A 56-year -old homeless man presents to ER with increased fatigability and
exertional dyspnea Physical examination shows significant lower extremity edema
and decreased sensation over the ankles and feet Further evaluation reveals
cardiac dilation and increased cardiac output . Which of the following nutrient
deficiency is most likely responsible for this patient's symptoms ?
7
8
n
n
u
u
C A . Vitamin A
18
18
17
13
’
B . Vitamin B.
O C . Vitamin B;
O D . Pyndoxine
O E. Niacin
C F . Ascorbic acid
O G . Vitamin B :
H
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Newt
*
Iakuldtor
Notes
4
5
6
7
£
to
11
n
A
-
A 56 year -old homeless man presents to ER with increased fatigability and
exertional dyspnea Physical examination shows significant lower extremity edema
and decreased sensation over the ankles and feet Further evaluation reveals
cardiac dilation and increased cardiac output Which of the following nutrient
deficiency is most likely responsible for this patient's symptoms ?
u
13
u
is
15
17
13
H
20
21
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25
?7
23
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37
O A . Vitamin A [1%]
* <§ B . Vitamin B [63%]
C C Vitamin B,[3%]
D . Pyridoxine [6%]
O E. Niacin [5%]
F. Ascorbic acid [1%]
O G . Vitamin B : [20%I
1
Explanation:
Thiamine deficiency is associated with infantile and adult beriberi , as well as
Wemicke-Korsakoff syndrome in alcoholics Manifestations of infantile benben
appear between the ages of two and three months and include a fulminant cardiac
syndrome with cardiomegaly . tachycardia , cyanosis, dyspnea and vomiting .
Adutt beriberi is categorized as dry or wet Dry beriberi describes a symmetrical
peripheral neuropathy accompanied by sensory and motor impairments , especially
of the distal extremities Wet beriberi includes this neuropathy as well as cardiac
involvement { eg. cardiomegaly , cardiomyopathy , congestive heart failure, peripheral
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I tslculdtor
Motes
C G Vitamin B : [20%J
A
1
Explanation:
Thiamine deficiency is associated with infantile and adult beriberi as well as
Wemicke-Korsakoff syndrome in alcoholics . Manifestations of infantile beriberi
appear between the ages of two and three months and include a fulminant cardiac
syndrome with cardiomegaly. tachycardia cyanosis dyspnea and vomiting .
,
1,
-
,
Aduft beriberi is categonzed as dry or wet Dry beriberi describes a symmetrical
peripheral neuropathy accompanied by sensory and motor impairments , especially
of the distal extremities Wet beriberi includes this neuropathy as well as cardiac
involvement ( eg , cardiomegaly , cardiomyopathy , congestive heart failure, peripheral
edema tachycardia ).
,
(Choice A) Vitamin A deficiency is characterized by night blindness xerophthalmia
and vulnerability to infection ( especially measles ).
,
,
(Choice C) Vitamin B (riboflavin ) deficiency is characterized by cheilosis, stomatitis
glossitis, dermatitis, corneal vascularization , and ariboflavinosis .
.
(Choice D) Pyridoxine (vitamin B, ) deficiency is charactenzed by cheilosis glossitis,
dermatitis, and peripheral neuropathy .
(Choice E) Niacin deficiency is characterized by pellagra ( dementia , dermatitis and
diarrhea ).
35
37
(Choice F) Ascorbic acid (vitamin C ) deficiency is characterized by scurvy
(hemorrhages , bleeding into joint spaces , gingival swelling , impaired wound healing,
and weakened immune response to local infections ).
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Notes
I dlculdtor
ft
involvement ( eg, cardiomegaly , cardiomyopathy congestive heart failure , peripheral
edema, tachycardia ).
7
a
to
11
(Choice A ) Vitamin A deficiency is characterized by night blindness xerophthalmia ,
and vulnerability to infection ( especially measles).
12
(Choice C) Vitamin B (riboflavin ) deficiency is characterized by cheilosis stomatitis
glossitis , dermatitis, comeal vascularization and ariboflavinosis .
IS
115
(Choice D) Pyridoxine (vitamin BJ deficiency is characterized by cheilosis , glossitis ,
dermatitis, and peripheral neuropathy.
u
u
17
13
19
20
(Choice E) Niacin deficiency
diauhea).
is
characterized by pellagra ( dementia, dermatitis and
21
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37
(Choice F) Ascorbic acid (vitamin C ) deficiency is characterized by scurvy
(hemorrhages , bleeding into joint spaces , gingival swelling , impaired wound healing,
and weakened immune response to local infections ).
(Choice G) Vitamin B,: (cobalamin ) deficiency is frequently associated with
pernicious anemia The classic presentation of pernicious anemia is an older ,
mentally slow woman of northern European descent who is "lemon colored" ( anemic
and icteric ), has a smooth shiny tongue indicative of atrophic glossitis , and
demonstrates a shuffling broad-based gait .
Educational Objective:
High-output congestive heart failure and neurological symptoms are strongly
suggestive of wet beriberi (thiamine deficiency ).
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Notes
I ole uhitor
4
b
6
7
a
9
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12
13
H
IS
115
17
13
19
20
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22
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29
A 34-year -old man is brought to the emergency department with severe headache
and blurry vision His symptoms began suddenly after having lunch at a new Itahan
deli in his neighborhood The patient says he “ ate a sandwich with lots of fancy
meats and cheeses ' and drank an iced tea. His past medical history is significant for
severe atypical depression ., He has no known medication or food allergies His
blood pressure is 210/130 mm Hg and heart rate is 110/min . On physical
examination he appears tremulous and diaphoretic . The medication used to treat
this patient's depression most likely affects which of the following steps of
monoamine neurotransmission ?
1
O A. Monoamine degradation
B . Binding to postsynaptic receptors
C Presynaptic non- selective monoamine uptake
C D . Presynaptic selective norepinephrine uptake
C E Presynaptic selective serotonin uptake
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*
Notes
t dlculdtor
4
A
&
6
A 34-year -old man is brought to the emergency department with severe headache
and blurry vision His symptoms began suddenly after having lunch at a new Italian
deli in his neighborhood . The patient says he "ate a sandwich with lots of fancy
meats and cheeses" and drank an iced tea. His past medical history is significant for
severe , atypical depression . He has no known medication or food allergies His
blood pressure is 210/130 mm Hg and heart rate is 110/min . On physical
examination he appears tremulous and diaphoretic . The medication used to treat
this patient's depression most likely affects which of the following steps of
monoamine neurotransmission?
7
a
9
i
11
12
13
U
IB
IS
17
13
14
b
A. Monoamine degradation [67%]
O B . Binding to postsynaptic receptors [3%]
C Presynaptic non- selective monoamine uptake [14%]
C D Presynaptic selective norepinephrine uptake [6%]
O E. Presynaptic selective serotonin uptake [10%]
**
20
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n
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29
Explanation:
30
Tyramine hypertensive crisis
31
32
33
34
* ;•
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37
Sausag«
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t f c
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NCKt
*
Notes
t ole uldtor
Explanation:
&
6
Tyramine hypertensive crisis
7
6
9
*
*i *
i
*
11
12
13
14
IS
IS
17
13
19
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rv
C/ w«
%
Draft beef
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transporter
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MAO- A inhibition
prevents Gl
deton
monoamine
Disptocenwit oi
.1
ephrme
by lyramirw
nofflpif
j.
A
-
'
21
22
I*
23
24
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27
23
29
u
rn
LJ
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-
7
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Reverse Uanspol of
excess fxnepinephnna
30
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31
32
33
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34
35
3&
37
t
Sympathetic activity & hypertension
NOfepmephrine
receptors
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r*
This patient is experiencing a hypertensive emergency ( severe hypertension ,
headache blurry vision ) and has signs of excessive sympathetic activity (tachycardia,
diaphoresis, tremors }, most likely as a result of ingesting tyramine containing
foods (eg aged cheeses cured meats draft beer), Tyramine is an indirect
sympathomimetic that is usually metabolized in the gastrointestinal tract by
monoamine oxidase (MAO ) . This mitochondrial enzyme is also responsible for
degradation of monoamine neurotransmitters within the presynaptic nerve terminal
MAO inhibitors ( eg , tranylcypromine, phenelzine ) function by increasing synaptic
monoamine levels and are used in some patients with atypical or treatment-resistant
depression . However, these medications also block the degradation of dietary
tyramine allowing it to enter the systemic circulation and cause severe hypertension
and other signs of sympathetic hyperactivity.
*
Notes
(
ilcufdtor
(
-
u
(Choice B) MAO inhibitors do not affect the binding of monoamines to their
receptors . Commonly used monoamine receptor blockers include antihistamines
and alpha and beta adrenergic receptor blockers.
(Choice C) MAO inhibitors do not affect monoamine reuptake Tricyclic
antidepressants (eg imipramine , amitriptyline, clomipramine ) inhibit reuptake of
monoamines (norepinephrine and serotonin), permitting them to remain for longer
periods at the receptor site .
(Choice D) Bupropion is a presynaptic selective dopamine -norepinephrine reuptake
inhibitor used to treat major depression and tobacco dependence Seizures are the
most significant side effect , and the drug is contraindicated in patients with bulimia or
anorexia nervosa
(Choice E) Selective serotonin reuptake inhibitors ( eg , fluoxetine sertraline,
paroxetine ) block presynaptic serotonin reuptake, thereby enabling more serotonin to
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*
Notes
(
alcufaior
tyrarnine , allowing it to enter the systemic circulation and cause severe hypertension
and other signs of sympathetic hyperactivity,
A
(Choice B) MAO inhibitors do not affect the binding of monoamines to their
receptors. Commonly used monoamine receptor blockers include antihistamines
and alpha and beta adrenergic receptor blockers,
(Choice CJ MAO inhibitors do not affect monoamine reuptake , Tricyclic
antidepressants (eg imipramine . amitriptyline, clomipramine) inhibit reuptake of
monoamines (norepinephrine and serotonin), permitting them to remain for longer
periods at the receptor site,
(Choice D) Bupropion is a presynaptic selective dopamine -norepmephnne reuptake
inhibitor used to treat major depression and tobacco dependence Seizures are the
most significant side effect , and the drug is contraindicated in patients with bulimia or
anorexia nervosa
b
(Choice E) Selective serotonin reuptake inhibitors ( eg , fluoxetine, sertraline,
paroxetine ) block presynaptic serotonin reuptake , thereby enabling more serotonin to
accumulate within the synaptic cleft They are first -line antidepressants and are also
useful for treating a number of anxiety disorders .
Educational objective:
Monoamine oxidase (MAO ) is a mitochondrial enzyme that degrades excess
monoamine neurotransmitters in presynaptic nerve terminals and detoxifies dietary
tyrarnine in the gastrointestinal tract . Tyrarnine hypertensive crisis can occur in
patients taking MAO inhibitors following the consumption of foods containing high
amounts of tyrarnine ( eg aged cheeses , cured meats, draft beer ).
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07 : 14
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'
Last updated (9 /24/2015 ]
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I ole u hit o r
4
&
6
7
8
9
10
1?
13
u
is
16
17
18
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23
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28
29
A 76- year-old Caucasian female is evaluated for painful lesions on her lips and at the
comers of her mouth She is mildly demented and lives alone Her urinary riboflavin
excretion is very low Activity of which of the following enzymes is most likely
decreased in this patient?
O A. Isocitrate dehydrogenase
C B Succinate dehydrogenase
O C . Succinate thiokinase
O D Malate dehydrogenase
O E . Fumarase
F . Glucose-6-phosphate dehydrogenase
O G. HMG-CoA reductase
30
31
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34
3$
36
37
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Lob Valuta
t alculdtor
Notes
A
6
7
a
9
n
12
13
U
IS
115
17
ia
20
21
n
21
2i
25
25
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28
29
30
31
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3
*
35
35
V
ft
A 76- year -old Caucasian female is evaluated for painful lesions on her lips and at the
corners of her mouth She is mildly demented and lives alone Her urinary riboflavin
excretion is very low Activity of which of the following enzymes is most likely
decreased in this patient?
C A Isocitrate dehydrogenase [13%J
^ * B. Succinate dehydrogenase [47%]
O C . Succinate thiokinase [8%]
O D. Malate dehydrogenase [9%]
O E. Fumarase [9%]
F . Glucose-6-phosphate dehydrogenase [7%]
O G. HMG-CoA reductase [7% ]
u
Explanation:
Marked riboflavin deficiency is rare in the United States, but can be seen in chronic
alcoholics and the severely malnourished Clinical manifestations of marked
riboflavin deficiency include angular stomatitis cheilitis glossitis seborrheic
dermatitis, eye changes (eg. keratitis , corneal neovascularization)r and anemia The
diagnosis is established with performance of the erythrocyte glutathione reductase
assay or evaluation of the urinary riboflavin excretion.
Metabolic modifications of nboflavin occur most frequently in the cells of the heart,
liver , and kidney . Typically , nboflavin is first phosphorylated to become the
coenzyme flavin mononucleotide ( FMN). It can then either be integrated into a
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Lab Value
V
*
Motes
t alculator
Explanation:
Marked riboflavin deficiency is rare in the United States , but can be seen in chronic
alcoholics and the severely malnourished . Clinical manifestations of marked
riboflavin deficiency include angular stomatitis , cheilitis glossitis seborrheic
dermatitis, eye changes { eg, keratitis , corneal neovascularization ), and anemia . The
diagnosis is established with performance of the erythrocyte glutathione reductase
assay or evaluation of the urinary riboflavin excretion .
u
IS
IS
17
13
19
20
21
n
21
n
25
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Metabolic modifications of nboflavin occur most frequently in the cells of the heart ,
liver , and kidney. Typically, riboflavin is first phosphorylated to become the
coenzyme flavin mononucleotide ( FMN), It can then either be integrated into a
coenzyme - flavin complex or can be further phosphorylated into flavin adenine
dinucleotide { FAD ) FMN and FAD are required cofactors for flavoproteins . which
are enzymes that participate in numerous reduction-oxidation reactions within the
human body . In the course of these reactions the FMN and FAD cofactors are
transformed into their reduced energy -carrying states (FMNH and FADH. ) through
the acceptance of electrons.
The riboflavin-containing coenzymes are key constituents of the electron transport
chain FMN is a component of complex I while FAD is a component of complex II.
FAD is an electron earner in the tricarboxylic acid cycle ( TCA ) and serves as a
cofactor for succinate dehydrogenase, which is an enzyme that mediates the
conversion of succinate into fumarate .
(Choices A , C, D, and E) Isocitrate dehydrogenase , succinate thiokinase , malate
dehydrogenase , and fumarase are enzymes that participate in the tricarboxylic acid
cycle (TCA ). These enzymes do not use FAD or FMN as cofactors
(Choice F) Glucose -6-phosphate dehydrogenase (G6PD ) is the rate -limiting
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The riboflavin-containing coenzymes are key constituents of the electron transport
chain FMN is a component of complex I, while FAD is a component of complex II .
FAD is an electron carrier in the tricarboxylic acid cycle ( TCA ) and serves as a
cofactor for succinate dehydrogenase , which is an enzyme that mediates the
conversion of succinate into fumarate.
20
(Choices A . C, D, and E) Isocitrate dehydrogenase , succinate thiokinase , malate
dehydrogenase and fumarase are enzymes that participate in the tricarboxylic acid
cycle ( TCA ) These enzymes do not use FAD or FMN as cofactors
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coenzyme -flavin complex or can be further phosphorylated into flavin adenine
dinucleotide ( FAD) FMN and FAD are required cofactors for flavoproteins , which
are enzymes that participate in numerous reduction-oxidation reactions within the
human body . In the course of these reactions the FMN and FAD cofactors are
transformed into their reduced energy -carrying states (FMNH, and FADH. ) through
the acceptance of electrons.
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(Choice F) Glucose -6-phosphate dehydrogenase (G6PD ) is the rate -limiting
enzyme in the pentose phosphate pathway. This pathway supplies NADPH for
glutathione reduction in RBCs .
(Choice G) HMG-CoA reductase is the rate -limiting enzyme in the cholesterol
synthesis pathway. FMN and FAD are not used as cofactors.
Educational Objective:
Riboflavin ( vitamin B2 ) is a precursor of the coenzymes FMN and FAD . FAD
participates m tricarboxylic acid cycle as a coenzyme of succinate dehydrogenase
which converts succinate into fumarate
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29
A mutation that leaves prokaryotes unable to replicate their DNA is induced in an
experimental setting . The ability to remove RNA primers during DNA replication is
affected by this experimental mutation Which of the following enzymes is most
likely nonfunctional?
b
O A Helicase
O B . Primase
O C Gyrase
< D DNA polymerase III
C E . DNA polymerase I
O F . Ligase
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ft
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6
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a
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11
A mutation that leaves prokaryotes unable to replicate their DNA is induced in an
experimental setting . The ability to remove RNA primers during DNA replication is
affected by this experimental mutation. Which of the following enzymes is most
likely nonfunctional?
O A. Helicase [2%]
O B. Pnmase [14%]
O C. Gyrase [3%]
O D . DNA polymerase III [16%]
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* • E. DNA polymerase I [58%]
O F. Ligase [6%]
n
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Explanation:
26
DNA polymerase 1 = 3' oxonucteasa atlmly
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Pnma&e - forms RNA prime
*
( DNA dtp RNA pOlyflMtfOSO )
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Explanation:
A
G
DMA polymwa« I = 5' euuotjeteaw acidly
7
:
a
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’ Primaso -
forms RNA prime
i ( DNA dtp RNA potymofatt)*
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Okazaki fragmen
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DNA polymerases are the main enzymes responsible for DNA replication. In E. cols
there are three major DNA polymerases : I, II, and III. Other enzymes that are
essential for the replication of DNA include: primase . helicase . Iigase and
topoisomerase l and II . During DNA replication new daughter strands are
synthesized in the 5‘to 3’ direction using the parent strands as complimentary
templates . Synthesis of DNA in the 3' to 5’ direction does not occur in this process .
Before the process of replication begins , the parent DNA double helix is unwound
and separated by the enzyme helicase Helicase binds DNA at the origin of
.
., 'i t, I L,
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I alculdlor
Notes
M l U I I J
A
Before the process of replication begins the parent DNA double helix is unwound
and separated by the enzyme helicase Helicase binds DNA at the origin of
replication with the assistance of DnaA protein and acts at the replication fork to
separate DNA. This separation and unwinding of the DNA produces positive
supercoiling that can lead to DNA fracture, if not relieved . Topoisomerases I and II
(II is also known as gyrase ) relieve unwinding tension.
b
DNA polymerases can not begin synthesis of daughter strands without a free
3'-hydroxyl group , which is provided by an RNA primer and synthesized by the
enzyme pnmase ( DNA dependent RNA polymerase ) . DNA replication then
proceeds with the leading strand being formed continuously in the 5 ' to 3' direction
toward the replication fork and the lagging strand being formed discontinuous in
the 5' to 3' direction, away from the replication fork Replication of the lagging strand
results in the formation of numerous short DNA segments called Okazaki fragments
and each separate DNA segment requires a new RNA primer upon which to initiate
synthesis . The fragments of the lagging strand are ultimately bound together by
ligase after numerous RNA primers have been removed and replaced with DNA .
The removal of RNA primers is accomplished by DNA polymerase I, the only
bacterial DNA polymerase with 5’ to 3' exonuclease activity This activity allows DNA
polymerase I to function both as an excision-repair enzyme and as the enzyme that
removes RNA primers.
^
Educational Objective;
Bacterial DNA polymerase I has 5‘to 3‘exonuclease activity , which is used to excise
RNA primers The gaps created after RNA excision are then replaced with DNA in
the 5' to 3’ direction by DNA polymerase I.
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A
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A 36-year -old male has orange yellow discoloration of his palmar creases and small
clusters of yellow papules on his elbows knees and buttocks Laboratory evaluation
suggests a lack of ApoE3 and ApoE4 in his circulating lipoproteins . Which of the
following is most likely impaired in this patient?
b
n
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IS
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O A Chylomicron secretion by the intestine
C B Chylomicron remnant uptake by liver cells
C C . LDL particle uptake by extrahepatic cells
O D . Lipoprotein lipase activation
C E . Cholesterol esterification in the blood
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u
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*v
<
A 36-year -otd male has orange-yellow discoloration of his palmar creases and small
dusters of yellow papules on his elbows , knees and buttocks Laboratory evaluation
suggests a lack of ApoE3 and ApoE4 in his circulating lipoproteins Which of the
following is most likely impaired in this patient?
v
*
A Chylomicron secretion by the intestine [6%]
B. Ch \ lom cron remnant uptake t \ liver ceils [50%]
C LDL particle uptake by extrahepatic cells [ 22% ]
k
O D . Lipoprotein lipase activation [14%]
O E. Cholesterol estenfication in the blood [8%]
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37
Explanation:
Familial dysbetalipoproteinemia (type III hyperlipoproteinemia ) is charactenzed
clinically by xanthomas and premature coronary and penpheral vascular disease .
The primary defects in familial dysbetalipoproteinemia are in ApoE3 and ApoE4 ,
apolipoproteins found on chylomicrons and VLDL that are responsible for binding
hepatic apohpoprotein receptors Without ApoE3 and ApoE4, the liver cannot
efficiently remove chylomicrons and VLDL remnants from the circulation , causing
their accumulation in the serum and resultant elevations in cholesterol and
triglyceride levels .
(Choice A) Chylomicrons particles composed primarily of triacylglyceroL are
synthesized on the RER and Golgi apparatus of small intestinal enterocytes, They
are released from enterocytes with only ApoB-48 apolipoprotein ; they subsequently
receive ApoC lt and ApoE from HDL particles .
-
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A
Familial dysbetalipoproteinemia ( type III hyperlipoproteinemia ) is characterized
clinically by xanthomas and premature coronary and peripheral vascular disease .
The primary defects in familial dysbetalipoproteinemia are in ApoE3 and ApoE4 ,
apolipoprotems found on chylomicrons and VLDL that are responsible for binding
hepatic apolipoprotein receptors . Without ApoE3 and ApoE4, the liver cannot
efficiently remove chylomicrons and VLDL remnants from the circulation, causing
their accumulation in the serum and resultant elevations in cholesterol and
triglyceride levels .
(Choice A) Chylomicrons particles composed primarily of triacylglycerol, are
synthesized on the RER and Golgi apparatus of small intestinal enterocytes. They
are released from enterocytes with only ApoB-48 apolipoprotein ; they subsequently
receive ApoC-ll and ApoE from HDL particles ,
(Choice C) ApoB -100 is present on LDL and is required for receptor -mediated
uptake of LDL by extrahepatic tissues.
(Choice D) Lipoprotein lipase is activated by ApoC - ll earned by chylomicrons and
VLDL. ApoC-ll deficiency results in hyperchylomicronemia (type 1
hyperlipoproteinemia ).
(Choice E) ApoA-l is required for esterification of free cholesterol in HDL particles
by lecithin-cholesterol acyltransferase ( LCAT). ApoA - l and LCAT deficiencies result
in low HDL and increased circulating free cholesterol levels.
Educational objective ;
The key functions of important apolipoproteins are as follows:
ApoA 4: LCAT activation (cholesterol esterification)
ApoB 48: Chylomicron assembly and secretion by the intestine
ApoB 100: LDL particle uptake by extrahepatic cells
-
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hepatic apolipoprotein receptors Without ApoE3 and ApoE 4 , the liver cannot
efficiently remove chylomicrons and VLDL remnants from the circulation, causing
their accumulation in the serum and resultant elevations in cholesterol and
triglyceride levels.
(Choice A) Chylomicrons particles composed pnmarily of triacylglycerol are
synthesized on the RER and Golgi apparatus of small intestinal enterocytes They
are released from enterocytes with only ApoB-48 apolipoprotein ; they subsequently
receive ApoC-ll and ApoE from HDL particles .
(Choice C) ApoB -100 is present on LDL and is required for receptor -mediated
uptake of LDL by extrahepatic tissues.
&
(Choice D) Lipoprotein lipase is activated by ApoC-ll carried by chylomicrons and
VLDL. ApoC-ll deficiency results in hyperchylomicronemia (type 1
hyperlipoproteinemia ).
(Choice E) ApoA- l is required for esterification of free cholesterol in HDL particles
by lecithin-cholesterol acyltransferase ( LCAT ). ApoA- l and LCAT deficiencies result
in low HDL and increased circulating free cholesterol levels.
Educational objective:
The key functions of important apolipoproteins are as follows:
ApoA-l: LCAT activation (cholesterol esterification)
ApoB-48: Chylomicron assembly and secretion by the intestine
ApoB-100; LDL particle uptake by extrahepatic cells
ApoC-ll: Lipoprotein lipase activation
ApoE 3 & 4: VLDL and chylomicron remnant uptake by liver cells
- -
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t olculdtor
4
5
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7
a
9
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13
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13
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n
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26
-
A 32 year-old man is recovering from extensive bums Fibroblasts near the site of
injury actively synthesize precursor mRNA to be used as templates for protein
synthesis After transcription , extensive processing of the precursor RNA occurs
form the finalized mRNA sequence . The finalized mRNA then exits the nucleus a
undergoes translation by nbosome complexes before being degraded . Which of
the following steps involving the processing and handling of mRNA occurs only
within the cytoplasm of cells?
%
C A S' -terminal guanosine triphosphate addition
C B. Methylation of the S'-terminal guanine
C. Multiple adenine nucleotide attachment to the 3' -end
.
D Interaction with snRNP
C E.
Removal of intervening sequences
C F. Interaction with P bodies
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t olc uhitor
4
ft
5
G
7
£
9
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12
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IS
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2&
26
A 32- year-old man is recovering from extensive burns Fibroblasts near the site of
injury actively synthesize precursor mRNA to be used as templates for protein
synthesis After transcription , extensive processing of the precursor RNA occurs to
form the finalized mRNA sequence . The finalized mRNA then exits the nucleus and
undergoes translation by nbosome complexes before being degraded . Which of
the following steps involving the processing and handling of mRNA occurs only
within the cytoplasm of cells?
A 5‘- terminal guanosine triphosphate addition [7 %]
B Methylation of the 5 '- terminal guanine [15%]
C Multiple adenine nucleotide attachment to the 3' - end [12%]
O D . Interaction with snRNP [15%]
O E . Removal of intervening sequences [9%]
v % F. Interaction with P bodies [42%]
,
?7
23
29
Explanation:
30
After transcription , the preliminary unprocessed mRNA is known as precursor
mRNA , or heterogeneous nuclear RNA (hnRNA ). Eukaryotic pre- mRNA undergoes
significant posttranscriptional processing before leaving the nucleus , including
5'-capping . poly A tail addition , and intron splicing .
31
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Once mRNA is finalized it leaves the nucleus bound to specific packaging proteins
Upon entering the cytoplasm , these mRNA complexes often associate with
ribosomes to undergo translation However , certain mRNA sequences instead
,
sicninHsafo with rtmteinc that aro found in P hnrlioQ
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fl”
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T
*
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t dlculdtor
Explanation:
After transcription, the preliminary, unprocessed mRNA is known as precursor
mRNA , or heterogeneous nuclear RNA (hnRNA ). Eukaryotic pre -mRNA undergoes
significant posttranscriptional processing before leaving the nucleus , including
5 '-cappingH poly A tail addition, and intron splicing .
Once mRNA is finalized , it leaves the nucleus bound to specific packaging proteins.
Upon entering the cytoplasm , these mRNA complexes often associate with
ribosomes to undergo translation However, certain mRNA sequences instead
associate with proteins that are found in P bodies P bodies are distinct foci found
within eukaryotic cells that are involved in mRNA regulation and turnover They play a
fundamental role in translation repression and mRNA decay , and contain numerous
proteins including RNA exonucleases, mRNA decapping enzymes , and constituents
involved in mRNA quality control and microRNA-induced mRNA silencing . P bodies
also seem to function as a form of mRNA storage , as certain mRNAs are
incorporated into P bodies only to be later released and utilized for protein translation
(Choices A and B) The 5‘end of all mRNA is capped with a 7-methylguanosine
residue by a unique 5' to 5' linkage which occurs in two stages . The first step is the
addition of guanine tnphosphate to the 5 ' end of mRNA in a reaction catalyzed by
guanylyltransferase. Methylation of the guanosine cap is then catalyzed by
guanine-7 -methyltransferase Capping of the precursor RNA occurs in the nucleus
as the RNA is being transcribed . This methylated cap protects mRNA from
degradation by cellular exonucleases , and allows it to exit the nucleus.
(Choice C) mRNA is polyadenylated at the 3' end by the polyadenylate polymerase
complex which recognizes a specific sequence (AAUAAA ), cleaves the pre-mRNA
molecule a few residues downstream from this sequence , and then adds a stretch of
20 - 250 adenine residues called the poly A tail. The addition of the poly A tail
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t dlculdtor
A
-
(Choices A and B) The 5‘end of all mRNA is capped with a 7 methylguanosine
residue by a unique 5' to 5' linkage , which occurs in two stages . The first step is the
addition of guanine tnphosphate to the 5’ end of mRNA in a reaction catalyzed by
guanylyltransferase . Methylation of the guanosine cap is then catalyzed by
guanme -7-methyltransferase Capping of the precursor RNA occurs in the nucleus
as the RNA is being transcribed . This methylated cap protects mRNA from
degradation by cellular exonucleases , and allows it to exit the nucleus.
(Choice C) mRNA is polyadenylated at the 3' end by the polyadenylate polymerase
complex , which recognizes a specific sequence (AAUAAA ) cleaves the pre-mRNA
molecule a few residues downstream from this sequence , and then adds a stretch of
20 - 250 adenine residues called the poly A tail. The addition of the poly A tail
occurs before the mRNA exits the nucleus . In the cytosol the poly A tail is gradually
shortened , eventually leading to mRNA degradation .
(Choices D and E) Since pre -mRNA contains both introns and exons , and only
exons code for proteins introns must be excised before translation through a
process known as splicing Splicing of pre -mRNA occurs within the nucleus and is
facilitated by the interaction of pre-mRNA with small ribonucleoprotein particles
called snRNPs ( or “ snurps " for short ).
Educational objective:
When mRNA is first transcribed from DNA it is in an unprocessed form called
pre - mRNA or heterogeneous nuclear mRNA (hnRNA ) Several processing steps
are required before finalized mRNA molecules can leave the nucleus including
5 '-capping poly A tail addition , and intron splicing . Cytoplasmic P bodies play an
important role in mRNA translation regulation and mRNA degradation.
References:
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Notes
t dlculdtor
A
&
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a
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11
A fragment of an mRNA molecule isolated from a Gram negative bacterium is shown
below .
5'
ACGCUACCAUUGCAAGUUAGCUAAAUA
128
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Which of the following recognizes codon 128?
O A. snRNA
O B . Uncharged tRNA
O C . Charged tRNA
O D Releasing factor 1
O E. Elongation factor 2
F . Transcription factor II D
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alculutor
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&
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A fragment of an mRNA molecule isolated from a Gram negative bacterium is shown
below .
ACGCUACCAUUGCAAGUUAGQJAAAUA
5'
128
12
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3'
Which of the following recognizes codon 128?
&
O A. snRNA [5%J
O B. Uncharged tRNA [13%]
O C . Charged tRNA [17%]
* Mi
D. Releasing factor 1 [54%]
O E. Elongation factor 2 [ 7%]
C F. Transcription factor II D [ 3%]
Explanation:
There are 64 codons in the genetic code , the majority of which code for amino
acids . Because there are only 20 amino acidst most amino acids have more than
one codon. For example , GUU , GUC , GUA and GUG all code for valine , In addition,
there are codons that call for the initiation and termination of protein synthesis AUG ,
which codes for methionine is the universal start codon UAA , UAG and UGA are
stop codons . The stop codons do not code for amino acids Instead, when the
ribosome encounters a stop codon, releasing factors bind to the ribosome and
stimulate release of the formed polypeptide chain and dissolution of the
ribosome-mRNA complex .
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acids Because there are only 20 ammo acids , most amino acids have more than
one codon. For example , GUUr GUC. GUA and GUG all code for valine. In addition,
there are codons that call for the initiation and termination of protein synthesis AUG.
which codes for methionine is the universal start codon. UAA , UAG and UGA are
stop codons . The stop codons do not code for amino acids Instead , when the
ribosome encounters a stop codon releasing factors bind to the ribosome and
stimulate release of the formed polypeptide chain and dissolution of the
nbosome-mRNA complex.
A
(Choice A) Transcription produces a pre-mRNA molecule containing both introns
and exons Posttranscriptional processing removes the introns Splicing is
accomplished via small nuclear ribonucleoproteins ( snRNPs).
b
(Choice B) Uncharged tRNA (without an amino acid ) would not interact with mRNA
and nbosomes during protein synthesis .
(Choice C) Charged tRNA delivers amino acids to the protein synthesis complex .
The anticodon on a tRNA molecule recognizes the corresponding codon on mRNA
assuring proper amino acid sequencing .
21
23
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31
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33
34
3$
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37
(Choices E and F) Initiation of gene transcription is governed by transcription factor
binding to the gene's regulatory region . Transcription factor II D is a transcription
factor that binds to the TATA promoter region located approximately 25 bp upstream
from the gene 's coding region Elongation factors facilitate tRNA binding and the
translocation steps of protein synthesis .
Educational Objective:
Releasing factors recognize the stop codons (UAA UAG and UGA ) to terminate
protein synthesis They facilitate release of the polypeptide chain from the ribosome
and dissolution of the ribosome -mRNA complex .
.
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Notes
t alculdtor
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*
An investigator is studying weight regulation using experimental mouse models
Knockout mice are created with a homozygous mutation in the gene coding forth ©
leptin receptor . This mutation prevents the receptor from binding leptin and initiating
its normal signaling cascade . The knockout mice are allowed to feed at will and their
body mass index (BMI ) and serum leptin levels are measured and compared with
control mice . On the graph below area C represents the normal relationship
between BMI and leptin in a control mouse.
,
Leptin vs. BMI
17
13
n
20
21
22
I
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28
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B
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21
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30
31
32
33
34
Serum leptin levels
Q U5MU
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35
37
Which of the following areas represents the expected relationship between BMI and
serum leptin levels in a leptin receptor mutant mouse?
V
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control mice . On the graph below area C represents the normal relationship
between BMI and leptin in a control mouse ,
7
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Notes
t alculdtor
I 3 0I C U W i l l
L I H 1C V d o Q I C I I I C 0 9 U I C U Q I I U 0 U I I |
ft
Leptin vs. BMI
8
9
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11
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1
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Serum leptin levels
OUtMHttbridUjC
27
28
29
30
31
32
33
34
3S
35
37
Which of the following areas represents the expected relationship between BMI and
serum leptin levels in a leptin receptor mutant mouse?
O A. A
O B. B
O c. c
O D. D
0 E. E
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control mice . On the graph below area C represents the normal relationship
between BMI and leptin in a control mouse ,
7
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Leptin vs. BMI
8
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37
D
E
Serum leptii levels
OlAMllltaU (K
Which of the following areas represents the expected relationship between BMI and
serum leptin levels in a leptin receptor mutant mouse?
O A . A [16%]
*
B . B [67%]
O c. C [1%1
OD. D [4 %l
O E . E [ 11%]
*
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Notes
(
aleufdtor
Leptm is a protein hormone that plays an important role in regulating appetite and
metabolism It is produced primarily in adipocytes , and large fat cells produce more
leptin than small ones Serum leptin concentrations are highly correlated with body
fat content
A
Leptin decreases food intake in the following important ways :
1. Leptin decreases the production of neuropeptide Y, a potent appetite
stimulant in the arcuate nucleus of the hypothalamus .
2. Leptin stimulates the production of proopiomelanocortin (POMC ) in the arcuate
nucleus . Alpha -melanocyte - stimulating hormone lalpha-MSH) is produced by
cleavage of POMC and inhibits food intake.
The knockout mouse described is homozygous for a mutation in the gene encoding
the leptin receptor ( db / db ), resulting in ineffective leptin signaling As a result these
mice become hyperphagic and profoundly obese . As leptin production is normal in
these mice leptin levels are elevated due to the increased lipocyte mass In
contrast , mice that are homozygous for a mutation resulting in impaired leptin
production ( ob ob ) also become hyperphagic and profoundly obese but their leptin
levels are low ( Choice A) .
b
.
Human obesity resulting from mutations in the leptm receptor and the leptin gene has
been described. However , most obese individuals do not have either of these
mutations . Instead , it is thought that the sustained elevation in leptin levels from the
enlarged fat stores results in leptin desensitization Thus obese individuals become
resistant to the effects of leptin in a manner similar to the development of insulin
resistance in type 2 diabetes.
(Choice D) Low leptin and low BMI correlate with low adipocyte stores and may be
seen after prolonged starvation.
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these mice , leptin levels are elevated due to the increased lipocyte mass In
contrast , mice that are homozygous for a mutation resulting in impaired leptin
production ( ob ob ) also become hyperphagic and profoundly obese , but their leptin
levels are low ( Choice A) .
*
Notes
(
akuhator
.
Human obesity resulting from mutations in the leptin receptor and the leptin gene has
been descnbed However , most obese individuals do not have either of these
mutations, Instead it is thought that the sustained elevation in leptin levels from the
enlarged fat stores results in leptin desensitization Thus obese individuals become
resistant to the effects of leptin in a manner similar to the development of insulin
resistance in type 2 diabetes.
(Choice D) Low leptin and low BMI correlate with low adipocyte stores and may be
seen after prolonged starvation.
I
-
(Choice E) With intact receptor signaling elevated leptin levels ( ie from exogenous
leptin administration ) would result in weight loss .
<
Educational objective:
Leptin is a protein hormone produced by adipocytes in proportion to the quantity of
fat stored Leptin acts on the arcuate nucleus of the hypothalamus to inhibit
production of neuropeptide Y ( decreasing appetite i and stimulate production of
alpha -MSH (increasing satiety ). Mutations in the leptin gene or receptor result in
hyperphagia and profound obesity .
References:
1. Leptin receptors.
37
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dkuldtor
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i
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*
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21
A 12-year -old boy is brought to the emergency department with severe chest pain .
He has had intermittent substernal chest pain for the past few months that typically
occurs after heavy activity . The boy's activities have been limited due to the chest
pam and he is no longer able to play on the soccer team. The patient does not use
tobacco or illicit drugs . His temperature is 36.7 C (98 F ) blood pressure is 130/60
mm Hgf pulse is 132/mm , respirations are 24/min , and pulse oximetry is 98% on
room air . BMI is 17 kg 'nr Physical examination shows an anxious - appearing boy
with a rapid but regular pulse No abnormalities are seen Troponin is elevated, and
ECG reveals ST segment elevations in leads II , 111, and aVF . After acute
stabilization and treatment , further laboratory workup shows an increased serum
methionine level Which of the following ammo acids is most likely essential in this
patient?
n
21
24
25
25
27
C A . Asparagine
C B. Cysteine
.
C. Isoleucine
23
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O D . Leucine
30
O E . Tyrosine
C F, Valine
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stabilization and treatment further laboratory workup shows an increased serum
methionine level. Which of the following amino acids is most likely essential in this
*
Notes
t ale uftilor
A
patient ?
a
9
n
n
12
13
U
IS
IS
13
19
20
21
A . Asparagine [5 %]
v » B . Cysteine [7Q %]
O C . Isoleucine [3%]
O D . Leucine [4%1
O E . Tyrosine [15%]
O F . Valine [2%]
Explanation:
n
Methionine cycle
23
24
25
26
THF
27
Methionine
23
29
30
Mefftfon/ne synthase
31
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33
34
Vitamin B t 2
5 -Methyi'THF
35
3S
3?
Homocysteine
Serine
Block Time Remaining:
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t alculdtor
4
A
5
6
THF
Methionine
7
a
9
Methionine synthase
n
Vitamin B 12
it
12
5 MethyMHF
13
U
is
is
Homocysteine
21
22
Methyt cobaiamin
S -adenosyt
methionine
Serine
13
19
20
b
Cysfatoionine
Sybase
23
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X
Vitamin Be
Methyttran sferase
S -adenosyl
homocysteine
Cystathionine
27
23
29
Methyl- X
30
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34
Cystathionase
Vitamin Be
35
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Cysteine
TU
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Notes
I alcufalor
The ammo acid methionine can be metabolized into S- adenosyl-methionine (SAM),
which acts as a methyl-donor for many methyltransferase reactions . After the
transfer of a methyl group SAM is converted into S-adenosyl-homocysteinet which is
broken down to form adenosine and homocysteine Subsequently the conversion
of homocysteine to cystathionine requires the enzyme cystathionine synthetase , the
amino acid serine , and the cofactor vitamin B,. Cystathionine is then converted to
cysteine by the enzyme cystathionase which also requires vitamin B as a
cofactor. Alternatively, the enzyme methionine synthase uses vitamin B,. as a
cofactor to revert homocysteine back to methionine.
(
This patient most likely has homocystinuria , a condition that leads to
hypercoagulability and thromboembolic occlusion Because homocysteine is
prothrombotic , individuals with complete cystathionine synthase deficiency can
develop premature acute coronary syndrome , as seen in this patient (based on his
troponin level and ECG findings ) Other clinical features include ectopia lentis
( ocular lens displacement ) and intellectual disability .
The most common cause of homocystinuria is a defect in cystathionine
synthase Affected patients cannot form cysteine from homocysteine : therefore
cysteine is essential m their diet . In addition homocysteine buildup results in
hypermethioninemia , as seen in this patient .
i
,
(Choice A) The enzyme asparagine synthase converts aspartate to asparagine , the
amino acid that is essential for rapidly dividing tumor cells that cannot produce it
quickly enough on their own . The chemotherapy drug asparaginase decreases
asparagine concentration in tumor cells and leads to lysis of these rapidly growing
cells.
(Choices C, D, and F) Maple syrup unne disease is an amino acid disorder caused
nase This deficiency leads
bv defici =JI CV * f branched-chait JIVGA •nacii
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Notes
/v
quickly enough on their own The chemotherapy drug asparaginase decreases
asparagine concentration in tumor cells and leads to lysis of these rapidly growing
cells
(Choices C, D, and F) Maple syrup urine disease is an amino acid disorder caused
by deficiency of branched-chain a-ketoacid dehydrogenase This deficiency leads
to toxic buildup of branched-chain amino acids (leucine isoleucine and valine ) and
their metabolites , resulting in feeding difficulties seizures, cerebral edema and a
sweet odor of the urine
(Choice E) Phenylalanine hydroxylase catalyzes the hydroxylation of the essential
amino acid phenylalanine to form tyrosine Deficiency of phenylalanine hydroxylase
is the most common cause of phenylketonuria , which results in severe intellectual
disability if left untreated .
&
Educational objective:
Homocystinuria is most commonly caused by a defect in cystathionine synthase ,
resulting in an inability to form cysteine from homocysteine . Cysteine becomes
essential in affected patients , and homocysteine buildup leads to elevated
methionine Homocysteine is prothrombotic resulting in premature thromboembolic
events ( eg . atherosclerosis, acute coronary syndrome) in these patients
30
31
32
33
34
35
36
References:
1. Hypermethioninemias of genetic and non genetic origin: A review.
*
2 . Overview of homocysteine and folate metabolism. With special
references to cardiovascular disease and neural tube defects.
37
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-
A 54 year-otd man with a history of cirrhosis is brought to the emergency department
by his wife who found him agitated and confused She reports that he vomited
bright red blood several times yesterday. His cirrhosis is secondary to chronic
hepatitis C infection , and he has received treatment for esophageal varices in the
past . Physical examination reveals abdominal distention, decreased liver span and
testicular atrophy A flapping tremor involving his hands is seen with wrist
extension . Serum studies show elevated ammonia levels. Which of the following
metabolic intermediates is most likely to be deficient in this patient' s brain?
tf
A o -ketoglutarate
O B Carnitine
O C Glutamine
O D . Lactate
O E Oxindoie
,
35
35
37
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t alculator
A
ft
5
6
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a
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10
n
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13
H
19
19
A 54-year -otd man with a history of cirrhosis is brought to the emergency department
by his wife who found him agitated and confused She reports that he vomited
bright red blood several times yesterday . His cirrhosis is secondary to chronic
hepatitis C infection , and he has received treatment for esophageal vances in the
past . Physical examination reveals abdominal distention decreased liver span and
testicular atrophy . A flapping tremor involving his hands is seen with wrist
extension . Serum studies show elevated ammonia levels. Which of the following
metabolic intermediates is most likely to be deficient in this patient' s brain?
17
A. a -ketog utarate [51% ]
v
19
2Q
O B . Camrtine [12%]
21
O C . Glutamine [33%J
23
2
29
25
O D, Lactate [4% ]
C E. Oxindole [1%J
n
*
27
23
29
Explanation:
30
31
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33
3
*
Effects of hyperammonemia on the
glutamate - glutamine cyde
}
v
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c * t**i tjp' iijfy
**
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t dlculdlor
E . Oxfndole [1%J
Explanation:
a
9
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IS
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I
Kr * to ( yflf
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dlculdtor
4
9
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7
£
9
10
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13
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39
This patient has hepatic encephalopathy , likely due to recent gastrointestinal
bleeding and a corresponding increase in ammonia and nitrogen absorption by the
gut . The pathogenesis of hepatic encephalopathy is related to increased circulatory
levels of ammonia and other neurotoxins due to failure of the liver to metabolize
waste products When ammonia levels rise acutely, astrocyte and neuron function
are affected .
Within the brain astrocytes and neurons interact to regulate the metabolism of
glutamate , glutamine and ammonia in a process known as the glutamate - glutamine
cycle Glutamate released by neurons during neurotransmission is taken up by
astrocytes and converted to glutamine a non-neuroactive compound . Glutamine is
then released by astrocytes and taken up by neurons , where rt is either converted
back to glutamate for use as a neurotransmitter or transaminated into a -ketoglutarate
for use in the Krebs cycle .
Ammonia toxicity results in part from depletion of glutamate and alpha -ketoglutarate
in the brain dunng the process of ammonia detoxification. Hyperammonemia
increases the conversion of glutamate into glutamine by glutamine synthetase within
astrocytes (Choice C) The resulting increase in glutamine leads to hyperosmolanty
and mitochondrial dysfunction, causing astrocytic swelling and
impairment . Increased glutamine formation also decreases total brain glutamate
stores , impairing excitatory neurotransmission (mediated by NMDA, AMPA . and
kainate receptors ) and neuronal energy production . In addition , ammonia can be
detoxified to glutamate via glutamate dehydrogenase depleting a -ketoglutarate and
further impairing energy metabolism in the brain.
3S
37
(Choice B) Carnitine is responsible for transport of fatty acids into the mitochondria
for beta - oxidation . Impaired beta-oxidation of fatty acids to acetyl-CoA can occur with
excessive alcohol consumption.
00 : 48
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O . Id
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O U U I U U I I , a i i i i i i u i i i a t a n u?
iuia / anu iicumnai eneiyy piuuu uun
m
^
^
*
I alt ufator
Notes
A
detoxified to glutamate via glutamate dehydrogenase depleting o -ketoglutarate and
further impairing energy metabolism in the brain.
(Choice B) Carnitine is responsible for transport of fatty acids into the mitochondna
for beta -oxidation . Impaired beta-oxidation of fatty acids to acetyl-CoA can occur with
excessive alcohol consumption .
(Choice D) Elevated lactate can be seen m patients with cirrhosis , particularly
conditions that cause elevated lactate production ( eg, sepsis , tissue
hypoperfusion). The fiver metabolizes lactate , and the lack of functioning
hepatocytes in cirrhosis decreases hepatic lactate disposal .
in
&
(Choice E) Oxindole is a tryptophan derivative formed by bacteria in the gut and
normally cleared by the liver It causes sedation muscle weakness hypotension
and coma Elevated levels of oxindole have been found in patients with hepatic
encephalopathy .
,
Educational objective:
Hyperammonemia in hepatic encephalopathy results in depletion of o -ketoglutarate
causing inhibition of the Krebs cycle Excess ammonia also depletes glutamate , an
excitatory neurotransmitter , and causes accumulation of glutamine , resulting in
astrocyte swelling and dysfunction
.
References:
1 , Ammonium metabolism in humans.
2. Roles of glutamine in neurotransmission.
37
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I akuldior
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A 34 year-old woman with a history of chronic urinary tract infections comes to the
office complaining of dysuria . A urine sample is obtained and sent for culture .
Gram-negative bacteria isolated from the urine are found to form pink colonies on
lactose -containing media Several days later, bacterial isolates from a new urine
sample do not ferment lactose due to the deletion of a single nucleotide from the
DNA sequence This genomic change is most consistent with which of the following
u
15
is
17
13
20
21
22
23
24
25
25
A Conservative mutation
B Missense mutation
C Nonsense mutation
C D Silent mutation
C Frameshrft mutation
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A 34-year -old woman with a history of chronic urinary tract infections comes to the
office complaining of dysuria . A urine sample is obtained and sent for culture .
Gram-negative bacteria isolated from the urine are found to form pink colonies on
lactose -containing media . Several days later bacterial isolates from a new urine
sample do not ferment lactose due to the deletion of a single nucleotide from the
DNA sequence This genomic change is most consistent with which of the following :
A Conservative mutation [2%]
C B . Missense mutation [18%]
C . Nonsense mutation [13%]
O D . Silent mutation [2%]
* » E . Frameshift mulatto - [65%]
Explanation:
A mutation is defined as any change in the DNA sequence When such changes
occur within exons or important non coding regions , altered protein quantity or
function may result Deletion of a single nucleotide causes a frameshift mutation , as
does deletion or insertion of any number of nucleotides that are not multiples of
three With a frameshift mutation, a shift in the reading frame by one or two base
pairs causes the production of an entirely different protein ( often with a premature
stop codon). In the above vignette , a nucleotide deletion alters lactose metabolism
by decreasing the formation of a required enzyme,
-
(Choice A) A conservative mutation is a specific type of missense mutation where
one amino acid is replaced with another amino acid that has similar biochemical
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Explanation:
A mutation is defined as any change in the DNA sequence When such changes
occur within exons or important non-coding regions , altered protein quantity or
function may result. Deletion of a single nucleotide causes a frameshift mutation , as
does deletion or insertion of any number of nucleotides that are not multiples of
three With a frameshift mutation , a shift in the reading frame by one or two base
pairs causes the production of an entirely different protein ( often with a premature
stop codon) In the above vignette , a nucleotide deletion alters lactose metabolism
by decreasing the formation of a required enzyme.
(Choice A) A conservative mutation is a specific type of missense mutation where
one amino acid is replaced with another ammo acid that has similar biochemical
characteristics . Conservative mutations do not alter protein length and may or may
not alter secondary structure protein stability , and protein function
(Choice B) A missense mutation is a point mutation ( single base substitution ) that
causes a codon change resulting in the incorporation of the wrong amino acid into
the polypeptide chain.
(Choice C) A nonsense mutation is a single base substitution that introduces a
premature stop codon (UAA , UAG , or UGA ) , resulting in the production of a
truncated, usually nonfunctional, protein . The mutation described in the question
stem is not a nonsense mutation because it is caused by a single base deletion, and
not by a single base substitution,
35
3S
37
(Choice ) A silent mutation is a single base substitution within a codon that does
not change the amino acid coded for by that codon ( due to redundancy of the code).
Silent mutations have no effect on protein formation or function .
V
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Ldb Valuer
Newt
Notes
(
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does deletion or insertion of any number of nucleotides that are not multiples of
three With a frameshift mutation , a shift in the reading frame by one or two base
pairs causes the production of an entirely different protein ( often with a premature
stop codon). In the above vignette , a nucleotide deletion alters lactose metabolism
by decreasing the formation of a required enzyme.
A
(Choice A) A conservative mutation is a specific type of missense mutation where
one amino acid is replaced with another amino acid that has similar biochemical
characteristics. Conservative mutations do not alter protein length and may or may
not alter secondary structure , protein stability, and protein function.
b
(Choice B) A missense mutation is a point mutation ( single base substitution ) that
causes a codon change resulting in the incorporation of the wrong amino acid into
the polypeptide chain
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(Choice C) A nonsense mutation is a single base substitution that introduces a
premature stop codon (UAA , UAG , or UGA } , resulting in the production of a
truncated , usually nonfunctional , protein The mutation descnbed in the question
stem is not a nonsense mutation because it is caused by a single base deletion , and
not by a single base substitution ,
(Choice D) A silent mutation is a single base substitution within a codon that does
not change the amino acid coded for by that codon ( due to redundancy of the code ) .
Silent mutations have no effect on protein formation or function.
Educational objective;
Single nucleotide deletions shift the reading frame , often creating a premature stop
codon or dramatically changing the protein structure
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The interaction of certain growth factors with their receptors leads to the following
sequence of events:
Binding of growth factor ligand
U
i
Autophosphorylation of tyrosine
residues
1
Interaction with SOS protein
n
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Activation of MX " protein
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Activation of Raf kinase
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Activation of MAP kinase
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Interaction with SOS protein
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Activation of " X ” protein
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Gene transcription
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Protein " X ' becomes activated when it binds which of the following :
1
C A ATP
C B . cAMP
O C. IP
j
:
D . GTP
C E . Ca
-
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Interaction with SOS protein
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Activation of " X" protein
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Activation of Raf kinase
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Activation of MAP kinase
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Protein " X " becomes activated when it binds which of the following :
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O A . ATP [15%]
O B . cAMP [19%]
O C . IP . [16%]
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*
D. GTP [40%]
O E. Ca: [10%]
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t alculutor
E . Ca- [10%]
A
Explanation:
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CYTOPLASM
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Binding proton complex fSOS )
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Transcription fcKtorB
VAP Kmiw
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l alculdior
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r%
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Growth factors can stimulate cell proliferation by altering the expression of certain
genes. This requires the use of signal transduction systems that can transfer the
signal to the nucleus. Examples of such systems include:
1. Ras -MAP kinase pathway
2. PI3K/Akt/mTOR pathway
3 Inositol phospholipid pathway
4. cAMP pathway
5. JAK/STAT pathway
.
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The Ras -MAP kinase pathway starts with a growth factor ligand binding to the
receptor tyrosine kinase , causing auto phosphorylation of the receptor .
Phosphotyrosine produced in this reaction interacts with a number of proteins ( such
as SH2-domain proteins and SOS protein ) leading to Ras activation Ras is a
G-protein that exists in active and inactive forms Inactive Ras contains GDP , while
the active form is bound to GTP . Activated Ras begins a phosphorylation cascade
starting with activation of Raf kinase This cascade results in the activation of MAP
(mitogen activated protein) kinase, which enters the nucleus to influence gene
transcription.
-
-
The Ras protein exists in a balance between its active and inactive forms Inactive
(GDP-containing ) Ras is activated by a signal originating from the receptor tyrosine
kinase Active (GTP-contaming ) Ras is inactivated by GAP (GTPase - activating
protein), which induces the hydrolysis of GTP into GDP, Mutation of Ras can lead to
an inability to split GTP; the resultant permanently activated Ras stimulates cell
proliferation and can lead to cancer,
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NPKt
Note
*
(
olculdtor
receptor tyrosine kinase causing auto-phosphorylation of the receptor .
Phosphotyrosine produced in this reaction interacts with a number of proteins ( such
as SH2-domain proteins and SOS protein), leading to Ras activation . Ras is a
G-protein that exists in active and inactive forms Inactive Ras contains GDP . while
the active form is bound to GTP. Activated Ras begins a phosphorylation cascade
starting with activation of Raf kinase This cascade results in the activation of MAP
(mitogen-activated protein) kinase , which enters the nucleus to influence gene
A
transcription.
The Ras protein exists in a balance between its active and inactive forms . Inactive
(GDP-containing) Ras is activated by a signal originating from the receptor tyrosine
kinase . Active (GTP-containmg ) Ras is inactivated by GAP (GTPase - activating
protein ), which induces the hydrolysis of GTP into GDP . Mutation of Ras can lead to
an inability to split GTP ; the resultant permanently activated Ras stimulates cell
proliferation and can lead to cancer.
(Choice A ) ATP is not a messenger in signal transduction pathways.
(Choice B) cAMP activates cAMP-dependent protein kinase (protein kinase A ) in
the cAMP transduction pathway.
(Choices C and E ) The inositol-iipid pathway uses IP . as a messenger IP causes
Ca'~ release from the endoplasmic reticulum.
Educational objective:
The MAP - kmase signal transduction pathway includes Ras protein , a G-protein that
exists in inactive (GDP-containing ) and active ( GTP -containing) forms Mutated
(permanently activated ) Ras is associated with the development of malignant tumors
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A 23-year -old Caucasian female with history of joint pains develops a facial rash .
Her serum tests are positive for antibodies against small nuclear ribonucleoprotein
particles ( snRNP ) . These snRNP particles participate in the function of which of the
following?
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u
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IS
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A. Peroxisomes
O B. Proteasomes
C . Spliceosomes
C D . Nucleosomes
I E . Ribosomes
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i
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13
1
*
/%
A 23-year-old Caucasian female with history of joint pains develops a facial rash .
Her serum tests are positive for antibodies against small nuclear nbonucleoprotein
particles ( snRNP ), These snRNP particles participate in the function of which of the
following?
C
15
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1?
n
H
20
n
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i-
A . Peroxisomes [3%J
B . Proteasomes [3%]
* # C, Spliceosomes [77%]
O 0 . Nucleosomes [8%]
O E. Ribosomes [9%]
Explanation:
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I alculalor
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*nRf, Ps
Irtlno n 1
5’
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RNA molecules that carry out functions without first being translated into proteins are
referred to as non-coding RNA . Some important forms of non-coding RNA include
small nuclear RNA ( snRNA ). ribosomal RNA ( rRNA ), and transfer RNA
( tRNA ) . Small nuclear RNA molecules are transcribed by RNA polymerase II or III
and are typically associated with specific proteins forming small nuclear
nbonucleoproteins ( snRNPs or “ snurps" ). A collection of snRNPs on pre-mRNA is
referred to as a spliceosome These spliceosomes remove intron sequences from
pre - mRNA by cleaving the 5 ' end of the intron and joining that end to the branch
point The 3' end is subsequently cleaved with the free ends of the remaining exon
01 : so
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Jii t \i tr A
Notes
*
t all tiUii or
*
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i..
RNA molecules that carry out functions without first being translated into proteins are
referred to as non- coding RNA . Some important forms of non- coding RNA include
small nuclear RNA ( snRNA ), ribosomal RNA (rRNA ), and transfer RNA
( tRNA ) Small nuclear RNA molecules are transcribed by RNA polymerase II or III
and are typically associated with specific proteins forming small nuclear
ribonucleoproteins ( snRNPs . or ‘snurps ' ). A collection of snRNPs on pre- mRNA is
referred to as a spliceosome . These spliceosomes remove mtron sequences from
pre - mRNA by cleaving the 5 ' end of the intron and joining that end to the branch
point . The 3' end is subsequently cleaved with the free ends of the remaining exon
mRNA and ligated with a phosphodiester linkage . Anti-snRNP antibodies are
present in mixed connective tissue disease
(Choice A) Peroxisomes are cytoplasmic organelles containing oxidative enzymes
such as catalase , D- amino acid oxidase and uric acid oxidase These organelles
are ubiquitous among eukaryotes and are most abundant in the liver and kidneys
where detoxification of ingested and environmental materials occurs In the liver
peroxisomes also play a role in the breakdown of fatty acids *
(Choice B) Degradation of proteins and polypeptides occurs mainly in
proteasomes and lysosomes . Proteasomes mainly degrade intracellular proteins,
while lysosomes degrade extracellular proteins.
(Choice D) Eukaryotic chromatin is composed of repeated subunits called
nucieosomes which consist of histone protein cores around which double stranded
DNA is wrapped . Nucieosomes are important for the compact packaging of dsDNA
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I dkuldtor
point The 3' end is subsequently cleaved with the free ends of the remaining exon
mRNA and ligated with a phosphodiester linkage . Anti-snRNP antibodies are
present in mixed connective tissue disease .
A
(Choice A) Peroxisomes are cytoplasmic organelles containing oxidative enzymes
such as catalase , D- amino acid oxidase and uric acid oxidase These organelles
are ubiquitous among eukaryotes and are most abundant in the liver and kidneys
where detoxification of ingested and environmental materials occurs . In the liver
peroxisomes also play a role in the breakdown of fatty acids,
17
13
(Choice B) Degradation of proteins and polypeptides occurs mainly in
proteasomes and lysosomes Proteasomes mainly degrade intracellular proteins ,
H
while lysosomes degrade extracellular proteins.
b
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(Choice D) Eukaryotic chromatin is composed of repeated subunits called
nudeosomes which consist of histone protein cores around which double stranded
DNA is wrapped . Nucleosomes are important for the compact packaging of dsDNA
into chromosomes with the aid of other packaging proteins.
(Choice E) Ribosomes are present in the cytoplasm and are required for the
translation of mRNA into protein The molecule rRNA along with ribosomal protein ,
comprises individual ribosomes . Within the cell rRNA engages mRNA and
facilitates the entry of tRNA during the formation of polypeptide chains .
Educational Objective;
Small nuclear ribonucleoprotein particles ( snRNPs) are important components of the
spliceosome a molecule which functions to remove introns from pre -mRNA during
processing within the nucleus .
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I tilculdior
A
b
6
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9
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12
Recent studies have suggested that increased serum levels of homocysteine may
predispose to thrombosis , and that decreasing one’s homocysteine levels may
benefit patients at risk for coronary or cerebral artery thrombosis . Folic acid and
vitamin B12 supplementation can decrease serum homocysteine levels according to
the following mechanism :
13
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3S
O A . Methionine
O B. Cysteine
O C . Serine
O D . Succinyl-CoA
E Methylmalonyl-CoA
F . Glutamic acid
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t <i l c u f t i l o r
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n
A
Recent studies have suggested that increased serum levels of homocysteine may
predispose to thrombosis , and that decreasing one’s homocysteine levels may
benefit patients at risk for coronary or cerebral artery thrombosis . Folic acid and
vitamin B12 supplementation can decrease serum homocysteine levels according to
the following mechanism :
13
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13
17
13
19
.
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Teirahydrofolaie
Cooommin
Matr.yicodalamin
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tv
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* <•A . Methionine [71%]
O B . Cysteine [16%]
O C . Serine [1%]
O D . Succinyl-CoA [3%]
E. Methylmalonyl-CoA [9%J
C F, Glutamic acid [1%]
33
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Explanation:
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5-Mflfflyl
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S-adanosyl
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Homocysteine
Serine
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Calculator
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Notes
t ottiiTTHno 4>nthaw
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Ldb Values
Newt
Explanation:
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<S
F . Glutamic acid [1%]
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Cystathionine
-
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homocysteine
Cystathionine
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Vitmm Be
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3S
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*
Cysteine
Methionine and tetrahydrofolate are formed when methyl- tetrahydrofolate donates a
methyl group to homocysteine. Tetrahydrofolate rapidly accepts one -carbon
moieties to re-enter the cycle The conversion of homocysteine to methionine
requires the cofactor vitamin B12 (cobalamin ). In vitamin B12 deficiency ,
tetrahydrofolate cannot be regenerated . Folate metabolism is consequently
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Notes
I dlcufdtor
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Vitamin 06
Cyst .-tWvc frjrf .n-
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Cysteine
Methionine and tetrahydrofolate are formed when methyl-tetrahydrofolate donates a
methyl group to homocysteine Tetrahydrofolate rapidly accepts one -carbon
moieties to re-enter the cycle The conversion of homocysteine to methionine
requires the cofactor vitamin B12 ( cobaiamm ). In vitamin B12 deficiency ,
tetrahydrofolate cannot be regenerated. Folate metabolism is consequently
impaired
Defective DNA synthesis and its resultant megaloblastic erythropoiesis is seen in
both vitamin B12 and folate deficiency . Homocysteine levels are elevated in both
conditions as well. Elevated homocysteine is a risk factor for arterial and venous
thrombosis .
Vitamin B12 is alone responsible for the conversion of methylmalonyl CoA to
succinyl CoA, Thus in vitamin B12 deficiency , methylmalonyl CoA levels are
elevated The result is the incorporation of nonphysiologic fatty acids into neuronal
lipids This contributes to the neurologic dysfunction in vitamin B12 deficiency as
well. Importantly, whereas homocysteine is elevated in both folic acid and vitamin
B12 deficiency, methylmalonyl CoA is elevated in vitamin B12 deficiency only.
Educational Objective:
Homocysteine is converted to methionine using methylcobalamin and methyl
tetrahydrofolate.
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tskuldior
1
5
6
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9
n
11
n
13
U
15
is
17
A 10-year -old boy suffers a laceration to his left knee while playing outside He is
brought to the emergency department where his wound is cleansed with sterile
saline solution and closed with sutures . In response to the injury, the patient' s
fibroblasts begin to increase protein synthesis locally . During this process an
aminoacyl- tRNA synthetase erroneously charges a proline -carrier tRNA with leucine.
Which of the following is most likely to happen with the leucine residue?
Q
18
19
C . It is randomly incorporated into the polypeptide chain , halting chain
elongation
21
n
27
leucine
B. It is incorrectly incorporated into the polypeptide chain in proline s place
20
21
25
26
A. It is properly incorporated into the polypeptide chain at a site requiring
-
D . It is never incorporated into the polypeptide chain and remains attached to
tRNA
O E . It is rapidly cleaved by glycosylase
23
29
30
31
32
33
31
35
3S
37
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A
ft
5
G
7
8
9
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n
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21
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37
A 10-year -old boy suffers a laceration to his left knee while playing outside He is
brought to the emergency department where his wound is cleansed with sterile
saline solution and closed with sutures . In response to the injury, the patient' s
fibroblasts begin to increase protein synthesis locally . During this process , an
aminoacyl- tRNA synthetase erroneously charges a proline -carrier tRNA with leucine,
Which of the following is most likely to happen with the leucine residue?
b
A. It is properly incorporated into the polypeptide chain at a site requiring
^ leucine [6%)
B t is incorrectly incorporated into the polypeptide chain in proline 's place
[56%]
C . It is randomly incorporated into the polypeptide chain , halting chain
elongation [5%]
D It is never incorporated into the polypeptide chan and remains attached to
“
tRNA [12%]
O E . It is rapidly cleaved by glycosylase [20%]
Explanation:
Ammo acid activation and attachment to the 3' end of tRNA is catalyzed by
aminoacyl- tRNA synthetases ( AA-tRNA synthetases ) . Each amino acid/tRNA pair
has a specific AA-tRNA synthetase that links them together . These enzymes are
responsible for ensuring that each amino acid binds to the tRNA with the proper
anticodon. Aminoacyl-tRNA synthetase activation and binding sites are highly
specific for their correct amino acids and tRNA molecules Additionally some
ifi *r fPMA mnlonil^ c anH huHrAliraa
AA
ci nfhofsicfic
"nroofroflH" fkoir c
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D It is never incorporated into the polypeptide chain and remains attached to
tRNA [12%]
C E. It is rapidly cleaved by glycosylase [20% ]
Explanation:
Amino acid activation and attachment to the 3 end of tRNA is catalyzed by
aminoacyl- tRNA synthetases ( AA -tRNA synthetases ) . Each amino acid/tRNA pair
has a specific AA-tRNA synthetase that links them together . These enzymes are
responsible for ensuring that each amino acid binds to the tRNA with the proper
anticodon . Aminoacyl- tRNA synthetase activation and binding sites are highly
specific for their correct amino acids and tRNA molecules Additionally, some
AA -tRNA synthetases can "proofread' their specific tRNA molecules and hydrolyze
the amino acid bond when their tRNAs are incorrectly charged The error rate for
AA-tRNA synthetases is thus very low at less than 1 error per 10 * charges.
r
b
1
During protein synthesis . tRNA acts as an adaptor molecule between the codons
found on mRNA and the amino acids being incorporated into the polypeptide chain .
The sequence of ammo acids in a polypeptide chain is dictated by binding of the
tRNA anticodon to its complementary codon on the mRNA molecule being
translated . Erroneous amino acidtRNA coupling by the AA - tRNA synthetase causes
the wrong amino acid to be incorporated into the growing polypeptide chain . In the
scenario descnbed in the question , when the ribosome encounters the proline codon
on mRNA ( e g, CCC ) , the complementary tRNA (GGG ) binds . If this tRNA is
improperly charged with leucine leucine gets incorrectly incorporated into the
growing polypeptide chain in proline ' s place .
(Choice C) During polypeptide chain elongation , ribosomes move from codon to
codon on mRNA in the 5 ' to 3 ' direction , sequentially adding amino acids from
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the ammo acid bond when their tRNAs are incorrectly charged The error rate tor
AA-tRNA synthetases is thus very low at less than 1 error per 10' charges
*
Notes
Calculator
During protein synthesis tRNA acts as an adaptor molecule between the codons
found on mRNA and the amino acids being incorporated into the polypeptide chain .
The sequence of amino acids in a polypeptide chain is dictated by binding of the
tRNA anticodon to its complementary codon on the mRNA molecule being
translated . Erroneous amino acid tRNA coupling by the AA - tRNA synthetase causes
the wrong amino acid to be incorporated into the growing polypeptide chain In the
scenario described in the question when the ribosome encounters the proline codon
on mRNA ( e g . , CCC ). the complementary tRNA (GGG ) binds. If this tRNA is
improperly charged with leucine leucine gets mcorrectly incorporated into the
growing polypeptide chain in proline ' s place .
,
20
21
22
*
2
25
26
*
27
(Choice C) During polypeptide chain elongation , ribosomes move from codon to
codon on mRNA in the 5 ' to 3' direction , sequentially adding amino acids from
aminoacyl- tRNA to the peptide chain. This continues until the ribosome encounters
a stop codon ( UAA , UAG , or UGA ). Releasing factors then assist in polypeptide
chain termination.
23
29
(Choice E) DNA glycosylases are enzymes involved in DNA base excision repair
30
Educational objective:
The sequence of amino acids in a growing polypeptide chain is dictated by the
interaction of the mRNA codon with the tRNA anticodon. tRNA that is mischarged
with the incorrect ammo acid ( and not corrected by AA -tRNA synthetase
proofreading ) will incorporate the wrong amino acid mto the growing polypeptide
chain as there is no amino acid proofreading during protein translation .
31
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35
3&
37
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A
&
6
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a
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A
A 78-year-old woman comes to the office due to tenderness and easy bleeding of
the gums when she brushes her teeth . The patient has brushed her teeth twice a
day for as long as she can remember and has not experienced these symptoms
before . Physical examination shows swollen gingiva that bleed on probing Her skin
findings are shown in the image below .
13
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0
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as
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3$
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37
Further questioning reveals that the patient lives alone and that her diet consists
primarily of tea and toast . Her symptoms are most likely caused by hypoactivity of an
Imeritc ?
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n
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13
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1-
15
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13
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20
21
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23
2S
26
27
23
29
Further questioning reveals that the patient lives alone and that her diet consists
primarily of tea and toast . Her symptoms are most likely caused by hypoactivity of an
enzyme found in which of the following compartments?
30
31
32
33
34
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C A . Extracellular space
O B Golgi apparatus
O C Lysosomes
O D . Mitochondria
O E. Nucleus
F . Rough endoplasmic reticulum
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t dkuldtor
A
5
6
7
a
9
10
11
1?
13
H
15
15
b
17
IB
20
:
21
22
25
25
27
28
29
*
l
19
23
5
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;
a
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.
-
Further questioning reveals that the patient lives alone and that her diet consists
primarily of tea and toast . Her symptoms are most likely caused by hypoactivity of an
enzyme found in which of the following compartments?
30
O A. Extracellular space [14%]
31
32
33
34
B . Golgi apparatus [13%]
O C . Lysasomes [6%]
3S
O D. Mitochondria [11%]
35
37
O E. Nucleus [4%]
^
* F. Rough endcp:a >r ; reticulum [52%]
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*
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I dkuldtsr
* F. Rough endoplasmic reticulum [52%]
A
Explanation:
9
10
11
Disorders caused by defective collagen synthesis
n
13
u
IS
15
17
Ii
Disorder
Efilers Danlos
syndrome
( types I & II )
* Hyperextensihle . fragile skm
* Most common form of EDS
Osteogenesis
imperfecta
* Spontaneous fractures
* Bone & tooth malformation
H
20
21
n
23
as
25
27
Scurvy
28
29
30
31
32
33
3
*
35
36
37
Defining characteristics
* Joint hypoirnobility
Impairment
Mutation in type V collagen
Mutation in Typo \ collagen
* Blue sclerae
* Bleeding gums
* Ecchymoses & petechtae
* Impaired wound healing
t ack of vitamin C
impairs
collagen hydroxylation
© eWorld
This patient likely has vitamin C deficiency ( scurvy ) . In the United States , vitamin
C deficiency is seen primarily among malnourished populations including alcoholics
the poor and the elderly The symptoms of scurvy reflect impaired formation of
collagen and include gingival swelling / bleeding , petechiae, ecchymoses. and
poor wound healing . Perifollicular hemorrhages and coiled ( corkscrew ) hairs
are also commonly seen .
,
,
V
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I
Notes
(
alculdlor
I
A
G
7
8
9
io
11
12
13
H
15
K
17
ij
H
20
21
22
*
23
24
25
25
This patient likely has vitamin C deficiency ( scurvy ). In the United States , vitamin
C deficiency is seen primarily among malnourished populations including alcoholics ,
the poor and the elderly The symptoms of scurvy reflect impaired formation of
collagen and include gingival swelling. bleeding, petechiae. ecchymoses and
poor wound healing Perifollicular hemorrhages and coiled (corkscrew ) hairs
are also commonly seen .
,
u
Collagen synthesis is a complex process that begins with the transcription of
collagen genes in the nucleus (Choice E) Collagen a-chains are then synthesized
by rough endoplasmic reticulum (RER }-bound ribosomes and directed into the
cisternae of the RER . Within the RER , specific proline and lysine residues are
post-transiationatly hydroxy lated to hydroxyproline and hydroxylysine by prolyl
hydroxylase and lysyl hydroxylase respectively Vitamin C is a required cofactor
for this post-translational modification Defective hydroxylation of these residues
severely diminishes the amount of collagen secreted by fibroblasts and impairs triple
helix stability and covalent crosslink formation.
27
23
29
30
31
32
33
34
35
35
37
(Choices A and B] After formation of the triple helix , procollagen molecules are
secreted from the cell via the Golgi apparatus . Propeptides at the N- and C termmals are cleaved by extracellular procollagen peptidase to form insoluble
tropocoilagen molecules These monomers then self -assemble into collagen fibrils
that are subsequently crosslinked via lysyl oxidase .
(Choices C and D) Lysosomes and mitochondria are not directly involved in the
synthesis of collagen .
Educational objective:
The hvdroxvlation of oroline and Ivsine residues in collaoen helps it attain its
04 : 29
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Ldb Values
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Notes
I rilcufdlor
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A
by rough endoplasmic reticulum ( RER ) bound ribosomes and directed into the
cisternae of the RER . Within the RER , specific proline and lysine residues are
post-translationally hydroxylated to hydroxyproline and hydroxylysine by prolyl
hydroxylase and lysyl hydroxylase respectively . Vitamin C is a required cofactor
for this post translational modification Defective hydroxylation of these residues
severely diminishes the amount of collagen secreted by fibroblasts and impairs triple
helix stability and covalent crosslink formation .
'
b
(Choices A and B) After formation of the triple helix , procollagen molecules are
secreted from the cell via the Golgi apparatus . Propeptides at the N- and C terminals are cleaved by extracellular procollagen peptidase to form insoluble
tropocollagen molecules These monomers then self -assemble into collagen fibrils
that are subsequently crosslinked via lysyl oxidase .
(Choices C and D) Lysosomes and mitochondria are not directly involved in the
synthesis of collagen .
Educational objective:
The hydroxylation of proline and lysine residues in collagen helps it attain its
maximum tensile strength This process occurs in the rough endoplasmic reticulum
and requires vitamin C as a cofactor . Impaired collagen synthesis resulting from
vitamin C deficiency ( scurvy ) can lead to fragile vessels predisposing to gingival
bleeding, ecchymosis, and petechia.
References:
1 . Be vigilant for scurvy in high- risk groups.
37
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I (ilcufdtor
A
b
6
7
a
9
10
11
12
13
U
15
15
1?
ia
19
20
-
A 22-year old Caucasian male develops recurrent skin blistering and prefers to work
as nocturnist . Laboratory evaluation shows elevated total plasma porphyrins . Which
of the following enzymes is most likely deficient in this patient?
C A 5 -Aminolevulinate synthase
B 5 -Aminolevulinate dehydrase
O C HMB synthase
D Uroporphynnogen decarboxylase
E Bilirubin glucuronyl transferase
21
72
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t <ilc uftilor
A
5
6
7
a
9
A
A 22-year -old Caucasian male develops recurrent skin blistering and prefers to work
as nocturnist . Laboratory evaluation shows elevated total plasma porphyrins Which
of the following enzymes is most likely deficient in this patient?
10
11
n
O A. 6 -Aminolevulinate synthase [14%]
13
U
15
IS
C B. 5'Aminolevutmate dehydrase [11%]
17
13
19
b
O C HMB synthase [&%]
*
D. Uroporphynnogen decarboxylase [67%]
E. Bilirubin glucuronyl transferase [2%]
20
2\
22
Explanation:
21
Alcohol
Barbiturates
Hypoxia
n
25
27
;
23
29
*
©
30
31
32
33
34
3S
ALA synthase
Afrundlevu inic acid
4
Sucboyl CoA
©
Glycine
ALA d&hyfratase
3S
37
Porphobilinogen < PBG )
Glucose
HUB synthase
.
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20
t akuldlor
ft
Alcohol
Barbiturates
Hypoxia
10
13
19
Notes
Explanation:
9
17
Lab Values
Newt
’J
©
Ammolevu
^ic Serf
Suoctnyl CoA
ALA synthase
4-
+
©
Glycine
ALA dehydratase
21
22
23
n
25
27
23
29
30
31
32
33
34
3S
3S
37
Porphobilinogen (PBG )
HMB sytilhjso
r Urt>poq)hynoogof \ I
Synthase)
1
GluCOSo
Heme
HMB
(Urutwphyriootjen |)
FefroeheWase
Uroporphyrinogen tit
synthase
Protoporphyrin IX
Uroporphyrinogen ill
l
Uroporphyrinogen
decarboxylase
f Protoporphyrmogen IX
Coproporphynnogen lit
v
C!* :* nt:*:*u Vn
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H
V
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o
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Lab Valuer
Notes
I aletifator
hi
A
FvtTQChQi&taSG
Uroporphyrinogen III
synthase
Protoporphyrin IX
Uroporphyrinogon III
1
Uropotph ymogen
CfocivboxyiasG
Copfopf>rphyrmoQon III
b
Prdoporphymogon ix
Copmporphynnogon
0 daso
"
17
13
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This patient most likely has porphyria cutanea tarda the most common disorder of
porphynn synthesis. Enzyme deficiencies in the early steps in porphyrin synthesis
cause abdominal pain and neuropsychiatric manifestations without photosensitivity
while late step derangements ( after the condensation of porphobilinogen ) cause
photosensitivity More specifically , defects in URO decarboxylase COPRO
oxidase , PROTO oxidase and Ferrochelatase result in photosensitivity
Photosensitivity induced by porphyria is thought to be mediated by the formation of
porphyrin- mediated superoxide free radicals from oxygen upon exposure to sunlight
,
,
25
27
28
29
30
31
32
33
34
35
35
37
(Choice A) Deficiency of ALA-synthase will result in a decrease in formation of all
porphyrins . This deficiency will not result in porphyria but will result in a decrease in
heme synthesis and concurrent hypochromic , microcytic anemia . Pyridoxal
phosphate ( Vitamin B6 ) is the cofactor required for activity of ALA synthase thus,
,
pyridoxine deficiency can result in microcytic hypochromic anemia secondary to
decreased heme synthesis
(Choices B and C) Deficiencies in ALA dehydrase and HMB synthase do not result
in photosensitivity because the metabolites that accumulate in these enzyme
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Notes
I dlculdtor
LJR
A
while late step derangements ( after the condensation of porphobilinogen ) cause
photosensitivity More specifically , defects in URO decarboxylase COPRO
oxidase . PROTO oxidase , and Ferrocheiatase result in photosensitivity .
Photosensitivity induced by porphyria is thought to be mediated by the formation of
porphyrin- med;ated superoxide free radicals from oxygen upon exposure to sunlight,
it
12
13
H
15
IS
17
13
19
(Choice A) Deficiency of ALA-synthase will result in a decrease in formation of all
porphyrins This deficiency will not result in porphyria but will result in a decrease in
heme synthesis and concurrent hypochromic , microcytic anemia . Pyridoxal
phosphate ( Vitamin Bg } is the cofactor required for activity of ALA synthase ; thus.
pyridoxine deficiency can result in microcytic hypochromic
decreased heme synthesis.
anemia
ti
secondary to
20
21
22
23
24
(Choices B and C) Deficiencies in ALA dehydrase and HM8 synthase do not result
in photosensitivity because the metabolites that accumulate in these enzyme
deficiencies are not porphyrinogens or porphyrins and are therefore unable to react
with oxygen upon excitation by ultraviolet light.
25
27
20
20
30
31
32
33
34
35
35
37
(Choice E) Bilirubin glucuronyltransferase is a hepatic enzyme responsible for the
conjugation of bilirubin with glucuronide , improving solubility for biliary excretion A
decrease in glucuronyltransferase results in unconjugated hyperbilirubinemia
Educational Objective:
Enzyme deficiencies of the early steps in porphyrin synthesis cause neuropsychiatnc
manifestations without photosensitivity, while late step derangements lead to
photosensitivity Photosensitivity in porphyria causes vesicle and blister formation on
sun- exposed areas as weil as edema , pruritus, pain and erythema
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Notes
I dlculdtor
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9
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A 24-year -old man comes to emergency department complaining of abdominal pain
vomiting and severe watery diarrhea . He recently returned from a camping trip and
admits to eating wild mushrooms that he collected in the woods His past medical
history is insignificant and he takes no medications . He does not use illicit drugs On
physical examination he is ill- appearing and jaundiced. His liver edge is soft, tender
and palpable 4 cm below the right costal margin Laboratory tests are significant for
elevated levels of alanine aminotransferase , aspartate aminotransferase , and
bilirubin Synthesis of which of the following is most likely to be directly inhibited by
the responsible toxin?
U
O A. DNA
8 Messenger RNA
O C Protein
C D Ribosomal RNA
C E . Transfer RNA
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A
-
A 24 year-old man comes to emergency department complaining of abdominal pain,
vomiting and severe watery diarrhea . He recently returned from a camping trip and
admits to eating wild mushrooms that he collected in the woods. His past medical
history is insignificant and he takes no medications . He does not use illicit drugs . On
physical examination he is ill- appearing and jaundiced His liver edge is soft, tender ,
and palpable 4 cm below the right costal margin Laboratory tests are significant for
elevated levels of alanine aminotransferase aspartate aminotransferase , and
bilirubin Synthesis of which of the following is most likely to be directly inhibited by
the responsible toxin?
&
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O A. DNA [G%]
* <• B .
Vessenger RNA (43%)
O C Protein [27%]
O D . Ribosomal RNA [20%]
O E. Transfer RNA [4%]
Explanation:
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Eukaryotic RNA polymerase
Major RNA product
RNA polymerase I
Ribosomal RNA
RNA polymerase II
Messenger RNA
HMA nnL
T
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III
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C D Ribosomal RNA [20%]
C E . Transfer RNA [4%]
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Explanation:
k
n
Eukaryotic RNA polymerase
Major RNA product
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RNA polymerase I
Ribosomal RNA
RNA polymerase II
Messenger RNA
RNA polymerase III
Transfer RNA
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© LWofld
Amatoxins are found in a variety of poisonous mushrooms ( eg , Amanita phailoides ,
known as death cap) and are responsible for the majority of mushroom poisoning
fatalities worldwide Ingestion of 1 or more amatoxin-containing mushrooms is a lifethreatening emergency . After absorption by the gastrointestinal tract, amatoxins are
transported to the liver via the portal circulation where active transport by organic
anion transporting polypeptide ( OATP ) and sodium taurochoiate co -transporter
( NTCP ) concentrates the toxin within the Irver cells There amatoxins bind to DNA dependent RNA polymerase type II and halt mRNA synthesis, ultimately resulting in
apoptosis , Other organ systems with rapid cellular turnover can also be affected in
amatoxin poisoning , including the gastrointestinal tract and proximal convoluted renal
tubules .
Symptoms typically start 6-24 hours after ingestion and include abdominal pain
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transported to the liver via the portal circulation where active transport by organic
anion transporting polypeptide ( QATP ) and sodium taurocholate co -transporter
( NTCP ) concentrates the toxin within the liver cells There amatoxins bind to DNA dependent RNA polymerase type II and halt mRNA synthesis, ultimately resulting in
apoptosis . Other organ systems with rapid cellular turnover can also be affected in
amatoxin poisoning , including the gastrointestinal tract and proximal convoluted renal
tubules .
Notes
( <alculdtor
A
b
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Symptoms typically start 6-24 hours after ingestion and include abdominal pain
vomiting , and severe , cholera-like diarrhea that may contain blood and
mucus Severe poisoning can lead to acute hepatic and renal failure Urine testing
for a-amanttin can confirm suspected amatoxin poisoning.
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3’
(Choice A) Acyclovir and related drugs ( eg , famciclovir and valacyclovir ) are
inhibitors of viral DMA polymerase .
(Choice C) Ricin ( from the castor oil plant Ricinus communis ) is a potent toxin that
inhibits protein synthesis by cleaving the rRNA component of the eukaryotic 60S
subunit .
(Choice D) The only function of RNA polymerase I is to transcribe the majority of
the eukaryotic ribosomal RNA components RNA polymerase I is insensitive to
amatoxins
(Choice E ) Eukaryotic RNA polymerase III transcribes transfer RNA , 5 S ribosomal
RNA , and other small RNA molecules It is only weakly affected by amatoxins
Educational objective:
Amatoxins are found in a variety of poisonous mushrooms { eg Amanita phalloides
known as death cap ) and are potent inhibitors of RNA polymerase II ( halting mRNA
synthesis )
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Microbiologists are investigating sugar metabolism in wild-type and mutant strains of
Escherichia coli Both strains are found to grow viable colonies on lactose containing media. Each strain is then cultured on a new growth medium containing
only glucose . Representative colonies of each strain from the new media undergo
Western blot processing using a fluorescently labeled probe specific for (3 galactosidase Wild-type bacterial colonies are found to contain only trace quantities
of pngalactosidase . However , the mutant colonies express significant amounts of (3galactosidase . Further analysis reveals that the variant strain contains a mutation that
inhibits the binding of a certain protein to its regulatory sequence In which of the
following locations did this mutation most likely occur?
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A Activator protein (CAP ) binding site
O B. Operator locus
O C. Promoter region
O D. Activator protein (CAP ) gene
O E . RNA polymerase cistron
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/%
Microbiologists are investigating sugar metabolism in wild-type and mutant strains of
Escherichia coli . Both strains are found to grow viable colonies on lactose containing media . Each strain is then cultured on a new growth medium containing
oniy glucose . Representative colonies of each strain from the new media undergo
Western blot processing using a fluorescently labeled probe specific for p galactosidase Wild-type bacterial colonies are found to contain only trace quantities
of p-galactosidase However the mutant colonies express significant amounts of (3galactosidase . Further analysis reveals that the variant strain contains a mutation that
inhibits the binding of a certain protein to its regulatory sequence . In which of the
following locations did this mutation most likely occur?
H
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A
* B.
O C.
O D.
O E.
Activator protein (CAP ) binding site [19%J
Operator locus [43%]
Promoter region [27%]
Activator protein (CAP ) gene [9%]
RNA polymerase cistron [2%]
Explanation:
CAP
37
Gene
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Lac I
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Structural
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RNA pel
LacZ
Luc Y
Lac A
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Explanation:
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Structural flaws
Regutalory gene
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CAP
P
Lac 1
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G n
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Lac Z
Lac Y
Lac A
Lac Z
LacV
Lac A
RNA polymornso
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Repressor protein
IS
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Lac I
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Repressor binds
the operator
preventing binding
of RNA polymerase
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cAMP
+
CAP
>)
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P
Lay 1
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Lacloss
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Inducer ( lactose ) causes a
conformational change
preventing rr- pm&sor from
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Lac 1
Lac 2
Uc Y
Lac A
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Lactose
( Inducer ]
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*
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Inducer ( locto&e i causes e
conform ttiiona i change
prove nting repressor Iron’
bribing operator
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onpron
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I a l l uldtor
Lactose
E.CoH cell
a
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*
The /ac operon consists of a regulatory gene ( / ac /), a promoter region ( lac p ), an
operator region ( lac o), and three structural genes ( lac Z lac Y , and lac A ) The lac
Z gene codes for [S-galactosidase , which is responsible for the hydrolysis of lactose
to glucose and galactose . The lac Y gene codes for permease which allows
lactose to enter the bacterium The lac p region is the binding site for RNA
polymerase dunng the initiation of transcription . The Lac I repressor protein is the
product of the lac I gene and is constitutiveJy expressed Repressor proteins , when
bound to the operator region prevent binding of RNA polymerase to the promoter
region, thus decreasing transcription of the lac Z . lac Y, and lac A genes Culture of
E coli in lactose -containing media causes a conformational change in the repressor
protein, preventing its attachment to the operator region and increasing transcription
of the lac operon structural genes.
b
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Culturing E coli in media containing glucose results m reduced expression of the lac
operon , even when the media contains lactose as well. This occurs because the lac
operon is positively regulated by the binding of catabolite activator protein (CAP ) to a
site slightly upstream from the promoter region. This only occurs when cAMP
concentrations are high . Since glucose decreases the activity of adenylyl cyclase
(reducing intracellular cAMP ), the lac operon is repressed in high-glucose
conditions . In summary, the lac operon is regulated by 2 distinct mechanisms :
.
1 Negatively by binding of the repressor protein to the operator locus
2 . Positively by cAMP-CAP binding upstream from the promoter region
Mutations impairing the binding of the repressor protein to its binding site at the
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operon is positively regulated by the binding of catabolite activator protein (CAP ) to a
site slightly upstream from the promoter region. This only occurs when cAMP
concentrations are high Since glucose decreases the activity of adenylyl cyclase
(reducing intracellular cAMP ), the lac operon is repressed in high-glucose
conditions In summary , the lac operon is regulated by 2 distinct mechanisms :
1. Negatively by binding of the repressor protein to the operator locus
2. Positively by cAMP-CAP binding upstream from the promoter region
Mutations impairing the binding of the repressor protein to its binding site at the
operator region will prevent repression of the genes of the lac operon in the
absence of lactose . This results in increased transcnption of the genes of the lac
operon in lactose-deficient media , although the presence of glucose will prevent
maximal transcriptional activity .
(Choices A and D) Mutations that impair the binding of cAMP -CAP to its regulatory
site upstream from the promoter will decrease transcnption of the lac operon as
cAMP-CAP is a posrtrve regulator.
(Choices C and E) Mutations impairing the binding of RNA polymerase to the
promotor region will also reduce transcription of the lac operon
Educational objective:
The lac operon is regulated by two distinct mechanisms : negatively by binding of the
repressor protein to the operator locus and positively by cAMP -CAP binding
upstream from the promoter region Constitutive expression of the structural genes
of the lac operon occurs with mutations that impair the binding of the repressor
protein ( Lac I ) to its regulatory sequence in the operator region.
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IS
-
A 2 year - old boy is brought to the local emergency room by his parents with
complaints of fever and diarrhea for several days Based on his clinical presentation
and the time of the year , the on-call pediatric resident admits the patient for
dehydration secondary to presumptive rotavirus -induced gastroenteritis . A few days
after discharge , the patient is seen by his pediatrician for abdominal distention and
diarrhea after each feeding . The symptoms improve significantly once dairy
products are eliminated from his diet . Which of the following steps in galactose
metabolism is most likely impaired in this patient?
17
U
Gatactrtol
H
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MM
©
Galactose
U
Galactna- 1 -phoaphato
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Galactosyl (J - 1.4 -glucose
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<Z
JDP - GiilaclOHt
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O A. A
O B. B
O c. c
ODD
O E. E
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1?
13
A
A 2-year - old boy is brought to the local emergency room by his parents with
complaints of fever and diarrhea for several days Based on his clinical presentation
and the time of the year , the on-call pediatric resident admits the patient for
dehydration secondary to presumptive rotavirus -induced gastroenteritis . A few days
after discharge , the patient is seen by his pediatncian for abdominal distention and
diarrhea after each feeding The symptoms improve significantly once dairy
products are eliminated from his diet . Which of the following steps in galactose
metabolism is most likely impaired in this patient?
Giikictitcl
®
®
H
Galactose- 1 -phosphate
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Gatatingy!
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jM . J -glucose
kj
<7
UDP - Galaclosa
O A. A [7%]
C B. B [42%]
O C . C [23%]
O D. D [7%]
**
E. E (21%]
Explanation:
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t alculdtor
O A. A [7%]
O B. B [42%]
O C C [23%]
ODD [7%I
* * E. E (21%]
A
Explanation:
—
GalacwoJ
( Causes catar acl )
Gtf/crcfofcw )tKe
-—
Galactose
Galactose I P
uOP-ghjeos**
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Galactose l ^ptwsphafe
*
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Ga aclosrdase
*
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undyl transferase
Lactose -*
(Galactosyl
Glucose
p- 1 .
Lactose
synthase
4 -glucose )
UOP -goiadose Glucose IP
Glucose 6'
phosphdie
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Pyru vale
Lactose ( galactosyl beta-1, 4-glucose or milk sugar ) is a disaccharide present in
milk It is synthesized in the mammary gland by formation of a 1 4 glycosidic linkage
between glucose and galactose . Lactose in the diet is catabolized into glucose and
galactose by an intestinal brush-border disaccharidase called lactase ( a type of
beta-galactosidase more specifically known as lactase-phlorizin hydrolase ). Lactose
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Pyruvate
0
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Lactose ( galactosyl beta -1, 4-glucose or milk sugar ) is a di&accharide present in
milk It is synthesized in the mammary gland by formation of a 1, 4 glycosidic linkage
between glucose and galactose Lactose in the diet is catabolized into glucose and
galactose by an intestinal brush-border disaccharidase called lactase ( a type of
beta -gaiactosidase more specifically known as lactase -phlorizin hydrolase ). Lactose
intolerance FS characterized by gastrointestinal upset upon ingestion of foods
containing lactose , such as dairy products , and is caused by deficiency of lactase
( Answer E ) Primary lactose intolerance is a very common disorder , particularly in
people of African and Asian descent . In contrast to most other races subjects of
Northern European descent maintain lactase activity throughout their life.
Secondary lactase deficiency occurs in association with a number of small intestinal
mucosal diseases such as celiac sprue and viral gastroenteritis The underlying
pathophysiology of this disorder is due to the fact that lactase is concentrated within
epithelial ceils in the microvilli of the small intestine (the brush border ). When these
ceils are damaged in gastroenteritis , the damaged cells slough off and are replaced
by immature cells that have low concentrations of lactase .
The other answer choice options are important in the metabolism of galactose to
either glucose or lactose Galactose is first phosphorylated to
gaiactose-1-phosphate by the enzyme galactokinase (Choice B) Next,
galactose-1-phosphate uridyl transferase ( GALT) catalyzes the conversion of
UDP -glucose and galactose -1-phosphate to UDP-galactose and
glucpse-1-phosphate ( Choice C) . UDP-galactose is then epimerized to
UDP -glucose by UDP -galactose -4 epimerase after which it can participate in the
_ _ iH , _
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t alculdtor
A
galactose-1-phosphate uridyl transferase ( GALT ) catalyzes the conversion of
UDP -glucose and galactose - 1-phosphate to UDP -galactose and
glucose-1-phosphate ( Choice C) . UDP -galactose is then epimerized to
UDP -glucose by UDP -galactoae -4 epimerase after which rt can participate in the
appropriate glucose-related metabolic pathways . Alternatively , UDP-galactose can
be converted to galactosyl beta -1 4-glucose ( lactose ) by lactose synthase within the
mammary glands as part of the formation of milk (Choice D)
,
Galactosemia is an illness that is distinct from lactose intolerance, and it is
characterized by symptoms that start soon after the initiation of breast - feeding .
Galactosemia can be caused by a deficiency of GALT (Type 1), galactokinase ( Type
2), or UDP -glucose 4-epimerase (Type 3 ), Excess galactose in patients with
galactosemia is converted to gaiactitol by aldose reductase ( Choice A ) , and high
levels of galactitol are responsible for many of the symptoms associated with
galactosemia ( especially cataract formation).
Educational objective:
Secondary lactase deficiency can occur after viral gastroenteritis or other diseases
that damage the intestinal epithelium . This disease causes abdominal distention
flatulence , and diarrhea after lactose ingestion.
References:
1 . Disaccharidase deficiency.
2 . Lactose intolerance
in
infants children and adolescents
,
3 . Lactose malabsorption
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t alculdtor
A nonfunctional protein expressed in a cell culture contains 156 amino acid residues
rather than the 130 amino acid residues normally seen in the functional protein. The
nonfunctional protein is still detected by specific antibodies against the functional
protein , Which of the following point mutations best explains the observed finding?
12
u
Notes
-
O A . Silent mutation in intron 2
O B . Missense mutation in exon 1
C . Frameshift mutation in codon 3 of exon 1
O D Nonsense mutation in exon 2
O E Splice site mutation
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A nonfunctional protein expressed in a cell culture contains 156 amino acid residues
rather than the 130 amino acid residues normally seen in the functional protein. The
nonfunctional protein is still detected by specific antibodies against the functional
protein Which of the following point mutations best explains the observed finding?
n
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3?
A. Silent mutation in intron 2 [7%]
O B. Missense mutation in exon 1 [9%]
v
C Frameshift mutation in codon 3 of exon 1 [19%]
O D. Nonsense mutation in exon 2 [8%]
* # E. Splice site mutation [56%]
Explanation:
Changes in the genetic code can result in the formation of altered proteins . For
instance , the protein formed in the cell culture described in the question stem is a
larger , nonfunctional protein The normal protein is shorter and functional This is
explained by mutations at the splice site ( Choice E) After transcription , mRNA
contains sequences from both introns and exons ; this type of RNA is called pre -RNA
or heteronuciear RNA (hnRNA ). The pre-RNA must be processed to mature mRNA
by posttranscriptional modifications including 5' methylguanosine capping, addition
of a 3’ polyadenine ( Poly A ) tail, and splicing Only exons contain the proper base
pairs in the correct order that will result in the formation of an appropriate functional
protein; therefore , introns are excised before translation by a process known as
splicing Mutation of splice sites result in the formation of larger proteins that are
usually nonfunctional but often retain the immunoreactivity of the normal protein
fhifiHinn
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or heteronuclear RNA (hnRNA ) The pre - RNA must be processed to mature mRNA
by posttranscriptional modifications including 5' methylguanosine capping addition
of a 3’ polyadenine ( Poly A ) tail and splicing Only exons contain the proper base
pairs in the correct order that will result in the formation of an appropriate functional
protein; therefore , introns are excised before translation by a process known as
splicing Mutation of splice sites result in the formation of larger proteins that are
usually nonfunctional but often retain the immunoreactivity of the normal protein
(binding to antibodies ).
27
the reading frame of the genetic code usually resulting in the formation of shorter,
nonfunctional proteins.
31
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37
t akulaior
A
b
(Choice B) Missense mutations are characterized by a change in the code through
base substitution, resulting in an amino acid change For instance changing UUU to
UCU changes the translated amino acid from phenylalanine (UUU) to serine (UCU),
The translated proteins may be dysfunctional but usually retain the same size .
(Choice C) Frameshift mutations occur with a deletion , or less commonly, an
insertion of base pairs which are not a multiple of three . Frameshift mutations alter
30
Notes
(Choice A) Silent mutations result m no changes in formed proteins The proteins
are functional and have the same size .
25
26
23
*
(Choice D) Nonsense mutations introduce a stop codon within a gene sequence
resulting in the formation of shorter , nonfunctional proteins .
Educational Objective:
Splice site mutations frequently result in the production of larger proteins with altered
function but preserved immune reactivity .
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14ilc uldlor
A 34- year -old man is found to have an LDL level of 310 mg ' dl and a normal serum
triglyceride level. His father suffered a myocardial infarction at age 39 . and his
paternal grandfather died of a heart attack at age 40 The patient's wife has a normal
lipid profile DNA samples are obtained from several family members for genetic
analysis ., Southern blotting of restriction fragments from a region containing the LDL
receptor gene shows the following pattern:
Patient
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Patient' s
wife
Patient ’s
Patient' s
Patient' s
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brother
father
A
20 kb
12 kb
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10 kb
n
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ClMOrtC
Which of the following statements best describes the DNA analysis results?
31
-
-
8 kb
34
36
37
A. The disease is transmitted in an X -linked recessive fashion
O B. The mutation is probably located in the 10 kb band
O C. The mutation is probably located in the 12 kb band
O D . The patient's brother most likely inherited the mutation
_
E. The patient’s son most likely inherited the mutation
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t alculator
A 34- year -old man is found to have an LDL level of 310 mg^ dL and a normal serum
triglyceride level . His father suffered a myocardial infarction at age 39 . and his
paternal grandfather died of a heart attack at age 40 The patient's wife has a normal
lipid profile DNA samples are obtained from several family members for genetic
analysis ., Southern blotting of restriction fragments from a region containing the LDL
receptor gene shows the following pattern:
n
15
IS
m'
Patient's
wife
Patient ’s
son
Patient's
brother
Patient' s
father
20 kb
13
H
20
12 kb
21
22
21
10 kb
n
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25
8 kb
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Which of the following statements best describes the DNA analysis results?
32
35
37
A. The disease is transmitted in an X -linked recessive fashion [4%]
B . The mutation is probably located in the 10 kb band [2%]
C C . The mutation is probably located in the 12 kb band [3%]
.
*•
D. The patient's brother most likely inherited the mutation [4%]
E. The pat e is son most ike y inherited the mutation |[08%]
-
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E. The patient's son most likely inhented the mutation [88%]
i*S
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37
Explanation:
This patient most likely has heterozygous familial hypercholesterolemia an
autosomal dominant LDL receptor defect that causes high LDL levels and
increases the risk of premature atherosclerosis . Homozygous familial
hypercholesterolemia ( a rarer and more severe form of the disease due to
inheritance of 2 defective LDL receptor alleles ) often presents with coronary heart
disease in childhood adolescence.
,
Southern blotting is a technique that can be used to detect DNA mutations . The
process involves the following steps:
1. DNA extraction from the individual' s cells
2. Restriction endonuclease digestion of the DNA sample into fragments
3. Gel electrophoresis to separate the various sizes of DNA fragments : larger
fragments move slowly and shorter fragments move faster
4. DNA probe ( a single - stranded segment of labeled DNA complementary to the
gene of interest ) to identify the target gene
Once the gene of interest is identified by the DNA probe , various family members'
Southern blots can be compared. Because both the patient and his father are
affected the common DNA segment between them ( 8 kb segment i most likely
represents the mutated gene . The patient's son also has the 8 kb segment,
meaning that he is probably affected as well.
(Choice A) Familial hypercholesterolemia is an autosomal dominant disorder
X -linked recessive mutations are transmitted from unaffected earner mothers to their
sons Father -to-son transmission does not occur
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Notes
(
aleuldtor
affected the common DNA segment between them | 8 Kb segment i most likely
represents the mutated gene The patient’s son also has the 8 kb segment ,
meaning that he is probably affected as well.
(Choice A) Familial hypercholesterolemia is an autosomal dominant disorder
X-hnked recessive mutations are transmitted from unaffected earner mothers to their
sons Father -to-son transmission does not occur
(Choices B and D) The patient ( affected by the disease ) does not possess the 10
kb segment so this segment does not correspond to the mutated gene . The
patient' s brother inherited the 10 kb segment from his father ( not the 8kb mutated
segment ), so he would not be affected
k
,
13
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2&
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*
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*
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(Choice C) The patient and his brother , but not their father , have a 12 kb segment
on Southern blot analysis . Therefore , this segment was likely inherited from the
mother and does not cany the mutation.
Educational objective:
Southern blotting is a technique used to identify DNA mutations. It involves
restnction endonuclease digestion of sample DNA , gel electrophoresis and gene
identification with a labeled DNA probe
References:
1 . Genetic defects causing familial hypercholesterolaemfa:
identification of deletions and duplications In the LDL-receptor gene
and summary of alJ mutations found in patients attending the
Hammersmith Hospital Lipid Clinic.
37
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E coti colonies grown on a lactose-containing medium up-regulate the production of
the enzymes p-galactosidase and galactoside permease Which of the following
best explains the synchronous production of both enzymes in response to lactose?
n
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A There are two activator binding sites for one activator protein
u
C B There are two operators for one repressor protein
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C C There are two repressors for one inducer
17
1B
D . There are two promoters in close proximity to each other
H
E. There is one mRNA coding for both enzymes
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a
9
E . coti colonies grown on a lactose-containing medium up-regulate the production of
the enzymes p-galactosidase and galactoside permease . Which of the following
best explains the synchronous production of both enzymes in response to lactose?
10
&
11
A. There are two activator binding sites for one activator protein [11%]
C B. There are two operators for one repressor protein [13%]
12
13
1J
15
15
17
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20
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22
O C. There are two repressors for one inducer [5%]
*
D . There are two promoters in close proximity to each other [15%]
® E. There is one mRNA coding for both enzymes [55%]
Explanation:
21
24
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CAP
Lac I
27
23
29
Structural genes
Regulatory gen*
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Lac Z
Lac r
Lac A
Lac Z
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RNA pa/ ymomc
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32
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Lacj
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Repressor bind
37
the operator
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or RNA potymefave
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Lactose
I Inducer
*
inducer ( lactose ) causes a
conformational change
preventing repressor from
biding operator
permease
37
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Repressor protein
7
a
*>
9
Lac Y
Lac A
10
It
b
Repressor binds
n
the operator
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( Inducer }
27
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Inducer ( lactose I causes ,1
permease
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conformational change
preventing re pressor from
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30
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The lac operon is the sequence of the E. colt genome which is required for the
metabolism of lactose The lac operon consists of a regulatory gene ( 0 promoter
region (p) , operator region ( o), and three structural genes (z. y, and a ). The z gene
codes for fJ-ga!actosidase ( (3-gal ), which is pnmarily responsible for the hydrolysis of
lactose to glucose and galactose . The y gene codes for permease, a
transmembrane enzyme that increases the permeability of the cell to lactose . The a
gene encodes a p-galactoside transacetylase , which transfers acetyl groups to pgalactosides and is unnecessary for lactose metabolism by E. coli .
u
In prokaryotes , one mRNA transcript contains the sequences for many proteins , and
a single mRNA molecule can be translated into multiple proteins or
polypeptides . For instance , all three proteins of the lac operon (P - galactosidase .
permease , and transacetylase } are synthesized from a single mRNA molecule
containing the z, y, and a gene sequences , respectively . Transcription and
translation of the genes of the lac operon is typically synchronous Remember that a
single mRNA molecule which codes for more than one protein JS referred to as a
polycistronic mRNA , and while most prokaryotic mRNA molecules are polycistronic,
eukaryotic mRNA is rarely polycistronic.
(Choices A - D ) The lac operon, which codes for all three aforementioned proteins
JS regulated by a single operator , a promoter, and a single group of regulatory
elements an inducer , repressor and catabolite activator protein Modulation of the
transcription of this operon through binding of the operator and action of the
repressor or other regulatory elements will change the transcnption of all three lacoperon structural genes ( z, y. and a). On the other hand there are no operators ,
repressors or inducers that can desynchronize the transcription of lac -operon
structural genes
V
il
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(Choices A
Notes
t alculalor
p
In prokaryotes , one mRNA transcript contains the sequences for many proteins , and
a single mRNA molecule can be translated into multiple proteins or
polypeptides For instance, all three proteins of the lac operon ( p - galactosidase
permease , and transacetylase } are synthesized from a single mRNA molecule
containing the z, y , and a gene sequences , respectively . Transcription and
translation of the genes of the lac operon is typically synchronous Remember that a
single mRNA molecule which codes for more than one protein is referred to as a
polycistronic mRNA , and while most prokaryotic mRNA molecules are polycistronic,
eukaryotic mRNA is rarely polycistronic.
n
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galactosides and is unnecessary for lactose metabolism by E coh
£
9
•
rf 1
A
- DJ The lac operon which codes for all three aforementioned proteins
is regulated by a single operator, a promoter , and a single group of regulatory
elements : an inducer, repressor , and catabolite activator protein . Modulation of the
transcription of this operon through binding of the operator and action of the
repressor or other regulatory elements will change the transcription of all three lacoperon structural genes ( z, y, and a) On the other hand , there are no operators
repressors , or inducers that can desynchronize the transcription of lac -operon
structural genes .
Educational Objective:
Bacterial mRNA can be polycistronic, meaning that one mRNA codes for several
proteins. An example of polycistronic mRNA is the bactenal lac operon , which
codes for the proteins necessary for lactose metabolism by E . coii \ the transcription
and translation of these bacterial proteins is regulated by a single promoter,
operator , and set of regulatory elements.
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A 15-year -old boy is found to have unexplained eryihrocytosis on routine laboratory
analysis . Evaluation of his immediate family shows that his father and sister also
have elevated red cell levels Genetic sequencing of the (S-globin gene is performed
in the affected family members . The results show a single base substitution at
amino acid position 82 that replaces the normal lysine residue with
methionine Further analysis shows that this amino acid replacement Empairs the
ionic interaction between the {3 -subunit and 2 , 3 -bisphosphoglycerate As a result of
this mutation , the patients hemoglobin will be most similar to which of the following
hemoglobin types?
C A. Hemoglobin A.
21
B . Hemoglobin C
23
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O C, Hemoglobin F
n
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k
O D Hemoglobin H
(
E . Hemoglobin S
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t aleuldtor
Notes
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a
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14
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18
17
A 15 *year -old boy is found to have unexplained erythrocytosis on routine laboratory
analysis . Evaluation of his immediate family shows that his father and sister also
have elevated red cell levels Genetic sequencing of the fJ- globin gene is performed
in the affected family members . The results show a single base substitution at
amino acid position 82 that replaces the normal lysine residue with
methionine Further analysis shows that this amino acid replacement impairs the
ionic interaction between the (3 -subumt and 2 , 3 -bisphosphoglycerate As a result of
this mutation , the patients hemoglobin will be most similar to which of the following
hemoglobin types ?
i
13
H
20
21
22
23
24
25
26
27
28
29
30
31
*
*
-
O A. Hemoglobin k [3%]
..
C B Hemoglobin C [13%J
v <# C. Hemoglobin F [64%]
C D . Hemoglobin H [5%]
O E. Hemoglobin S [15%]
Explanation:
2 , 3 bisphosphoglycerate and hemoglobin binding
33
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Explanation:
6
A
2 , 3‘bisphosphoglycerate and hemoglobin binding
7
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Hemoglobin F
Hemoglobin A
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-
Replacement Of hm! dme by Sfir ne in fetal
hemoglobin ( educes ponton charge of the
binding pocket limiting 2 3 ' BPG aiiachment
Positrvcly charged histidine and tywno
residues secure negahvely charged
2 3 BPG to the binding pocket
OLWoflfl
This patient most likeiy has familial erythrocytosis due to a p -globin mutation
resulting in reduced binding of 2,3-bisphosphoglycerate ( 2 , 3 - BPG ) 2 , 3-BPG is
synthesized from glycolytic intermediates and binds strongly to deoxyhemoglobin in
^
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i
A
&
6
Mark
A
Replacement of hi&l dtne by senme in fetaJ
hemoglobin reduce positive charge of the
binding pocket limiting 2 3 BPG anechfflent
Positively changed histidme and lysine
residues secu e netjutvely charged
'
2 3 BPG to Ihe binding pocfcei
*
CUWorHJ
This patient most likely has familial erythrocytosis due to a P -globin mutation
resulting in reduced binding of 2,3 *bisphosphoglycerate ( 2 , 3- BPG ). 2 , 3- BPG is
synthesized from glycolytic intermediates and binds strongly to deoxyhemoglobin in
a pocket formed between the 2 beta chains This binding reduces the oxygen
affinity of hemoglobin allowing more oxygen to diffuse into the peripheral
tissues. The hemoglobin 2.3- BPG binding pocket contains positively charged amino
acids ( eg . histidine and lysine ) that attract the negatively charged phosphate groups
in 2 , 3- BPG . Mutations that decrease the positive charge of the binding site
decrease 2 , 3-BPG binding and increase hemoglobin oxygen affinity
Fetal hemoglobin (hemoglobin F ) is synthesized primarily during fetal development
('6 weeks until term) and consists of the usual 2 alpha chains with 2 gamma chains
in place of beta chains. The gamma chains do not bind effectively to 2, 3 - BPG due
to replacement of a histidine residue with serine. As a result , fetal hemoglobin has
significantly higher oxygen affinity than adult hemoglobin A . This allows fetal
hemoglobin to extract more oxygen from the mother ' s adult hemoglobin in the
placenta providing the developing fetus with an adequate supply of oxygen.
(Choice A) Hemoglobin A is formed by nomenzymatic glycosylation of hemoglobin
A Glycosylation can interfere with the binding of 2 , 3-BPG to hemoglobin by altering
the physical structure of the binding pocket which is compensated for by increased
red cell 2.3- BPG levels in patients with diabetes . However the reduced 2 , 3- BPG
binding affinity of this patient s mutated hemoglobin more closely resembles that of
,
kanunrilplniin
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21
n
23
24
25
26
27
2d
29
30
31
* 32
-
*
33
34
*
35
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36
37
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rormea oy non- enzymatic giycosyiation or nemogiooin
A Giycosyiation can interfere with the binding of 2 , 3 -BPG to hemoglobin by altering
the physical structure of the binding pocket which is compensated for by increased
red cell 2,3-BPG levels in patients with diabetes . However , the reduced 2 , 3 -BPG
binding affinity of this patient s mutated hemoglobin more closely resembles that of
hemoglobin F.
{
^ noice AJ
Memogiooin A IS
*
Notes
t alculator
/v
,
b
'
(Choice B) Hemoglobin C results from a mutation in the p -g!obin chain that causes
glutamate to be replaced by lysine Hemoglobin C forms hexagonal crystals and
promotes red cell dehydration , causing a mild chronic hemolytic anemia 2, 3 - BPG
binding and tissue oxygen delivery are not significantly altered.
(Choice D) A defect in the synthesis of alpha chains results in varying degrees of
alpha thalassemia , which is characterized by the formation of p-globin and y -globin
tetramers { hemoglobin H and Barts, respectively ). These abnormal tetramers have
extremely high oxygen affinity (resembling myoglobin ) and are ineffective at
delivering oxygen to tissues.
(Choice E) Hemoglobin S is the predominant form of hemoglobin in sickle cell
disease and is caused by replacement of a glutamate by valine in the (3-globin
chain This results in formation of hemoglobin polymers with reduced oxygen affinity .
Educational objective:
2 , 3 -bisphosphoglycerate ( 2 , 3-BPG ) normally forms ionic bonds with the beta
subunits of deoxygenated hemoglobin A facilitating oxygen release in the peripheral
tissues Mutations that result in loss of the 2,3 -BPG binding pocket' s positive charge
cause hemoglobin A to resemble fetal hemoglobin, which binds oxygen with a higher
affinity due to its inability to interact with 2 3-BPG .
f
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Notes
t alculdior
4
b
6
7
a
9
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13
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15
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17
13
A 52-year -old woman comes to the office with a 2-month history of skin rash that
worsens with sun exposure . Her family says that lately she has become irritable ,
hostile , and has had episodes of disorientation . The patient does not use tobacco or
illicit drugs but has been dnnking half a bottle of gm daily . Further questioning
reveals poor nutritional intake and intermittent diarrhea The patient restricts her diet
for weight control Body mass index is 17 kg nU On examination , she has a
well-demarcated, hyperpigmented, scaly rash on the hands , forearms , and upper
chest . The cause of most of her symptoms is determined to be a lack of the
precursor vitamin for synthesis of NAD- coenzyme The compensatory pathway to
synthetize this coenzyme uses which of the following as a precursor?
'
14
20
21
O A Arginine
22
O B Carotene
23
n
25
25
27
28
29
30
C Cholesterol
O D Orotic acid
C E . Phenylalanine
G F. Tryptophan
31
32
-
34
*
3S
*
35
37
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Newt
Notes
t alculaior
4
5
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7
a
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15
17
1j
A
A 52-year-old woman comes to the office with a 2-month history of skin rash that
worsens with sun exposure Her family says that lately she has become irritable
hostile , and has had episodes of disorientation The patient does not use tobacco or
illicit drugs but has been drinking half a bottle of gm daily Further questioning
reveals poor nutritional intake and intermittent diarrhea. The patient restricts her diet
for weight control Body mass index is 17 kgTn:. On examination , she has a
well-demarcated, hyperpigmentedf scaly rash on the hands , forearms , and upper
chest . The cause of most of her symptoms is determined to be a lack of the
precursor vitamin for synthesis of NAD- coenzyme . The compensatory pathway to
synthetize this coenzyme uses which of the following as a precursor?
14
20
O A Arginine [14%J
21
22
O B. Carotene [6%]
O C . Cholesterol [4%]
O D . Orotic acid [8%]
23
n
25
25
27
28
29
30
31
32
-
34
*
3S
-
-
35
37
^
•
£. Phenylalanine [8%]
F . Tryptophan [60%]
Explanation:
Pellagra ( Tough skin * in Italian vernacular ) is due to niacin deficiency and is
characterized by the "3 Ds": dermatitis, diarrhea, and dementia:
* Dermatitis is bilateral and symmetric on sun- exposed areas of the body and is
characterized by rough, thick, scaly skin.
epithelium
)
* Diarrhea is due to atrophy (and occasional ulceration of columnar
08 : *13
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Lab Values
Explanation:
n
11
n
13
U
19
19
17
ia
H
t alculdtor
I
a
9
Notes
K I ryptophan [bU%|
A
b
6
0i
Pellagra ("rough skin" in Italian vernacular ) is due to niacin deficiency and is
characterized by the "3 Ds": dermatitis, diarrhea, and dementia:
Dermatitis is bilateral and symmetric on sun-exposed areas of the body and is
characterized by rough , thick, scaly skin .
* Diarrhea is due to atrophy (and occasional ulceration) of columnar epithelium
of the gastrointestinal tract .
•Dementia is due to neuronal degeneration in the brain and spinal cord, with
lesions similar to those associated with pernicious anemia .
#
20
21
n
2}
n
29
25
27
23
29
30
31
32
*
-
34
39
35
37
Niacin [ nicotinic acid , or vitamm B;) is an essential component of the coenzymes
nicotinamide adenine dinucleotide (NAD ) and nicotinamide adenine dinucleotide
phosphate ( NADP ). which participate in redox metabolism Specifically, NAD
functions as a coenzyme for dehydrogenases involved in the metabolism of fats ,
carbohydrates and amino acids ; NADP is crucial in the hexose -monophosphate
shunt of glucose metabolism and for biosynthesis of cholesterol and fatty acids .
Niacin can be obtained through dietary intake or synthesized endogenously from
tryptophan In developing countries, niacin deficiency is seen in populations that
subsist primarily on corn products (niacin in corn occurs in a bound unabsorbable
form ). In developed countries it is pnmarily seen in patients with impaired nutritional
intake ( eg, alcoholism, chronic illness ). Pellagra can also be seen occasionally in
those with carcinoid syndrome prolonged isomazid therapy , or Hartnup disease
(Choice A) Arginine is the precursor of nitric oxide urea ornithine and agmatine It
is also necessary for the formation of creatine .
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Newt
Notes
t <ilt ulator
4
&
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£
9
10
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12
13
U
15
15
17
13
20
21
/%
Niacin can be obtained through dietary intake or synthesized endogenously from
tryptophan In developing countries niacin deficiency is seen in populations that
subsist primarily on corn products (niacin in corn occurs in a bound unabsorbable
form). In developed countries , rt is primarily seen in patients with impaired nutritional
intake ( eg , alcoholism , chronic illness ). Pellagra can also be seen occasionally in
those with carcinoid syndrome prolonged isoniazid therapy , or Hartnup disease
b
(Choice A) Arginine is the precursor of nitric oxide , urea ornithine and agmatine !t
is also necessary for the formation of creatine .
(Choice B) Carotene is the precursor to vitamin A.
(Choice C) Cholesterol is the precursor to steroid hormones
(Choice D) Orotic acid is a precursor of pyrimidine .
22
23
24
25
25
27
23
29
30
31
32
-
34
*
35
-
35
(Choice E) Phenylalanine is the precursor to tyrosine an amino acid necessary for
the formation of catecholamines .
Educational objective:
Niacin (vitamin B; ) can be synthesized endogenously from tryptophan and is an
essential component of nicotinamide adenine dinucleotide (NAD ) and nicotinamide
adenine dinucleotide phosphate ( NADP ), A deficiency of this vitamin results in
pellagra , which is characterized by dermatitis , diarrhea , and dementia
References:
1 . Pellagra and alcoholism: a biochemical perspective.
37
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13
A 46-year -old obese man is referred to a dietitian for evaluation of his food intake.
He has been trying to lose weight but has been unsuccessful . The patient is 172.7
cm (5 ft 0 in ) tall and weighs 113 kg ( 250 lb ). Analysis of his food intake shows that
he is consuming 3600 Calories a day . The dietitian recommends increasing physical
activity and implementing a dietary plan. In the first phase, the patient is advised to
reduce his daily dietary intake to 3, 000 Calories , with 30% coming from protein How
much protein per day will this patient consume on the new dietary plan?
Q
A. 130 g
H
20
21
. 160 g
QB
22
n
25
25
c
C . 100 g
D 225 g
21
23
29
30
g
Q E. 250
31
32
33
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*
Notes
t dlculdtor
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5
6
7
a
9
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11
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19
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22
21
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30
31
32
33
*
35
36
-
37
A
-
A 46 year-old obese man is referred to a dietitian for evaluation of his food intake
He has been trying to lose weight but has been unsuccessful . The patient is 172.7
cm (5 ft 8 in ) tall and weighs 113 kg ( 250 lb ). Analysis of his food intake shows that
he is consuming 3600 Calories a day . The dietitian recommends increasing physical
activity and implementing a dietary plan. In the first phase , the patient is advised to
reduce his daily dietary intake to 3, 000 Calories , with 30% coming from protein How
much protein per day will this patient consume on the new dietary plan?
L
O A 130 g [9%1
O B . 160 g [6%l
C C 180 g [18%]
^ m 0. 225 g [57%1
O E. 250 g [9%]
Explanation:
Dietary energy comes predominantly from protein , carbohydrate , and fat .
Metabolism yields 4 Calories (Cal) per gram of protein or carbohydrate and 9 Cal per
gram of fat . Ethanol yields 7 Cal per gram .
This patient is instructed to consume 3000 Cal per day , 900 ( 30% ) of which are to
be from protein Because 1 g of protein yields 4 Cal of energy , this patient should
consume (900 Cal /4 Cal) = 225 g/day of protein .
Educational objective:
Metabolism of 1 g of protein or carbohydrate produces 4 Calories of energy .
V
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nas been trying to lose weight out has oeen unsuccessful I he patient is 1 ( Z /
cm (5 ft 8 inj tall and weighs 113 kg ( 250 lb ) Analysis of his food intake shows that
he is consuming 3600 Calories a day . The dietitian recommends increasing physical
activity and implementing a dietary plan. In the first phase , the patient is advised to
reduce his daily dietary intake to 3, 000 Calories with 30% coming from protein . How
much protein per day will this patient consume on the new dietary plan?
He
*
Notes
t dlculdtor
,
A
t
,
*
12
O A. 130 g [9%]
13
14
15
IS
O B . 160 g [6%]
O C 180 g [18%]
17
13
19
20
v
*D
225 g [57%1
O E. 250 g [9%]
21
22
21
24
25
25
27
23
29
30
31
32
33
*
35
36
37
Explanation:
Dietary energy comes predominantly from protein , carbohydrate , and fat .
Metabolism yields 4 Calories (Cal) per gram of protein or carbohydrate and 9 Cal per
gram of fat . Ethanol yields 7 Cal per gram
This patient is instructed to consume 3000 Cal per day, 900 (30% ) of which are to
be from protein Because 1 g of protein yields 4 Cal of energy this patient should
consume (900 Cal / 4 Cal) = 225 g/day of protein .
Educational objective:
Metabolism of 1 g of protein or carbohydrate produces 4 Calories of energy ;
metabolism of 1 g of fat produces 9 Calories .
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*
*
I alt uldtor
SuspefKi
tnd Blurt
Note
A
5
6
i
a
9
10
11
12
Biochemistry researchers are investigating the speed at which various carbohydrates
are metabolized within the liver . They hypothesize that different monosaccharides
delivered to the liver have different rates of intracellular metabolism . Which of the
following substances is most likely to have the fastest rate of metabolism in the
glycolytic pathway?
13
14
15
15
1?
13
H
20
21
22
A. Fructose-1-phosphate
B . Galactose-1-phosphate
C Glucose - 1-phosphate
’
D Glucose-6-phosphate
E Mannose - 6-phosphate
23
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Lob Values
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Notes
t tilculdtor
A
A
5
6
7
8
9
10
11
12
13
U
15
1&
17
13
14
20
2\
Biochemistry researchers are investigating the speed at which various carbohydrates
are metabolized within the liver . They hypothesize that different monosaccharides
delivered to the liver have different rates of intracellular metabolism . Which of the
following substances is most likely to have the fastest rate of metabolism in the
glycolytic pathway?
'
A . Fructose-1-phosphate [41%]
O B . Galactose-1-phosphate [2%]
C C Glucose-1-phosphate [11%]
C D Glucose-6-phosphate [45%]
E. Mannose -6-phosphate [2%]
22
23
n
Explanation:
25
25
Non -glucose monosaccharides & glycolysis
27
23
29
Glucose
30
31
32
33
34
-
Hcxokjnase/
giiJCOkmase
Galactose —
Galactose 1P
Glucose 1P
Glucose 6- phosphate
36
Phosphoghjco-
37
isowetase
Sucrose
i
Mannose
Block Time Remaining:
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09 : 30
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Fructose 6 phosphate
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t alculdtor
Notes
explanation:
rs
5
6
Non -glucose monosaccharides & glycolysis
7
a
9
Glucose
10
11
12
13
u
|r
Galactose —
Galactose 1P
Glucose 1P
*-
Glucose 6- phosphate
15
15
Phosptyoqluco
17
isomerase
13
19
Mannose
Mannose 6phosphate
20
21
25
25
Fructose 6 - phosphate
Fructose - 1. 6
btsphos4 )h8 t ;tse
22
21
24
PFK - 1
FrucfoAinase
Fructose 1-phosphate
Fructose - 1 , B- bisphosphate
27
2H
b
Hexoktnasa/
gfucohnase
'
Aktotase 0
Atdc : dse A 4 B
29
30
31
32
33
34
-
DHAP
Tnok ;nase
Gfyceraldehyde
Giyceraldehyde
3- phosphate
i
36
37
4
i
Fructose bypasses a major
regulatory step in glycolys« s .
a Mowing faster metabolism
DHAP
Triose
phosphate
tsomerase
i
Pyruvate
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Notes
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Non-glucose monosaccharides ( eg , galactose mannose , fructose ) enter the
glycolytic pathway at different points as Intermediates of glycolysis Of these ,
fructose is the only one whose metabolites bypass phosphofructokinase , one of
the key enzymes involved in regulating the rate of glycolysis As a result , fructose is
metabolized by the liver faster than the other monosaccharides and is rapidly cleared
from the bloodstream following dietary absorption.
a
9
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ii
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IB
IS
17
13
20
21
22
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25
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37
b
Metabolism of fructose in the liver begins with phosphorylation by fructokmase to
fructose- 1- phosphate ( F 1 P) Aldolase B can use both fructose -1,6-bisphosphale
and F1P as substrates; it converts F1P into dihydroxy acetone phosphate
( DHAP) and glyceraidehyde Glyceraldehyde can be either phosphorylated to
glyceraldehyde-3 phosphate by triokinase or converted to DHAP DHAP is
converted by triose phosphate isomerase to glyceraldehyde-3-phosphate, which
continues down the glycolytic pathway .
-
(Choices B, C, D, and E) Galactose-1-phosphate , glucose-1-phosphate ,
glucose-6 -phosphate and mannose-6 -phosphate enter glycolysis upstream of
phosphofructokinase , a major rate-limiting enzyme of glycolysis This slows down
the rate of their metabolism relative to fructose and its metabolites ( eg, F1P),
Educational objective:
Dietary fructose is phosphorylated in the liver to F1P and is rapidly metabolized
because it bypasses PFK - 1 , the major rate -limiting enzyme of glycolysis Other
sugars ( eg. glucose galactose mannose ) enter glycolysis prior to PFK - 1 and as a
result are metabolized more slowly .
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Notes
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7
a
9
10
11
n
13
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15
IS
1?
13
H
20
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24
A 2-year - old boy is evaluated for easy bruising His parents report that he develops
marked bruising and open wounds following minor trauma . The skin is difficult to
suture due to its extreme fragility . Physical examination reveals hyperextensible skin,
multiple ecchymoses over the forearms and pretibial regions, and an umbilical
hernia . A skin biopsy is performed , and histochemica! evaluation of the biopsy
reveals a defect in extracellular processing of collagen Which of the following steps
of collagen synthesis is most likely impaired in this patient?
A Glycosylation of hydroxylysine residues
O B . Interchain C - termmal disulfide bond formation
O C N-terminal propeptide removal
O D . Proline residue hydroxylation
O E. Triple helix formation
25
25
27
28
29
30
31
32
33
34
35
*
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*
(Notes
Calculator
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ft
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A 2-year - old boy is evaluated for easy bruising His parents report that he develops
marked bruising and open wounds following minor trauma . The skin is difficult to
suture due to its extreme fragility . Physical examination reveals hyperextensible skin,
multiple ecchymoses over the forearms and pretibial regions, and an umbilical
hernia . A skin biopsy is performed , and histochemica! evaluation of the biopsy
reveals a defect in extracellular processing of collagen Which of the following steps
of collagen synthesis is most likely impaired in this patient?
A Glycosylation of hydroxylysine residues [10%]
B Interchain C - terminal disulfide bond formation [18%]
v @ C. N-terminal propeptide removal £27%]
O D . Proline residue hydroxylation [17%]
O E. Triple helix formation [26%]
Explanation:
Collagen synthesis
30
31
32
33
34
35
Signal sequence directs
growing polypeptide chain
into endoplasmic reticulum
Prepro -a- chains
Signal sequence is cleaved
Pro -O Chains
i
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“J
*
Notes
t alculdtor
/%
Explanation:
a
Collagen synthesis
9
10
11
12
13
u
15
IS
Signal sequence directs
growing polypeptide chain
into endoplasmic reticulum
Preproa- chams
Signal sequence is cleaved
Pro-a chains
I
I
17
13
H
20
21
22
23
24
25
25
27
Hydroxylation of selected
prohne & lysine residues
{ vitamin C dependent )
OH
Glycosylation of
selected hydroxylysme
residues
Galactose
Glucose
OH
OH
OH
OH
23
29
30
31
32
33
34
35
37
i
f
OH
•
Assembly of pro -a - chains into
procollagen triple helix
Procollagen
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i
ta
•Glucose
Assembly of pro- a - chains into
procollagen triple helix
(Notes
*
(
dlculdlor
Procollagen
9
10
b
It
12
13
U
15
15
Procollagen transferred to
Golgi apparatus & secreted
into extracellular matrix
17
13
H
20
21
22
23
n
25
26
27
23
29
30
31
32
33
34
Terminal propeptides cleaved
by N- & C~ procoilagen
peptidases
Collagen molecules
spontaneously
assemble
Tropocollagen
i
Collagen fibrils
a
: QSZS&ZSSXZSS:
i
35
37
Covalent cross links
formed by lysyl oxidase
v
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Notes
(
dlcutdtor
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7
Ldb Values
J
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o
Sent
© U World
This child likely has Ehlers-Danlos syndrome a group of rare hereditary disorders
characterized by defective collagen synthesis . The condition can be caused by a
deficiency in procollagen peptidase the enzyme that cleaves terminal
propeptides from procollagen in the extracellular space Impaired propeptide
removal results in the formation of soluble collagen that does not properly
crosslink Consequently, patients often have joint laxity hyperexlensible skin J
fragile tissue with easy bruising , and poor wound healing
,
,
Each collagen molecule consists of 3 polypeptide a -chains held together by
hydrogen bonds forming a triple helix . Collagen assumes this conformation
because each of the a-chains has a simple repetitive amino acid sequence
represented as (Gly-X -Y )„ The smallest amino acid glycine (Gly ) , is necessary at
every third position to ensure compact coding of the helix . Many of the amino acids
represented by X and Y are proline residues , which kink the polypeptide chain and
enhance the rigidity of the helical structure due to their ring configuration
Mature collagen is synthesized by fibroblasts osteoblasts and chondroblasts
through the following steps:
1. As translation begins in the cytoplasm , an ammo acid signal sequence at the Nterminus of the o -chain facilitates ribosomal binding to the rough endoplasmic
reticulum ( RER ) and passage of the growing polypeptide chain (pre -pro -o chain ) into the RER,
2. Inside the RERr the hydrophobic signal sequence is cleaved to yield the pro-a cham Proline and lysine at the Y positions of the pro-a chain are hydroxylated
to hydroxyproline and hydroxylysine respectively (Choice DJ Glycosylation of
select hydroxylysine residues also occurs within the RER (Choice A )
-
,
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-
t olcul iior
chain ) into the RER.
2 . Inside the RER the hydrophobic signal sequence is cleaved to yield the pro-achain Proline and lysine at the Y positions of the pro-a-chain are hydroxylated
to hydroxyprohne and hydroxylysine . respectively (Choice D). Glycosylalion of
select hydroxylysine residues also occurs within the RER (Choice A )
3 . The central helical region of the pro-a-chain is flanked by N- and C- terminal
propeptides Disulfide bond formation between the C-terminal propeptide
region of 3 c -chains brings the chains into an alignment favorable for assembly
into a triple helix (procollagen molecule ) ( Choices 6 and E) .
4 . Procollagen molecules are then transported through the Golgi apparatus into
the extracellular space The N- and C- terminal propeptides are cleaved by
procollagen peptidases, converting procollagen into less soluble
tropocollagen .
5. Tropocollagen monomers self -assemble into collagen fibrils Finally lysyl
oxidase helps create covalent crosslinks between collagen fibrils to form
strong collagen fibers.
/\
.
Educational objective:
Ehlers- Danlos syndrome is a group of rare hereditary disorders characterized by
defective collagen synthesis . It can be caused by procoliagen peptidase deficiency,
which results in impaired cleavage of terminal propeptides in the extracellular
space Patients often have joint laxity , hyperextensible skin, and tissue fragility due
to the formation of soluble collagen that does not properly crosslink
References:
1. Defect in conversion of procoliagen to collagen in a form of EhlersDanlos syndrome.
2 . Defects in the biochemistry of
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collagen in diseases
v
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.
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IB
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A 26-year-old man develops myalgia , nasal congestion , and cough . His temperature
is 38.3 C (101 F ). Examination shows nasal and pharyngeal hyperemia Rapid
influenza antigen testing of his nasopharyngeal secretions is positive tn the infected
cells of the respiratory tract , viral proteins are degraded and attached to major
histocompatibility l molecules that are then expressed on the cell surface for
presentation to cytotoxic CD8+ lymphocytes Which of the following enzymes is
most likely involved in this process ?
f&m
A Acid phosphatase
O B . Caspase
C Guanylate cyclase
D Myeloperoxidase
C E . Ubiquitin ligase
25
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A
A 26-year -old man develops myalgia , nasal congestion , and cough His temperature
is 38.3 C (101 F ). Examination shows nasal and pharyngeal hyperemia Rapid
influenza antigen testing of his nasopharyngeal secretions is positive. In the infected
cells of the respiratory tract, viral proteins are degraded and attached to major
histocompatibility I molecules that are then expressed on the cell surface for
presentation to cytotoxic CD8+ lymphocytes Which of the following enzymes is
most likely involved in this process ?
A . Acid phosphatase (14%]
O B Caspase [25%]
C Guanylate cyclase [7%]
D. Myeloperoxidase [17%J
E. Libia. t
gase [ 38%]
^
*
Explanation:
23
29
Ubiqultln dependent protein catabolism
30
31
32
33
34
Ubiquitin clpjvpd
from target proton
4 recycled
—
}
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Uhiqutfifi
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Explanation:
a
&
Ubiquitin dependent protein catabolism
9
10
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12
from target proton
& recycled
13
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23
29
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Hydrolyzed
QiNjopepudei
30
31
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© UWodd
'
J
The ubiquitin proteasome pathway (UPP ) is essential for breakdown of
intracellular proteins, both native and foreign and helps recycle them into the amino
acid building blocks Ubiquitin functions as a tag that is attached to proteins to mark
them for destruction . This process is performed by ubiquitin ligases . enzymes that
recognize specific protein substrates and catalyze ubiquitin attachment . These
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12
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*
UWorld
a
9
(Notes
b
The ubiquitin proteasome pathway (UPP ) is essential for breakdown of
intracellular proteins both native and foreign and helps recycle them into the amino
acid building blocks Ubiquitin functions as a tag that is attached to proteins to mark
them for destruction This process is performed by ubiquitin ligases enzymes that
recognize specific protein substrates and catalyze ubiquitin attachment . These
tagged proteins are then taken up by the proteasome , where they are broken down
into their constituent oligopeptides and , eventually , amino acids
,
,
U
H
20
21
22
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24
2&
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34
3S
3S
The role of the UPP in the immune response is related to its ability to degrade
foreign intracellular proteins , such as viral particles These proteins are
degraded to an appropriate size coupled to major histocompatibility class I
protein complex in the endoplasmic reticulum , and then presented on the cell
surface for recognition by cytotoxic CD8 + lymphocytes . Once the cytotoxic
lymphocytes recognize non- native ( eg viral ) proteins on infected cells the
presenting cells are destroyed as part of the immune response
(Choice A) Acid phosphatase is found in the lysosome of most cells and
hydrolyzes organic phosphates , but it is not involved in protein breakdown
(Choice B) Caspases are a family of proteases that are essential for apoptosis .
They also play a role in necrosis and inflammation but are not involved in the antigen
presenting system .
(Choice C) Guanylate cyclase is involved in activation of the cyclic guanosine
monophosphate messenger system If plays a crucial role in the activation and
regulation of diverse physiological processes such as smooth muscle relaxation and
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Notes
t iiiculdtor
(Choice B ) Caspases are a family of proteases that are essential for apoptosis.
They also play a role in necrosis and inflammation but are not involved in the antigen
presenting system.
b
(Choice C) Guanylate cyclase is involved in activation of the cyclic guanosine
monophosphate messenger system It plays a crucial role in the activation and
regulation of diverse physiological processes such as smooth muscle relaxation and
retinal phototransduction .
(Choice D) Myeloperoxidase is a hey component of the respiratory burst of
neutrophils and is released into the phagocytic vacuole and extracellular space as
part of the immune response to bacteria and other pathogens.
Educational objective:
A key step in the ublquitin proteasome pathway is controlled by ubiquitin ligases
which recognize specific protein substrates and attach a ubiquitin tag. The proteins
are then degraded to an appropriate size and coupled with the major
histocompatibility I protein complex in the endoplasmic reticulum Finally , they are
presented on the cell surface for recognition by cytotoxic CD8+ lymphocytes .
*
27
28
29
30
31
32
33
34
35
35
References:
1 Protein degradation by the ubiqultln-proteasome pathway in normal
and disease states *
*
2 . Cbl- and Nedd4 - family ubiquitin ligases: balancing tolerance and
immunity *
3 . Ubiquitin makes its mark on immune regulation *
Time Spent 6 seconds
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11 : IS
Turor
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I dlculdtor
A 3-year -old girl is brought to the clinic due to several months of fatigue and difficulty
walking . She ambulates normally at first but rapidly becomes weak and tired The
patient has not been ill recently and is usually happy and playful She has a history of
mild motor delays but is otherwise developmental normal. Vital signs are within
normal limits . Examination shows mildly decreased power in all extremities but no
ataxia. Cardiac auscultation reveals a 1/6 systolic murmur and an S3 gallop.
Laboratory results are as follows:
Serum chemistry
37 mg /dL
Glucose
304 UfL
Creatine kinase
A
^
Urinalysis
Protein
none
Glucose
negative
Ketones
negative
Leukocyte esterase negative
Nitrites
negative
Muscle biopsy shows a very low carnitine content Which of the following
substances has deficient synthesis in this patient‘s disease?
30
31
32
33
34
O A. Acetoacetate
C B . Arachidonic acid
O C . Glutathione
O D . Homocysteine
O E Lactate
O F . Palmitate
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A 3-year - old girl is brought to the clinic due to several months of fatigue and difficulty
walking . She ambulates normally at first but rapidly becomes weak and tired. The
patient has not been ill recently and is usually happy and playful . She has a history of
mild motor delays but is otherwise developmental! / normal Vital signs are within
normal limits . Examination shows mildly decreased power in all extremities but no
ataxia. Cardiac auscultation reveals a 1/6 systolic murmur and an S3 gallop
Laboratory results are as follows:
Serum chemistry
37 mg /dL
Glucose
304 U/ L
Creatine kinase
Notes
l alculdior
A
b
Urinalysis
Protein
none
Glucose
negative
Ketones
negative
Leukocyte esterase negative
Nitrites
negative
Muscle biopsy shows a very low carnitine content Which of the following
substances has deficient synthesis in this patient's disease?
30
31
32
33
34
* <§
A. Acetoacetate [42%]
C B . Arachidonic acid [5%]
. C . Glutathione [8 %]
C D. Homocysteine [9%]
O E. Lactate [8%]
Q F. Palmitate [29%]
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I akulator
A
&
Fatty acid oxidation
E
9
10
11
Fatty acid
12
13
14
is
is
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13
Primary carnitine deficiency
* Muscle weakness
Acyl CoA synthase
Acyl CoA
*
Cardiomyopathy
*
Hypoketotic hypoglycemia
-
Elevated muscle triglycerides
H
CAT I
20
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Carnitine
Acyl- carnitine
Cytoplasm
CAT il
Carnitine *
Acyl- carnitme
•
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Mitochondrial
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Acyl CoA
Acyl CoA
dehydrogenase
FADH2
Trans- enoyl CoA
I
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Medium chain acyl CoA
dehydrogenase (MCAD)
deficiency
Hypoglycemia
* Hypoketotic hypoglycemia
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dehydrogenase (MCAD)
deficiency
*
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Hypoglycemia
Hypoketotic hypoglycemia
u
3 - Hydroxyacyl CoA
N
20
21
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NADH
21
24
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3 - Ketoacyl CoA
I
i
Acetyl CoA
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Ketone bodies
TCA
Cycle
© UWorld
This patient has myopathy ( eg . elevated creatine kinase weakness ), cardiomyopathy
—
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9
10
ii
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34
This patient has myopathy teg . elevated creatine kinase weakness ) cardiomyopathy
( eg S 3 gallop), and hypoketotic hypoglycemia in the setting of decreased muscle
carnitine content which is consistent with primary carnitine deficiency This
condition is caused by a defect in the protein responsible for carnitine transport
across the plasma membrane Without sufficient carnitine , fatty acids cannot be
transported from the cytoplasm into the mitochondria as acyl-carnitine ( carnitine
shuttle ). The mitochondria therefore cannot |3 - oxidize the fatty acids into acetyl CoA ,
the carbon substrate for the citric acid cycle As a result, cardiac and skeletal
myocytes cannot generate ATP from fatty acids and the liver is unable to synthesize
ketone bodies (ie , acetoacetate ) when glucose levels are low, Hypoketotic
hypoglycemia is also seen in other fatty acid oxidation disorders (eg , medium- or
very-long-chain acyl CoA dehydrogenase deficiency).
,
(Choice B) Arachidonic acid can be ingested or synthesized from phospholipids in
the cell membrane Its eicosanoid derivatives (eg, prostanoids , leukotnenes ) are
important modulators of inflammation It is not affected by carnitine levels .
(Choice C) Glutathione is a tripeptide that can be synthesized from amino acids
( glutamate cysteine , and glycine ) , it is an important antioxidant and plays a role in
DMA synthesis and repair.
} Homocysteine is an amino acid that is synthesized from
methionine Using vitamin cofactors , it can be converted to cysteine (pyridoxine ) or
recycled into methionine ( folate ).
(Choice
(Choice E) Lactate is produced from pyruvate under anaerobic conditions. Patients
J
i8
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Next
a n u u M u a m OIMJ
piays a
Notes
esa
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IUIC M
(Choice D) Homocysteine is an amino acid that is synthesized from
methionine Using vitamin cofactors , it can be converted to cysteine (pyridoxine ) or
recycled into methionine ( folate ).
(Choice E) Lactate is produced from pyruvate under anaerobic conditions Patients
with carnitine deficiency synthesize lactate normally but may produce increased
lactate during times of catabolic stress due to inability to utilize fatty acids for energy
production.
&
(Choice F) Palmitate is a fatty acid that can be ingested or synthesized from
carbohydrates Palmitate synthesis occurs in the cytosol and would not be affected
by carnitine deficiency .
Educational objective:
Carnitine deficiency impairs fatty acid transport from the cytoplasm into mitochondna
preventing (3-oxidation of fatty acids into acetyl CoA. This leads to cardiac and
skeletal myocyte injury (lack of ATP from citric acid cycle ) and impaired ketone body
production by the liver during fasting periods .
References:
1 . Primary carnitine deficiency: novel mutations and insights Into the
cardiac phenotype *
2 Systemic primary carnitine deficiency: an overview of clinical
manifestations * diagnosis, and management.
,
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t alt ufalor
A 14-month-old boy is evaluated for failure to thrive and developmental delay . His
mother reports that at 12 months he could barely lift his head and had difficulty sitting
unsupported . The toddler has not started babbling or forming words. He is at the
10 th percentile for height and 5th percentile for weight . Laboratory results are as
follows:
8.6 g /dL
Hemoglobin
Mean corpuscular
114 fL
volume
1%
Reticulocytes
42 pg/dL normal 40-80
Ammonia , plasma
pg/dL
10
ii
12
13
14
IS
16
17
13
14
20
21
Urine specimens contain large amounts of orotic acid crystals Supplementation with
which of the following substances would most likely benefit this patient?
n
23
24
2&
25
(
27
A. Ascorbic acid
O B. Folic acid
O C Guanine
23
29
30
D. Iron
C E Pyndoxine
O F . Uridine
31
32
33
34
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30
31
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31
A 14 month-old boy is evaluated for failure to thrive and developmental delay . His
mother reports that at 12 months he could barely lift his head and had difficulty sitting
unsupported . The toddler has not started babbling or forming words. He is at the
10 th percentile for height and 5th percentile for weight . Laboratory results are as
follows:
8.6 g /dL
Hemoglobm
Mean corpuscular
114 fL
volume
1%
Reticulocytes
42 pg/dL normal 40-80
Ammonia plasma
pg/dL
Ct
Urine specimens contain large amounts of orotic acid crystals Supplementation with
which of the following substances would most likely benefit this patient?
O A. Ascorbic acid [3%]
O B . Folic acid [24%]
O C . Guanine [7%]
O D . Iron [4%]
O E. Pyndoxme [25%]
* =•F. Undine [38%]
Explanation:
De novo pyrimidine synthesis
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(Notes
[ 38% ]
A
Explanation:
De novo pyrimidine synthesis
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*
Note
*
l alcutalor
dTMP
/%
This patient likely has hereditary orotic aciduria a rare autosomal recessive
disorder of de novo pyrimidine synthesis that results in physical and mental
retardation ( eg. low height/ weight delayed developmental milestones ),
megaloblastic anemia ( eg elevated mean corpuscular volume, low reticulocyte
count ), and elevated urinary orotic acid levels Increased unnary orotic acid may
also be seen in ornithine transcarbamylase deficiency however, patients with this
condition classically have failure to thrive and hyperammonemic encephalopathy
within the first few weeks of life (due to impaired urea synthesis).
Hereditary orotic aciduria occurs due to a defect in uridine 5 *monophosphate
( UMP) synthase a polypeptide containing 2 enzymatic domains ( orotate
phosphonbosyltransferase and OMP decarboxylase ) that catalyze the final
conversion of orotic acid to UMP . Impaired conversion of orotic acid to UMP results
in the excretion of large amounts of orotic acid in the urine and the clinical features
described above . Uridine supplementation can bypass this enzymatic defect and
improve symptoms as uridine is converted to UMP via nucleoside kinases.
,
(Choice A ) Ascorbic acid (vitamin C } is required for hydroxylation of proline and
lysine residues in collagen synthesis ; therefore , it plays an important role in
connective tissue maintenance and wound healing,
(Choice B) Folate participates jn single carbon transfer reactions as in the de novo
synthesis of purines and thymidine Folate supplements will improve megaloblastic
anemia resulting from folate deficiency but will not improve the anemia in orotic
aciduria
(Choice C) Guanine and adenine are purine bases present in DNA and RNA
Orotic aciduria is a defect in the synthesis of pyrimidine bases so supplementation
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A
(Choice B) Folate participates m single carbon transfer reactions , as in the de novo
synthesis of purines and thymidine Folate supplements will improve megaloblastic
anemia resulting from folate deficiency but will not improve the anemia in orotic
aciduria
(Choice C) Guanine and adenine are punne bases present in DNA and RNA .
Orotic aciduria is a defect in the synthesis of pyrimidine bases , so supplementation
with purines would not affect orate synthesis
[,
(Choice D) Iron supplementation improves iron deficiency anemia classically a
microcytic hypochromic anemia,
(Choice E) Pyridoxine (vitamin B ) supplementation is indicated during treatment
with isomazid . Pyridoxine is a cofactor in transamination , deamination,
decarboxylation , and condensation reactions .
{
Educational objective;
Orotic aciduna is a rare autosomal recessive disorder of de novo pynmidine
synthesis that occurs due to a defect in undine 5 '- monophosphate ( UMP ) synthase
Children typically present with physical and mental retardation megaloblastic anemia ,
and large amounts of urinary orotic acid Uridine supplementation can improve
symptoms as uridine is converted to UMP via nucleoside kinases.
References:
1 . Orotic aciduria and uridine monophosphate synthase:
^ reappraisal,
2. Inborn errors of pyrimidine metabolism: clinical update and therapy.
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i
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9
10
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15
1?
13
19
2Q
21
n
23
21
2&
25
A patient with newly diagnosed type 2 diabetes mellitus comes to your office asking
for more information about her disease. You recall that the physiology of glucose
homeostasis is complex, involving multiple intercellular signaling pathways mediated
through transmembrane receptor proteins . Binding of intracellular GTP to a specific
membrane - associated protein causes rapid metabolic changes m hepatocytes
These metabolic changes include a decrease in intracellular glycogen stores and the
release of glucose into the blood. Which of the following is the most likely mediator
responsible for these effects?
&
A . cGMP -dependent protein kinase
B . Tyrosine-specific protein kinase
O C . Protein kinase A
O D . Phosphodiesterase
O E. Janus tyrosine kinase ( JAK )
27
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t akuldlor
ft
&
£
S
10
11
12
13
14
IS
15
A patient with newly diagnosed type 2 diabetes mellitus comes to your office asking
for more information about her disease. You recall that the physiology of glucose
homeostasis is complex , involving multiple intercellular signaling pathways mediated
through transmembrane receptor proteins. Binding of intracellular GTP to a specific
membrane - associated protein causes rapid metabolic changes in hepatocytes .
These metabolic changes include a decrease in intracellular glycogen stores and the
release of glucose into the blood . Which of the following is the most likely mediator
responsible for these effects?
17
13
H
20
21
n
23
24
2&
25
71
23
29
30
31
32
33
34
A. cGMP -dependent protein kinase [16%]
C B . Tyrosine - specific protein kinase [24%]
* # C. Protein kinase A [50%]
D. Phosphodiesterase [4%]
O E. Janus tyrosine kinase ( JAK ) [7%]
Explanation:
Gluugon
TSH
PTH
iin
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*
Note
*
t alcufdtor
E. Janus tyrosine kinase ( JAK ) [ 7%]
Explanation:
Giucagon
ISM
PTH
9
10
11
&
A
12
13
U
IS
15
1?
13
H
20
21
Idea) laic
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23
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ATP
r \ rli r
'
\ tlhr priorin hinaw V
71
\
23
29
Physiologic atf «cTS
30
31
32
33
34
.
cAMP
The scenario described in the question stem is characteristic of the interaction of a
hormone with the G-protein-mediated adenylate cyclase second messenger
system. A number of hormones such as glucagon, thyroid- stimulating hormone
( TSH ). and parathyroid hormone (PTH } exert their effect through transmembrane
receptors in coordination with G- proteins and cyclic AMP.
G-protein is a heterotrimer consisting of alpha beta , and gamma subunits
*1,
i
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Physiologic » 1 fecit
A
B
9
10
11
o
Mnft
*
Notes
t dlculdtor
A
The scenario described in the question stem is characteristic of the interaction of a
hormone with the G-protein-mediated adenylate cyclase second messenger
system . A number of hormones such as glucagon , thyroid- stimulating hormone
( TSH ). and parathyroid hormone ( PTH } exert their effect through transmembrane
receptors in coordination with G-proteins and cyclic AMP.
G-protein is a heterotrimer consisting of alpha beta , and gamma subunits
associated with the intracellular domains of cell membrane - associated
receptors. The alpha subunit of the inactivated G-protein is bound to GDP Upon
activation of the receptor, the alpha subunit undergoes a conformational change and
GDP is released. Subsequent binding of GTP then allows for the dissociation of the
alpha subunit from the remainder of the G-protein complex . There are multiple
subtypes of alpha G-proteins , each with different secondary effects A specific
alpha subunit known as Gt (present in the glucagon , TSH , and PTH receptor
complexes ) activates adenylate cyclase when released from the G-protein
complex Once formed from ATP , cyclic AMP activates a family of enzymes known
as the cAMP-dependent protein kinases , or protein kinase A,
Protein kinase A phosphorylates specific serine or threonine residues in some
enzymes , thereby leading to their activation or deactivation. Protein kinase A also
phosphorylates several proteins that bind to regulatory regions of genes on the DNA
molecule itself
(Choice A) Cyclic GMP activates protein kinase G , which mediates a number of
intracellular effects including relaxation of smooth muscles , platelet activation , sperm
metabolism, and cell division. Cyclic GMP is inactivated by cGMP
phosphodiesterase, an enzyme specifically involved in cleaving cyclic GMiP in the
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*
Note
*
t ale ulator
phosphorylates several proteins that bind to regulatory regions of genes on the DNA
molecule itself
(Choice A) Cyclic GMP activates protein kinase G , which mediates a number of
intracellular effects including relaxation of smooth muscles , platelet activation, sperm
metabolism and cell division . Cyclic GMP is inactivated by cGMP
phosphodiesterase, an enzyme specifically involved in cleaving cyclic GMP in the
corpus cavemosum of the penis and regulating penile erection . Drugs such as
sildenafil inhibit cGMP-specific phosphodiesterase and are used in the treatment of
erectile dysfunction.
,
(Choice B) Hormones such as insulin act through cell surface - associated receptors
that have intrinsic tyrosine kinase domains.
(Choice D) Cyclic AMP is cleaved by the enzyme cAMP phosphodiesterase to its
inactive form 5‘- AMP . Drugs that inhibit cyclic AMP phosphodiesterase lead to the
prolongation of the actions of cyclic AMP, An example of this is the use of
theophylline in bronchial asthma .
(Choice E) Growth hormone , erythropoietin and cytokines such as interferon act
through cell membrane - associated receptors in the JAK-STAT pathway. JAK is a
specific intracellular tyrosine kinase associated with receptors that dimerize upon
ligand binding .
Educational objective:
Protein kinase A is primarily responsible for the intracellular effects of the G -protein
adenylate cyclase second messenger system. Some hormone receptors that use
this mechanism include the TSH , glucagon, PTH , and beta - adrenergic receptors
Time Spent 4 seconds
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t ale ufator
9
9
io
11
n
13
H
19
19
1?
13
A 46-year -old man comes to the emergency department due to recurrent
nosebleeds When interviewed for additional history he becomes belligerent and
uncooperative. The patient has a history of alcohol abuse and chronic mental
illness. He has been placed m homeless shelters on multiple occasions but has not
remained there for any prolonged penods . Physical examination shows swollen
gums , scattered ecchymoses , and hyperkeratosis He also has a chronic ulcer on
the left lower extremity that does not appear to be infected . Which of the following
mechanisms accounts for this patient's examination findings?
A Abnormal oxidative decarboxylation of ketoacids
20
21
O B . Abnormal proline hydroxylation
O C . Abnormal transamination
23
24
29
26
27
O D Deficient methionine synthesis
n
&
O E . Diminished synthesis of purines
26
29
30
31
32
33
34
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10
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IS
15
17
13
H
20
21
22
21
21
2&
25
27
28
29
A 46-year -old man comes to the emergency department due to recurrent
nosebleeds When interviewed for additional history he becomes belligerent and
uncooperative. The patient has a history of alcohol abuse and chronic mental
illness. He has been placed in homeless shelters on multiple occasions but has not
remained there for any prolonged periods. Physical examination shows swollen
gums , scattered ecchymoses , and hyperkeratosis He also has a chronic ulcer on
the left lower extremity that does not appear to be infected . Which of the following
mechanisms accounts for this patient's examination findings?
IT
A. Abnormal oxidative decarboxylation of ketoacids [5%]
* B . Abnormal proline hydroxylation [80%]
O C. Abnormal transamination [4%]
O D. Deficient methionine synthesis [6%]
O E. Diminished synthesis of purines [5%]
•
Explanation:
30
-
Water soluble vitamins
31
32
33
31
Vitamin
Bi
( thiamine )
Primary function
DecarboxylHi ion of a-keto
acids (carlwhydtMte
motdbohsm )
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Deficiency
* Beriberi (peripheral
*
neuropathy , heart failure )
Wernicke-Korsakoff syndrome
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M M V I IV
I U M M 1 4 MII «
*
(Note
*
t aleuldtor
A
5
Water -sotuble vitamins
9
to
11
Vitamin
Decarboxylalion of a-heto
12
13
14
is
115
1?
13
Primary function
8 , ( thiamine )
B; (riboflavin)
H
acids (carbohydrate
metabolism)
Mitochondrial hydrogen
earner ( f MN FAD)
20
Deficiency
Beriberi (peripheral
neuropathy, heart failure )
• Wernicke Korsakoff syndrome
• Angular cheilosis, stomatitis,
glossitis
• Normocytic anemia
21
22
21
24
2&
25
27
Hydrogen acceptor
(NAD)NADH )
• Pellagra (dermatitis, dementia,
B<j (pyridoxme )
Transamination of amino
acids (ammo acid synthesis )
• Cheilosis
B ., (folate , folic
acid)
HydroxymethyLformyl
carrier (purine & thymine
synthesis)
* Megaloblastic anemia
• Neural tube delects ( fetus)
Bi ( niacin )
23
29
30
31
32
33
34
diarrhea )
stomatitis , glossitis
Isomerase &
Bi ? ( cobalamrn)
methyftransferase cotactor
(DNA & methionine
synthesis)
* Megaloblastic anemia
* Neurologic deficits
V
JS _ f
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*
Notes
t alculator
4
A
5
Hydroxymet hyIf ormyI
carrier (purine & thymine
synthesis )
Mega IohId Stic anemia
• Neural lube defects ( felus )
Bu ( cobalamin)
Isomerase &
methyttransterase cofactor
(DNA & methionine
synthesis )
• Megaloblastic anemia
C (ascorbic acid)
Hydroxytation of proltne &
tysme { collagen synthesis )
• Scurvy
Bo ( folate, folic
acid)
9
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IS
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* Neurologic deficits
OUWlxld
In the United States vitamin C deficiency ( scurvy ) is most often seen in severely
malnourished individuals (eg homeless alcohol or drug abusers ) , Symptoms of
vitamin C deficiency are the result of decreased connective tissue strength. The
capillary walls are especially fragile , leading to easy bruising , mucosal bleeding,
and perifollicular petechial hemorrhages . Patients may also suffer from periodontal
disease ( gum swelling, loosening of the teeth, and infection) and poor wound
healing , and have hyperkeratotic follicles with corkscrew hairs Scurvy is even
more severe in children and manifests with hemorrhages , bony deformities , and
subperiosteal and joint hematomas.
Vitamin C is necessary for the hydroxytation of proline and lysine residues
during collagen synthesis . This reaction is executed by prolyl and lysyl
hydroxylases, with vitamin C serving as a reducing agent Hydroxyproline and
hydroxylysine are essential for cross linking collagen molecules . In scurvy, collagen
cross -linking is compromised thereby greatly reducing its tensile strength
-
r
^
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Lab Values
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(Notes
t dkulator
J'
/%
(Choice A) Thiamine (vitamin B i serves as a coenzyme in the decarboxylation
reactions mediated by several dehydrogenase enzymes It is necessary for the
conversions of pyruvate to acetyl-CoA and of alpha -ketoglutarate to succinyl-CoA in
the citric acid cycle Vitamin B deficiency can cause peripheral neuropathy, heart
failure , and central nervous system dysfunction (Wernicke - Korsakoff syndrome ).
(Choice C) Vitamin B, (pyridoxinel serves as a cofactor in many reactions that
involve amino acids ( eg , transamination, decarboxylation, deamination ). Pyridoxine
deficiency manifests with seborrheic dermatitis , glossitis , and peripheral neuropathy
(Choice D) Vitamin B . is necessary for the synthesis of methionine from
homocysteine and for the synthesis of succinyl-CoA from methylmalonyl-CoA .
Deficiency of vitamin B . causes megaloblastic anemia and subacute combined
degeneration of the spinal cord.
(Choice E) Purine and thymidine synthesis is diminished in patients with folate
deficiency . The resultant decreased ability of erythropoietic cells to form DNA
causes megaloblastic anemia .
Educational objective;
Vitamin C is necessary for the hydroxylation of proline and lysine residues in
pro-collagen . Vitamin C deficiency ( scurvy) is most often seen in severely
malnourished individuals and leads to capillary bleeding, poor wound healing, and
periodontal disease In children bony deformities and subperiosteal hemorrhages
are also characteristic .
References:
1. Scurvy masquerading as leukocytoclastic vasculitis: a case report
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SPKt
C*
Note
*
t olr uldlor
A 44- year -old homeless man is brought to the emergency department after police
officers found him agrtated and confused. During transport to the hospital, he is
started on intravenous fluids with dextrose . On arrival the patient is disorientated but
cooperative Physical examination shows bruises on his forehead , forearms , and
shins Extraocular findings include bilateral horizontal nystagmus and decreased
lateral eye movements He also has an unsteady gait with widely - spaced legs and
short steps . The ambulance personnel state that the patient's extraocular
movements were intact when they picked him up A review of the medical record
shows that the patient has been admitted to the hospital with alcohol intoxication
several times before Which of the following reactions is likely to be the most
impaired in this patient ?
A
n
19
Oxaloatetate
20
21
22
23
24
2S
25
Citrate
Malate
27
23
29
30
31
32
33
34
Fumarate
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Notes
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4
r\
B
9
10
11
12
F Lima rate
Isocitrate
13
H
IS
15
17
13
H
20
Succinate
21
aKetoglutarate
22
23
24
Succinyl CoA
25
26
27
23
29
30
31
32
33
34
O A. A
OB B
OC C
ODD
O E. E
O F. F
O G. G
O H. H
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4
/%
6
8
9
10
11
F jma rate
12
Isocitrate
13
u
is
V5
1/
13
H
20
Succinate
21
a Ketoglutarate
22
21
24
SuccinyltoA
?s
CuU«IV »441.U.
*
25
27
O A. A [11%1
28
29
O B . B [2%J
O C C [ 12%]
* <® D . D [59%]
30
31
32
33
34
E E [5%J
O F. F [3%]
O G . G [2%J
O H. H [7%1
I
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(Notes
t dlculdtor
C H . H [7%]
A
6
8
9
10
11
Explanation:
Inhibition of the citric acid cycle by ethanol
12
Pyruvate
13
H
IB
V5
1/
13
1
I
Pyruvate
NAD *
i
dehydrogenase
( Thiannine dependent)
NAOH
I
H
I
20
Acetyl CoA
Ethanol
21
22
23
24
TNADH
?B
25
27
28
29
Malate
Oxatoacetate
dehydrogenase
30
31
32
33
34
NAOH
Malate
u
Citrate synthase
Citrate
Fumorose
Acomtase
Fumarate
Mitochondrion
r
Isocitrate
FAOH,
Succtnate
dehvdroaenase
FAD
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Newt
A
NAD
6
NADH
ryruvdie
i
j
I
*
Note
*
lm i l u f d l o r
A
dehydrogenase
(Thiamine dependent )
i
s
Ethanol
n
n
12
#
Acetyl CoA
TNADH
13
u
IS
16
1?
13
Malore
Qxaloacetate
dehydrogenose s^C*
^ NADH
f \
Malate AD
14
20
21
Aconitase
'
Foma rate
Succinate
dehydrogenase
Mitochondrion
FA on,
Isocitrate
NAP'
FAD
NADH -
30
31
32
33
31
Citrate
Fumarase
22
23
21
2S
25
2J
23
29
Citrate synthase
Succinate
Succinate thiokinase
(SaCCinyl CoA Synthetase)
isocitrate
dehydrogenase
,
GTP
CO
GOP
NAI
Ketoglutarate
0‘
NAUM
Suconyl- CoA
CO,
a-Ketoglutarate
dehydrogenase complex
(Thiamine dependent)
© UWorld
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(Notes
(
alculdlor
synmerasej
A
Succiiiyl-CoA
6
8
9
ID
Lab Value!
Next
Previous
,
CO
a Ketoglutarate
dehydrogenase complex
( Thiamine dependent)
© UWortd
Patients with chronic alcoholism are frequently deficient in thiamine a necessary
cofactor for pyruvate dehydrogenase, a-ketoglutarate dehydrogenase and
transketolase Administration of glucose to thiamine - deficient patients can cause
rapid depletion of the small amount of thiamine remaining in the circulation . This can
result in neuronal injury within highly metabolic brain regions , leading to acute
Wernicke encephalopathy.
,
u
H
20
21
n
23
24
2S
26
27
23
29
30
31
32
33
34
The metabolism of ethanol by alcohol dehydrogenase and aldehyde dehydrogenase
consumes NAD and increases the NADH to NAD ratio . This skewed ratio inhibits
all pathways requiring NAD *: as a result , the entire citric acid cycle is
inhibited . However , in the setting of Wernicke encephalopathy , thiamine -dependent
enzymes are especially affected due to the lack of NAD and thiamine (Choices A ,
B , C, E, F, G . and H)
'
'
-
Educational objective:
Pyruvate dehydrogenase and a -ketoglutarate dehydrogenase require thiamine as a
cofactor Administration of glucose to thiamine -deficient patients ( eg alcoholics i
can result in Wernicke encephalopathy (eg , acute confusion , ophthalmoplegia and
ataxia ) due to increased thiamine demand.
References:
1 . Thiamine pyrophosphate- requiring enzymes are altered during
pyrithiamine-induced thiamine deficiency in cultured human
I*
V
u ^u
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The following vignette applies to the next 2 items The items in the set must be
answered in sequential order Once you click Proceed to Next Item , you will not be
able to add or change an answer .,
An 87- year -old nursing home resident is brought to the emergency department with a
two day history of fever vomiting and progressive lethargy His past medical history
is significant for advanced dementia and stroke His temperature is 38.3 C (101°
F ) blood pressure is 78 /62 mm Hg , and pulse is 125/min and regular. On
examination, he is lethargic but arousable Coarse rhonchi are heard over the right
midlung. Laboratory results are as follows :
23.000 ceils /pL
Leukocyte count
Platelets
210.000 /pL
T
.
21
n
21
2A
29
26
27
28
29
30
31
32
33
34
Sodium
Potassium
Chloride
Bicarbonate
Creatinine
Glucose
Lactic acid
141 mEq/L
4.2 mEq/L
100 mEq/L
14 mEq/L
0,9 mg/dL
121 mg/dL
5.2 mmol/ L (normal , < 2
mmol/L )
Item 1 of 2
Which of the following is the most likely cause of this patient 's increased anion gap9
A . Impaired hepatic gluconeogenesis
B Increased lipolysis and ketogenesis
03 : 17
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*
(Notes
t dkuldter
4
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8
9
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12
13
U
IS
115
1?
r
*
An 87- year - old nursing home resident is brought to the emergency department with a
two day history of fever vomiting , and progressive lethargy . His past medical history
is significant for advanced dementia and stroke His temperature is 38.3 ' C (101°
F ) blood pressure is 78 /62 mm Hg and pulse is 125/min and regular On
examination he is lethargic but arousabie Coarse rhonchi are heard over the right
midlung. Laboratory results are as follows:
23 000 ceils/pL
Leukocyte count
Platelets
210,000 /pL
20
Sodium
Potassium
Chloride
21
Bicarbonate
22
Creatinine
Glucose
Ij
19
21
24
2&
26
27
28
29
30
31
32
33
34
b
.
Lactic acid
141 mEq/L
4.2 mEq/L
100 mEq/L
14 mEq/L
0.9 mg/dL
121 mg/dL
5.2 mmol/L ( normal, < 2
mmol/L)
Item 1 of 2
Which of the following is the most likely cause of this patient ’s increased anion gap?
O A. Impaired hepatic gluconeogenesis
C B. increased lipolysis and ketogenesis
O C. Increased protein breakdown
C D . Impaired renal tubular bicarbonate reabsorption
C E . Decreased oxidative phosphorylation
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*
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lo l c u f a l o r
ft
An 87- year - old nursing home resident is brought to the emergency department with a
two day history of fever, vomiting and progressive lethargy . His past medical history
a
is significant for advanced dementia and stroke His temperature is 38 3 C ( 101*
F ) blood pressure is 78 /62 mm Hg and pulse is 125/min and regular On
examination he is lethargic but arousabie Coarse rhonchi are heard over the right
midlung . Laboratory results are as follows:
23.000 cells/pL
Leukocyte count
Platelets
210 ,000 / JJL
,
16
17
13
Sodium
Potassium
Chloride
Bicarbonate
Creatinine
Glucose
14
20
21
22
23
24
2S
25
Lactic acid
141 mEq/L
4.2 mEq/L
100 mEq/L
14 mEq/L
0.9 mg/dL
121 mg /dL
5.2 mmol /L ( normal, < 2
mmol/L )
,
27
28
29
30
Item 1 of 2
Which of the following is the most likely cause of this patient ’ s increased anion gap?
31
32
33
34
O A . Impaired hepatic gluconeogenesis [6 %]
C B . Increased tipolysis and ketogenesis [ 23%]
O C . Increased protein breakdown [9%]
*
C D . Impaired renal tubular bicarbonate reabsorption [19% ]
® E. Decreased oxidative phosphorylation [43%]
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O . Id
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*
Notes
lolcufdlor
A
•Enhanced metabolic rate ( eg, seizures
and exercise )
•Reduced oxygen delivery (eg cardiac
or pulmonary failure , shock, and tissue
infarction)
Diminished lactate catabolism due to
hepatic failure or hypoperfusion
•Decreased oxygen utilization ( eg
cyanide poisoning)
•Enzymatic defects in glycogenolysis or
gluconeogenesis
13
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Causes of lactic acidosis
12
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33
34
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VP
Explanation:
10
27
o
* I E. Decreased Oxidative phoaphorylat on [43%]
9
28
29
<1
L,
This patient' s signs and symptoms ( eg fever , leukocytosis hypotension , and
tachycardia ) suggest that he is in septic shock His decreased bicarbonate level and
increased anion gap are indicative of amon-gap metabolic acidosis , and his elevated
lactic acid level suggests that these metabolic derangements are secondary to lactic
acidosis .
Lactic acidosis is an anion-gap metabolic acidosis that results from overproduction
and/or impaired clearance of lactic acid . In septic shock , impaired tissue
oxygenation decreases oxidative phosphorylation leading to the shunting of pyruvate
to lactate after glycolysis . Hence , there is an increase in lactic acid
Iftrmatinn
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Newt
Notes
t dkuldtor
Lactic acidosis is an anion-gap metabolic acidosis that results from overproduction
and /or impaired clearance of lactic acid. In septic shock , impaired tissue
oxygenation decreases oxidative phosphorylation , leading to the shunting of pyruvate
to lactate after glycolysis. Hence , there is an increase in lactic acid
formation Hepatic hypoperfusion also contnbutes to the buildup of lactic acid as
the liver is the primary site of lactate clearance .
(Choice A) Impaired hepatic gluconeogenesis can result in the buildup of lactic
acid as seen in congenital pyruvate carboxylase deficiency . However, this patient s
serum glucose level , lack of specific history suggestive of enzymatic deficiency, and
symptoms of septic shock make this option unlikely.
U
(Choice B) Increased lipolysis and ketogenesis occur in patients with diabetic
ketoacidosis , which also presents with an anion-gap metabolic acidosis
(Choice C } Increased protein breakdown can occur in the setting of chronic
metabolic acidosis . However, it is not a direct cause of acidosis.
(Choice D) Impaired renal tubular bicarbonate reabsorption produces type 2
(proximal ) renal tubular acidosis Poor bicarbonate reabsorption can occur in a
variety of inherited or acquired conditions , including multiple myeloma and drug
toxicity ( eg, acetazolamide ).
Educational objective:
Lactic acidosis occurs in patients with septic shock because of tissue hypoxia , which
results in impaired oxidative phosphorylation and the shunting of pyruvate to lactate
following glycolysis. Hepatic hypoperfusion also contributes to the buildup of lactic
acid, as the liver is the primary site of lactate clearance .
Time Spent 13 seconds
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*
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t dlculdior
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s
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17
13
The patient is treated with intravenous fluids and antibiotics with improvement in his
fever, leukocytosis and hypotension. On the third day of hospitalization , his
temperature is 36 7 * C ( 98° F ), blood pressure is 122/78 mm Hg and pulse is
86< min His BMl is 19.5 kg / m: . On examination he is mildly lethargic and his voice is
soft and breathy . There is left- sided hemiparesis from his previous stroke Chest
x-ray demonstrates dense airspace opacities in the superior region of the lower
lobes . Which of the following is the most likely cause for this patient' s current
hospitalization?
,
,
b
H
20
21
n
23
2
2S
25
*
27
28
29
A Reduced intercostal muscle strength
C B Swallowing muscle dysfunction
O C . Extended immobility
O D Decreased enteral feeding
O £ . Impaired cellular immunity
30
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t tiiculdior
4
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Item 2 of 2
9
10
11
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13
u
15
15
1?
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The patient is treated with intravenous fluids and antibiotics with improvement in his
fever leukocytosis , and hypotension . On the third day of hospitalization , his
temperature is 36.7 C (98° F ), blood pressure is 122/78 mm Hg. and pulse is
86 /min His BMI is 19.5 kg / m:. On examination he is mildly lethargic and his voice is
soft and breathy . There is left- sided hemiparesis from his previous stroke Chest
x-ray demonstrates dense airspace opacities tn the superior region of the lower
lobes. Which of the following is the most likely cause for this patient' s current
hospitalization?
C
b
H
20
21
22
21
24
25
25
27
23
29
30
31
32
33
34
A . Reduced intercostal muscle strength [12%]
* <§ B . Swallowing muscle dysfunction [52%]
O C . Extended immobility [13%J
O D Decreased enteral feeding [4%]
O E. Impaired cellular immunity [18%]
Explanation:
Predisposing conditions for aspiration
pneumonia
•Altered consciousness impairing cough
reflex and glottic closure teg . dementia
and drug intoxication }
•Dysphagia due to neurologic deficits
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I alculalor
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Explanation:
9
ID
12
13
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IS
IS
17
13
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20
21
22
23
24
25
25
27
23
29
30
31
32
33
34
Predisposing conditions for aspiration
pneumonia
•Altered consciousness impairing cough
reflex and glottic closure (eg . dementia
and drug intoxication }
•Dysphagia due to neurologic deficits
( eg . stroke and neurodegeneratrve
disease )
•Upper gastrointestinal tract disorders
( eg , GERD)
Mechanical compromise of aspiration
defenses (eg , nasogastric and
endotracheal tubes)
Protracted vomiting
Large -volume tube feedings in
recumbent position
b
This patient’s fever, leukocytosis , and radiographic lung opacities are most
consistent with pneumonia Given his history of dementia and stroke with residual
hemiparesis. the pneumonia is most likely due to aspiration. The chest x -ray findings
further support a diagnosis of aspiration pneumonia because the opacities are
located in the superior regions of the lower lobes , which (in addition to the posterior
regions of the upper lobes ) are the most dependent locations in the lungs of supine
individuals Anaerobic bacteria { Peptostreptococcus , Bacteroides . Prevotelta and
Fusobactenum ) are the dominant organisms in the upper airway and may be
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This patientVrever , leukocytosis and radiographic lung opacities are most
consistent with pneumonia Given his history of dementia and stroke with residual
hemiparesis , the pneumonia is most likely due to aspiration. The chest x -ray findings
further support a diagnosis of aspiration pneumonia because the opacities are
located in the superior regions of the lower lobes , which (in addition to the posterior
regions of the upper lobes ) are the most dependent locations in the lungs of supine
individuals. Anaerobic bacteria ( Peptostreptococcus, Bacteroides , Prevotetla , and
Fusobacterium ) are the dominant organisms in the upper airway and may be
isolated from cultures in patients with anaerobic pneumonia .
*
Notes
I iilculdlor
&
(Choices A and C) Reduced intercostal muscle strength can result in atelectasis
and hypoxia Extended immobility can produce atelectasis in the postenor lungs if
the patient remains in a supine position These conditions can also predispose to
the development of pneumonia if pulmonary hygiene is inadequate However,
aspiration pneumonia secondary to dysphagia is more likely given the patient's
residual hemiparesis and soft , breathy voice (indicative of vocal cord paralysis and
likely swallow muscle dysfunction )
.
(Choice D) Decreased enteral feeding would result in starvation and weakness,
which do not manifest as airspace opacities on chest x ray .
-
(Choice E) Impaired celiular immunity usually results in recurrent viral and fungal
infections.
Educational objective:
Elderly patients with dementia or hemiparesis may also have dysphagia , which is a
risk factor for aspiration pneumonia Dependent lung consolidation is commonly
seen in aspiration pneumonia .
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t dieulator
4
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6
7
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IB
16
A group of investigators is studying the regulation of catecholamine synthesis in
response to severe stress . In the experiments , subject rats are randomly assigned
to either an experimental or control group The experimental rats undergo resection
of the pituitary gland , and the control rats undergo craniotomy without pituitary
resection , The experimental animals are found subsequently to have decreased
production of epinephrine by the adrenal medulla compared with the control animals
Decreased activity of which of the following enzymes is most likely responsible for
the observed effect ?
17
ia
H
20
21
n
23
24
2B
26
C A . Catechol-O-methyl transferase
O 6 . Dopa decarboxylase
O C Dopamine beta-hydroxylase
O D Monoamine oxidase
,
O E. Phenylalanine hydroxylase
27
F . Phenylethanollamine -N-methyltransferase
23
29
G . Tyrosine hydroxylase
30
31
32
33
34
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t dlcufdtor
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7
9
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n
21
24
25
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r>
A group of investigators is studying the regulation of catecholamine synthesis in
response to severe stress . In the experiments , subject rats are randomly assigned
to either an experimental or control group The experimental rats undergo resection
of the pituitary gland, and the control rats undergo craniotomy without pituitary
resection . The experimental animals are found subsequently to have decreased
production of epinephrine by the adrenal medulla compared with the control animals
Decreased activity of which of the following enzymes is most likely responsible for
the observed effect ?
C A. Catechol-O-methyl transferase [14%]
O B . Dopa decarboxylase [16%]
O C . Dopamine beta-hydroxylase [16%]
C D . Monoamine oxidase [5%]
E Phenylalanine hydroxylase [4%]
* F Pheny ethane am nr. N- - transferase (33%]
C G. Tyrosine hydroxylase [12%]
.
.
—
30
31
32
33
34
Explanation:
Catecholamine synthesis
Tyrosine
Tyrosine hydroxylase
V
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t <ilc uhitor
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A
5
Explanation:
6
7
Catecholamine synthesis
9
n
Tyrosine
n
12
I
13
u
is
16
17
13
fyros/ne rtytfroxytese
DOPA
H
20
Dopa rfeca/boxy/ase
21
22
23
24
25
26
27
28
29
Dopamine
I
Dopamtfie fl-fiycfroxyiase
Norepinephrine
30
31
32
33
34
I
PJVMP
—
+)
Cortisol
Epinephrine
I
' Phen yiethan olamme- iV-methy it ran sfe ( ase
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*
Notes
t <ilc uldlor
T
Epinephrine
n
11
12
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115
1?
13
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34
' Phenyieth&ryotemine- N - methyltran $fer&$e
b
CUWOrM
The 3 main circulating catecholamines are dopamine norepinephrine and
epmephnne Norepmephnne and dopamine are produced in the central as well as
the peripheral nervous system , whereas epinephrine is predominantly produced in
the adrenal medulla The first step in the synthesis of catecholamines is the
conversion of tyrosine to dihydroxyphenylalanine (DOPA ) by tyrosine hydroxylase
{ Choice G) This is the rate -limiting step in the synthesis of catecholamines DOPA
is converted to dopamine by dopa decarboxylase (Choice B) , which is then
converted to norepinephrine by dopamine beta -hydroxylase (Choice CJ In the
adrenal medulla, norepinephrine is rapidly converted to epinephrine by
phenylethanolamine-N-methyltransferase (PNMT).
,
,
Expression of PNMT in the adrenal medulla is upregulated by cortisol. Because the
venous drainage of the adrenal cortex passes through the adrenal medulla cortisol
concentrations in the medulla can be very high , and PNMT is expressed at a high
level. However, following pituitary resection , ACTH secretion and subsequent
cortisol production would be low The result would be decreased PNMT activity
and reduced conversion of norepinephrine to epinephrine.
(Choices A and D) Catechol- O -methyltransferase (COMT) and monoamine
oxidase iMAO ) are the enzymes responsible for breakdown of catecholamines
COMT converts epinephrine to metanephnne and norepinephrine to
normetanephnne . MAO converts metanephnne and normetanephrine to
vanillylmandelic acid.
Block Time Remaining:
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17
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25
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Sent
Notes
t alt ufalor
1
>
/
Expression of PNMT in the adrenal medulla is upregulated by cortisol Because the
venous drainage of the adrenal cortex passes through the adrenal medulla cortisol
concentrations in the medulla can be very high , and PNMT is expressed at a high
level However , following pituitary resection . ACTH secretion and subsequent
cortisol production would be low . The result would be decreased PNMT activity
and reduced conversion of norepinephrine to epinephrine .
b
(Choices A and D) CatechoJ- O -methyltransferase (COMT ) and monoamine
oxidase ( MAO ) are the enzymes responsible for breakdown of catecholamines .
COMT converts epmephnne to metanephnne and norepmephnne to
normetanephrine. MAO converts metanephrine and normetanephnne to
vanillylmandelic add.
(Choice E) Tyrosine required for the synthesis of catecholamines is obtained from
either dietary intake or by conversion of phenylalanine by phenylalanine hydroxylase
in the liver . Deficiency of this enzyme causes phenylketonuria .
Educational objective:
Cortisol increases the conversion of norepinephrine to epinephrine in the adrenal
medulla by increasing the expression of phenylethanolamine -N-methyltransferase
30
31
32
33
34
References:
1 . Stress-triggered changes in peripheral catecholaminergic systems.
2 . Epinephrine biosynthesis: hormonal and neural control during
stress .
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c
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*
Notes
t all ufalor
A 76-year-old man is brought to the emergency department due to excessive fatigue
and altered mental status . He was recently discharged home from the emergency
department with a diagnosis of pneumonia The patient has a long history of
diabetes meliitus . His blood pressure is 10Q/60 mm Hg and pulse is 100/min .
Physical examination shows a dry mouth, cracked lips, and severe cataract
formation Laboratory tests show a serum glucose level of 750 mg /dL and a normal
serum ketone level The pathophysiology of this patient 's cataract formation involves
certain enzymes within the lens An enzyme called aldose reductase produces
sorbitol, a substance that cannot readily exit the lens cells Which of the following is
the most likely end product of sorbitol metabolism in the lens of healthy individuals?
'
A, Glucose
C B . Fructose
C C . Galactose
D Galactitol
C ) E . Xylulose
30
31
32
33
34
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Notes
l tilculalor
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IS
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17
lj
A 76 year-old man is brought to the emergency department due to excessive fatigue
and altered mental status He was recently discharged home from the emergency
department with a diagnosis of pneumonia The patient has a long history of
diabetes mellrtus, His blood pressure is 1GQ /60 mm Hg and pulse is 100/min .
Physical examination shows a dry mouth, cracked lips, and severe cataract
formation Laboratory tests show a serum glucose level of 750 mg/dL and a normal
serum ketone level The pathophysiology of this patient 's cataract formation involves
certain enzymes within the lens An enzyme called aldose reductase produces
sorbitol, a substance that cannot readily exit the lens cells Which of the following is
the most likely end product of sorbitol metabolism in the lens of healthy individuals?
&
H
20
21
n
21
24
25
25
27
23
29
30
31
32
33
34
O A . Glucose [13%}
* •B
se [44%}
O C . Galactose [8%]
C D Galactrtol [25%]
,
O E Xylulose [11%]
,
Explanation:
This patient has diabetes mellitus and presents with hyperosmolar hyperglycemia , a
metabolic derangement often precipitated by infection (pneumonia in this case ) and
characterized by dehydration , hyperglycemia and hyperosmolarity without significant
ketoacidosis His cataracts likely formed from oversaturation of the polyol pathway
secondary to long- term hyperglycemia.
Polyol pathway
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*
l tilculalor
4
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Explanation:
£
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11
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13
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17
This patient has diabetes meliitus and presents with hyperosmolar hyperglycemia a
metabolic derangement often precipitated by infection (pneumonia in this case ) and
characterized by dehydration, hyperglycemia , and hyperosmolahty without significant
ketoacidosis His cataracts likely formed from oversaturation of the polyol pathway
secondary to long-term hyperglycemia .
Polyol pathway
lj
H
20
Glucose
21
n
Sorbitol
dehydrogenase
Aldose
reductase
21
24
25
25
27
23
29
NADPH
\
NADP+
Fructose
Sorbitol
NAD+
NAOH
30
31
32
33
34
Glycolysis
(DUWofW
Aldose reductase converts glucose into sorbitol during the first step in the polyol
pathway of glucose metabolism . Sorbitol cannot readily cross cell membranes and
is therefore trapped inside the cells within which it is formed If the enzyme sorbitol
rlohvdrnnon^ce
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(Notes
l tilculalor
OJWofU
A
&
B
9
10
11
n
13
14
IS
15
17
13
H
2Q
21
22
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Aldose reductase converts glucose into sorbitol during the first step in the polyol
pathway of glucose metabolism. Sorbitol cannot readily cross cell membranes and
is therefore trapped inside the cells within which it is formed If the enzyme sorbitol
dehydrogenase ( sometimes referred to as polyol dehydrogenase ] is also present in
the cell , it can convert sorbitol into fructose. This pathway , known as the polyol
pathway, is especially active in the seminal vesicles , as sperm use fructose as their
primary energy source . Other tissues , such as the retina , renal papilla , and Schwann
cells , have much less sorbitol dehydrogenase activity .
&
The human lens does contain significant levels of sorbitol dehydrogenase , ( see
references ), which allows for the production of fructose However , this enzyme has
a significantly lower Vmax in the sorbitol-to -fructose direction than in the reverse
direction. When glucose levels are low. the limited forward activity of this enzyme is
sufficient to convert enough sorbitol into fructose to prevent sorbitol accumulation.
In contrast , states of long-standing hyperglycemia lead to the production of an
excessive amount of sorbitol that is trapped in the celis. This increases the osmotic
pressure and facilitates the influx of water into the lens cells , leading to the
development of hydropic lens fibers that degenerate Eventually this results in lens
opacification and cataract formation . In addition to osmotic cell injury, oxidative
stress resulting from the depletion of NADPH contributes to cataract formation and
other diabetic complications such as neuropathy and retinopathy.
(Choice A) The end product of sorbitol metabolism is fructose, not glucose
Glucose is formed in celis by the processes of glycogenolysis and
gluconeogenesis .
(Choices C and D) Another function of aldose reductase is conversion of
galactose into galactitol | ie , this enzyme converts sugars into their corresponding
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Glucose is formed in cells by the processes of glycogenolysis and
gluconeogenesis
C.7
*
Note
*
t alculalor
A
(Choices C and D) Another function of aldose reductase is conversion of
galactose into galactitol | ie , this enzyme converts sugars into their corresponding
sugar alcohols ) Galactitol production via this pathway is normally insignificant . In
galactosemia ( galactose 1-phosphate uridyltransferase deficiency), an increased
amount of galactitol is produced, resulting in cataract formation .
&
(Choice E) Xylulose is an end product of glucuronic acid metabolism and an
intermediate in the pentose phosphate pathway .
Educational objective:
Aldose reductase converts glucose into sorbitol , which is further metabolized into
fructose by sorbitol dehydrogenase. This pathway is most active in the seminal
vesicles , The lens also contains significant levels of sorbitol dehydrogenase , which
become overwhelmed in the setting of hyperglycemia Other tissues , such as the
retina , renal papilla , and Schwann cells, have much less sorbitol dehydrogenase
activity .
.
References:
1, The sorbitol pathway in the human lens: aldose reductase and
polyol dehydrogenase.
2 , Osmotic stress induced oxidative damage: possible mechanism of
cataract formation in diabetes
.
-
3, Contribution of polyol pathway to diabetes induced oxidative stress.
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*
Notes
4 ii
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b
B
9
I
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13
A protein believed to play a role in signal transduction and the cellular response to
thyroid stimulating hormone is studied Special attention is paid to a region of this
protein that contains several alpha -helical regions each composed of approximately
20 amino acid residues— consisting primarily of valine , alanine , and
isoleucine Which of the following functions does this particular region most likely
perform?
&
u
15
15
1?
13
H
20
O A . Phosphorylating tyrosine residues
O B Binding to intranuclear DNA
O C Spanning the cellular membrane
21
C 0 . Interacting with signaling substances
23
24
25
25
C
n
Interacting with metal ions in transporting proteins
27
28
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30
31
32
33
34
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I alcufator
4
/%
&
6
9
I
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n
13
M
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16
17
1B
20
2i
n
23
24
2&
25
27
23
29
30
A protein believed to play a role in signal transduction and the cellular response to
thyroid stimulating hormone is studied Special attention is paid to a region of this
protein that contains several alpha- helical regions each composed of approximately
20 amino acid residues— consisting primarily of valine , alanine , and
Isoleucine Which of the following functions does this particular region most likely
perform?
O A . Phosphorylating tyrosine residues [12%]
O B . Binding to intranuclear DNA [20%]
C. Spanning the cellular membrane [46%]
O D . interacting with signaling substances [17%]
E . Interacting with metal ions in transporting proteins [5%]
Explanation:
Extracellular domain
tor hormone binding
Coll membrane
composed of a
phospholipid bifayer
3t
32
33
34
CYTOPLASM
v
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£ dlculdtor
Notes
A
A
&
Explanation:
B
9
Cell membrane
i
11
12
13
14
15
15
1?
13
H
20
21
22
23
24
composed of a
phospholipid hilayer
30
31
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33
34
11
IfllfllM
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/
/
CYTOPLASM
* -helical segments
Trantmembrine domain with 7 alpha
each containing approximately 20 amino acids
with hydrophobic , nonpolar ft -groups Intracellular domain with
i
tyrosine kinaee
ability OR the ability to activate
second-messenger systems such as G -proteins.
25
25
27
28
29
Extracellular domain
for hormone binding
Non-polar , hydrophobic amino acids such as valine alanine, isoleucine , methionine ,
and phenylalanine are generally located interiorly on globular proteins where they are
shielded from direct contact with water The classic pfasma membrane - spanning
proteins are executors for glycoprotein hormones such as TSH, LH , and FSH,
These G-protein-coupled membrane -bound receptors for glycoprotein hormones
contain three major domains extracellular (responsible for ligand binding ),
transmembrane (consisting of hydrophobic amino acids ) , and intracellular ( coupled
with G-proteins ),
The question stem describes repeating alpha -helical segments each composed of
approximately 20 hydrophobic ammo acids this description is characteristic of the
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transmembrane (consisting of hydrophobic ammo acids ) , and intracellular ( coupled
with G-proteins ).
B
9
The question stem desenbes repeating alpha -helical segments each composed of
approximately 20 hydrophobic amino acids this description is characteristic of the
transmembrane region( s ) of a protein (Choice C) .
i
11
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IS
IS
17
13
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Lab Value
Newt
*
u
(Choice A) The receptors for insulin IGF - 1 . and several cytokines also have three
domains: extracellular (ligand binding ] , transmembrane (composed of hydrophobic
amino acids) and intracellular The intracellular ( cytosolic ) domain of these
membrane - associated receptor proteins contains a tyrosine kinase that is activated
upon extracellular ligand binding Once activated, tyrosine kinase will phosphorylate
available tyrosine residues
I dlculaior
Notes
.
(Choice B) Cyclic AMP formed during the activation of a Gt protein can activate
protein kinase A . Protein kinase A has the unique ability to phosphorylate and
thereby activate proteins that are capable of translocating into the nucleus. Once
within the nucleus, these special proteins bind to the promoter regions of DNA and
modulate transcription. (The proteins that actually bind to DNA are not membrane
proteins.)
(Choice D) The binding of a ligand to the extracellular domain of a transmembrane
protein causes indirect activation of the adenylate cyclase system the opening of ion
channels , direct activation of tyrosine kinase activation of the calcium- calmodulin
system, and activation of the inositol triphosphate system. The region of a
transmembrane protein that interacts with the ligand is found in the extracellular
space and therefore could not possibly be composed mainly of hydrophobic amino
acids
(Choice E) This choice describes the intracellular iron-containing proteins
.
i
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Nest
avaiiaoie tyrosine resiaues .
*
t alcufdtor
Notes
A
&
(Choice B) Cyclic AMP formed during the activation of a G protein can activate
protein kinase A . Protein kinase A has the unique ability to phosphorylate and
thereby activate proteins that are capable of translocating into the nucleus Once
within the nucleus , these special proteins bind to the promoter regions of DNA and
modulate transcription ( The proteins that actually bind to DNA are not membrane
proteins . }
;
B
9
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21
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34
b
(Choice D) The binding of a ligand to the extracellular domain of a transmembrane
protein causes indirect activation of the adenylate cyclase system the opening of ion
channels, direct activation of tyrosine kinase , activation of the calcium- calmodulin
system , and activation of the inositol tnphosphate system The region of a
transmembrane protein that interacts with the ligand is found in the extracellular
space and therefore could not possibly be composed mainly of hydrophobic amino
acids .
(Choice E) This choice describes the intracellular iron-containing proteins
(hemeproteins) such as hemoglobin myoglobin, and cytochrome oxidase Heme is
a complex of protoporphyrin IX and iron Hemoglobin A , the most common
hemoglobin in adults consists of two alpha and two beta globin chains held together
by non-covalent interactions. Each subunit has stretches of alpha helices and a
crevice lined by nonpolar amino acids , where heme binding occurs.
Educational Objective:
Integral membrane proteins contain transmembrane domains composed of alpha
helices with hydrophobic amino acid residues such as valine , alanine isoleucine ,
methionine , and phenylalanine.
Time Spent 5 seconds
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Notes
t alcuLilor
A
5
a
An agent that specifically blocks the interaction of inositol triphosphate with its
intracellular receptor would most likely decrease the activity of :
9
n
12
13
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15
115
17
10
b
A. Phospholipase C
O B . Lipoxygenase
C. Protein kinase C
O D . Phosphodiesterase
E . Adenylate cyclase
H
20
21
22
23
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25
21
23
29
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32
33
3
*
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Notes
(
olc uhnor
4
A
6
B
9
An agent that specifically blocks the interaction of inositol triphosphate with its
intracellular receptor would most likely decrease the activity of :
n
O A . Phospholipase C [31%]
C B . Lipoxygenase [2%]
12
13
14
IS
1S
1?
13
*
*
C
C . Protein kinase C [56%]
D . Phosphodiesterase [5% ]
_ E . Adenylate cyclase [6%]
H
20
21
22
23
24
2&
25
2T
28
29
30
31
32
33
34
Explanation:
<
Epinephrine ai )
A
MW
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Phospholipose -c
Cell Membrane
Phospholipids
DAG
IP :
SR
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*
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I alc uhnor
4
6
A
Explanation:
Epinephrine (a , )
4
B
9
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13
14
IS
1S
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13
ifflii (I piiiiW
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20
21
Phospholipase -c
Cell Membrane
Phospholipids
DAG
22
23
24
2&
25
2T
28
29
30
31
32
33
34
&
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The G-protem-coupled-receptors have a very characteristic structure with seven
transmembrane regions an extracellular domain and an intracellular domain coupled
with the Irimeric G- protein . In their inactivated state , G-proteins exist as
hotfimtHfnort r Ancictinn r %f alnhai heta anH
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115
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Notes
£ alculator
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6
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The G-protem-coupled-receptors have a very characteristic structure with seven
transmembrane regions , an extracellular domain and an intracellular domain coupled
with the trimenc G -protein In their inactivated state , G -proteins exist as
heterotnmers consisting of alpha beta and gamma subunits with guanosine
diphosphate (GDP ) tightly bound to the alpha subunit . G proteins are activated after
binding of hormone to the extracellular domain. The first step in activation of a
G-protein occurs when GDP is exchanged for GTP on the alpha subunit Once
bound to GTP, the alpha subunit dissociates from the beta and gamma subunits and
exposes its catalytic domain for either adenylate cyclase or phospholipase C
depending on the ligand
If the G-protein alpha subunit activates phospholipase C then the degradation of
phosphatidylinositol 4 , 5-bisphosphate to inositol 1 4.5 -triphosphate ( IP ) and
diacylglycerol (DAG ) occurs Diacylglycerol stimulates protein kinase C , which is
responsible for some intracellular effects Inositol 1, 4 , 5 -tnphosphate ( IP ) produces
most of the intracellular effects of this pathway by increasing intracellular calcium ,
and elevated intracellular calcium activates protein kinase C. If the action of IP, were
blocked as described m the question stem , then decreased activation of protein
kinase C would occur upon hormone binding ( Choice C ).
(Choice A) The activity of phospholipase C would be unchanged if IP3 were
blocked because phospholipase C exerts its effect before IP. in the calcium I
phosphatidylinositol second messenger system .
(Choice B) Lipoxygenase is an enzyme responsible for formation of leukotrienes
from arachidonic acid It is not involved in intracellular signaling
(Choice D) Termination of the effects of hormones that act by cAMP or cGMP
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lespuiisnjiBiurauMicf
^ . n i u s n u i i, ** , 0 -111^11 IUSJJJ I O I C v i r ; p i u u u t c s
most of the intracellular effects of this pathway by increasing intracellular calcium ,
and elevated intracellular calcium activates protein kinase C . If the action of IP. were
blocked as described m the question stem , then decreased activation of protein
kinase C would occur upon hormone binding ( Choice C),
umaLCHUioi eiie
is
*
Note
Iolcufdtor
*
A
TO
12
13
H
IB
16
ir
13
H
20
21
n
23
21
2&
25
2T
23
29
30
31
32
33
34
(Choice A) The activity of phospholipase C would be unchanged if IP3 were
blocked because phospholipase C exerts its effect before IP in the calcium /
phosphatidyl!nositol second messenger system.
b
(Choice B) Lipoxygenase is an enzyme responsible for formation of leukotrienes
from arachidonic acid. It is not involved in intracellular signaling
(Choice D) Termination of the effects of hormones that act by cAMP or cGMP
G-protein second messenger systems is carried out by the enzyme
phosphodiesterase Phosphodiesterase has no effect on the IP , second
messenger system*
(Choice E ) Activation of adenylate cyclase leads to the formation of cyclic AMP and
the subsequent activation of protein kinase A . Protein kinase A activates the
proteins that produce the intracellular effects of hormones
Educational Objective:
After a hormone binds a G-protein coupled receptor that activates phospholipase C,
the initial step of the IP , second messenger system involves degradation of
membrane lipids into diacylglycerol (DAG ) and inositol triphosphate ( IP; ) by that
enzyme Protein kinase C is activated by DAG as weli as calcium released from
sarcoplasmic reticulum under the influence of IP,*
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Notes
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alculator
A
b
B
9
10
11
b
Hepatocytes exposed to an external stimulus demonstrate a rapid increase in
intracellular glycogen stores and a decrease in glucose release into the blood
Which of the following most likely promotes the effects described above ?
A. Protein phosphatase -1
13
u
IB
16
13
H
20
O B . Protein kinase A
O C Phospholipase C
D . Janus protein kinase { JAK )
O E Lipoxygenase
21
n
23
21
25
25
?T
23
29
30
31
32
33
34
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*
Notes
t dlcufdtor
/%
Hepatocytes exposed to an external stimulus demonstrate a rapid increase in
intracellular glycogen stores and a decrease in glucose release into the blood
Which of the following most likely promotes the effects described above ?
tf
* ® A. Protein phosphatase - 1 [32%]
O B. Protein kinase A [29%]
O C. Phospholipase C [13%]
O D . Janus protein kinase i JAK ) [22%]
O E. Lipoxygenase [3%]
20
21
Explanation:
23
2i
The response of the hepatocytes to the stimulus described in the question stem is
characteristic of the response of these cells to insulin . Insulin is an anabolic
hormone that promotes the synthesis of glycogen , triacylglycendes , nucleic acids ,
and proteins. Insulin inhibits glycogenolysis and gluconeogenesis . Insulin acts via a
tyrosine kinase mechanism . The insulin cell surface receptor is a transmembrane
protein that also has cytosolic tyrosine kinase activity . The tyrosine kinase causes
phosphorylation of a poorly characterized class of proteins known as insulin receptor
substrates leading to activation of protein phosphatase . Protein phosphatase
dephosphorylates glycogen synthase thereby activating that protein and promoting
glycogen synthesis, Protein phosphatase also dephosphorylates fructose
1, 6 -bisphosphatase thereby inactivating that enzyme and inhibiting
gluconeogenesis . This is also a good example of how phosphorylation and
dephosphorylation of enzymes by second-messenger proteins can cause activation
n
as
25
71
28
29
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uaiiauiciiiuioiic
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protein that a ^ so has cytosolic tyrosine kinase activity The tyrosine kinase causes
phosphorylation of a poorly characterized class of proteins known as insulin receptor
substrates leading to activation of protein phosphatase . Protein phosphatase
dephosphorylates glycogen synthase thereby activating that protein and promoting
glycogen synthesis. Protein phosphatase also dephosphorylates fructose
1 6-bisphosphatase thereby inactivating that enzyme and inhibiting
gluconeogenesis This is also a good example of how phosphorylation and
dephosphorylation of enzymes by second- messenger proteins can cause activation
of some enzymes and inactivation of others.
B
9
10
11
M I l o s e M 115
1
i
in
Note
*
t alcufalor
r\
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13
U
IS
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1?
13
19
20
21
n
23
24
2&
25
(Choice B) Protein kinase A is the primary intracellular affector enzyme in the
G-protein ! adenylate cyclase second messenger system . Increased levels of
cAMP stimulate protein kinase A to activate the necessary enzymes to carry out the
intracellular actions of the hormone that bound the cell and activated adenylate
cyclase in the first place .
(Choice C ) Phospholipase C is active in the G- protein / Inositol tnphosphate ( IP3 ) /
30
Calcium second messenger system Hormone binds its receptor and activates a
G-protein that in turn activates phospholipase C to degrade phospholipids into
inositol triphosphate and diacylglycerol . Both diacylglycerol and the increased
intracellular calcium caused by IP will activate protein kinase C.
31
32
33
34
(Choice D) Janus protein kinase ( JAK ) is a part of the second messenger system
for peptide hormones such as some cytokines in a pathway referred to as
?r
28
29
^
JAK-STAT ( signal transducers and activators of transcription ) JAK has tyrosine
kinase activity .
(Choice E) Lipoxygenase is an enzyme involved in arachidonic acid metabolism and
is responsible for the arm of that pathway that synthesizes leukotrienes
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o,
*
*
Notes
l ukuhitor
/%
(Choice B ) Protein kinase A is the primary intracellular affector enzyme in the
G-protein i adenylate cyclase second messenger system Increased levels of
cAMP stimulate protein kinase A to activate the necessary enzymes to carry out the
intracellular actions of the hormone that bound the cell and activated adenylate
cyclase in the first place .
b
(Choice C ) Phospholipase C is active in the G- protein l Inositol triphosphate ( IP ) f
^
Calcium second messenger system . Hormone binds its receptor and activates a
G-protein that in turn activates phospholipase C to degrade phospholipids into
inositol triphosphate and diacylglycerol . Both diacylglycerol and the increased
intracellular calcium caused by IP 3 will activate protein kinase C
,
20
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23
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25
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34
(Choice D) Janus protein kinase ( JAK ) is a part of the second messenger system
for peptide hormones such as some cytokines in a pathway referred to as
JAK- STAT ( signal transducers and activators of transcription ) . JAK has tyrosine
kinase activity .
(Choice E) Lipoxygenase is an enzyme involved in arachidonic acid metabolism and
is responsible for the arm of that pathway that synthesizes leukotrienes
Educational Objective:
Insulin is an anabolic hormone that acts via a tyrosine kinase second messenger
system to stimulate the synthesis of glycogen proteins , fatty acids and nucleic
acids Tyrosine kinase leads to the activation of protein phosphatase within cells,
and protein phosphatase directly modulates the activity of enzymes in the metabolic
pathways regulated by insulin,
,
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A
b
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9
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19
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27
A 35-year-otd man comes to the office due to fatigue , dark urine , and back pain . His
wife returned from a business trip in Egypt several days ago and brought home a
type of large flat bean that he had never seen before. The patient ate some of the
beans for dinner last night and first noticed the symptoms when he woke up In the
morning . Physical examination shows jaundice and pallor Laboratory studies show
a hemoglobin level of 8 g /dL Further evaluation reveals deficiency of an enzyme
involved in the conversion of glucose to ribulose - 5 -phosphate Impairment in the
activity of this enzyme is most likely to inhibit which of the following biochemical
pathways?
I
© A. ADP phosphorylation
B Cholesterol synthesis
O C Glycogen storage
O D . Ketone body synthesis
C E . Protein degradation
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Notes
Iolcuhitor
A
ft
&
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to
11
12
H
IS
15
1?
13
A 35-year-old man comes to the office due to fatigue , dark urine , and back pain . His
wife returned from a business trip in Egypt several days ago and brought home a
type of large flat bean that he had never seen before The patient ate some of the
beans for dinner last night and first noticed the symptoms when he woke up In the
morning. Physical examination shows jaundice and pallor. Laboratory studies show
a hemoglobin level of 8 g /dL Further evaluation reveals deficiency of an enzyme
involved in the conversion of glucose to ribulose-5 -phosphate Impairment in the
activity of this enzyme is most likely to inhibit which of the following biochemical
pathways?
14
20
21
22
23
2i
2&
25
27
23
29
i
C A. ADP phosphorylation [36%]
* • B . Cholesterol synthesis [29%)
C Glycogen storage [16%J
O D . Ketone body synthesis [10%]
O E. Protein degradation [9%]
Explanation:
30
Pentose phosphate pathway
31
32
33
3
*
OXIDATIVE
( IRREVERSIBLE )
Glucose - 6 ’ phosphate
NADP *
NADPH
*
Cholesterol & fatty
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( rate limiting step )
1 .
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Explanation:
Lab Values
r
Notes
(
ulcuf -i l o r
A
Pentose phosphate pathway
OXIDATIVE
( IRREVERSIBLE )
to
11
n
Glucoae-6 - phosphate
NADP *
Gtvcose-S -phosphate dehydrogenase
( rate limiting step )
NADPH
H
is
1S
1?
13
Cholesterol & fatty
acid synlhesis
* Glutathione antioxidant
mechanism
*
14
21
30
*
6-phosphoghucondte dehydrogenase
Rib u I ose- 5- phosphate
22
31
32
33
3
NADP *
NADPH
20
23
2i
2S
25
27
23
29
6- phosph ©gluconate
NONOXIDATIVE
(REVERSIBLE )
Ri buiose - 5- phosphate
Xylulose - 5 - phosphate
R ibose- 5-phospha te
Sedoheptulose- 7'phosphate
Nucleotide
synthesis
Glyceraldehyde- 3 -phosphate
f
Tmnsakfofase
\
\
.±
M
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Xy lutose 5 - p hosphate
Lab Value
Newt
\
Rfboseaphosphate
—
*
(Notes
t alculdtor
Nucleotide
synthesis
Transketolase
;
9
10
Sedoheptulose- 7 - phospha te
11
Gl ycer aIdefi yd e- 3 -phospha te
i,
12
Transaldolase
14
IS
is
Ery t hrose - 4 pho spha t e
IT
13
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.
F ructo se-6-pho s p ha te
Glycolysis
© UWodd
This patient is presenting with acute hemolytic anemia, which can occur in those with
glucose 6 phosphate dehydrogenase (G6 PD) deficiency after certain infections
( eg pneumonia , viral hepatitis ), consumption of fava beans or use of specific
medications leg. primaquine and certain sulfa drugs ) . G6PD catalyzes the first
(rate -limiting ) step in the pentose phosphate pathway (PPP ). In the oxidative
portion of the pathway, glucose-6-phosphate is converted to ribulose - 5 -phosphate
and 2 molecules of NADPH are produced . The nonoxidative reactions of the PPP
are used primarily to interconvert ribose - 5-phosphate (used for nucleotide synthesis )
into glycolytic intermediates that can be used for energy production.
--
Because the PPP is the main source of NADPH the pathway is particularly active in
1. Cells experiencing high oxidative stress ( eg , erythrocytes), where NADPH is
used to regenerate reduced glutathione
2. Organs ( eg , liver , adrenals ) involved in reductive biosynthesis leg fatty
acids, cholesterol, steroids ) and cytochrome P 450 metabolism
3. Phagocytic cells generating a respiratory burst (NADPH oxidase )
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*
Notes
I ale ulator
/s
Because the PPP is the main source of NADPH the pathway is particularly active in :
1. Cells experiencing high oxidative stress ( eg . erythrocytes), where NADPH is
used to regenerate reduced glutathione
2. Organs ( eg . liver adrenals ) involved in reductive biosynthesis ( eg fatty
acids , cholesterol, steroids ) and cytochrome P 450 metabolism
3. Phagocytic cells generating a respiratory burst (NADPH oxidase )
u
20
'
are used primarily to interconvert nbose - 5-phosphate (used for nucleotide synthesis )
into glycolytic intermediates that can be used for energy production .
12
15
15
1?
13
M
b
(Choice A) ADP phosphorylation does not occur in the PPP . Although NADPH
itself is a form of energy it cannot be used to convert ADP into ATP (in contrast to
NADH),
21
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(Choice C) Glycogenesis stores glucose for later use via formation of the glucose
polymer glycogen from glucose-1-phosphate However , this process does not
require NADPH as a cofactor.
(Choice D) Ketone bodies are formed mainly in the liver during states of excessive
fat degradation . Cytosolic HMG-CoA synthase is the starting point of cholesterol
synthesis , whereas the mitochondnal version of the enzyme is the rate - limiting step
in ketone body synthesis . Unlike cholesterol synthesis , ketone body production
does not require NADPH.
(Choice E) Protein catabolism begins with the hydrolysis of polypeptides into amino
acids , which are then degraded , starting with the removal of the amine nitrogen by
transamination reactions. Nitrogen is funneled from different amino acids into a
small number of compounds (mainly glutamate ) which in turn are oxidatively
deaminated. producing ammonia The urea cycle then converts ammonia into urea
V
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a respiratory Durst ( NAUKM oxiaase )
*
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*
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akufdlor
5
B
9
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15
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13
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(Choice A) ADP phosphorylation does not occur in the PPP. Although NADPH
itself is a form of energy , it cannot be used to convert ADP into ATP (in contrast to
NADH).
(Choice C) Glycogenesis stores glucose for later use via formation of the glucose
polymer glycogen from glucose -1-phosphate. However , this process does not
require NADPH as a cofactor ,
(Choice D) Ketone bodies are formed mainly in the liver during states of excessive
fat degradation . Cytosolic HMG-CoA synthase is the starting point of cholesterol
synthesis whereas the mitochondrial version of the enzyme is the rate -limiting step
in ketone body synthesis . Unlike cholesterol synthesis ketone body production
does not require NADPH.
(Choice E) Protein catabolism begins with the hydrolysis of polypeptides into amino
acids , which are then degraded , starting with the removal of the amine nitrogen by
transamination reactions Nitrogen is funneied from different amino acids into a
small number of compounds (mainly glutamate ), which in turn are oxidatively
deaminated producing ammonia . The urea cycle then converts ammonia into urea
for elimination in the urine .
Educational objective:
Glucose-6-phosphate dehydrogenase is the rate -limiting enzyme in the pentose
phosphate pathway, the major source of cellular NADPH This reducing molecule is
necessary for reducing glutathione (preventing oxidative damage ) and for the
biosynthesis of cholesterol , fatty acids, and steroids.
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Notes
l alculator
4
5
-
B
9
10
11
12
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13
A 24 year -old woman is diagnosed with gestational diabetes mellitus during her first
pregnancy Although her glycemic status improves markedly after delivery , her
fasting glucose levels remain modestly elevated The patient ' s past medical history
is otherwise unremarkable, but her mother and younger sister had "high blood
sugars' during pregnancy. If this patient’s gestational hyperglycemia is genetically
predisposed she is most likely to have decreased activity in which of the following
enzymes'?
1
O A . Aldolase
21
O B . Enolase
O C. Glucokinase
22
C
H
20
21
24
25
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i
D Lactate dehydrogenase
C E. Phosphofructokinase
F . Pyruvate carboxylase
t G . Pyruvate kinase
30
31
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*
Note
*
I dlcufaior
4
ft
&
-
6
9
n
11
n
13
A 24 year-old woman is diagnosed with gestational diabetes mellrtus during her first
pregnancy . Although her glycemic status improves markedly after delivery , her
fasting glucose levels remain modestly elevated The patient' s past medical history
is otherwise unremarkable, but her mother and younger sister had 'high blood
sugars" during pregnancy. If this patient's gestational hyperglycemia is genetically
predisposed she is most likely to have decreased activity in which of the following
enzymes?
15
15
1T
ia
O A Aldolase [ 4%]
O B Enolase [1%]
H
20
21
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*
C . Glucokinase [57%]
D . Lactate dehydrogenase [3%J
O E. Phosphofructokinase [21%]
F. Pyruvate carboxylase [7% ]
G . Pyruvate kinase [7%]
Explanation:
31
32
33
34
Pancreatic beta cell function
Glucose
#
GLUT 2
V
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Explanation:
Pancreatic beta cell function
n
Glixov?
11
#
b
12
13
GLUT 2
*
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ia
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F
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'
r
Calcium
••* * •
GLP 1 receptOf
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*
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Calcium
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_•* •*•
GLP 1 receptor
•
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© U World
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B
9
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A
Exocytosis
© U World
Glucose is the major stimulant of insulin secretion. Glucose -induced insulin release
from the beta cells requires the following steps:
-
1. Glucose enters the beta cell through glucose transporter type 2 (GLUT 2)
2. Glucose is metabolized by glucokinase to glucose-6-phosphate
3. Glucose-6-phosphate is further metabolized by glycolysis and the Krebs cycle
to produce ATP
4. A high ATP to ADP ratio within the beta cell results in the closure of
ATP-sensitive potassium ( K c h a n n e l s
5 Depolarization of beta cells results in opening of voltage - dependent calcium
channels
6. High intracellular calcium causes insulin release
.
27
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31
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34
Glucokinase has a lower glucose affinity than other hexokinases . This allows it to
function as a glucose sensor in beta cells by varying the rate of glucose entry into the
glycolytic pathway based on blood glucose levels Heterozygous mutations of the
glucokinase gene cause a decrease in beta cell metabolism of glucose , less ATP
formation and diminished insulin secretion . This produces a type of maturity onset
diabetes of the young which is characterized by mild nonprogressive
hyperglycemia that often worsens with pregnancy-induced insulin resistance.
Homozygous mutations lead to fetal growth retardation and severe hyperglycemia at
birth.
-
—
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(Choices A , B, E, and G) Glycolytic enzyme deficiencies ( eg aldolase A enolase,
phosphofructokinase , pyruvate kinase ) generally present with hemolytic anemia , as
red blood cells rely completely on anaerobic glycolysis for energy production.
20
(Choice D) Lactate dehydrogenase is present in most cells and catalyzes
conversion of pyruvate to lactate during anaerobic glycolysis Deficiency can cause
21
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29
30
31
32
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*
(Notes
t <i l r ufalor
function as a glucose sensor in beta cells by varying the rate of glucose entry into the
glycolytic pathway based on blood glucose levels. Heterozygous mutations of the
glucokinase gene cause a decrease in beta cell metabolism of glucose , less ATP
formation and diminished insulin secretion This produces a type of maturity-onset
diabetes of the young which is characterized by mild, nonprogressive
hyperglycemia that often worsens with pregnancy-induced insulin resistance
Homozygous mutations lead to fetal growth retardation and severe hyperglycemia at
birth .
1
H5
1T
13
19
22
Ldb Value
Next
A
decreased exercise tolerance and muscle stiffness .
(Choice F) Pyruvate carboxylase one of the gluconeogenetic pathway enzymes in
the mitochondria catalyzes the conversion of pyruvate to oxaloacetate Deficiency
of pyruvate carboxylase causes lactic acidosis and fasting hypoglycemia
Educational objective:
Insuiin release by pancreatic beta cells is stimulated by increased ATP production
Glucokinase functions as a glucose sensor in pancreatic beta cells by controlling the
rate of glucose entry into the glycolytic pathway . Mutations in the glucokinase gene
are a cause of maturity-onset diabetes of the young.
References:
1 . Glucokinase MODY and implications for treatment goals ot common
V
forms of diabetes.
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(
ulcufdtor
4
b
B
9
10
It
12
to
u
IS
17
10
H
20
21
22
23
24
2b
25
A 45- year-old man is referred to an endocrinologist for newly diagnosed diabetes
mellitus A week ago . his primary care physician noted an elevated fasting serum
glucose level . The endocrinologist discusses the different treatment options
available Including oral and Injectable medications . He recommends treatment with
a medication that alters glucose metabolism within the liver by increasing the
concentration of fructose 2 , 6-bisphosphate within hepatocytes Which of the
following conversions will be inhibited by high intracellular concentrations of this
metabolite ?
b
O A . Acetyl CoA -+ fatty acids
O B Alanine
glucose
C Fructose-6-phosphate — fructose-1 , 6-bisphosphate
O D. Glucose — glycogen
O E NAD NADH
*
'
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(
ulculdtor
A
A
&
B
9
to
It
12
13
U
16
17
13
H
20
21
n
23
21
2&
25
27
23
29
A 45 - year-old man is referred to an endocrinologist for newly diagnosed diabetes
mellitus A week ago. his primary care physician noted an elevated fasting serum
glucose level . The endocrinologist discusses the different treatment options
available including oral and injectable medications . He recommends treatment with
a medication that alters glucose metabolism within the itver by increasing the
concentration of fructose 2 , 6-bisphosphate within hepatocytes Which of the
following conversions will be inhibited by high intracellular concentrations of this
metabolite ?
0
A. Acetyl CoA ~+ fatty acids [5%]
v # B. Alanine glucose [35%1
. C . Fructose -6 -phosphate -» fructose -1, 6-bisphosphate [ 42% ]
. D. Glucose
glycogen [13%]
O E. NAD NADH [6%]
—
-
Explanation:
Fructose 2 ,6 - tnsphosphoto and glucose metabolism
30
31
32
33
34
insulin
Glucose
( CAMP )
?
Fructose
Phosphofrvaokinase 2
2 , 6-blipho phaie
*
It
T
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Fructose 6- pnosphate
07 : 13
T UtCK
Glucagon
11 CAMP
fructose
2.6 txsphospt' atas
*
Fructow
1 2 , 6-bispbosphate
/
V
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Foedn .i cK
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Calculator
NADH [6%]
u
Fructose 2,6 -bisphosphate and glucose metabolism
n
it
12
(
13
H
N
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Explanation:
9
15
17
13
'
Fructose
2, S - tHB phosphate
20
-
Insulin
tcAMP )
Glucose
Glutsgon
i
9
+ cAMP
Fri/CiOS
Fructose 6- phosphate *
*
2.6 tHspfrospfijtJse
L
t.
Fructose
2 $-b spt>osphate
'
21
22
Fructose f
Pnosphatn
21
24
2&
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Citrate
t - bpfcspnalase
fructose T .6 - bsphosphate
27
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Pyruvate
© UWortr)
-
Fructose 2 ,6 bisphosphate helps control the balance between giuconeogenesis
and glycolysis through inverse regulation of phosphofmctokinase-1 t PFK -1 ) and
fructose 1, 6-bisphosphatase . Fructose 2 6-bisphosphate activates PFK -1, the main
regulatory enzyme involved in glycolysis which converts fructose 6-phosphate to
fructose 1,6-bisphosphate . The opposite reaction ( fructose 1, 6 -bisphosphate to
fructose -6-phosphate ) occurs in giuconeogenesis and is catalyzed by the enzyme
fructose -1,6-bisphosphatase (inhibited by fructose 2,6-bisphosphate )
07 : 17
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Lab Values
NCKt
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Notes
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alculator
'
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fructose 1,6-bisphosphatase . Fructose 2, 6-bisphosphate activates PFK -1, the mam
regulatory enzyme involved in glycolysis , which converts fructose 6 -phosphate to
fructose 1, 6 -bisphosphate . The opposite reaction ( fructose 1, 6 -bisphosphate to
fructose -6-phosphate ) occurs in gluconeogenesis and is catalyzed by the enzyme
fructose -1, 6-bisphosphatase (inhibited by fructose 2 , 6-bisphosphate ) ,
i
-
The interconversion of fructose-6 -phosphate and fructose 2,6 -bisphosphate is
achieved by a bifunctional enzyme complex composed of PFK - 2 ( increases fructose
2 ,6 -bisphosphate levels ) and fructose 2, 6-bisphosphosphatase ( decreases fructose
2 , 6 -bisphosphate levels ) Insulin causes activation of PFK -2 , leading to increased
fructose 2,6 -bisphosphate levels and augmented glycolysis High
concentrations of fructose 2 , 6 -bisphosphate also inhibit gluconeogenesis , leading
to decreased conversion of alanine and other gluconeogenic substrates to
glucose .
(Choice A) Fatty acid synthesis is upregulated by insulin and high citrate levels
( which increase when acetyl-CoA is abundant as with active glycolysis ) Therefore ,
fatty acid synthesis is likely to be upregulated in metabolic states in which fructose
2 ,6 -bisphosphate concentration is increased.
(Choice C) The conversion of fructose-6 -phosphate to fructose-1,6-bisphosphate
is catalyzed by the enzyme PFK -1. This enzyme is allostencally activated by high
levels of fructose-2 , 6-bisphosphate and so conversion would be increased .
(Choice D) Glycogen formation is stimulated by increased levels of insulin and
glucose-6 -phosphate Because elevated insulin levels also increase fructose
2 ,6 -bisphosphate formation the rise of fructose 2,6-bisphosphate levels in
hepatocytes is typically concurrent with increased glycogen synthesis .
(Choice E ) The rise in fructose-2 , 6-bisphosphate accelerates glycolysis , leading to
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uc upicyuioicu
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Lab Value
Sent
OIUIC
^
u i vvi nu
11 U
^
&
2, 6 -bisphosphate concentration is increased.
E
9
(Choice C) The conversion of fructose - 6 -phosphate to fructose -1, 6-bisphosphate
is catalyzed by the enzyme PFK -1, This enzyme is allostencally activated by high
levels of fr\Jctose-2,6-bisphosphate and so conversion would be increased .
n
11
n
13
14
IS
17
ia
H
20
21
n
23
24
2&
25
27
23
29
30
31
32
33
34
Notes
*
I dlcutdtor
IUSC
b
(Choice D) Glycogen formation is stimulated by increased levels of insulin and
glucose-6 -phosphate Because elevated insulin levels also increase fructose
2 , 6 -bisphosphate formation , the rise of fructose 2 , 6-bisphosphate levels in
hepatocytes is typically concurrent with increased glycogen synthesis.
(Choice E) The rise in fructose - 2 , 6 -bisphosphate accelerates glycolysis , leading to
increased conversion of NAD * to NADH.
Educational objective:
Fructose 2 , 6-bisphosphate activates phosphofructokinase-1 (increasing glycolysis )
and inhibits fructose 1.6-bisphosphatase ( decreasing gluconeogenesis ) . Fructose
2.6 -bisphosphate concentration is regulated by a bifunctional enzyme complex :
phosphofructokinase-2 increases levels in response to insulin , and fructose
2.6 -biSphosphatase decreases levels in response to glucagon.
References:
1. Glucagon and regulation of glucose metabolism.
2. Elucidation of the role of fructose 2,6-bisphosphate in the regulation
of glucose fluxes in mice using in vivo (13)C NMR measurements of
hepatic carbohydrate metabolism.
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t dlcutdtor
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9
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u
IB
1?
ia
A 30-year -old woman comes to the office with a 3- week history of mild fatigue cold
intolerance and constipation . The patient has also experienced dry skin , weight
gam and myalgias . Her mother had similar symptoms and was diagnosed with a
thyroid disorder at a similar age . The patient has no history of significant past
medical problems and takes no medications . She has been pregnant once and
gave birth to a healthy infant 3 years ago . Examination reveals dry skin , delayed
ankle jerk reflexes and bradycardia. Her thyroid gland shows mild diffuse
enlargement , with no tenderness or nodules Laboratory evaluation of this patient is
likely to show which of the following patterns?
Free T4
Total T3
Decreased
Normal
O B . Increased
Normal
Normal
O C Normal
Normal
Normal
O D. Normal
Normal
Increased
O E. Increased
Increased
Increased
O F. Decreased
Decreased
Decreased
C G . Decreased
Increased
Increased
Norma'
Decreased
TSH
I
20
21
n
23
24
2&
25
27
23
29
30
31
32
33
34
_
A. Increased
,
H. Normal
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Notes
I olc uhnor
4
A
&
6
9
n
11
12
13
1
19
*
17
13
19
A 30-year -old woman comes to the office with a 3-week history of mild fatigue , cold
intolerance and constipation The patient has also experienced dry skin , weight
gam . and myalgias . Her mother had similar symptoms and was diagnosed with a
thyroid disorder at a similar age . The patient has no history of significant past
medical problems and takes no medications . She has been pregnant once and
gave birth to a healthy infant 3 years ago, Examination reveals dry skin, delayed
ankle jerk reflexes, and bradycardia Her thyroid gland shows mild diffuse
enlargement , with no tenderness or nodules Laboratory evaluation of this patient is
likely to show which of the following patterns?
Free T4
Total T 3
Decreased
Normal
[74% ]
O B . Increased
Normal
Normal
[ 3%]
C C Normal
Normal
Normal
[D%]
D. Normal
Normal
Increased
[1%J
E. Increased
Increased
Increased
[ t %]
Decreased
Decreased
[12%]
Increased
Increased
13 1
Norma;
Decreased
TSH
b
20
21
22
21
24
25
25
21
28
29
* <§
'
,
30
31
32
33
34
A . Increased
.
.
O F Decreased
G. Decreased
O H. Normal
*
[5%]
V
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*
Notes
(
alculdior
/s
Explanation:
Thyroid hormone regulation
n
it
i?
13
1
b
Hypothalamus
*
IS
M
13
H
T4
20
21
Pituitary gland
22
23
24
2S
26
71
23
29
T3
Thyroid gland
30
31
32
33
34
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MI Mift
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Notes
I alculaior
T 3 ^“
&
A
E
9
TO
It
Thyroid gland
b
13
H
IS
17
13
T9
20
2t
We a five feedback
^
fjnflCfrve)
22
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34
Physiologic
effects
Curtorid
Regulation of thyroid hormone secretion involves feedback inhibition along the
hypothalamus -prtuitary -thyroid axis . The hypothalamus releases thyrotropin-releasing
hormone ( TRH ) which tnggers release of thyroid- stimulating hormone (TSHT
thyrotropin) from the pituitary TSH stimulates release of thyroid hormone from the
thyroid , which in turn inhibits release of both TRH and TSH. Thyroid hormone exists
in 2 active forms T4 (thyroxine ), the primary form secreted by the thyroid and T 3
( triiodothyronine ) , the more active form produced primarily by deiodination ofT4 in
peripheral tissues . Hypothyroidism may occur due to dysfunction involving the
thyroid giand (primary hypothyroidism ) or, less commonly, due to disorders of the
pituitary ( secondary ) or hypothalamus (tertiary ) .
r
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Notes
I dlcufdtor
hypothalamus-prtuitary-thyroid axis . The hypothalamus releases thyrotropin-releasing
hormone ( TRH ), which tnggers release of thyroid- stimulating hormone (TSH,
thyrotropin) from the pituitary TSH stimulates release of thyroid hormone from the
thyroid , which in turn inhibits release of both TRH and TSH. Thyroid hormone exists
in 2 active forms T4 (thyroxine ), the primary form secreted by the thyroid and T 3
(triiodothyronine ) , the more active form produced primarily by deiodination of T4 in
peripheral tissues . Hypothyroidism may occur due to dysfunction involving the
thyroid gland (primary hypothyroidism ) or, less commonly , due to disorders of the
pituitary ( secondary ) or hypothalamus (tertiary ).
A
This patient, with fatigue , cold intolerance and a diffusely enlarged thyroid ( goiter ),
has common features of hypothyroidism , In the United States , the most common
cause of hypothyroidism is Hashimoto thyroiditis which is characterized by
autoimmune destruction of the thyroid gland. As this autoimmune destruction of the
thyroid progresses, thyroid hormone production declines This leads to loss of
feedback inhibition of TSH secretion, with low T4 and elevated TSH (Choices B
and C). Because most T 3 is produced in peripheral tissues under the control of
multiple factors. T3 levels usually remain normal until relatively late - stage
hypothyroidism (Choice H) ,
(Choice D) Thyroid hormones circulate mostly in protein-bound form, primarily to
thyroxine -binding globulin ( TBG ) transthyretin and albumin. When TBG levels are
elevated ( eg , pregnancy or oral contraceptive use ), total thyroid hormone levels are
high but the free hormone levels are normal These patients are euthyroid and have
t
normal TSH.
(Choice E) Elevated levels of TSH , T4 , and T 3 are consistent with hyperthyroidism
due to a TSH-secreting pituitary adenoma ,
(Choice F) Secondary hypothyroidism due to pituitary failure is characterized by
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B
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Notes
t ale ufator
) Thyroid hormones circulate mostly in protein-bound form , pnmarily to
thyroxine -binding globulin ( TBG), transthyretin and albumin When TBG levels are
elevated ( eg , pregnancy or oral contraceptive use ), total thyroid hormone levels are
high but the free hormone levels are normal These patients are euthyroid and have
(Choice
normal TSH,
(Choice E) Elevated levels of TSH T4 , and T 3 are consistent with hyperthyroidism
due to a TSH-secreting pituitary adenoma
(Choice F) Secondary hypothyroidism due to pituitary failure is characterized by
decreased levels of both thyroid hormones and TSH. This is significantly less
common than primary hypothyroidism and usually associated with loss of other
pituitary hormones
(Choice G) Suppressed TSH with elevated thyroid hormone levels is characteristic
of thyrotoxicosis ( eg , Graves disease ). This patient has no clinical features of
hyperthyroidism.
Educational objective:
Primary hypothyroidism is characterized by decreased thyroxine (T4 ) levels and
increased thyroid-stimulating hormone (TSH ). Triiodothyronine (T 3) is primarily
produced by conversion from T 4 in peripheral tissues and levels may be normal in
early hypothyroidism
.
References:
1 . Pitfalls in the measurement and interpretation of thyroid function
tests
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Last updated: [11/12/2015 ]
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t akufdtor
A
MUSCLE GIYCOGFN
$
19
20
21
n
23
24
25
25
GLUCOSE- 1- PHOSPHATE
§
§
27
23
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31
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33
34
Mdrt
A 15 - year -old boy is being evaluated for poor exercise endurance He recently
started weightlifting with some of his fhends and is disappointed that he is ' the
weakest one by far " He says he tries as hard as possible but his "arms feel like
jelly after just a few repetitions " The patient also says he sometimes experiences
severe muscle cramping and urine discoloration after penods of intense straining.
Further evaluation reveals that his exercise tolerance can be greatly improved by
drinking an oral glucose solution before beginning a strenuous activity . This patient
is most likely deficient in an enzyme that catalyzes which of the following
conversions?
u
30
'
6 PHOSPHO
GLUCONATE
GLUCOSE 6 PHOSPHATL
GLUCOSE
FRUCTOSi - 6 - PHOSPHATF
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Newt
*
Notes
(
ulcuf -i t o r
/v
is most likely deficient in an enzyme that catalyzes which of the following
conversions?
MUSCLE GLYCOGEN
TO
$
It
12
13
u
GLUCOSE 1 PHOSPHAK
IS
IS
$
13
20
21
22
21
24
2S
25
6 - PHOSPHO GLUCONATE
GLUCOSE - 6- PHOSPHATF
GLUCOSE
§
27
28
29
FRUCTOSE 6 PHOSPHATE
30
31
32
33
34
C A. A
C BB
O C. C
O D. D
O E. E
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Item: 17 of 34
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is most likely deficient in an enzyme that catalyzes which of the following
conversions?
B
9
MUSCLE GLYCOGEN
10
*
11
n
13
14
15
IS
§
$
H
20
n
6 - PHOSPHQ
GLUCONATE
GLUCOSE - 6- PHOSPHATF
24
25
25
27
23
29
*
Notes
I akulaior
ft
GLUCOSE - 1- PHOSPHATE
n
2\
Lab Value
Newt
b
GLUCOSE
FRUCTOSE - 6 PHOSPHATE
30
31
32
33
34
* ® A. A [68%]
O B . B [10%1
O C . C [ 18%]
O D. D [2% J
O E. E [2%1
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E . E [2%]
*
Note
*
( dlculdtor
A
Explanation:
Impairments in glycogenotysis
to
11
12
13
14
15
16
Ebguitrt&nt by
18
19
pftGUpfrOfyliMe
Duco» 1-P S
20
22
*
*
Type
• Mu$de photpiwyl»»
+ Normal gluooee level
* Sever cardrarnCHjay
detafcnc ,
- frali ne^i A 1 rtfju.
* *> UooO lactate
*
• No to
II: PomM tiieeaee
*
*
* Glycogen accumulation
n fyvosomo
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Limrt
dextrin
Dctrtrtchr
^-
entyme
(a- 1 4 -* o 1f 4rfrarcsJfer Wl
*
27
28
29
Tvpt Til: Con diuini
*
Hepatomegaly
Ketotc 'ftypogfycem i
* Myooton a & weak neu
- Abnormal glycoqeo wtti
*
30
31
32
33
34
Acv a-gtvcoitf s&
Tvio# V: MeArdla diww
*
21
23
24
2&
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U
* G k coi*
0< iysrK **Hirt(j emyw
(0 1 6 gkjcQsxiawt
-*
G'UCOSP
P
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very abort outer ebom
Gtyicogcn
• pfiospfjatylasm
'
Glucose l-P
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Item: 17 of 34
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Newt
Mdrk
C rifflulffiTtfNir tty
Lab Value
^
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TVM V: Me Ardl di < J c
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* *
*
MuscJ phoapftoryfcwo
*
defioonc /
10
11
• AfliiluvrtA
12
13
dfrjclMn
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,
Type (H:
*
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Glycogen necumu^ t'fcpfi
t\ lyiMomu
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Limit
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15
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Notes
Aetf Q htCosrtM
.
Gycogen
ptaurtioryttee
*
fa 1 6
Lj,Tucas jjjff
*
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^
* Ab'tarrul giycOfl«r> **4 #i
wy hort cMof chom
*
*
23
24
25
25
"'N C^wee'
27
23
29
j
Gluooi IP
*
30
1
31
32
33
34
GIltLOB* 6 P
Tvod I: von GiftrK* d lflitl
64i&w
60toj£>nar£5r
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gahy K tTPitOM
*
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* Lactic acfcfcwtt
*
*
* HyTwruftOSTP ft &
hyperiipidefma
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Item: 17 of 34
3
O Id
B
9
10
11
12
13
U
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i M „f t
<1
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Lab Value
Newt
llMCtW
4
5
M
*
Notes
(
ulrufdlor
A
QUWorld
This patient most likely has McArdle disease ( glycogen storage disease type
V ) This condition is caused by deficiency of myophosphorylase , an isoenzyme of
glycogen phosphorylase present in muscle tissue Deficiency of this enzyme leads
to decreased breakdown of glycogen during exercise resulting in poor exercise
tolerance muscle cramps , and rhabdomyolysls . The prognosis is generally
good and symptoms can be improved by consuming simple sugars before
beginning physical activity .
t
*
During glycogenolysis , glycogen phosphorylase shortens glycogen chains by
cleaving a -1 4 -glycosidic linkages between glucose residues , liberating glucose 1
phosphate in the process. This occurs until 4 residues remain before s branch point
( the " limit” dextrin ). At this point , the debranching enzyme performs 2 enzymatic
functions :
-
1. Glucosyltransferase cleaves the 3 outer glucose residues of the 4 that are left
by glycogen phosphorylase and transfers them to a nearby branch
2. The enzyme a-1 6-glucosidase removes the single remaining branch residue
producing free glucose and a linear glycogen chain that can be further
shortened by glycogen phosphorylase
(Choice B ) The glucose 1-phosphate generated by glycogenolysis is converted by
phosphoglucomutase to glucose 6 -phosphate , which can then undergo glycolysis .
(Choice C) Within the liver and kidney , glucose 6-phosphatase converts glucose 6phosphate to glucose to help maintain blood glucose levels during periods of
fasting . However , muscles Jack glucose 6-phosphatase and so must utilize glucose
6-phosphate for glycolysis during muscle contraction.
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Lab Value
Sent
W B F 4
*
Notes
I alculaior
A
2. The enzyme a -1, 6- glucosidase removes the single remaining branch residue ,
producing free glucose and a linear glycogen chain that can be further
shortened by glycogen phosphorylase
(Choice B) The glucose 1-phosphate generated by glycogenolysis is converted by
phosphoglucomutase to glucose 6-phosphate , which can then undergo glycolysis .
&
(Choice C) Within the liver and kidney , glucose 6-phosphatase converts glucose 6phosphate to glucose to help maintain blood glucose levels during periods of
fasting However , muscles lack glucose 6-phosphatase and so must utilize glucose
6 -phosphate for glycolysis during muscle contraction,
(Choice D) The first step in the pentose phosphate pathway (hexose
monophosphate pathway ) is conversion of glucose 6-phosphate to 6phosphogJuconate by glucose-6-phosphate dehydrogenase. This pathway
maintains cellular NADPH levels and produces pentose sugars for nucleotide
synthesis ,
(Choice E) Glucose 6-phosphate is converted into fructose 6-phosphate by
glucose 6-phosphate isomerase during the second step of glycolysis
Educational objective:
Glycogen serves as a source of glucose during fasting and as an energy store that
can be mobilized quickly during strenuous muscle contraction Myophosphorylase
deficiency iMcArdle disease or glycogen storage disease type V) causes failure of
muscle glycogenolysis , resulting in decreased exercise tolerance , muscle pain and
cramping and myoglobinuria with physical activity
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(Notes
(
ulcufdlor
A
b
6
9
n
11
12
13
14
15
15
17
19
A 4-month-old boy is brought to the office for his first visit since arriving in the United
States . The patient was recently adopted and his adoptive mother says the boy is
tremulous compared to her biological children Over the past week the boy has also
had episodes of upward eye deviation and bilateral arm and leg shaking for
approximately 2 minutes at a time . Biological family history is not available His
temperature is 36.7 C ( 98 1 F ) , blood pressure is 90 /40 mm Hgt pulse is 120/min
and respirations are 30 /min . Examination shows a fair - skinned infant with blue eyes
and a musty body odor . Which of the following amino acids is most likely essential in
this patient?
,
,
20
O A . Cysteine
21
O B. Isoleucine
23
24
O C . Leucine
C D . Phenylalanine
C E . Tyrosine
n
25
25
27
23
29
O F . Valine
30
31
32
33
34
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Ldb Vdlutf
*
(Notes
I iilcufdlDr
4
A
5
B
9
n
11
12
13
14
15
1&
1?
19
A 4-month-old boy is brought to the office for his first visit since arriving in the United
States . The patient was recently adopted , and his adoptive mother says the boy is
tremulous compared to her biological children. Over the past week , the boy has also
had episodes of upward eye deviation and bilateral arm and leg shaking for
approximately 2 minutes at a time . Biological family history is not available His
temperature is 36.7 C (98 1 F ) , blood pressure is 90/40 mm Hgf pulse is 120/min,
and respirations are 30 /min . Examination shows a fair -skinned infant with blue eyes
and a musty body odor . Which of the following amino acids is most likely essential in
this patient?
20
O A. Cysteine [ 4%]
21
O B. Isoleucine [1%]
22
21
24
25
26
27
23
29
30
31
32
33
34
O C Leucine [2%]
C D . Phenylalanine [20%]
V m E. Tyrosine [71%]
O F. Valine [1%]
Explanation:
Phenylketonuria
Dthytfrofltenchfyt
reductase
OUe
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Ldb Value
NPKt
Previous
*
Note
*
I a k uldlor
Explanation:
Phenylketonuria
10
It
Dih / droptendine
n
reductase
u
u
IS
is
Melanin
17
19
Phenylalanine
20
Tyrosine
*
>
OOPA
Phcnytaiantne
hydroxytase
21
Catecholamines
22
23
24
2S
26
Phenylpyruvate
Homogentisate
23
29
I
HomoQeriU&c eoo
27
Phenylacetate
Phenyllactate
30
dtoxygenase
MaIey laceto acetate
31
32
33
34
FumaryIacetoacetate
|
—
*
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/
TCA
\
Fumarate i Cyde
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Fumarate
TCA
Cycle
b
GUWortd
This patient' s clinical presentation suggests phenylketonuria ( PKU) . Classic PKU
( the most common form ) occurs most frequently in persons of Scandinavian
descent . Individuals who are homozygous for this autosomal recessive disorder
have a severe deficiency of phenylalanine hydroxylase , the enzyme responsible
for conversion of phenylalanine to tyrosine Tyrosine becomes an essential amino
acid in these patients as it cannot be synthesized from phenylalanine , although most
patients receive adequate amounts in their diet and do not require supplementation.
Furthermore the excess phenylalanine inhibits tyrosinase, which normally results in
the downstream production of melanin , The lack of melanin causes the fair
complexion seen in this patient.
Although development is initially normal, most untreated patients develop
intellectual disability by age 6 months Other classic findings include seizures
eczema , light hair /skin pigmentation, and a "musty , " odor . However , cognitive
impairment can be lessened by early detection (newborn screening ) and restriction
of phenylalanine intake
(Choice A) Cystmuria is due to impaired transport of renal cystine ( a homodimer of
cysteine ) , which leads to cystine renal stone formation , Clinical manifestations
typically include flank pain hematuria , and possible stone passage in children or
adolescents
O P anH F I Tka h r p n r
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eczema , light hair / skin pigmentation and a musty " odor , However , cognitive
impairment can be lessened by early detection ( newborn screening) and restriction
of phenylalanine intake .
*
Notes
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alculator
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A
(Choice A) Cystinuria is due to impaired transport of renal cystine ( a homodimer of
cysteine } , which leads to cystine renal stone formation . Clinical manifestations
typically include flank pain hematuria , and possible stone passage in children or
b
adolescents
(Choices B, C, and F) The branched-chain amino acids (leucine isoleucine , valine )
are elevated in maple syrup urine disease , an amino acid disorder marked by
deficiency of branched-cham alpha-ketoacid dehydrogenase complex Buildup of
the branched-cham amino acids and their metabolites is toxic, leading to feeding
difficulties seizures , cerebral edema and a sweet odor of the urine
(Choice D) Phenylalanine levels are significantly elevated in PKU, leading to
neurologic defects Therapy depends on dietary restriction of phenylalanine
Educational objective:
Phenylketonuna (PKU ) results from an inabilrty to convert phenylalanine to tyrosine by
the phenylalanine hydroxylase system , making tyrosine an essential amino acid in
these patients Classic clinical features of untreated PKU include intellectual
disability , seizures light pigmentation, and a "musty" odor.
References:
1 . Phenylalanine hydroxylase deficiency: diagnosis and management
guideline.
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A 69- year -old woman with Alzheimer disease is brought to the emergency
department after her son found her wandering in a local park after being unable to
contact her for the last day . The patient says that she got lost while taking a walk and
has not eaten or drunk anything for over 24 hours . On examination, she is mildly
confused and dehydrated Laboratory studies show a blood glucose level within the
normal range despite her prolonged fasting Which of the following hormones
contributes to this patient's laboratory findings by binding to an intracellular receptor ?
IB
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O A. Cortisol
O B. Epinephrine
O C. Glucagon
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D Growth hormone
O E. Insulin
F . Norepinephrine
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Motes
Iakufdtor
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ia
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A 69-year -old woman with Alzheimer disease is brought to the emergency
department after her son found her wandering in a local park after being unable to
contact her for the last day. The patient says that she got lost while taking a walk and
has not eaten or drunk anything for over 24 hours . On examination, she is mildly
confused and dehydrated. Laboratory studies show a blood glucose level within the
normal range despite her prolonged fasting Which of the following hormones
contributes to this patent's laboratory findings by binding to an intracellular receptor ?
*®
A . Cortisol [61%]
B, Epinephrine [3%]
O C Glucagon [30%]
D . Growth hormone [3%]
O E . Insulin [3%]
F. Norepinephrine [0% ]
Explanation:
Cortisol mechanism of action
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Diffusion through
1
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F Norepinephrine [0% ]
Explanation:
n
Cortisol mechanism of action
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Diffusion through
Modification of
gone transcription
COll membrane
Sg
Cortisol
Receptor
dimerization
&
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Nuclear
translocation
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ftiRNAj
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Comsoi receptor
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Heat shock protein
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Nudeat DMA
Blood glucose in normal subjects does not fall in the hypoglycemic range with fasting
due to decreased insulin secretion and the protective actions of multiple
counterregulatory hormones Glucagon is the primary hormone secreted in
response to a rapid drop in blood glucose levels , with epinephrine acting as the
major backup hormone. Cortisol and growth hormone contribute to glucose
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(
alcufator
A
Blood glucose in normal subjects does not fall in the hypoglycemic range with fasting
due to decreased insulin secretion and the protective actions of multiple
counterregulatory hormones Glucagon is the primary hormone secreted in
response to a rapid drop in blood glucose levels , with epinephnne acting as the
major backup hormone Cortisol and growth hormone contribute to glucose
homeostasis during prolonged fasting by altering transcription of many key enzymes
b
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In the inactivated state cortisol receptors are found within the cytoplasm in
association wrth heat shock proteins . Binding of cortisol to the carboxy terminal
portion of the receptor causes the release of the heat shock proteins and receptor
dimerization The activated homodimers are then transported to the nucleus
where they control gene expression by binding to hormone -responsive DNA
elements in the promoter region of target genes . Cortisol increases the
transcription of enzymes involved in gluconeogenesis ( formation of glucose from
fat and protein substrates ) as well as those involved in lipolysis and proteolysis
(Choices B , C, and F) Epinephrine norepinephrine , and glucagon exert their
metabolic effects via membrane-bound G protein-coupled receptors that activate
adenyl cyclase and increase cyclic AMP production , Glucagon increases hepatic
glycogenolysis and gluconeogenesis . Epinephrine ( and norepinephrine to a lesser
extent ) increases hepatic and renal glycogenolysis and gluconeogenesis: it also
increases the release of gluconeogenic substrates from muscle and fat,
(Choice D) Growth hormone receptors are membrane-bound receptors that result
in activation of the JAK -STAT pathway . Growth hormone antagonizes insulin action ,
increases gluconeogenesis. and promotes lipolysis (provides gluconeogenic
substrates ).
(Choice E) In addition to the production of counterregulatory hormones , inhibition of
insulin release from oancreatic beta cells Dlavs a primary role in preventina
09 : 54
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where they control gene expression by binding to hormone -responsive DNA
elements in the promoter region of target genes Cortisol increases the
transcription of enzymes involved in gluconeogenesis ( formation of glucose from
fat and protein substrates ) as well as those involved in lipolysis and proteolysis .
(Choices B , C, and F) Epinephrine , norepinephrine , and glucagon exert their
metabolic effects via membrane bound G protein-coupled receptors that activate
-
adenyl cyclase and increase cyclic AMP production Glucagon increases hepatic
glycogenolysis and gluconeogenesis . Epinephrine land norepinephrine to a lesser
extent ) increases hepatic and renal glycogenolysis and gluconeogenesis: rt also
increases the release of gluconeogenic substrates from muscle and fat .
-
(Choice D) Growth hormone receptors are membrane bound receptors that result
in activation of the JAK -STAT pathway. Growth hormone antagonizes insulin action
increases gluconeogenesis, and promotes lipolysis (provides gluconeogenic
substrates).
(Choice E) In addition to the production of counterregulatory hormones , inhibition of
insulin release from pancreatic beta ceils plays a pnmary role in preventing
hypoglycemia during fasting . Insulin acts on a receptor tyrosine kinase present on
23
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the cell surface
30
Educational objective:
Unlike the other counterregulatory uet insulin- opposing ) hormone receptors , cortisol
receptors are located within the cytoplasm and translocate to the nucleus after
binding to their substrate . In the nucleus the cortisol-receptor complex binds to
hormone -responsive DNA elements altering gene transcription to enhance hepatic
glucose production and limit peripheral glucose utilization.
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l alculdlor
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A heafthy 34-year -old coal mine worker is trapped underground following partial
collapse of an access shaft Rescue efforts are directed toward clearing the
obstructed tunnel, but it takes 2 days to reach him . While being taken to the surface
the miner tells rescuers that he feels dizzy and weak . He had an emergency supply
of water but has not eaten anything for over 30 hours. Fingerstick blood glucose
concentration is 78 mg /dL Which of the following biochemical reactions is most
likely responsible for maintaining this patient's current blood glucose levels?
b
—
A Acetoacetyl CoA * 3 -hydroxy -3-methylglutaryl-CoA
O B . Acetyl CoA palmitic acid
—
O C . Fructose 6-phosphate — * fructose 1, 6-bisphosphate
O D . Glycogen glucose-1-phosphate
C E . Oxaloacetate phosphoenolpyruvate
—
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<*S
5
B
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ir
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•
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n
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30
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A healthy 34-year -old coal mine worker is trapped underground following partial
collapse of an access shaft Rescue efforts are directed toward clearing the
obstructed tunnel , but it takes 2 days to reach him . While being taken to the surface
the miner tells rescuers that he feels dizzy and weak. He had an emergency supply
of water but has not eaten anything for over 30 hours. Fingerstick blood glucose
concentration is 78 mg /dL. Which of the following biochemical reactions is most
likely responsible for maintaining this patient's current blood glucose levels?
li
—
A Acetoacetyl CoA * 3 -hydroxy -3-methylglutaryl-CoA [14%]
B, Acetyl CoA palmitic acid [7%]
C Fructose 6- phosphate » fructose 1, 6-bisphosphate [4%]
O 0. Glycogen glucose-1-phosphate [25%}
* # E. Oxaloacetate phosphoenolpyruvate [50%]
—
—
Explanation:
Important gluconeogenic substrates & steps
Glucose
Gly cc rol-3 -ptiosphatc
II
Ph osphoen ol py ruvale
Pyruvate kinase
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Explanation:
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Important gluconeogenic substrates & steps
n
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Glucose
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Pyruvate kmase
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* Lactate
* Alanine
* Other glucogenic
amino acids
Pyruvate
n
23
2t
Phosphoenolpyru va te
25
carboxyktnase
25
*
27
Pyruvate
28
29
Pyruvate
30
31
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3
*
carboxylase
Oxaioaceiate
I
Malate
MaSate
I
Oxaioaceiate
Z'
Acetyl CoA
/
Cytosol
\
;
Mitochondrion
dehydrogenase
/
/
Malate
Malate
shuttle
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Citric acid cycle
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Pyruvate kinase
4
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• Lactate
Pyruvate
£
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• Alanine
• Other glucogenic
Phosphoenolpyruva te
carbQxykinase
amino acids
n
12
Pyruvate
13
U
Pyruvate
carboxytese
15
is
17
Oxaloacetate
13
Maiate
I
H
21
23
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Mitochondrion
\
Maiate
Maiate
shuttle
/
25
I
Oxaloacelate
b
Cytosol
Acetyl CoA
dehydrogenase
Maiate
n
25
•Glutamine
27
•Other
28
29
glucogenic
amino acids
30
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c*
CUWorid
The 2 major processes that maintain plasma glucose between meals are
glycogenolysis and gluconeogenesis . Glycogenolysis is the primary source of
glucose for the first 12-18 hours of fasting Once hepatic glycogen stores become
depleted , gluconeogenesis becomes the major process used by the body to keep
blood glucose levels within the normal range . During gluconeogenesis glucose is
if A m
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t alculaior
yiuwyciTH#
amino acids
© uworid
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The 2 major processes that maintain plasma glucose between meals are
glycogenolysis and gluconeogenesis . Glycogenolysis is the primary source of
glucose for the first 12-18 hours of fasting Once hepatic glycogen stores become
depleted , gluconeogenesis becomes the major process used by the body to keep
blood glucose levels within the normal range . During gluconeogenesis , glucose is
formed from lactate glycerol , and glucogenic amino acids . This process uses many
of the enzymes involved in glycolysis However , hexokinase phosphofructokinaseT
and pyruvate kinase are unidirectional and must be bypassed by distinct
gluconeogenic enzymes.
The first committed step of gluconeogenesis is the biotin-dependent carboxylation
of pyruvate to oxaloacetate by mitochondrial pyruvate carboxylase Oxaloacetate
is subsequently converted to malate by malate dehydrogenase to facilitate exit from
the mitochondria , and then is converted back to oxaloacetate by cytosolic malate
dehydrogenase (malate shuttle ). In the cytosol, phosphoenolpyruvate
carboxykinase ( PEPCK ) converts oxaloacetate to phosphoenolpyruvate .
Therefore , pyruvate carboxylase and PEPCK work together to bypass pyruvate
kinase . The 2 other unique gluconeogenic enzymes are fructose
1,6 bisphosphatase (bypasses phosphofructokinase ) and
glucose 6 phosphatase (bypasses hexokinase ),
-
--
(Choice A) Conversion of acetoacetyl-CoA to 3-hydroxy- 3 - methylglutaryl-CoA
occurs during the synthesis of cholesterol and ketone bodies. Ketone body
synthesis is increased in starvation situations however, ketone bodies cannot be
used to synthesize glucose
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(Choice B) Palmitic acid is the first fatty acid produced from acetyl CoA during
lipogenesis in the fed state . However during prolonged fasting, lipolysis
predominates and leads to generation of glycerol and fatty acids .
H
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Iu k u l d t o r
A
(Choice A) Conversion of acetoacetyl- CoA to 3-hydroxy- 3- methylglutaryl- CoA
n
u
u
Notes
kinase . The 2 other unique gluconeogenic enzymes are fructose
1,6 -bisphosphatase (bypasses phosphofructokinase ) and
glucose-6-phosphatase (bypasses hexokinase ).
occurs during the synthesis of cholesterol and ketone bodies. Ketone body
synthesis is increased in starvation situations , however ketone bodies cannot be
used to synthesize glucose .
it
*
ti
,
(Choice C) Conversion of fructose 6- phosphate to fructose 1 , 6- bisphosphate
occurs during giycotysis and is catalyzed by phosphofructokinase Dunng starvation ,
glycolysis is minimized and gluconeogenesis predominates
(Choice D) The first step of glycogenolysis is breakage of 1-4 glycosidic linkage to
form giucose- 1- phosphate After 24 hours of fasting maintenance of blood glucose
levels is achieved mostly through gluconeogenesis , not by glycogenolysis .
,
Educational objective:
After 12-18 hours of fasting gluconeogenesis becomes the principal source of
blood glucose . Gluconeogenesis uses many glycolytic enzymes but hexokinase .
phosphofructokinase , and pyruvate kinase need to be bypassed as they are
unidirectional. The initial steps of gluconeogenesis involve the conversion of
pyruvate to oxaloacetate and oxaloacetate to phosphoenolpyruvate by pyruvate
carboxylase and phosphoenolpyruvate carboxykinase respectively.
,
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*
Notes
(
ukuhitor
A
b
B
9
n
n
13
14
15
15
1?
U
14
20
Nutrition researchers investigating the relationship between fructose consumption
and cardiovascular disease conduct a prospective cohort study on a population of
randomly selected young adults . Study participants undergo semiannual
measurement of waist circumference , blood pressure, and serum cholesterol and
triglyceride concentrations . Dietary fructose consumption is assessed through the
use of questionnaires and by measuring unnary fructose excretion, A 23 - year old
man enrolled in the study is found to excrete large amounts of fructose in his urine
compared to other study participants despite maintaining a moderate fructose
intake Further evaluation shows a hereditary defect in fructose metabolism, but he
is asymptomatic and has no other medical problems. This patient most likely
remains able to metabolize fructose due to the compensatory activity of which of the
following enzymes?
-
n
23
2
*
25
25
O A. Aldolase B
O B Aldose reductase
21
O C. Fructokinase
23
29
O D . Hexokinase
30
31
32
33
34
E UDP -galactose-4- epimerase
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(Notes
I olcuhifor
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A
5
B
9
n
it
n
13
14
15
IS
1?
13
H
20
Nutrition researchers investigating the relationship between fructose consumption
and cardiovascular disease conduct a prospective cohort study on a population of
randomly selected young adults . Study participants undergo semiannual
measurement of waist circumference , blood pressure, and serum cholesterol and
triglyceride concentrations . Dietary fructose consumption is assessed through the
use of questionnaires and by measuring urinary fructose excretion, A 23 -year old
man enrolled in the study is found to excrete large amounts of fructose in his urine
compared to other study participants despite maintaining a moderate fructose
intake Further evaluation shows a hereditary defect in fructose metabolism, but he
is asymptomatic and has no other medical problems. This patient most likely
remains able to metabolize fructose due to the compensatory activity of which of the
following enzymes?
b
-
22
21
24
25
26
21
28
29
30
31
32
33
34
O A. Aldolase B [28%]
*•
B . Aldose reductase [16%]
C . Fructokinase [22%]
D . Hexokinase [29%]
E UDP-gaiactose-4- epimerase [4%]
Explanation:
Disorders of fructose metabolism
Glucose
V
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Newt
Notes
(
dlcutdtor
A
Disorders of fructose metabolism
Glucose
Hexaktnase
10
n
Giucos© 6- phosphat ©
Sucrose
n
D>
Phosphogluco-
13
H
IB
1G
17
13
Essential
fructosuria
isometase
(benign condition )
Fructose
Hexokmase
- Fructose 6-phosphate
Fructose - 1. 6-
Ftuetakinase
H
P? K ~ 1
btsphos&twtas *
*
20
22
Fructose - 1. $ -bisphosphate
Fructose 1 phosphate
23
24
2B
25
27
23
29
Aldoiase B
DHAP
30
31
32
33
34
Aldolase A A 8
Giyceraklehyd©
Hereditary fructose intolerance
Hypogtycemta & vomiting after
- fructose
ingestion
- Faiure lo ttmve liver A renal failure
,
Tnohnasc
DHAP
Gtycaraldohyde
phosphate
^
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phosphate
tscmerase
1
Pyruvate
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Fructose is obtained in the diet primarily from fruits and food sweeteners such as
I
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Lab Values
Newt
Notes
(
alcufator
1
F allure to thrive liver & renal failure
A
Pyruvate
CUWwM
B
9
to
11
12
13
14
15
16
17
13
Fructose is obtained in the diet primarily from fruits and food sweeteners such as
table sugar ( sucrose ) and high-fructose com syrup Fructose is absorbed in the
proximal intestine through the GLUTS fructose transporter . It is normally
phosphorylated by fructokinase in the liver , yielding fructose-1-phosphate which
is converted by aldolase B to dihydroxyacetone phosphate (DHAP ) and
glyceraldehyde (Choice C) . Glyceraldehyde and DHAP can be converted to
glyceraldehyde-3-phosphate , which can then be metabolized in the glycolytic
pathway,
i
H
20
22
23
24
25
26
27
23
29
30
31
32
33
34
Fructokinase deficiency ( essential fructosuria ) is an asymptomatic , autosomal
recessive disorder that causes dietary fructose to be excreted unchanged in the
urine . In fructokinase deficiency, hexoklnase takes over the role of fructose
metabolism, converting dietary fructose into fructose-6-phosphate
Fructose-6 -phosphate can be metabolized in the glycolytic pathway or converted to
glucose-6 -phosphate or glucose- 1-phosphate , which can be used in the pentose
phosphate pathway or for glycogen synthesis, respectively.
,
(Choice A) Aldolase B plays a nonessentiai role in glycolysis due to the redundant
function of aldolase A. However , it is particularly important during fructose
metabolism as deficiency of this enzyme leads to toxic accumulation of
fructose -1-phosphate (hereditary fructose intolerance ) This life -threatening disorder
presents in infancy after the introduction of fructose-containing foods .
(Choice B) Aldose reductase is the enzyme that converts glucose to sorbitol .
Aldose reductase has a low affinity for glucose , and normally only very small
amounts of glucose are metabolized by this enzyme . The amount of glucose
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5
FructoKmase deficiency ( essential fructosuria } is an asymptomatic , autosomal
recessive disorder that causes dietary fructose to be excreted unchanged in the
urine . In fructokinase deficiency, hexokinase takes over the role of fructose
B
9
to
It
n
metabolism, converting dietary fructose into fructose-6-phosphate
Fructose-6 -phosphate can be metabolized in the glycolytic pathway or converted to
glucose-6 -phosphate or glucose-1-phosphate , which can be used in the pentose
phosphate pathway or for glycogen synthesis , respectively.
13
U
15
16
17
13
(Choice A) Aldolase B plays a nonessential role m glycolysis due to the redundant
function of aldolase A. However , it is particularly important during fructose
metabolism as deficiency of this enzyme leads to toxic accumulation of
fructose -1-phosphate ihereditary fructose intolerance / This life -threatening disorder
4
Lab Value
Sent
*
Notes
t dlcutalor
presents in infancy after the introduction of fructose -containing foods .
20
n
23
24
25
25
(Choice B) Aldose reductase is the enzyme that converts glucose to sorbitol .
Aldose reductase has a low affinity for glucose , and normally only very small
amounts of glucose are metabolized by this enzyme . The amount of glucose
metabolized by the aldose reductase pathway increases significantly in diabetes
mellitus and contributes to chronic complications such as neuropathy and retinopathy.
?7
28
29
30
31
32
33
34
(Choice E) UDP-galactose - epimerase is involved in the metabolism of
galactose ; it does not play a role in fructose metabolism .
^
Educational objective:
Essential fructosuria is a benign disorder of fructose metabolism caused by
fructokinase deficiency In patients with essential fructosuria some of the dietary
fructose load is converted by hexokinase to fructose -6-phosphate , which can then
enter glycolysis ; this pathway is not significant in normal individuals
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*
Notes
I «i( cu ( dlor
4
5
£
9
to
An infant suffering from severe neurological symptoms and lactic acidosis is
diagnosed with pyruvate dehydrogenase complex deficiency . When placed on a
special diet , the patient demonstrates some decrease in the blood lactate level.
Which of the following substances can be most safely supplemented to this patient?
It
O A Glycerol
13
U
15
IS
17
13
O B . Alanine
O C . Galactose
O D . Lysine
20
O F . Serine
n
21
’l
23
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.
O E. Asparagine
-
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*
(Note
*
(
olcuhHor
A
£
9
n
11
12
13
14
IB
15
1?
1j
H
20
An infant suffering from severe neurological symptoms and lactic acidosis is
diagnosed with pyruvate dehydrogenase complex deficiency . When placed on a
special diet , the patient demonstrates some decrease in the blood lactate level.
Which of the following substances can be most safely supplemented to this patient?
u
O A. Glycerol [11%]
O B . Alanine [26%J
O C. Galactose [11%]
* <S> D. Lysine [35%]
E. Asparagine [12%]
O F . Senne [5%]
21
21
24
2B
25
27
28
29
30
31
32
33
34
Explanation:
Pyruvate dehydrogenase ( PDH) is an allosteric enzyme that converts pyruvate into
acetyl-CoA in the presence of oxygen (i. e dunng aerobic metabolism ). In the
absence of oxygen or with a deficiency of PDH. pyruvate is alternatively converted to
lactate by the enzyme iactate dehydrogenase . Excessive lactate production in these
states results in lactic acidosis .
PDH deficiency has a wide spectrum of presentations . Because carbohydrates may
aggravate lactic acidosis , a ketogenic diet is recommended in these patients Amino
acid catabolism following removal of the amino group results in formation of
intermediates that are either glucogenic ( producing intermediates of the citnc acid
cycle or pyruvate ) or ketogenic (producing acetoacetate or its precursors) Some
amino acids such as phenylalanine , isoleucine and tryptophan are both glucogenic
as well as ketogenic. Leucine and lysine (Choice D) are exclusively ketogenic and
|
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5
B
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it
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id
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Newt
Note
*
t dlculdtor
F . Serine [ 5%]
A
Explanation:
Pyruvate dehydrogenase ( PDH) is an allosteric enzyme that converts pyruvate into
acetyl-CoA in the presence of oxygen (i. e during aerobic metabolism). In the
absence of oxygen or with a deficiency of PDH pyruvate is alternatively converted to
lactate by the enzyme lactate dehydrogenase. Excessive lactate production in these
states results in lactic acidosis.
i
*
PDH deficiency has a wide spectrum of presentations . Because carbohydrates may
aggravate lactic acidosis , a ketogenic diet is recommended in these patients Amino
acid catabolism following removal of the amino group results in formation of
intermediates that are either glucogenic (producing intermediates of the citric acid
cycle or pyruvate ) or ketogenic (producing acetoacetate or its precursors). Some
amino acids such as phenylalanine , isoleucine and tryptophan are both glucogenic
as well as ketogenic . Leucine and lysine (Choice D) are exclusively ketogenic and
would not lead to increased formation of lactic acid Lysine is an essential amino
acid that is totally ketogenic .
(Choice A ) Glycerol forms the backbone to triglycerides , and in the fasting state
when triglycendes are degraded for energy , glycerol is released to be used in
gluconeogenesis . Glycerol can aggravate lactic acidosis in patients with pyruvate
dehydrogenase deficiency because it can be converted to dihydroxyacetone
phosphate, which in turn can form pyruvate and subsequently lactate
(Choices B and F) Alanine is a non- essential glucogenic amino acid that can be
transaminated to form pyruvate Similarly, senne is a non- essential amino acid that
can be converted to pyruvate by serine hydratase
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Notes
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A
5
6
9
n
11
12
13
H
15
16
1?
ia
A
(Choice A) Glycerol forms the backbone to triglycerides , and in the fasting state
when triglycerides are degraded for energy, glycerol is released to be used in
gluconeogenesis . Glycerol can aggravate lactic acidosis in patients with pyruvate
dehydrogenase deficiency because it can be converted to dihydroxyacetone
phosphate, which in turn can form pyruvate and subsequently lactate
(Choices B and F ) Alanine is a non- essential glucogenic amino acid that can be
transaminated to form pyruvate Similarly , senne is a non-essential amino acid that
can be converted to pyruvate by serine hydratase
20
(Choice C) Galactose , when combined with glucose as a disacchande , forms the
milk sugar lactose Galactose is metabolized in the glycolytic pathway thus
galactose feeding will result in aggravation of lactic acidosis in patients with pyruvate
2\
dehydrogenase deficiency .
v.
23
24
25
25
27
28
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32
33
34
(Choice E) Asparagine is a non- essential glucogenic amino acid that is catabolized
Initially to aspartate by the enzyme asparaginase Aspartate is transaminated with
alpha ketoglutarate to produce glutamate and oxaloacetate a gluconeogenic
intermediate and an intermediate in the TCA cycle.
Educational Objective:
Pyruvate dehydrogenase deficiency is a disease with multiple possible
presentations ranging from neonatal death to mild episodic symptoms in adulthood .
By preventing the conversion of pyruvate to acetyl CoA pyruvate is shunted to lactic
acid resulting in lactic acidosis in these patients Lysine and leucine are exclusively
ketogenic and would not increase the blood lactate level in patients suffering from
pyruvate dehydrogenase deficiency.
.
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Notes
I dkufalor
A
5
E
9
Inhibition of lactate dehydrogenase in strenuously exercising skeletal muscles would
eventually lead to an inhibition of glycolysis due to intracellular depletion of which of
the following substances?
10
It
12
t3
14
15
15
1?
13
19
20
21
22
O A . AMP
C B . FADH.
O C NAD*
D Carnitine
E . Pyruvate
.
OF Citrate
24
25
25
27
23
29
30
31
32
33
34
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*
Note
*
(
akuldior
4
*v
<
&
E
9
Inhibition of lactate dehydrogenase in strenuously exercising skeletal muscles would
eventually lead to an inhibition of glycolysis due to intracellular depletion of which of
the following substances?
10
b
It
O A . AMP [5%]
O B. FADH, [3%|
12
t3
14
15
16
1?
13
19
20
21
*
C . NAD [70%J
O D . Carnitine [1%I
E . Pyruvate [18% ]
O F . Citrate [3%J
n
Explanation:
24
Glucose
25
25
27
23
29
30
Fructose 1.6-
31
32
33
34
BipTiospnate
:
Glyceraktehyrk: 3
Phosptiate
NAD"
-
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DihydroKyacetonB
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Glycerol
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Ldb Values
Notes
I dlculdior
J
/s
F . Citrate [3% ]
Explanation:
9
Glucose
10
b
11
13
U
15
16
1?
Fructose 1.6
.
8 phosphate
lj
H
20
GtyoefaKkfhyde -3Phosphate
21
n
24
25
26
:
4 S
-
Dihydroxy acetone
Phosphate
T
MAD'
Gfycerot
NADU
21
1.3-BPG
23
29
30
31
32
33
34
Pyruvate
Lactate
v
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C7
Lab values
Notes
t dlculdtsr
u« \i
&
Lactate
TO
£
9
i
,i. l *
10
b
11
12
13
14
15
IS
17
13
H
20
21
22
24
25
25
21
23
29
30
31
32
33
34
In glycolysis glucose is metabolized to pyruvate and lactate Under aerobic
conditions , the dominant product in most tissues is pyruvate , which is converted to
acetyl CoA to enter the TCA cycle. When oxygen is depleted, such as in exercising
muscle , the dominant product is lactate . This is referred to as anaerobic glycolysis .
During glycolysis, glyceraldehyde- 3-phosphate is converted to
1-3-bisphosphoglycerate by the enzyme glyceraldehyde-3-phosphate
dehydrogenase This enzyme uses NAD-dependent oxidation , and in this reaction
NAD- is converted into NADH NAD' is present in limited amounts in most cells , and
it must be regenerated from NADH for glycolysis to continue Under aerobic
conditions , NAD' is converted to NADH in the TCA cycle and NADH is then
converted back to NAD ' in the electron transport chain as the energy in NADH is
utilized to synthesize ATP In anaerobic glycolysis, NAD' is regenerated from NADH
when pyruvate is converted to lactate by lactate dehydrogenase. In strenuously
exercising muscle glycolysis can be inhibited by limited regeneration of NAD from
NADH (choice C)
,
*
(Choice A) During muscle contraction, glycogen is broken down by glycogen
phosphorylase for energy production via the glycolytic pathway. Epinephrine through
cyclic AMP causes phosphorylation of glycogen phosphorylase making it active
Non- phosphorylation dependent activation of glycogen phosphorylase can occur
during muscle contraction by increased intracellular calcium concentrations and by
AMP under extreme conditions .
V
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I
I
*
Note
*
t alt ufator
A
cyclic AMP causes phosphorylation of glycogen phosphorylase making it active
Non-phosphorylation dependent activation of glycogen phosphorylase can occur
during muscle contraction by increased intracellular calcium concentrations and by
AMP under extreme conditions
b
(Choice B) FADH is not produced in glycolysis FADH is produced from FAD
during the conversion of succinate to fumarate in the TCA cycle by the enzyme
succinate dehydrogenase
U
IS
IS
1?
13
(Choice D) Carnitine is the amino acid responsible for transport of fatty acids into
the mitochondria for beta-oxidation . It is synthesized from lysine and methionine .
Vitamin C is essential for the synthesis of carnitine .
H
20
21
22
24
2&
25
27
28
29
30
31
32
33
34
(Choice E) In glycolysis pyruvate is formed from phosphoenolpyruvate by a
unidirectional enzyme called pyruvate kinase . Pyruvate is usually present in large
quantities in cells ; therefore , it is not the limiting factor for glycolysis ,
(Choice F) Citrate is formed from the condensation of acetyl CoA with oxaloacetate
in the first step of the TCA cycle. Citrate is a very powerful allosteric inhibitor of
phosphofructokinase -1 Increased citrate concentrations thereby decrease
glycolysis. In exercising muscles under anaerobic conditions , oxidative
phosphorylation of glucose through the citnc acid cycle is not a dominant pathway
and hence excess citrate is not produced.
Educational Objective:
Under anaerobic conditions NADH transfers protons to pyruvate to form lactate and
to regenerate NAD *. NAD* is required to convert glyceraldehyde 3 -phosphate to
1-3-bisphosphoglycerate in glycolysis .
'
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Lob Values
Notes
t alcufdtor
4
5
6
9
10
11
n
A 78-year-old female suffers from nausea , anorexia , fatigue , and skin rash on her
legs. She is mildly demented and is believed to have severe malnutrition.
Laboratory findings include an organic aciduria You suspect that biotin deficiency
may contribute to this patient’s symptoms. Which of the following conversions is
impaired in those with biotin deficiency?
13
U
15
16
1?
O A . Pyruvate to alanine
O B. Pyruvate to oxaloacetate
C Glucose to ribose-5-phosphate
U
(
H
O D . Pyruvate to acetyt-CoA
20
2\
22
E . Succinate to oxaloacetate
23
2S
26
21
23
29
30
31
32
33
3
*
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*
Note
*
I olc ulator
A
&
B
9
10
11
12
13
H
15
is
1?
13
H
20
21
A 78-year-old female suffers from nausea , anorexia , fatigue , and skin rash on her
legs. She is mildly demented and is believed to have severe malnutrition
Laboratory findings include an organic aciduria You suspect that biotin deficiency
may contribute to this patient s symptoms , Which of the following conversions is
impaired in those with biotin deficiency?
O A . Pyruvate to alanine [6%]
B . pyruvate to oxaloacetate [58%]
C Glucose to ribose - 5 -phosphate [5%]
O D . Pyruvate to acetyl-CoA [23%]
E , Succinate to oxaloacetate [9%)
22
23
Explanation:
25
25
27
23
29
30
31
32
33
34
Biotin acts as a CO; earner on the surface of the carboxylase enzyme which
encompasses the enzymatic subtypes acetyl-CoA carboxylase (ACC), pyruvate
carboxylase ( PC ), propionyf carboxylase (PCC ), and beta-methylcrotonyl CoA
carboxylase (MCC ), All of these enzymes play roles in carbohydrate and lipid
metabolism. In the tissues responsible for gluconeogenesis, for instance, pyruvate
carboxylase iand therefore biotin) is necessary for the conversion of pyruvate to
oxaloacetate In biotin-deficient individuals, the level of pyruvate rises and the
pyruvate is converted to lactic acid instead Metabolic acidosis results .
,
Another example of metabolic derangement secondary to biotin deficiency is the
need for propionyt CoA carboxylase ( and therefore biotin ) to synthesize succinyl
CoA from amino acids such as valine In biotin-deficient individuals the propiionvl
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A
5
B
9
10
11
12
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H
20
21
22
23
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25
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Ldb Value
Newt
*
Note
*
I olc ulator
Explanation:
A
Biotin acts as a CO, earner on the surface of the carboxylase enzyme which
encompasses the enzymatic subtypes acetyLCoA carboxylase (ACC) pyruvate
carboxylase ( PC ), propionyl carboxylase (PCC ), and beta -methylcrotonyl CoA
carboxylase ( MCC ). All of these enzymes play roles in carbohydrate and lipid
metabolism. In the tissues responsible for gluconeogenesis , for instance , pyruvate
carboxylase ( and therefore biotin) is necessary for the conversion of pyruvate to
oxaloacetate In biotin-deficient individuals , the level of pyruvate rises and the
pyruvate is converted to lactic acid instead Metabolic acidosis results .
Another example of metabolic derangement secondary to biotin deficiency is the
need for propionyl CoA carboxylase land therefore biotin ) to synthesize succinyl
CoA from amino acids such as valine In biotin-deficient individuals the propionyl
CoA builds up and is instead metabolized into a surplus of odd-chain fatty acids
Deficiencies in this cofactor are rare but can occur secondary to poor diet ,
excessive raw egg white consumption ( due to the high levels of biotin-binding avidin
in egg whites ) and congenital disorders of biotin metabolism.
T
27
23
29
30
31
32
33
34
(Choices A . C, D, and E ) These
conversions
are not impaired in patients with biotin
deficiency.
Educational Objective;
Biotin acts as a CO . earner on the surface of the carboxylase enzyme and is
necessary for numerous conversions , including pyruvate to oxaloacetate Excessive
ingestion of avidm (which is found in egg whites ) has been associated with biotin
deficiency.
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Newt
*
I alculaior
Suspend
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Note
A
b
£
9
10
11
12
13
14
IB
A previously healthy 8- year-old boy is brought to the emergency department due to
vomiting and lethargy The patient had been on an overnight hiking trip with his
family. During the trip the family lost their food pack while canoeing . They had to
hike back to the car . The child became weak and was carried the last mile . No one
has eaten for approximately 24 hours . On examination , he appears listless. Mild
hepatomegaly is noted . Laboratory results are as follows:
30 mg /dL
Glucose
,
is
1?
Acetoacetate
ij
Aspartate
H
20
21
22
aminotransferase
Alanine
aminotransferase
not
detected
341 U/ L
412 U/ L
23
24
26
The child begins seizing as the intravenous line is placed . Which of the following
enzymes is most likely deficient in this patient?
27
23
29
30
31
32
33
34
A Acetyl-CoA carboxylase
B AcyJ - CoA dehydrogenase
( C Acid a - glucosidase
C D Glucose 6-phosphatase
I E Glycogen phosphorylase
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Sent
*
Notes
t alculdtor
A
r\
b
£
9
10
11
12
13
14
IB
is
1?
ij
H
20
A previously healthy 8 -year-old boy is brought to the emergency department due to
vomiting and lethargy The patient had been on an overnight hiking trip with his
family. During the tnp . the family lost their food pack while canoeing . They had to
hike back to the car. The child became weak and was carried the last mile . No one
has eaten for approximately 24 hours . On examination , he appears listless. Mild
hepatomegaly is noted Laboratory results are as follows:
30 mg/dL
Glucose
.
Acetoacetate
not
detected
Aspartate
aminotransferase
341 U / L
21
Alanine
22
aminotransferase
21
24
25
412 U/ L
The child begins seizing as the intravenous line is placed . Which of the following
enzymes is most likely deficient in this patient?
27
23
29
30
31
32
33
34
A Acetyl-CoA carboxylase [13%]
* # B. Acyl-CoA dehydrogenase [28%]
O C . Acid ct -glucosidase [4%]
D . Glucose 6-phosphatase [30%]
E. Glycogen phosphorylase [25%J
Explanation:
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Newt
I olc uhHor
Notes
> KJ
i
i
A
E Glycogen phosphorylase [25%]
e
Explanation:
9
io
11
Fatty acid oxidation
12
13
U
15
1&
1?
ii
H
Fatty acid
Primary carnitine deficiency
weakness
- Muscle
Cardiomyopathy
Acyl CoA synthase
20
•
21
23
24
25
-
Acyl CoA
22
U
Hypoketotic hypoglycemia
Elevated muscle triglycerides
CATf
Acyl- carnitine
Carnitine
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Cytoplasm
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CAT II
Carnitine *
Acyl - carnitine
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Mitochondrial
matrix
Acyl CoA
Medium chain acyl CoA
Acyl CoA
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dehydrogenase {MCADJ
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Acyl CoA
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FADH;
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dehydrogenase {MCADJ
deficiency
* Hypoglycemia
* Hypoketotic hypoglycemia
3 ’Hydroxyacyl CoA
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Acyl CoA
dehydrogenase
I
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b
Medium chain acyl CoA
Trans-enoyl CoA
20
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3 - Ketoacyl CoA
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Acetyl CoA
Ketone bodies
TCA
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© UWorld
This child has hypoketotic hypoglycemia after a period of fasting . These findings
are consistent with a defect in fatty acid 0 oxidation in the mitochondria The most
common enzymatic defect leading to impaired 0- oxidation is acyl*CoA
dehydrogenase deficiency . Normal 0-oxidation of fatty acids yields FADR and
NADH for ATP production and generates acetyl-CoA for the citric acid cycle and
ketone bodies. During periods of fasting , patients with acyl-CoA deficiency cannot
oxidize fatty acids for energy or produce ketone bodies This results in the
characteristic hypoketotic hypoglycemia.
-
There are several variants of acyl-CoA dehydrogenase that metabolize different length fatty acid chains ( very long-chaint long-chaint medium-chain, or shortchain ) . However , the clinical findings are similar regardless of the type of acyl-CoA
dehydrogenase deficiency . Carnitine deficiency prevents fatty acids from being
transported into the mitochondria for p-oxidation and causes similar
features. Affected patients may remain asymptomatic for long periods until they
experience a significant fast : therefore , fatty acid oxidation disorders and carnitine
deficiency have been added to newborn screening
Treatment of acyl-CoA dehydrogenase deficiencies requires prevention of
catabolism . This means avoiding prolonged fasting as well as promptly supplying
glucose during periods of illness .
(Choice A) Acetyl-CoA carboxylase is the rate -limiting enzyme that catalyzes the
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A
There are several variants of acyl-CoA dehydrogenase that metabolize differentlength fatty acid chains ( very long-chain, long-chain, medium-chain , or shortchain ) However , the clinical findings are similar regardless of the type of acyl-CoA
dehydrogenase deficiency . Carnitine deficiency prevents fatty acids from being
transported into the mitochondna for p-oxidation and causes similar
features Affected patients may remain asymptomatic for long periods until they
experience a significant fast : therefore , fatty acid oxidation disorders and carnitine
deficiency have been added to newborn screening
b
Treatment of acyl-CoA dehydrogenase deficiencies requires prevention of
catabolism This means avoiding prolonged fasting as well as promptly supplying
glucose during periods of illness .
(Choice A) Acetyl-CoA carboxylase is the rate -limiting enzyme that catalyzes the
first step in fatty acid synthesis . During prolonged fasting acetyl-CoA carboxylase is
suppressed as fatty acids are catabolized.
(Choices C, D , and E) Acid o - glucosidase , glucose 6 -phosphatase, and glycogen
phosphorylase are involved in glycogenolysis. Deficiency of any of these
enzymes can lead to glycogen storage disease However , patients with glycogen
storage disease have normal fatty acid oxidation and produce ketones during
periods of fasting.
Educational objective:
Impaired (3 -oxidation of fatty acids causes hypoglycemia after prolonged fasting and
inappropriately low levels of ketone bodies . Acyl-CoA dehydrogenase catalyzes the
first step in the p-oxidation pathway and is the most commonly deficient enzyme .
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Notes
I olc uhHor
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A
&
Researchers analyzing eukaryotic genome structure and function perform an
experiment to extract DNA from exocrine pancreatic cells During the purification
process , they isolate small circular DNA molecules that resemble a bacterial
chromosome . Further analysis shows that these molecules code for proteins ,
transfer RNA and ribosomal RNA . From which of the following cellular structures
did these DNA molecules most likely originate ?
B
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O A. A
OB B
O C. C
ODD
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Newt
Notes
I olc uhHor
A
A
&
Researchers analyzing eukaryotic genome structure and function perform an
experiment to extract DNA from exocrine pancreatic cells During the purification
process , they isolate small circular DNA molecules that resemble a bacterial
chromosome . Further analysis shows that these molecules code for proteins ,
transfer RNA and ribosomal RNA . From which of the following cellular structures
did these DNA molecules most likely originate ?
B
9
n
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k
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O A. A [5%]
OB. B [29% ]
OC . C [ 10%]
J
I
* <•D. D [53%|
OEE [3%1
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olc uhHor
E . E [3%]
&
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9
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Explanation:
n
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CllWuM
Nuclear chromosomes contain most of the DNA found in human cells However,
mitochondna also contain their own DNA called mitochondrial DNA (mtDNA ). This
DNA exists as a small circular chromosome with a slightly different genetic code
than that of nuclear DNA consistent with the endosymbiotic theory that mitochondria
originated as prokaryotic cells that were later engulfed by ancient eukaryotes . Over
time most of the genes coding for mitochondrial proteins have migrated to nuclear
DNA . However. mtDNA stilt codes for about 14 proteins ( some involved in oxidative
metabolic pathways ) and the ribosomal and transfer RNA needed for mitochondrial
protein synthesis . Each mitochondrion contains 1-10 copies of maternally derived
mtDNA . As a result diseases arising from mutations in mtDNA are transmitted from
v
the mother to all nf her nffenrinn Mitor honrlria ran he identifier! on elortron
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Notes
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(
ukuhitor
time most of the genes coding for mitochondrial proteins have migrated to nuclear
DNA . However , mtDNA still codes for about 14 proteins ( some involved in oxidative
metabolic pathways ) and the ribosomal and transfer RNA needed for mitochondrial
protein synthesis . Each mitochondrion contains 1-10 copies of maternally derived
mtDNA. As a result diseases arising from mutations in mtDNA are transmitted from
the mother to all of her offspring Mitochondria can be identified on electron
microscopy by their characteristic double membrane and wavy cristae
A
(Choice A) The rough endopiastic reticulum has a stippled appearance secondary
to the presence of numerous ribosomes bound to its membranes . These
nbosomes are involved in the synthesis of integral membrane proteins and proteins
destined for export or packaging into granules or organelles.
b
(Choice B) The dark region identified within the nucleus is the nucleolus the site of
synthesis and assembly of ribosomal components . There is no lipid membrane
separating the nucleolus from the rest of the nucleus.
(Choice C) The lighter ' electron-lucent" regions within the nucleus signify
euchromatin (unpackaged DNA actively being transcribed),
27
23
29
M'
-
(Choice E) This electron dense membrane -bound spherical structure represents an
exocrine granule containing enzymes and other proteins packaged for secretion .
Educational objective:
Mitochondrial DNA (mtDNA ) is the most common non -nuclear DNA found in
eukaryotic cells It resembles prokaryotic DNA and is maternally derived Mutations
involving mtDNA or nuclear DNA that codes for mitochondrial proteins can cause a
variety of mitochondrial disorders including Leigh syndrome and MELAS.
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*
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22
A 55-year-old man is found to have abnormal serum chemistries during routine
laboratory testing He has a history of hypertension for which he is being treated
with pharmacologic therapy, weight reduction and decreased sodium intake
Laboratory results are as follows:
134 mEq/L
Serum sodium
3.8 mEq / L
Serum potassium
Blood urea nitrogen 18 mg /dL
0.8 mg / dL
Serum creatinine
Calcium
11 0 mg/dL
Blood glucose
98 mg/dL
Parathyroid hormone Decreased
Which of the following is most likely responsible for these findings?
23
n
25
25
*
-
--
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32
C A. Familial hypocalciunc hypercalcemia
O B Hypothyroidism
O C Medication
C D
Primary hyperparathyroidism
C E . Secondary hyperparathyroidism
33
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I alculdlor
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it
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1
15
is
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*
20
21
22
A 55-year -old man is found to have abnormal serum chemistries during routine
laboratory testing He has a history of hypertension , for which he is being treated
with pharmacologic therapy, weight reduction and decreased sodium intake
Laboratory results are as follows:
134 mEq/L
Serum sodium
3.8 mEq /L
Serum potassium
Blood urea nitrogen 18 mg/dL
0.8 mg / dL
Serum creatinine
Calcium
11 Omg/dL
Blood glucose
98 mg /dL
Parathyroid hormone Decreased
I
Which of the following is most likely responsible for these findings9
23
n
25
26
-
-
--
2S
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C A. Familial hypocalciunc hypercalcemia [14%]
B . Hypothyroidism [2%]
v ® C. Medication [71%]
O D. Primary hyperparathyroidism [4%]
O E. Secondary hyperparathyroidism [1Q%]
Explanation:
Effect of thiazide diuretics on distal tubular
calcium reabsorption
v
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I Hjlc uldtor
oeconaary riypeiparamyruTaism [ IUVOJ
A
Explanation:
s
9
Effect of thiazide diuretics on distal tubular
calcium reabsorption
10
n
n
13
14
15
is
Early distal tubule cells
Interstitial
Tubular
lumen
fluid
13
H
20
22
23
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-
--
2K *
- Na *
- - Cl
Thiazides
Ca *
30
32
-
J, INa - I
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<<-
3 Na *
21
Ca;+
3Naf
33
34
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-
t ale uf H or
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IB
15
Ql > SM ( F World . UC
Thiazide diuretics ( eg chlorthalidone hydrochlorothiazide ) are some of the most
common agents used for treatment of primary hypertension. The use of high-dose
thiazides is associated with a variety of metabolic and electrolyte complications
including hypercalcemia , hyperglycemia, hypercholesterolemia , hyperuricemia ,
hyponatremia and hypokalemia However , the incidence of these side effects is
relatively low and generally only 1 or 2 will develop in any particular patient
,
13
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n
21
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2B
25
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--
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Thiazide diuretics inhibit the Na /C ) cotransporter in the distal tubule leading to
increased excretion of Na - and H O as well as K - and H* ions . This also increases
distal tubular Ca reabsorption which causes both hypercalcemia and
hypocalciuria The increase in serum Ca - will result in appropriate suppression of
parathyroid hormone (PTH ) levels,
,
(Choice A ) Familial hypocalciuric hypercalcemia (FHH ) is a benign autosomal
dominant disorder caused by defective Ca: - sensing by parathyroid and renal tubule
cells . This defect prevents PTH from being suppressed in response to an increase
in serum Ca - * , resulting in hypercalcemia with normal-to -high serum PTH levels.
Although both FHH and primary hyperparathyroidism present with elevated Ca - and
PTH levels , FHH will also have low urinary Ca: - excretion (ie hypocalciuna ) due to
impaired renal Ca:* sensing.
(Choice B) Hypothyroidism is usually not accompanied by significant changes in
serum Ca levels . In contrast , hypercalcemia can be seen in some patients with
thyrotoxicosis and is typically caused by increased osteoclastic bone resorption.
;
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increased excretion of Na * and H:0 as well as K and H* ions . This also increases
distal tubular Ca reabsorption, which causes both hypercalcemia and
hypocalciuria The increase in serum Ca * will result in appropnate suppression of
parathyroid hormone (PTH ) levels.
A
(Choice A ) Familial hypocalciunc hypercalcemia ( FHHJ is a benign autosomal
&
dominant disorder caused by defective Ca:~ - sensing by parathyroid and renal tubule
cells This defect prevents PTH from being suppressed in response to an increase
in serum Ca; , resulting in hypercalcemia with normal- to -high serum PTH levels.
Although both FHH and primary hyperparathyroidism present with elevated Ca • and
PTH levels , FHH will also have low urinary Ca - excretion (ie hypocalciuria ) due to
impaired renal Ca:* sensing .
(Choice B) Hypothyroidism is usually not accompanied by significant changes in
serum Ca - levels In contrast , hypercalcemia can be seen in some patients with
thyrotoxicosis and is typically caused by increased osteoclastic bone resorption .
(Choice D) Patients with pnmary hyperparathyroidism have increased Ca:- and
elevated PTH levels. This patient has decreased PTH levels,
(Choice E) Secondary hyperparathyroidism is due to chronic renal failure and
malabsorption of vitamin D and is characterized by hypocalcemia and elevated PTH
levels
Educational objective :
Thiazide diuretics cause hypercalcemia by increasing the distal tubular reabsorption
of filtered Ca * . The increased serum Ca: levels result in suppression of parathyroid
hormone , which distinguishes the side effect from primary hyperparathyroidism.
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t dlculdtor
A
B
9
10
12
13
u
A 6-month-old boy is brought to the emergency department by his mother because
of recent onset of vomiting irritability , and jaundice . The infant was born at term and
had been healthy until the onset of these symptoms . All of his vaccinations are
up - to - date He had been breast- fed exclusively until 1 week ago , when cereals and
fruit juices were introduced into his diet Further evaluation reveals hepatomegaly
and abnormal liver function tests Which of the following enzymes is most likely to
be deficient in this patient?
IS
IS
1?
13
20
2\
72
23
2i
2&
25
O A. Galactose-1-phosphate uridyl transferase
O B. Aldolase B
O C . Fructokinase
O D Galactokinase
O E Acid a-glucosidase
21
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*
t dlculdtor
A
A
5
B
9
n
11
12
13
14
15
15
1?
13
H
20
21
22
21
24
25
25
27
*
-
b
A Galactose- 1 -phosphate uridyl transferase [11%]
* B . Aldolase B [63%J
O C . Fructokinase [21%]
O D Galactokinase [3%]
O E . Acid a-glucosidase [2%]
.
*
Explanation:
Glucose
29
30
31
*
32
-
33
34
-
A 6-month-old boy is brought to the emergency department by his mother because
of recent onset of vomiting irritability , and jaundice The infant was born at term and
had been healthy until the onset of these symptoms . All of his vaccinations are
up- to- date He had been breast- fed exclusively until 1 week ago , when cereals and
fruit juices were introduced into his diet Further evaluation reveals hepatomegaly
and abnormal liver function tests Which of the following enzymes i s most likely to
be deficient in this patient?
Glucose 6-phosphate
Sucrose/ S orbitol
Diet
l
Fructose 6- phosphate
Fructose
bispfrosphate
Fructose
FrurJnUmas
PFK 1
V
*
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(Notes
t ole uLilor
J
/\
Explanation:
Glucose
io
it
n
Glucose 6-phosphate
13
M
is
is
1?
13
Sucrose/ Sorttltol
I
Fructose 6- phosphate
Fruaose
Fructose
Diet
bisphosphato
p/tosp/wfase- 7
H
20
Fructokinaso
21
22
-
--
Fructose
Fructose t - phosphate
23
24
26
25
27
Aldolase B
29
DHAP
PFK 1
1.6 - Bisphosphate
A /do/ase A & B
Glycer aldehyde
Triokmase
* GlyceraIde hyde
DHAP
3 - phosphate
30
31
*
32
*
33
34
[
T
Pyruvate
Dietarv fructose is obtained mainlv from fruits veaetables . honey table suaar
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[
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/%
i
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z
9
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17
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19
20
21
22
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25
27
-
-
*
29
30
31
32
33
31
Dietary fructose is obtained mainly from fruits, vegetables , honey table sugar
( sucrose ), and processed foods . Fructose is rapidly absorbed in the proximal small
bowel by the hexose transporter GLUT 5 . Initial metabolism of fructose involves
three enzymes fructokinase , aldolase B, and ( riokinase Fructose is phosphoryiated
on the first carbon by hepatic fructokinase, yielding fructose -1’
phosphate . Metabolism of fructose - 1-phosphate by aldolase B generates
dihydroxyacetone phosphate (DHAP ) and glyceraldehyde . Glyceraldehyde is then
phosphoryiated to glyceraldehyde - 3-phosphate ( G3P ), an intermediate of glycolysis ,
by tnose kinase . DHAP can also be converted to G3P by tnose phosphate
b
isomerase.
Aldolase B deficiency causes the potentially life-threatening disorder known as
hereditary fructose intolerance . Patients typically present when fructose - containing
foods are introduced into the diet. The primary manifestations are vomiting and
hypoglycemia about 20-30 minutes after fructose ingestion Hypoglycemia results
from intracellular accumulation of fructose-1-phosphate and depletion of inorganic
phosphate, which inhibit glycogenolysis and gluconeogenesis . Failure to thrive
hepatomegaly , and jaundice can also occur Undiagnosed individuals may eventually
develop liver and renal failure. Elimination of dietary fructose is the mainstay of
treatment and results in symptom improvement with a good long - term prognosis .
(Choices A and D) Galactose -1-phosphate uridyl transferase deficiency (classic
galactosemia ) is an autosomal recessive disorder characterized by vomiting feeding
intolerance neonatal jaundice , hepatomegaly , and death if untreated Symptoms
start soon after breastfeeding is initiated . Galactokinase deficiency is a more benign
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hypoglycemia about 20-30 minutes after fructose ingestion Hypoglycemia results
from intracellular accumulation of fructose-1-phosphate and depletion of inorganic
phosphate, which inhibit glycogenolysis and gluconeogenesis . Failure to thrive
hepatomegaly, and jaundice can also occur . Undiagnosed individuals may eventually
develop liver and renal failure . Elimination of dietary fructose is the mainstay of
treatment and results in symptom improvement with a good long-term prognosis
*
*
Notes
l alculdtor
A
(Choices A and D) GaJactose - 1-phosphate uridyl transferase deficiency (classic
galactosemia ) is an autosomal recessive disorder characterized by vomiting feeding
intolerance neonatal jaundice , hepatomegaly and death if untreated Symptoms
start soon after breastfeeding is initiated. GaJactokmase deficiency is a more benign
disorder of galactose metabolism that results in the formation of neonatal cataracts
b
(Choice C) Fructokinase deficiency causes essential fructosuria . a benign
autosomal recessive disorder. Fructose from the diet JS absorbed and secreted
freely in the urine due to impairment of the first step in fructose metabolism .
(Choice E) Glycogenolysis is accomplished mainly by glycogen phosphorylase and
debranching enzyme , but a small amount is also broken down by the lysosomal
enzyme a-1.4-glucosidase Alpha - glucosidase ( or acid maltase ) deficiency causes
Pompe disease This disease presents not with hypoglycemia , but with
cardiomyopathy and hypotonia .
Educational objective:
Aldolase B deficiency causes hereditary fructose intolerance . This disease
manifests after introduction of fructose into the diet with vomiting and hypoglycemia
about 20-30 minutes after fructose ingestion . These infants can present with failure
to thrive , jaundice , and hepatomegaly .
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l olc uf i l o r
A
&
9
n
11
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u
A 35-year -old woman comes to the emergency department with nausea vomiting
and fever Her symptoms began 24 hours ago, and she has been unable to eat or
drink anything since. She has a 3 - year -old daughter who had similar symptoms 2
days earlier but is now fine Laboratory studies show a blood glucose level of 82
mg / dl despite her 24-hour fast . Maintenance of this patient's blood glucose levels
is facilitated by hepatic conversion of pyruvate into glucose Which of the following
substances is an allosteric activator of the first step of this process ?
_
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ia
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O A. Acetyl-CoA
O B . Alanine
O C . Citrate
D Fructose 2 6 -bisphosphate
O E . Lactate
F . Oxaloacetate
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14
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29
26
A 35-year-old woman comes to the emergency department with nausea , vomiting
and fever Her symptoms began 24 hours ago , and she has been unable to eat or
drink anything since. She has a 3 - year -old daughter who had similar symptoms 2
days earlier but is now fine Laboratory studies show a blood glucose level of 82
mg /dl despite her 24-hour fast , Maintenance of this patient's blood glucose levels
is facilitated by hepatic conversion of pyruvate into glucose Which of the following
substances is an allosteric activator of the first step of this process?
_
b
A. Acetyl-CoA [41%]
O B. Alanine [7%]
O C , Citrate [12%]
D . Fructose 2.6-bisphosphate [18%]
O E, Lactate [6%]
O F. Oxaloacetate [16%]
27
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Explanation:
Metabolic fate of pyruvate
30
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Pyruvate
C02
Pyruvate
carhoxvlas
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Explanation:
A
Metabolic fate of pyruvate
B
9
Pyruvate
10
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13
U
19
is
17
13
COi
Pyruvate
carboxylase
Pyruvate
dehydrogenase
co2
1 -1
20
21
22
Oxaloacetate
Acetyl- CoA
Gluconeogenesis
Citnc acid cycle
Glucose
Energy
23
24
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—
Fatty acid
oxidation
27
28
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31
-
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A
5
Energy
Glucose
B
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OUWortct
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During gluconeagentrsis , substances such as lactate and alanine are converted to
pyruvate. However , pyruvate cannot be converted to phosphoenolpyruvate directly
as pyruvate kinase is unidirectional To convert pyruvate to phosphoenolpyruvate,
pyruvate first undergoes biotm- dependent carboxylahon to oxaloacetate in the
mitochondria This reaction is catalyzed by pyruvate carboxylase The activity of
pyruvate carboxylase is increased by acetyl-CoA . an allosteric activator of the
enzyme This critical regulatory step diverts pyruvate to pyruvate dehydrogenase
when acetyl-CoA levels are too low. preventing the cell from becoming energy
starved . When acetyl-CoA levels are high (indicating energy excess }, pyruvate
carboxylase can operate at full capacity and convert most of the pyruvate into
oxaloacetate for use in gluconeogenesis .
&
25
27
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30
-
31
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33
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31
(Choice B) Muscle converts pyruvate to alanine via transamination, which is then
transported to the liver where it is converted back to pyruvate for use in
gluconeogenesis . Alanine allostencally inhibits pyruvate kinase , preventing
phosphoenolpyruvate from be ng consumed by glycolysis during the gluconeogenic
state .
(Choice C) Citrate is formed within mitochondria in the first reaction of the Krebs
cycle and elevated levels act as an indicator of high cellular energy stores and
abundant biosynthetic intermediates . Citrate is therefore an important positive
regulator of acetyl-CoA carboxylase and fructose -1, 6 -bisphosphatase key enzymes
invofved in fatty acid synthesis and gluconeogenesis , respectively
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I alculaior
(Choice C) Citrate is formed within mitochondria in the first reaction of the Krebs
cycle, and elevated levels act as an indicator of high cellular energy stores and
abundant biosynthetic intermediates Citrate is therefore an important positive
regulator of acetyl-CoA carboxylase and fructose -1, 6 -bisphosphatase key enzymes
involved in fatty acid synthesis and gluconeogenesis respectively .
.
(Choice D) Regulation of glycolysis and gluconeogenesis occurs mainly through
the inverse regulation of phosphofructokmase -1 and fructose 1,6 -bisphosphatase by
fructose 2.6-bisphosphate High levels of fructose 2 , 6-bisphosphate activate
phosphofructokinase-1 and accelerate glycolysis ; low levels disinhibit fructose
1, 6 -bisphosphatase and promote gluconeogenesis .
(Choice E) Lactate is an important source of carbon atoms for glucose synthesis
during gluconeogenesis. During anaerobic glycolysis in skeletal muscle , pyruvate is
reduced to lactate by lactate dehydrogenase Lactate formed in the contracting
muscles is released into the bloodstream and transported to the liver, where it is
converted back into glucose .
Educational objective:
Acetyl-CoA is an allosteric activator of gluconeogenesis that acts by increasing the
activity of pyruvate carboxylase when acetyl-CoA is abundant. This regulatory step
allows pyruvate to be shunted toward acetyl-CoA production when acetyl-CoA levels
are low, preventing the cell from becoming depleted of energy
.
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sent
References:
1. Regulation of the structure and activity of pyruvate carboxylase by
acetyl CoA.
2 . Allosteric regulation of the biotin-dependent enzyme pyruvate
carboxylase by acetyl-CoA.
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5
£
9
10
11
n
13
u
Some transmembrane receptors employ Janus kinase ( JAK ) to stimulate enzymes in
the cytoplasm . Which of the following substances is most likely to utilize this
pathway?
C A. Insulin
B . Plateiet-derived growth factor ( PDGF )
15
IS
O C . Growth hormone
17
O 0 Atrial natnuretic peptide
13
H
20
2\
O E . Progesterone
F . Gamma-aminobutyric acid (GABA )
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(Notes
t alculdtor
A
A
5
6
9
Some transmembrane receptors employ Janus kinase ( JAK ) to stimulate enzymes in
the cytoplasm Which of the following substances is most likely to utilize this
pathway?
n
11
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1?
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O A. Insulin [19%]
B . Platelet-derived growth factor ( PDGF ) [26%]
* # C. Growth hormone [37%]
D. Atrial natnuretic peptide [8%]
O E. Progesterone [4%]
C F. Gamma- aminobutyric acid ( GABA ) [ 4%]
21
n
23
n
23
29
Receptors are subdivided into four major categories : steroid ion channel,
enzyme -linked , and G-protem-linked . Steroid receptors are located in the cell
cytoplasm or nucleus while the other three receptor types are found on the cell
surface
31
Enzyme -linked receptors are proteins that span the cell membrane , extruding an
25
25
27
-
--
Explanation:
32
33
3t
,
extracellular terminal that binds to the corresponding growth factor Once bound the
receptor protein configuration is changed , which triggers a cascade of events.
Some of the enzyme -linked receptors are enzymes themselves , and some activate
enzymes present in the cytosol The differences between these two types of
enzyme-linked receptors are as follows :
Receptor [ With intrinsic enzyme activity
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I Without intrinsic
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Newt
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I dlculdlor
surface
Enzyme -linked receptors are proteins that span the cell membrane , extruding an
extracellular terminal that binds to the corresponding growth factor . Once bound , the
receptor protein configuration is changed which triggers a cascade of events.
Some of the enzyme -linked receptors are enzymes themselves , and some activate
enzymes present in the cytosol The differences between these two types of
enzyme-linked receptors are as follows :
Receptor
With intrinsic enzyme activity
kinase )
( receptor tyrosine
17
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Stmc Lure
H
Extracellular domain ( binds the
growth factor)
&
Without intrinsic enzyme
activity (tyrosine-kinase
associated receptor )
L> .tracellular domain
20
21
Transmembrane domain
Trarsmemtirane domain
cytosolic domain ( enzyme)
Cytosolic domain ( lacks enzymatic
22
23
21
25
25
activity )
sigr a ng
pathway
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JAK/STAT
Receptor autophospnorylates
i
in j jars : 1 ' St h< jrylatior f
23
29
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‘
- 33
- *
MAP - kinase
L
Ras protei n
Examples
Growth factor receptors EGF
PDGF , PGR , etc
Receptor activates Janus kl nases
( JAKs ) which phosphorylate
.
STATs {signaltransducers and
activators of transcription)
Receptors for cytoknes. growth
hormone prolactin, IL- 2
(Choices A and B) The insulin receptor and platelet- derived growth factor receptor
have extracellular ligand- bmding domains , transmembrane segments , and
cytoplasmic tails that demonstrate intrinsic enzyme activity . The JAK/STAT signal
transduction system is not used by these receptors.
V
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Signaling
pathway
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Examples
o
.
Lab Value
'-
MAP kinase
JAK/STAT
Rec epto r a uto phosp fi oryiates
and diggers phosphorylation of
Ras prote: n
Receptor activates Janus kl nases
( JAKs ) , which phosphory late
STATs {sigoal transducers and
activators of transcription)
: t v tr r yt A HOS grc / "i
ji e . prolactin, IL - 2
r
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it
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activity }
E
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Growth factor receptors £Gf .
FDGP FCfLeK
.
*
Notes
lalculator
A
u
(Choices A and B) The insulin receptor and platelet- derived growth factor receptor
have extracellular ligand - bmding domains , transmembrane segments , and
cytoplasmic tails that demonstrate intrinsic enzyme activity. The JAK/STAT signal
transduction system is not used by these receptors.
(Choice D) The atrial natriuretic peptide receptor has intrinsic guanylate cyclase
activity , The JAK/STAT signal transduction system is not used by this receptor.
(Choice E) The progesterone receptor requires the ligands to diffuse through the
cell membrane before binding can occur Transcription is then actuated. The
JAK/STAT signal transduction system is not used by this receptor ,
(Choice F) The gamma - aminobutyric acid (GABA ) receptor uses either a
ligand-gated ion channel or G protein coupling , The JAK /STAT signal transduction
system is not used by this receptor.
Educational Objective:
Colony-stimulating factors prolactin, growth hormones and cytokines utilize tyrosine
kinase -associated receptors and the JAK/STAT signaling pathway .
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Note
4
5
E
9
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It
12
13
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15
is
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H
20
21
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21
24
A 12-year -old girl is involved in a motor vehicle accident and brought to the
emergency department . An abdominal CT scan is negative for acute trauma
However , she is incidentally found to have an abnormal intra - abdominal mass
Biopsy of the mass reveals ectopic tissue . On electron microscopy , these cells
have a highly developed smooth endoplasmic reticulum . These cells most likely
receive stimulation from which of the following agents?
A Parathyroid hormone
C B . Progesterone
O C . ACTH
OD
Aldosterone
O E Dopamine
25
25
2T
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30
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(Notes
t alculdlor
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1&
1?
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H
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22
21
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25
25
A 12-year -old girl is involved in a motor vehicle accident and brought to the
emergency department . An abdominal CT scan is negative for acute trauma
However , she is incidentally found to have an abnormal intra - abdominal mass
Biopsy of the mass reveals ectopic tissue . On electron microscopy , these cells
have a highly developed smooth endoplasmic reticulum. These cells most likely
receive stimulation from which of the following agents?
b
O A Parathyroid hormone [5%J
O B . Progesterone [18%]
,
v m C . ACTH [69%]
D . Aldosterone [4%]
E, Dopamine [4%]
Explanation:
21
Rough & smooth endoplasmic reticulum
23
29
30
--
Nucleus
32
Nuclear envelope
33
34
Rough ER
u
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(Notes
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A
Explanation:
9
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Nucleus
n
Nuclear envelope
13
14
15
15
1?
Rough ER
13
H
20
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22
Smooth ER
23
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25
21
23
29
30
Transport
vesicles
‘ 32
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34
Ribosomes
I
V
cmtWnrtd
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4
ft
© UWorid
£
9
10
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U
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IS
1?
13
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22
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2S
25
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-
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The endoplasmic reticulum (ER ) is a continuous system of folded membranes that
enclose a single cisternal space The rough ER (RER ) has ribosomes attached and
is the site of synthesis of secretory , lysosomal , and integral membrane proteins
Once synthesized, many proteins undergo post-translational modification inside the
RER and most are targeted for export to the Golgi apparatus . These proteins pass
from the RER to the Golgi apparatus within COP II coated transport vesicles The
Golgi apparatus sorts and distributes proteins to the cell membrane organelles , and
secretory granules .
b
In contrast to the RER , the smooth ER (SER ) contains enzymes for steroid and
phospholipid biosynthesis. All steroid-producing cells ( eg , cells in the adrenals,
gonads and liver ) contain a well-developed SER . ACTH primarily stimulates the
cells in the adrenal cortex to produce glucocorticoids , but it also increases adrenal
production of mineralocorticoids and androgens to a lesser extent These are all
cholesterol- derived steroid hormones , and the cells that synthesize them contain
prominent SERs . The SER is also involved in the detoxification of numerous drugs,
especially in hepatocytes. In addition, a specialized SER functions as the
sarcoplasmic reticulum in striated muscle cells .
,
(Choice A) Parathyroid hormone helps regulate serum calcium levels by increasing
osteoclastic bone resorption , increasing the distal tubular absorption of calcium , and
increasing 1,25 - dihydroxyvitamin D conversion through upregulation of the renal
enzyme 1- alpha hydroxylase Parathyroid hormone does not upregulate steroid or
lipid synthesis.
(Choice B) Progesterone is a steroid hormone secreted from the corpus luteum It
prepares the endometrium for conception and helps maintain the new pregnancy , It
»
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B
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21
n
23
21
25
25
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--
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Neat
*
Notes
I olc uhHor
prominent SERs The SER is also involved in the detoxification of numerous drugs,
especially in hepatocytes . In addition, a specialized SER functions as the
sarcoplasmic reticulum in striated muscle cells
A
(Choice A) Parathyroid hormone helps regulate serum calcium levels by increasing
osteoclastic bone resorption increasing the distal tubular absorption of calcium , and
increasing 1, 25-dihydroxyvitamin D conversion through upregulation of the renal
enzyme 1-alpha hydroxylase Parathyroid hormone does not upregulate steroid or
lipid synthesis.
(Choice B) Progesterone is a steroid hormone secreted from the corpus luteum. it
prepares the endometrium for conception and helps maintain the new pregnancy It
acts via a type 1 nuclear receptor that on binding progesterone translocates to the
nucleus and binds directly to DNA to influence protein production The ovarian lutein
cells that produce progesterone contain abundant smooth ER , but the endometrial
cells on which progesterone acts do not.
(Choice D) Aldosterone acts mainly on the distal tubules and collecting ducts of the
nephron to increase absorption of sodium and secretion of potassium, increasing
blood volume and blood pressure It does not affect steroid or lipid synthesis
(Choice E) Dopamine is a powerful vasoactive and inotropic molecule derived from
tyrosine . It functions via a G-protein-coupled receptor .
Educational objective:
In contrast to the rough endoplasmic reticulum (ER ) , the smooth ER contains
enzymes for steroid and phospholipid biosynthesis All steroid-producing cells ( eg
cells in the adrenals , gonads and liver ) contain a well-developed smooth ER .
Time Spent 6 seconds
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I dlculdlor
A
B
9
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11
1?
13
A
A 23-year-old man is brought to the emergency department by paramedics following
a motor vehicle accident He was an unrestrained passenger in the front seat
Several days after hospitalization , his fluid volume and plasma osmoiarity are
measured and illustrated in the image below ( solid line normal , dotted line , patient ) .
c>
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1?
13
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E
20
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21
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23
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29
30
inrj JticltAar
31
*
33
34
flu 4 ( f C F )
flxtijcclhlar
nu J (ECF )
*
Fluid volume (L)
0
OLTWond
Which of the following conditions is most likely to cause the findings shown in the
image ?
A
..
A
4
*
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Notes
I olc uhnor
A
/%
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B
9
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11
12
13
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17
18
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inftritefcl* flu l ( ( CF )
23
24
Fluid volume < L )
25
25
27
28
29
30
FKir <KMUdr nutd (ECF )
0
*
eiMlvfd
Which of the following conditions
image ?
is
most likely to cause the findings shown in the
31
-
-
33
u
A Acute gastrointestinal hemorrhage
O B . Adrenal insufficiency
O C . Diabetes insipidus
O D . Hypertonic saline infusion
C E . Primary polydipsia
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t dkuldtor
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.
miiltn !
B
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E
13
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6
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inff < eluldr flu«j ( ICF )
*
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F <lf «Mdf IU J ( ECF )
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Fluid volum (L)
*
0
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Which of the following conditions is most likely to cause the findings shown in the
image ?
3t
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3
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'
A . Acute gastrointestinal hemorrhage [17%]
B . Adrenal insufficiency [8%]
C . Diabetes insipidus [60%]
C D. Hypertonic saline infusion [13%]
E. Primary polydipsia [3%]
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Iolc ulotor
E . Primary polydipsia [3 % j
&
A
Explanation:
£
9
n
Diabetes insipidus (Dl )
11
Hyperosmotic volume contrucUon
12
13
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17
13
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20
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24
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31
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liwrjte iHd( nud ( fCF )
Ejoraccnudr
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34
Fluid volume (L )
0
Dud ( EC F )
*
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The figure shows the fluid volume ( x - axis ) within the intracellular ( ICF ) and
extracellular ( ECF ) fluid compartments and the osmolanty ( y- axis | of the fluid in those
compartments The changes seen in this patient are indicative of hyperosmotic
I
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O Id
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'
Mif
*
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Notes
*
t alcufdtor
A
&
B
9
10
11
n
13
U
15
IS
1?
13
19
20
2\
22
23
24
25
26
27
28
29
30
31
*
33
*
34
A
til/rtnnd
The figure shows the fluid volume ( x - axis ) within the intracellular ( ICF ) and
extracellular ( ECF ) fluid compartments and the osmolarity ( y - axis ) of the fluid in those
compartments. The changes seen in this patient are indicative of hyperosmotic
volume contraction . This occurs when the loss of free water exceeds the loss of
electrolytes, resulting in increased osmolarity and contracted volumes in the ICF and
ECF compartments Hyperosmotic volume contraction classically occurs in the
setting of diabetes insipidus (likely secondary to head trauma in this patient ) but
can also occur with profuse sweating (due to the hypotonic nature of sweat )
b
(Choice A) Acute gastrointestinal hemorrhage ( or diantiea ) would cause an
isotonic loss of ECF volume with no effects on osmolarity or ICF volume This is
referred to as isosmotic volume contraction
(Choice B) Adrenal insufficiency primarily causes hypertonic loss of NaCI with
some extracellular volume loss (hyposmotic volume contraction ). The low osmolarity
of the ECF results in shifting of free water into the ICF compartment, causing ICF
expansion .
(Choice D) Infusion of large amounts of hypertonic saline leads to hypertonic
volume expansion. Both the volume and osmolarity of the ECF are increased . The
high osmolarity of the ECF leads to shifting of water from the ICF, further increasing
the ECF volume .
(Choice E) Primary polydipsia is a condition in which patients feel compelled to
consume massive amounts of free water . This causes expansion of both the ECF
and ICF compartments and a decrease in the osmolarity of both compartments
(hyposmotic volume expansion). SIADH is another example of this type of volume
expansion,
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O Id
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5
B
9
10
11
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19
M<
Mdrk
<
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(Choice B ) Adrenal insufficiency primarily causes hypertonic loss of NaCI with
some extracellular volume loss (hyposmotic volume contraction ). The low osmolarity
of the ECF results in shifting of free water into the ICF compartment , causing ICF
expansion.
22
the ECF volume .
25
26
27
28
29
30
31
*
33
*
34
Notes
t alcufdtor
(Choice A) Acute gastrointestinal hemorrhage ( or diarrhea ) would cause an
isotonic loss of ECF volume , with no effects on osmolarity or ICF volume . This is
referred to as isosmotic volume contraction
2\
23
24
*
setting of diabetes insipidus (likely secondary to head trauma in this patient ) but
can also occur with profuse sweating ( due to the hypotonic nature of sweat }
(Choice D) Infusion of large amounts of hypertonic saline leads to hypertonic
volume expansion Both the volume and osmolarity of the ECF are increased The
high osmolarity of the ECF leads to shifting of water from the ICF, further increasing
20
Lab Value
Newt
b
(Choice E) Primary polydipsia is a condition in which patients feel compelled to
consume massive amounts of free water This causes expansion of both the ECF
and ICF compartments and a decrease in the osmolarity of both compartments
(hyposmotic volume expansion ) SIADH is another example of this type of volume
expansion .
Educational objective:
Volume contraction and expansion can be divided into isosmotic , hyposmotic , and
hyperosmotic states The loss of free water (with retention of electrolytes ) is seen in
diabetes insipidus and in states of excessive sweating without fluid replacement .
These conditions cause hyperosmotic volume contraction .
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*
Note
*
t ale ufalor
A
A
&
5
9
n
11
12
13
U
15
15
17
ia
19
Researchers are investigating the relationship between glucose transporters and
insulin concentration in various cells and tissues Data are collected and plotted on
the graph below . The graph shows the number of glucose transporters found on the
surface of 2 types of cells ( circles versus triangles ) compared to serum insulin
concentration .
A
c
20
o
21
(A
(A
22
23
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25
to
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to
27
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23
29
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25
30
31
32
CL
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A
A
A
A
A A AA
• •
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AAA
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Lab Value
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*
Notes
I dlculdlor
.2
<A
</>
gj
a
e
*>
a
9
10
11
r
12
tA
13
14
15
is
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13
H
a;
5.
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to
tft
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u
3
o
A
A
A
A A AAA
A
•A •
AA
20
21
22
21
24
2&
25
27
2H
29
30
Insulin concentrations
OUWtxKJ
Which of the following cell types are most likely represented by the circles and
triangles , respectively?
31
32
-
34
O A. Adipocytes and skeletal muscle cells
B . Hepatocytes and cortical pyramidal cells
O C . Hepatocytes and renal tubular cells
_ D . Pancreatic p -cells and intestinal epithelial cells
E . Skeletal muscle cells and renal tubular cells
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A
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o
Ldb Value
Newt
*
Notes
I dlculdlor
.2
>
l/
&
<X )
a
e
x
a?
9
10
11
r
12
(A
&
13
m
u
01
15
is
1?
13
o
u
3
o
H
A
•A •
AA
A A AAA
AA
20
21
22
23
24
2&
25
27
2H
29
30
Insulin concentrations
DUWortd
(
Which of the following cell types are most likely represented by the circles and
triangles, respectively?
31
32
-
34
O A. Adipocytes and skeletal muscle cells [15%]
B . Hepatocytes and cortical pyramidal cells [12%]
O C . Hepatocytes and renal tubular cells [11%]
. D. Pancreatic p -celis and intestinal epithelial cells [8%]
E* Skeletal muscle cells and rena tubular ceils [53%J
•
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Lob Value
Newt
*
Note
*
t alculdtor
O 6 . Hepatocytes and cortical pyramidal cells [12%J
O C Hepatocytes and renal tubular cells [11%]
O D Pancreatic {J-cells and intestinal epithelial cells [8%]
10
^
E . Skeletal muscle ce Is ann renal tubular ce s [53%J
11
12
13
1J
15
is
Explanation:
GLUT transporters
u
17
13
19
Type
Major cellular distribution
Distinctive features
GLUT - 1
Erythrocytes, blood brain barrier
Basal glucose
transport
GLUT- 2
Hepatocytes, pancreatic p-cells .
renal tubular cells small intestine
Regulation of insulin
release
GLUT - 3
Placenta & neurons
Placental & neuronal
glucose transport
GLUT-4
Skeletal muscle celts
& adipocytes
Insulin-mediated
glucose uptake
GLUT- 5
Spermatocytes
& gastrointestinal tract
Fructoso transport
20
21
22
23
n
25
25
27
23
29
30
31
32
- *
3
C eWorld
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Newt
*
Notes
t rile uldtor
A
A
5
GLUT -5
I
9
n
11
12
13
1J
15
IS
17
13
21
22
23
n
25
25
F ructose transport
© UWofid
Of the 5 major facilitate glucose transporters (GLUTs ), only GLUT-4 is
responsive to insulin GLUT-4 is expressed predominantly in muscle cells and
adipocytes In the absence of insulin, GLUT-4 is sequestered in the
cytoplasm However, as insulin concentrations rise , the receptors translocate to the
plasma membrane , facilitating glucose transport down the concentration gradient into
the cell. In the absence of insulin, muscle cells and adipocytes are impermeable to
glucose .
14
20
Spermatocytes
A gastrointestinal tract
-
-
ct
In contrast to GLUT 4 , GLUT 1 2 , 3, and 5 are always present on the plasma
membrane and constitutively transport glucose { insulin-independent } .
t
(Choice A) Adipocytes and skeletal muscle cells have insulin-responsive GLUT -4
transporters. The corresponding graph would show increasing GLUT expression
on both cells with rising insulin concentrations .
21
23
29
30
31
32
- *
3
(Choices B , C, and D) Hepatocytes , pyramidal neurons , renal tubular cells ,
pancreatic p-cells . and intestinal epithelial cells all express insulin-independent
GLUTs . The corresponding graph would show unchanging GLUT expression on
both cells with rising insulin concentrations .
Educational objective:
GLUT 4 is expressed primarily in muscle cells and adipocytes and is the major
glucose transporter that is responsive to insulin.
-
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C7
Lab Value
Newt
A 43-year-old man prospecting for gold in Arizona becomes stuck in the desert after
his truck breaks down. He brought a large supply of water with him but only a few
granola bars as food. After 3 days, he is able to flag down a passing vehicle and
obtain transportation to the nearest settlement. During this ordeal, his liver begins to
synthesize large quantities of glucose from source molecules such as alanine ,
lactate , and glycerol As part of this process , phosphoenolpyruvate is formed from
oxaloacetate in a reaction that requires a specific nucleoside triphosphate as a
cofactor . Which of the following reactions dtrectly synthesizes this cofactor?
*
Notes
(
alcufdtor
U
15
is
17
13
Oxaloacetate
H
Citrate
Malate
20
21
22
21
2i
25
25
27
23
29
Fumarate
Isociirate
30
31
32
33
Succinate
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17 : 01
TUttlT
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*
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Newt
Lob volute
Notes
( dkuldtsr
n
11
n
Fumarate
Isocitrate
13
14
15
is
w
13
H
20
Succinate
2\
a - Ketoglutarate
22
23
2t
25
Suctinyf CoA
>
( 4Ak4 ltori4 ltl
25
27
2H
29
30
31
32
33
O A. A
OB B
O C. C
ODD
O E. E
O F. F
O G. G
O H. H
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Note!
I alculaior
A
10
11
F jma rate
n
Isocitrate
13
n
is
16
1?
13
b
H
20
Succinate
21
a Ketoglutarate
22
23
24
2S
26
SuccinyltoA
.
IK
V
O A . A [4 %]
O B. B [3% ]
2H
29
30
O c . C [ 10%1
O D . D [21%)
' < * E. E [40%]
31
32
33
O F. F [10%]
C G. G [5%]
O H. H [6%]
I
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i
Lob Values
Notes
(
olc ufdtor
H . H [6% ]
rs
Explanation:
Citric acid cycle
10
12
Glucose
13
11
15
15
*
i
17
*
> PtiosphoenoIpyruvate
13
H
20
Pyruvate kinase
21
22
23
21
* <JOP * CO,
Pyruvate
PfPOf
25
<jTP
25
27
Pyruvate
28
29
Pyruvate
30
31
32
33
Cytosol
carboxylase
Oxaloacetate
Matate
dehydrogenas
f \
Malate +
Malate HAD
/
Fumarase
Block Time Remaining:
Oxaloacetate
HADrt
Pyruvate
dehydrogenase
Acetyl CoA
Citrate synthase
Citrate
\
Aconitase
17: 15
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GDP
o
Ldb Value
Newt
*
Notes
(
alcuhilor
I'
- CO,
Pyruvate
Cytosol
PtPCK
GTP
so
Pyruvate
Pyruvate
It
dehydrogenase
12
13
14
15
is
Pyruvate
Acetyl CoA
carboxylase
Oxaioacetate
Motor ?
dehydrogen os
17
13
Oxaloacetate
H
KADH
20
Malate
Malate
21
Fumarase
n
24
Acomtase
Fumarate
25
25
Mitochondrion
Isocitrate
FADH
27
Succinate
28
29
NAD*
dehydrogenase
f AD
30
NAOH t CO,
Succinate
31
32
33
i-
Citrate
NAD
f
22
Citrate synthase
isocitrate
dehydrogenase
\ r
\ / GDP
MAD * a - Ketoglutarate
Sucanate thiokmase ^\ y
HAW * CO
^
(Succinyl - CoA synthetase)
SuccinyKoA
a- Ketoglutarate
dehydrogenase complex
^
Tkft
ATD
i
4
Block Time Remaininq :
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A
AIII
\1 : 26
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SucanGte ffitofo/ jose
/Sucdny / C o A synthetase )
Lob Value
7
GOP
NADHt
coI
NAD *
*
Notes
t alculdtor
A
a - Ketogiutarate
Suainyf -CoA
B
9
10
<3
Previous
cMydfogpriflip comp/p x
oimtxv]
It
n
13
14
15
IS
17
13
H
20
21
22
23
24
25
25
27
28
29
30
31
32
33
The majority of ATP used for cellular processes is generated by the oxidation of
acetate in the tricarboxylic acid (TCA ) cycle The enzymes of the TCA cycle are
located in the mitochondria and generate reduced nicotinamide adenine dinucleotide
( NADH ) and flavin adenine dmucleotide (FADH. ) (Choices C, D, F, and H) These
molecules drive the process of oxidative phosphorylation, which converts their
reducing potential into high-energy ATP via the electron transport chain
ATP can also be generated by substrate-level phosphorylation which involves
the direct transfer of a phosphate group from a reactive intermediate to a nucleotide
diphosphate ( eg ADP , GDP ) Succinyl-CoA synthetase converts succinyl-CoA to
succinate and uses the high- energy thioester present in succinyl- CoA to drive GTP
synthesis. This GTP can then be used to transphosphorylate ADP to ATP, or it
may be utilized by specific GTP-hydrolyzing enzymes , such as
phosphoenolpyruvate carboxykinase (converts oxaloacetate to
phosphoenolpyruvate during gluconeogenesis).
Educational objective:
GTP is synthesized by succinyl-CoA synthetase during the conversion of
succinyl-CoA to succinate in the citric acid cycle. During gluconeogenesis
phosphoenolpyruvate carboxykinase uses GTP to synthesize phosphoenolpyruvate
from oxaloacetate
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O
Ne « t
Lab Vdiue
*
Notes
I dlruldtor
Suspend
tnd Biotic
A
5
G
7
8
9
10
u
15
IS
17
1j
H
20
21
22
Laboratory animals deprived of folic acid experience a marked increase in marrow
erythroid precursor cell production . Subsequently , many of these erythroid
precursor cells undergo apoptosis without further maturation . Provision of which of
the following supplements would reduce erythroid precursor cell apoptosis in these
animals ?
A Homocysteine
O B . Cytosine
C C. Thymidine
O 0 Cobalamin
E . Glutamine
21
24
25
26
21
23
29
30
31
32
33
34
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Lab Value
sent
*
Note
*
( alculator
4
A
5
G
7
8
9
10
u
is
is
17
18
14
20
21
Laboratory animals deprived of folic acid experience a marked increase in marrow
erythroid precursor cell production . Subsequently , many of these erythroid
precursor cells undergo apoptosis without further maturation . Provision of which of
the following supplements would reduce erythroid precursor cell apoptosis in these
animals?
A Homocysteine [13%]
O B . Cytosine [ 4%J
' C. Thymidine [37%]
O D . Cobalamin [41%]
O E. Glutamine [4%]
*
22
21
24
2&
25
27
28
29
30
31
32
33
34
Explanation:
Folate derivatives are crucial in the synthesis of DNA and in the conversion of vitamin
B, to one of its coenzyme forms As summarized in the following chemical reaction ,
thymidylate synthetase catalyzes the methylation of dUMP (deoxyuridine
monophosphate ) to dTMP ( deoxythymidine monophosphate ) while converting the
folate derivative 5,10 -methylenetetrahydrofolate to dihydrofolate
5,10-methylenetetrahydrofolate + dUMP - dihydrofolate + dTMP.
This is the only de novo pathway for dTMP production. Thymidylate synthetase is
therefore essential in its regulation of the supply of the four nucleotide precursors of
DNA replication. DNA synthesis is also impaired when a deficit of
5 , 10-methylenetetrahydrofolate develops One of the common consequences of
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i
Lab Value
NCKt
*
Notes
esa
t alculdtor
O D . Cobalamin [41%]
E Glutamine [4%J
ft
Explanation:
n
Folate derivatives are crucial in the synthesis of DNA and in the conversion of vitamin
B. to one of its coenzyme forms As summarized in the following chemical reaction .
thymidylate synthetase catalyzes the methyJation of dUMP (deoxyuridine
monophosphatei to dTMP ( deoxythymidine monophosphate ) while converting the
folate derivative 5,10 -methylenetetrahydrofolate to dihydrofolate :
^
14
16
15
17
18
19
20
21
22
21
24
2S
25
27
23
29
30
31
32
33
34
b
5,10-methylenetetrahydrofolate + dUMP = dihydrofolate + dTMP.
This is the only de novo pathway for dTMP production Thymidylate synthetase is
therefore essential in its regulation of the supply of the four nucleotide precursors of
DNA replication, DNA synthesis is also impaired when a deficit of
5 ,10-methylenetetrahydrofolate develops . One of the common consequences of
reduced DNA synthesis is megaloblastosis . There is a salvage pathway using
thymidine kinase that normally accounts for 5 *10% of dTMP synthesis , Therefore ,
activation of this pathway with thymidine supplementation can partially compensate
for diminished dTMP synthesis .
(Choice A) Homocysteine levels are elevated in folate deficiency . Folate
supplementation can reduce the homocysteine levels .
(Choice B) Although cytosine is a pyrimidine base , it is not a dTMP precursor.
Therefore , cytosine supplementation would not be expected to increase available
dTMP levels .
(Choice D) Although cobalamin supplementation can mitigate the impact of folate
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Nent
*
Notes
I alcufdtor
thymidine kinase that normally accounts for 5 -10% of dTMP synthesis . Therefore ,
activation of this pathway with thymidine supplementation can partially compensate
for diminished dTMP synthesis
A
(Choice A) Homocysteine levels are elevated in folate deficiency. Folate
supplementation can reduce the homocysteine levels
14
IB
115
17
(Choice B) Although cytosine is a pyrimidine base , it is not a dTMP precursor
Therefore cytosine supplementation would not be expected to increase available
dTMP levels .
13
(Choice D) Although cobalamm supplementation can mitigate the impact of folate
deficiency on erythropoiesis . it would not be as effective as directly increasing the
20
available levels of dTMP,
21
22
23
24
2B
26
27
23
29
30
31
32
33
34
&
(Choice E) Glutamine is the major source of nitrogen in the synthesis of
nucleotides . As such it contributes a nitrogen atom to the biosynthesis of dUMP,
Later, 5.10-methylenetetrahydrofolate donates one carbon group to dUMP . resulting
in the formation of dTMP In those with severe folate deficiency , however , any
increases in dUMP stimulated by glutamine supplementation would typically not
increase the available dTMP levels as effectively as thymidine supplementation
would.
Educational Objective:
Folate deficiency inhibits the formation of deoxythymidine monophosphate ( dTMP ) ,
which limits DNA synthesis and promotes megaloblastosis and erythroid precursor
cell apoptosis . Because thymidine supplementation can moderately increase dTMP
levels, it can reduce erythroid precursor cell apoptosis .
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Mn t
Lab Values
Notes
t alculator
Some human cells are unable to generate NADPH from glucose metabolism but are
able to synthesize ribose from fmctose-6 -phosphate Which of the following
enzymes is essential for the latter finding?
A . Glucose-6-phosphate dehydrogenase
u
15
15
1?
ia
19
2Q
O B Glutathione reductase
O C. Enolase
O D . Aconrtase
E Transketolase
21
n
21
24
25
25
27
23
29
30
31
32
33
34
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Lab Valuer
Newt
*
r
Notes
l <i k ulator
A
Some human cells are unable to generate NADPH from glucose metabolism but are
able to synthesize ribose from fructose-6 -phosphate Which of the following
enzymes is essential for the latter finding?
b
C A. Glucose- 6-phosphate dehydrogenase [20%]
O B. Glutathione reductase [11%]
O C. Enolase [13%]
u
is
15
13
O D. Aconrtase [5%J
v
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*
E- Transketolase [52%J
20
2\
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21
21
2&
25
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29
Explanation:
Glucose
^bosphate
Glucose '6 - Phosphate Dehydrogenase
(Fate limiting step )
NADPH+H*
OXIDATIVE
( IRREVERSIBLE )
30
E- ph osphogluconate
NATP*
3t
32
33
34
::: : .r.ohogkiconete
Dehydrogenase
Ribulose - 5 - phosphate
Ribuioss - 5-phosphate
NONOXIDATIVE
J — xl : v
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Explanation:
—
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t olc ufdtor
/%
--
I
n
+
'
Qtucose -6 - Phosphate Dehydrogenase
(Rate limiting step )
NADPH*H*
13
OXIDATIVE
u
( IRREVERSIBLE )
15
1
17
18
19
B- phosphogluconate
NflOP*
5- phosphogtuconate Dehydrogenase
UADPH4H'
Ribuiose - 5- phosphate
20
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Lab Valuer
Newt
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Glucose 6 phosphate
8
9
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Pibuiose - 5- phosphate
NONOXIDATIVE
( REVERSIBLE )
--
Rbose - 5 - phosphate
Xylulose 5 phosphate
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SecoheptJ ose - 7-ptiosphato
G fy cera I dehy de - 3- ph osph ate
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l ale ufalor
Ributose - 5-phosphaie
A
6
G
7
8
9
10
Ribuiosfl -5-phosphaie
*
NONOXIDATIVE
( REVERSIBLE )
F&bose - 5- phosphare
Xylul ose - £-phasp hate
u
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IS
17
ia
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20
Transh&oteae
/
/
Giyceraldehyde
SedohepnJ ose - 7-phosphate
phosphate
A
2\
n
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Trunstrldatax
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Erythn.- -.- r -4-phosphate
Fructose - 6- phosph ate
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Glycolysis
NADPH, the reduced form of NADP , is produced mainly when glucose is
metabolized via the hexose monophosphate shunt {HMP shunt , pentose phosphate
pathway ). NADPH, in contrast to NAD is primarily used as reducing equivalent in the
cytosol rather than as a source of energy for ATP synthesis in the electron transport
chain The HMP shunt is also responsible for the production of ribose 5- phosphate
needed for the synthesis of nucleotides From one molecule of glucose the HMP
OI : 10
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u
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NADPH, the reduced form of NADP , is produced mainly when glucose is
metabolized via the hexose monophosphate shunt (HMP shunt , pentose phosphate
pathway ). NADPH, in contrast to NAD; is primarily used as reducing equivalent in the
cytosol rather than as a source of energy for ATP synthesis in the electron transport
chain The HMP shunt is also responsible for the production of ribose 5 -phosphate
needed for the synthesis of nucleotides From one molecule of glucose the HMP
shunt forms a five -carbon sugar , two molecules of NADPH and C02. The HMP
shunt consists of two different types of reactions : oxidative (irreversible ) and
nonoxidative (reversible )
.
All reactions of the HMP shunt occur exclusively in the cytoplasm . The primary
enzymes involved in the non- oxidative steps of the HMP shunt are transaldolase and
transketolase. Transketolase transfers two -carbon groups between substrates of
the HMP shunt and requires thiamine pyrophosphate as a cofactor , and
transaldolase transfers three - carbon groups between substrates of the HMP shunt
All cells can synthesize nbose from the glycolysis intermediates fructose
6 -phosphate and glyceraldehyde 3-phosphate with the help of transketolase and
transaldolase even rf the oxidative reactions of the HMP pathway are not active in
those cells .
(Choices A and B) In the oxidative reactions of the HMP shunt glucose
6 -phosphate is first converted to 6-phosphoglucono!actone , forming one molecule
of NADPH This reaction is catalyzed by glucose 6-phosphate dehydrogenase the
rate-limiting enzyme of the HMP shunt Next 6 -phosphogluconolactone is
hydrolyzed to ribulose 5 phosphate by the enzyme 6-phosphogluconate
rlphyrlrr>npnflgp
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l alculdtor
6 -phosphate and glyceraldehyde 3-phosphate with the help of transketolase and
transaldolase even tf the oxidative reactions of the HMP pathway are not active in
those cells .
u
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(Choices A and B) In the oxidative reactions of the HMP shunt , glucose
6 -phosphate is first converted to 6-phosphogluconolactone forming one molecule
of NADPH This reaction is catalyzed by glucose 6-phosphate dehydrogenase, the
rate -limiting enzyme of the HMP shunt Next . 6-phosphogluconolactone is
hydrolyzed to ribulose 5 -phosphate by the enzyme 6-phosphogluconate
dehydrogenase producing a second molecule of NADPH.
The oxidative portion of the HMP shunt pnmanly occurs in tissues active in reductive
biosynthesis. For example , NADPH is consumed in fatty acid and steroid synthesis
and is also used in the cytochrome p45 G pathway and for the generation of
superoxide in phagocytes Thus , the oxidative reactions of the HMP shunt are active
in the liver, adrenal cortex , gonads , adipose tissue and erythrocytes.
(Choice C) Enoiase catalyzes the conversion of 2 -phosphoglycerate to
phosphoenolpyruvate in glycolysis.
27
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(Choice D) Aconitase catalyzes the isomerization of citrate to isocitrate m the TCA
cycle The TCA cycle is not involved in generation of ribose 5 -phosphate
Educational Objective:
Transketolase and transaldolase carry out the nonoxidative reactions of HMP shunt .
Some cells do not use the oxidative phase reactions to produce cytosolic NADPH
but all cells can synthesize ribose from fructose -6-phosphate using the nonoxidative
reactions .
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t a l l uldtor
A 40-year -old woman comes to the office with a 3-month history of progressive
limitation of physical activity due to fatigue She says . Ml could barely walk from my
car to your office " Past medical history is significant for a positive tuberculin skin
test 7 months ago with a norma! chest radiograph . She has been compliant with the
prescribed treatment despite its bitter taste . Physical examination shows a
tired-appearing woman with conjunctival and palmar pallor . Results of complete
blood count are as follows:
9 g/dU
Hemoglobin
Hematocrit
28%
Mean corpuscular volume
72 fL
Bone marrow aspirate revealed the following representative sample under Prussian
blue stain .
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>»
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> O| *>
/%
7
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1S
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© UWorld
<>-
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34
Decreased activity of which of the following enzymes most likely explains the anemia
found in this patient?
A 6 - aminolevulinate dehydratase
C B . 5 - aminolevulinate synthase
O C, Cystathionine synthase
_
D . Glucose-6-phosphate dehydrogenase
O E . Pyruvate kinase
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© UWorld
f*4
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Decreased activity of which of the following enzymes most likely explains the anemia
found in this patient?
C A . 6 - aminolevulinate dehydratase [18%]
*
B . fi-ammolevulinate synthase [55%]
O C . Cystathionine synthase [7%]
D . Glucose-6-phosphate dehydrogenase [16 %J
E . Pyruvate kinase [3%]
*
.
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Acute intermittent porphyria
&
6
Lab Value
Sent
CYP450 inducers
Barbrluiates
• A/itiepilepUcs
* FlOhl & smoking
7
8
9
n
t
u
1
19
19
*
Succmy!
CoA
ALA synthase
6 Aminolevulirm: arid ( ALA )
+
17
Glycine
13
19
ALA dehyritatase
21
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23
2X
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k
Acute Intermittent
porphyria
20
Abdominal pain
Neurological
*
Porphobilinogen (PBG)
Glucose
*
manifestations
PBG deamtnase
* No photosensitivity
Pon - wine
colored unne
* PBG & ALA in unne
Heme
Hydraxymethyibiiane (HMB)
Perrochelatase
Fe3*
Uroporphyrinogen III
synthase
Protoporphyrin IX
Uroporphyrinogen III
Uroporphynnogt rt
decarboxylase
1
Cop no porphyrinogen III
*
Pnoioporphynnogen IX
V
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t alculdtor
/%
Ft; '
*
Uropcwp/ jynnogen ffl
syntfiase
Protopwphynn IX
Uroporphynrogen III
Uf oporphyt tnogen
decarboxylase
Coproporphynnogen III
1
15
16
*
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n
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** Protoptxphynnogen IX
Coproporph yrt noyen
oxidase
© u*crid .co*n
This patient with latent tuberculosis has laboratory values and a bone marrow
aspirate consistent with sideroblastic anemia due to isoniazid use Sideroblastic
anemia is diagnosed by bone-marrow examination with Prussian blue stain Causes
include X -linked sideroblastic anemia idue to an 5 - aminolevulinate synthase
mutation ) , myelodysplastic syndrome , alcohol abuse , copper deficiency , and certain
medications (eg , isoniazid, chloramphenicol, Ihnezolid ) .
Isoniazid directly inhibits the enzyme pyridoxine phosphokinase , which normally
converts pyridoxine (vitamin B,) to its active form pyridoxai 5' phosphate Pyridoxal
5 ' phosphate is a cofactor for 6-aminolevulinic acid ( ALA) synthase , the enzyme
that catalyzes the rate-limiting step in heme synthesis Inhibition of this enzyme
produces a microcytic, hypochromic anemia Iron is transported to developing
erythrocytes that cannot form heme and its granules accumulate circumferentially
around the nucleus , forming ring sideroblasts .
Because pyridoxal 5' phosphate is a cofactor for numerous enzymes, pyridoxine
deficiency can also lead to dermatitis stomatitis , neuropathy , and
•
-fcilw *.
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t dlculdlor
erythrocytes that cannot form heme , and its granules accumulate circumferentially
around the nucleus , forming ling sideroblasts
A
Because pyridoxal 5‘phosphate is a cofactor for numerous enzymes pyridoxine
deficiency can also lead to dermatitis , stomatitis , neuropathy and
confusion. Therefore , pyridoxine is typically prescribed with isoniazid.
(Choice A) 5- aminolevulinate dehydratase is also involved in heme synthesis Lead
poisoning can cause anemia by directly inhibiting this enzyme
(Choice C } Cystathionine synthase is a pyridoxine -dependent enzyme that
catalyzes the formation of cystathionine from homocysteine . Cystathionine synthase
deficiency results m homocystinuria . an autosomal recessive disorder characterized
by a marfanoid body habitus and hypercoagulability.
[
,
(Choice D) G!ucose-6 -phosphate dehydrogenase ( G6PD ) deficiency results in
increased red blood cell susceptibility to oxidative stress ( eg , infection , medication,
fava beans ), which triggers hemolysis This results in a normocytic normochromic
anemia with increased reticulocyte count and decreased haptoglobin.
(Choice E ) Pyruvate kinase deficiency, an autosoma! recessive disorder , results in
a hemolytic anemia characterized by a normocytic normochromic anemia,
reticulocytosis , and elevated indirect bilirubin .
Educational objective:
Isoniazid inhibits pyridoxine phosphokinase leading to pyridoxine (vitamin B,)
deficiency Pyndoxine s active form is the cofactor for 6 - aminolevulmate synthase,
the enzyme that catalyzes the rate - limiting step of heme synthesis Inhibition of this
step can result in sideroblastic anemia .
'
V
References *
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t dlculdlor
A 34-year-old male who is accidentally exposed to nitrites at work presents to the ER
with anxiety , weakness , dyspnea , and headaches Physical examination reveals
cyanosis that is not corrected by oxygen supplementation. Which of the following is
most likely to be normal in this patient?
k
u
G A Oxygen content of the arterial blood
IB
1S
1?
ia
C B . Oxygen carrying capacity of the arterial blood
H
O D Bound fraction of oxygen in the arterial blood
O E. Oxygen delivery to peripheral tissues
20
21
n
C . Partial pressure of oxygen in the arterial blood
23
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2B
26
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*
t alculdlor
/v
A 34-year -old male who is accidentally exposed to nitrites at work presents to the ER
with anxiety , weakness dyspnea , and headaches Physical examination reveals
cyanosis that is not corrected by oxygen supplementation. Which of the following is
most likely to be norma! in this patient?
u
is
IS
17
13
H
2Q
b
O A Oxygen content of the arterial blood [15%]
B Oxygen carrying capacity of the arterial blood [17%]
* <# C Partial pressure of oxygen in the arterial blood [57%]
D Bound fraction of oxygen in the arterial blood [7%]
O E. Oxygen delivery to peripheral tissues [4%]
.
2\
n
23
2X
2S
26
Explanation:
Drugs
(Nitrites)
27
28
29
30
*
Fe ‘
( Hemoglobin)
31
32
33
34
.
Fe1*
(Met hemoglobin)
t
V
i
Binds hghtly to cyanide
Causes dusky discoloration io skin
Cannot carry oxygen
Iron bound to heme is normally in the reduced ferrous [Fe( ll )J state Oxidation of the
ferrous iron of hemoglobin to ferric iron leads to the formation of methemoglobm
With iron in the oxidized ferric state , methemoglobin is unable to bind oxygen . In
arlrlrtinn
thp
affinrtv nf a ny
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t aleuldtor
ft
increased , causing a leftward shift of the oxygen- dissociation curve . Nitrites actually
cause poisoning by oxidizing the heme iron to the ferric state .
In patients with methemoglobinemia , the partial pressure of oxygen in blood (Choice
C) is normal because the amount of oxygen dissolved in the plasma is the same
4
Methemoglobinemia causes dusky discoloration to the skin ( similar to cyanosis ) and
because methemoglobin is unable to carry oxygen , a state of functional anemia is
induced .
21
27
1 M rk
Iron bound to heme is normally in the reduced ferrous [Fe( ll )] state . Oxidation of the
ferrous iron of hemoglobin to ferric iron leads to the formation of methemoglobin
With iron in the oxidized ferric state , methemoglobin is unable to bind oxygen In
addition, the affinity of any residual ferrous iron in the hemoglobin tetramer is
% saturation
(Oj p&r gm Hb )
PO;
( dissolved oxygen )
n
23
24
25
26
M
CO poisoning
Anemia
,
0 content
(bolti dissolved and O. attached (o Hb )
Decreased
Decreased
Normal
( CO competes with Oj )
riowHb)
Normal
Normal
Decreased
Polycythemia
(high Hb )
Normal
Normal
Increased
30
31
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34
(Choices A . B , D and E ) Methemoglobin cannot bind to oxygen and therefore the
oxygen content and oxygen canying capacity of the arterial blood will decrease The
bound fraction of oxygen and oxygen delivery to the peripheral tissues will also
decrease because of the inability of methemoglobin to carry and deliver O -
^
Educational Objective:
Methemoglobinemia causes dusky discoloration to the skin ( similar to cyanosis) and
^
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*
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t alculdior
/%
&
6
7
s
9
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U
15
15
17
13
14
20
2\
In patients with methemoglobinemia the partial pressure of oxygen in blood ( Choice
C) is normal because the amount of oxygen dissolved in the plasma is the same .
Methemoglobinemia causes dusky discoloration to the skin ( similar to cyanosis ), and
because methemoglobin is unable to carry oxygen , a state of functional anemia is
induced .
Po,
(dissolved OKygen )
CO poisoning
Anemia
0? content
% saturation
(0 / per gm Hb )
.
(boiti dissolved and O attached to Hb >
Decreased
Normal
Decreased
(CO competes wrth O
flowHbl
Normal
Normal
Polycythemia
( high Hbj
Normal
Normal
^J
Qecreased
Increased
22
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23
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34
(Choices A , B, D and E ) Methemoglobin cannot bind to oxygen and therefore the
oxygen content and oxygen carrying capacity of the arterial blood will decrease . The
bound fraction of oxygen and oxygen delivery to the peripheral tissues will also
decrease because of the inability of methemoglobin to carry and deliver O*.
Educational Objective:
Methemoglobinemia causes dusky discoloration to the skin ( similar to cyanosis ) and
because methemoglobin is unable to carry oxygen , a state of functional anemia is
induced . The blood partial pressure of 02 however, will be unchanged in this
condition because oxygen's partial pressure is a measure of 0? dissolved in the
plasma and is not related to hemoglobin function.
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(Notes
I air uftilor
4
6
7
s
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10
u
IS
1v
17
13
20
A 43-year-old man is evaluated for progressive neuropsychiatric symptoms A year
ago, he began feeling depressed and having hallucinations Five months later, he
developed intermittent paresthesias and progressively worsening choreiform
movements , myoclonus , and ataxia . These symptoms have not improved despite
multiple hospitalizations ; an extensive workup has been unrevealing. The patient is a
slaughterhouse worker with extensive exposure to bovine offal As part of the
evaluation for prion disease , a tissue sample digested with protease is processed
via gel electrophoresis and transferred to fitter paper Antibodies to a specific prion
protein are added to the filter . Next a marked protein that combines with the
antibody -protein complex is used to determine whether the test is positive . Which of
the following best descnbes this test'?
21
22
23
2i
25
25
27
23
29
O A Microarray
O B Northern blot
C C Southern blot
D Southwestern blot
C £ . Western blot
30
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*
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I ale uftilor
4
6
7
s
9
10
u
IS
1v
17
13
20
ft
A 43-year-old man is evaluated for progressive neuropsychiatric symptoms A year
ago, he began feeling depressed and having hallucinations Five months later, he
developed intermittent paresthesias and progressively worsening choreiform
movements myoclonus , and ataxia These symptoms have not improved despite
multiple hospitalizations ; an extensive workup has been unrevealing. The patient is a
slaughterhouse worker with extensive exposure to bovine offal As part of the
evaluation for prion disease , a tissue sample digested with protease is processed
via gel electrophoresis and transferred to filter paper Antibodies to a specific prion
protein are added to the filter . Next a marked protein that combines with the
antibody -protein complex is used to determine whether the test is positive . Which of
the following best descnbes this test'?
u
,
21
O A . Microarray [6%]
O B . Northern blot [3%]
O C Southern blot [5%]
22
23
2i
25
25
27
23
29
30
31
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33
3
*
*
D . Southwestern blot [6%J
® E. Western blot [80%]
Explanation:
Western blotting is used to detect a target polypeptide or protein from within a
mixed sample . Potential target proteins are separated by gel electrophoresis . The
separated proteins are then transferred to a nitrocellulose membrane and probed
with a primary antibody specific for the protein of interest . The membrane is then
washed and treated with a ( secondary ) marked antibody that binds to the primary
antjKf%Hv / anH r
sin
KojHoforfqH / so
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Ne «t
Ldb Value
*
Notes
(
akufator
4
A
Explanation:
b
7
8
9
n
1
is
1S
17
13
*
20
2\
22
23
24
2S
26
27
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30
31
32
33
34
Western blotting is used to detect a target polypeptide or protein from within a
mixed sample . Potential target proteins are separated by gel electrophoresis . The
separated proteins are then transferred to a nitrocellulose membrane and probed
with a primary antibody specific for the protein of interest . The membrane is then
washed and treated with a ( secondary ) marked antibody that binds to the primary
antibody and can be detected (eg . by colorimetry }.
For example , a serum sample from a patient with suspected HIV infection can be
analyzed via Western blot to detect antibodies directed against specific viral
proteins . Following separation of viral proteins by gel electrophoresis and protein
transfer to a nitrocellulose membrane the membrane is treated with the patient's
serum. Patients who are HIV positive are likely to have antibodies that react with viral
p 24. gp 4l and gp120 16 Q . If 2 of these 3 bands are positive the test is considered
positive .
'
-
Western blotting is similar to the enzyme linked immunosorbent assay (ELISA )
technique ; however, in ELISA the patient' s serum is tested directly , whereas in
Western blotting the proteins are first separated by electrophoresis.
(Choice B) Northern blots analyze mRNA A sample containing a large number of
mRNA molecules is separated via gel electrophoresis Separated bands are then
transferred to a membrane and hybridized with a probe containing a nucleotide
sequence complementary to the mRNA of interest .
(Choice C) Southern blotting is used to analyze DNA sequences. DNA that is
fragmented using restnction endonucleases is separated by gel electrophoresis and
transferred to a nitrocellulose membrane A radiolabeled DNA probe containing a
seauence complementary to an area of interest is then used for hybridization
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(
ulcul -i t o r
Western blotting the proteins are first separated by electrophoresis,
A
(Choice B) Northern blots analyze mRNA A sample containing a large number of
mRNA molecules is separated via gel electrophoresis Separated bands are then
transferred to a membrane and hybridized with a probe containing a nucleotide
sequence complementary to the mRNA of interest .
i
is
115
1?
*
n
H
(Choice CJ Southern blotting is used to analyze DNA sequences DNA that is
fragmented using restriction endonucleases is separated by gel electrophoresis and
transferred to a nitrocellulose membrane A radiolabeled DNA probe containing a
sequence complementary to an area of interest is then used for hybridization.
Restriction site mutations can be detected by Southern blotting because they alter
DNA fragment lengths, thereby altering electrophoresis migration patterns .
20
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?7
23
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Microarray analysis is similar to Southern and Northern blotting but involves
hybridization of a large number of probes at once (Choice A), The genomic DNA or
cDNA being analyzed is labeled with a fluorescent tag and placed on a gene chip
containing complementary sequences for a large number of genes The degree of
fluorescence corresponds to the mRNA expressed in the particular sample
(Choice D) Southwestern blotting is a technique that analyzes DNA - binding proteins
using principles of the Southern and Western blot techniques. DNA -binding proteins
are recognized by their ability to bind specific oligonucleotide probes .
Educational objective:
Western blotting is used to identify proteins , Northern blotting identifies specific RNA
sequences and Southern blotting identifies specific DNA sequences in an unknown
sample .
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1B
H
20
21
n
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?T
23
29
A 54 year-old man with a history of chronic kidney disease due to hypertension
develops anemia Evaluation reveals that the cause of anemia is erythropoietin
deficiency . Erythropoietin increases the numbers of erythroid precursor cells in the
bone marrow and induces heme production in erythrocyte precursors , in this patient
mature erythrocytes are found that are unable to synthesize heme even though they
contain detectable levels of cytoplasmic enzymes involved in heme synthesis . Lack
of which of the following cellular organelles best explains this phenomenon?
.
A. Endoplasmic reticulum
O B . Golgi apparatus
O C Mitochondria
O D . Nucleus
O E . Peroxisomes
.
F . Proteasomes
30
31
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Notes
l tilculalor
4
A
&
7
8
9
10
14
IS
115
17
1B
H
20
21
22
21
24
2&
25
27
23
29
-
A 54 year-old man with a history of chronic kidney disease due to hypertension
develops anemia Evaluation reveals that the cause of anemia is erythropoietin
deficiency . Erythropoietin increases the numbers of erythroid precursor cells in the
bone marrow and induces heme production in erythrocyte precursors Sn this patient
mature erythrocytes are found that are unable to synthesize heme even though they
contain detectable levels of cytoplasmic enzymes involved in heme synthesis . Lack
of which of the following cellular organelles best explains this phenomenon?
A. Endoplasmic reticulum [15%1
O B Golgi apparatus [6%]
* * C Mitochondria [63%]
O D Nucleus [10%]
O E. Peroxisomes [4%]
C ' F, Proteasomes [2%]
.
Explanation:
Heme synthesis
30
31
32
33
34
Mitochontfrta
Cytoplasm
Succinyl
ALA syntftate
6- Aminolevuhnic actd ( ALA )
*
iV
* *
Ai A dehydratase
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Ment
Notes
l okuFdtor
/%
Explanation:
&
Heme synthesis
7
a
Mitochondria
Cytopliwn
10
Suconyl
CoA
ALA synthase
13
H
is
16
17
13
6'Aminolevulinic acid ( ALA ) *
+
0
Porphotutinogen {PBGl
H
Heme
20
21
Glycine
i\
ALA dehydratase
Glucose
PBG deaminase
22
2\
2
2&
26
*
Ferroch&atase
Hydroxymet hylbtlane (HMB )
Fe
'
Urofxxphynnogen ill
27
synthase
23
29
Protoporphyrin IX
i
Uroporphyrinogen III
30
31
32
33
34
Umporphynnogert
decarboxylase
Coproporphyrjnogen III
* Protoporphynnogen
IX
CoptopOfphywyoQen
oxidase
© UWortd
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\JI ly 111 i\/y CM i A i
*
Notes
l alculator
CopfOpotphyrinogGn
Oxidase
OUWorid
n
is
IS
17
ia
19
2Q
J
Heme synthesis occurs partly in the mitochondria and partly in the cytoplasm of
erythrocytes. Mitochondria are necessary for the first and the final 3 steps
Erythrocyte precursors divide a number of times before finally losing their nuclei and
mitochondria and forming mature red blood cells that survive for about 120 days (4
months) When erythrocytes lose their mitochondria , they lose the ability to generate
heme and therefore hemoglobin .
u
synthesized in virtually every organ but the principal sites of synthesis are
erythrocyte precursor cells (located in the bone marrow ) and hepatocytes (use heme
in microsomal cytochrome P450 system ).
Heme
is
21
n
23
2A
26
26
27
28
29
30
31
32
33
3
*
(Choices A , B, and D) Although mature erythrocytes do not contain a nucleus , a
Golgi apparatus , or an endoplasmic reticulum, their cytoplasm still contains residual
amounts of the enzymes necessary for heme synthesis . Therefore the lack of
mitochondria ( and their associated heme biosynthetic enzymes ) is a better
explanation for the lack of heme synthesis .
(Choices E and F ) Proteasomes are involved in protein recycling and peroxisomes
are involved in fatty acid catabolism . These organelles typically disappear during
erythrocyte development .
Educational objective:
Maturing erythrocytes lose their ability to synthesize heme when they lose their
mitochondria which are necessary for the first and final 3 steps of heme synthesis
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*
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*
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alcufalor
A
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9
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14
IS
15
1?
13
19
20
21
n
23
24
25
25
A 65 *year -old woman with chronic obstructive pulmonary disease and type II
diabetes mellitus comes to the emergency department due to profound fevers and
malaise . After initial evaluation , she is hospitalized for septicemia . Blood cultures
plated on lactose - containing media grow rapidly dividing gram-negative bacteria
Bidirectional DNA replication in these microbial cells requires synthesis of 2
daughter strands of DNA , each using one of the parent strands as a template
Which of the following processes will differ the most between the 2 daughter strands
during their synthesis?
t
O A. Enzymatic function of DNA helicase
B Interaction with single - stranded DNA-binding proteins
O C . Joining of DNA fragments by ligase
O D . Proofreading of the newly synthesized DNA
O E. Relief of supercoils by topoisomerase
27
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34
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I tilculdtor
A
&
6
8
9
in
u
19
19
n
H
20
21
n
23
24
2S
25
27
28
29
/%
A 65- year-old woman with chronic obstructive pulmonary disease and type II
diabetes mellitus comes to the emergency department due to profound fevers and
malaise . After initial evaluation , she is hospitalized for septicemia . Blood cultures
plated on lactose- containing media grow rapidly dividing gram-negative bacteria
Bidirectional DNA replication in these microbial cells requires synthesis of 2
daughter strands of DNAr each using one of the parent strands as a template
Which of the following processes will differ the most between the 2 daughter strands
during their synthesis?
i
O A . Enzymatic function of DNA helicase [2%]
C B . Interaction with single- stranded DNA-binding proteins |10%]
* # C . Joining of DNA fragments by ligase [62%]
C D . Proofreading of the newly synthesized DNA [21%]
O E . Relief of supercoils by topoisomerase [4 %]
Explanation:
DNA replication fork
30
31
32
33
34
3'
3'
DNA
helicase
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lUlQ- T
5'
Leading strand
DNA polymerase
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I alculaior
A
5
6
A
Explanation:
E
9
to
DNA replication fork
3'
S'
u
19
19
1?
DNA
helicase
n
JF
-
Leading strand
DNA polymerase
Jf
19
Movement of replication fork
20
21
22
%
23
24
2S
25
2/
28
29
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31
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34
RNA primer
l agging strand
Primase
© UWorld
Single stranded
DNA binding protein
Oka / aki fragments
DNA replication is similar in prokaryotes and eukaryotes , with DNA polymerases l
and III being the main polymerase enzymes involved in prokaryotic DNA
replication . For DNA replication to begin DNA helicase must first unwind the DNA
double helix and separate the parent strands (Choice A ) . The unwound singlestranded DNA is stabilized by the binding of single - stranded DNA -binding proteins to
prevent spontaneous reanneahng (Choice B).
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prevent spontaneous reannealing tunoice
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(Notes
t dlculdtor
BJ .
A
9
6
S
9
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u
19
19
17
n
2Q
Synthesis of the daughter strands occurs simultaneously from both parent
3' direction one
strands Because DNA synthesis can occur only in the 5’-‘
daughter strand is synthesized continuously toward the replication fork (leading
strand). However , the other strand must be synthesized discontinuously in a
direction away from the replication fork (lagging strand ) , with more and more
segments being added as the replication fork moves across the DNA double
helix , This results in the formation of Okazaki fragments short stretches of newly
synthesized DNA that are separated by RNA primers . These primers are removed
and replaced with DNA , and the Okazaki fragments are subsequently joined together
by DNA ligase. Because of the discontinuous nature of DNA synthesis on the
lagging strand, DNA ligase acts many more times on the lagging strand than on the
leading strand.
,
b
21
n
21
29
26
27
23
29
30
31
32
33
34
(Choice D) DNA polymerases I and III have proofreading ability (ie , 3’— * 5 '
exonuclease activity ) and the proofreading function of these polymerases is not
restricted to either the leading or lagging strand .
(Choice E) Topoisomerase II produces negative supercoiling in the DNA helix
ahead of the replication fork to reduce the strain produced by unwinding , which
causes positive supercoiling .
Educational objective:
DNA replication occurs in the 5' — ^ 3' direction on both strands. In contrast to the
continuous synthesis of the leading strand, lagging strand synthesis occurs
discontinuous and is composed of short stretches of RNA primer plus newly
synthesized DNA segments (Okazaki fragments). As a result , lagging strand
synthesis requires the repetitive action of DNA primase and DNA ligase
^
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Red blood cells from a 24-year-old African American male who suffers from periodic
hemolysis demonstrate a low activity of glucose-6-phosphate dehydrogenase
Deficiency of which of the following erythrocyte enzymes has the same
pathophysiology as this patient’s condition?
u
IB
16
17
13
H
20
21
Notes
I alt uldtor
b
O A Pyruvate kinase
O B . Glutathione reductase
C . Bisphosphoglycerate mutase
O D . Hexokinase
C E. Transketolase
n
23
24
2B
26
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Notes
t dkuldtor
4
&
G
7
8
9
to
14
15
IS
17
18
H
20
/%
Red blood cells from a 24- year-old African American male who suffers from periodic
hemolysis demonstrate a low activity of glucose-6- phosphate dehydrogenase
Deficiency of which of the following erythrocyte enzymes has the same
pathophysiology as this patient’ s condition?
O A. Pyruvate kinase [24 %]
v ® 8- Glutathione reductase [60%]
C Bisphosphoglycerate mutase [3%J
O D . Hexokinase [5%]
O E. Transketolase [8%]
21
n
23
24
25
26
27
28
29
30
31
32
33
34
Explanation:
Red blood cells do not have mitochondria or a nucleus ; therefore, metabolism of
glucose in these cells occurs via glycolysis and the hexose monophosphate ( HMP )
shunt Glycolysis provides energy for erythrocyte survival; whereas , the HMP shunt
provides the reducing agent NADPH to prevent oxidant damage In the initial
oxidative portion of the HMP shunt , glucose 6-phosphate is converted to 6phosphogluconolactone and one molecule of NADPH is formed . This reaction is
catalyzed by glucose 6-phosphate dehydrogenase the rate limiting enzyme of the
HMP shunt . In the second reaction of the oxidative portion of the HMP shunt , 6phosphogiuconolactone is hydrolyzed to ribulose 5-phosphate by the enzyme 6phosphogluconate dehydrogenase producing a second molecule of NADPH .
,
NADP'
4
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Glutathkine
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Hydrogen
Peroxide
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17
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21
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Ldb Vdlurs
Newt
Red blood cells do not have mitochondria or a nucleus ; therefore, metabolism of
glucose in these cells occurs via glycolysis and the hexose monophosphate ( HMP )
shunt . Glycolysis provides energy for erythrocyte survival; whereas , the HMP shunt
provides the reducing agent NADPH to prevent oxidant damage In the initial
oxidative portion of the HMP shunt , glucose 6-phosphate is converted to 6phosphogluconolactone and one molecule of NADPH is formed This reaction is
catalyzed by glucose 6-phosphate dehydrogenase , the rate limiting enzyme of the
HMP shunt . In the second reaction of the oxidative portion of the HMP shunt 6phosphogluconolactone is hydrolyzed to ribulose 5 -phosphate by the enzyme 6 phosphogiuconate dehydrogenase producing a second molecule of NADPH.
Reduced
Glutathione
NADP*
Glutathione
NADPH * H
27
IPJAOPH purwratvd KI ir,e i M*
30
Previous
Notes
t dlculdtor
Explanation:
23
24
2&
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31
32
33
34
o
3
*
n
23
29
-
M rt
reductase
A
‘
-
&
Hydrogen
\
f
Peroxide
Glutathione
peroxidase
Oxidized
Glutathione
*
¥
Water
tfunl hdpfc 10 ke p (jlk/dtftlona m
a reduced ilaral
*
In erythrocytes hydrogen peroxide produced by partial reduction of molecular
oxygen is detoxified by glutathione peroxidase Glutathione is oxidized during this
reaction. The regeneration of reduced glutathione is carried out by the enzyme
glutathione reductase using NADPH as an electron donor NADPH in red blood cells
is produced solely by the HMP shunt , and this is how the HMP shunt contributes to
protecting red blood cells from oxidative stress Defective generation of NADPH
due to defects in the oxidative portion of the HMP shunt will increase the
susceptibility of the RBCs to oxidative damage , and glutathione reductase deficiency
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(Notes
*
I alculdtor
*
A
is produced solely by the HMP shunt , and this is how the HMP shunt contributes to
protecting red blood cells from oxidative stress . Defective generation of NADPH
due to defects in the oxidative portion of the HMP shunt will increase the
susceptibility of the RBCs to oxidative damage , and glutathione reductase deficiency
will lead to similar clinical picture ( choice B ) .
(Choices A , C and D) Enzymatic defects in the glycolytic pathway lead to poor RBC
survival because the decreased availability of ATP as an energy source The RBC
membrane becomes stiffened and deformed, and these RBCs are prematurely
cleared from the circulation by the spleen or by intravascular hemolysis . The
mechanism of hemolysis due to defective RBC glycolysis is different from
hemolysis resulting from defects in the HMP shunt in that hemolysis due to glycolytic
defects causes a chronic hemolytic anemia while hemolysis due to HMP shunt
defects causes an episodic hemolytic anemia Pyruvate kinase deficiency
constitutes more than 95% of cases of hemolytic anemia due to defective RBC
glycolysis,
b
(Choice E) Transketolase and transaldolase catalyze the non- oxidatrve reactions of
the HMP shunt and are primarily designed to generate ribose 5 -phosphate from
fructose 6 -phosphate and glyceraldehyde 3 phosphate or the reverse NADPH is
not generated in the non-oxidative portion of the HMP shunt .
—
Educational Objective;
Glucose 6-phosphate dehydrogenase deficiency is a defect in the HMP shunt that
impairs glutathione reduction due to failure to produce NADPH, Glutathione
reductase deficiency causes a similar clinical picture and is pathophysiologically
similar to G6PD deficiency.
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t akutdtor
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&
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10
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IS
115
17
13
H
20
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24
25
26
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28
29
A 2-day-old boy is being examined in the newborn nursery prior to discharge from
the hospital . He was born at 38 weeks gestation by vaginal delivery . The pregnancy
and delivery were uncomplicated , and the boy has been breastfeeding, stooling and
urinating normally . The patient's mother has beta - thaiassemia trait , and his father has
a normal hemoglobin electrophoresis . Vital signs and physical examination are
normal Which of the following hemoglobin compositions is most likety predominant
in this infant?
C A . o 2(32
O B . a 2y 2
C a 262
O D p4
O E. y 4
C f £2t 2
C G . £2y2
30
31
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Notes
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4
&
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7
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10
u
IS
18
17
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25
26
27
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30
A
A 2-day-old boy is being examined in the newborn nursery prior to discharge from
the hospital He was bom at 38 weeks gestation by vaginal delivery . The pregnancy
and delivery were uncomplicated , and the boy has been breastfeeding , stooiing and
urinating normally . The patient's mother has beta -thalassemia trait , and his father has
a normal hemoglobin electrophoresis . Vital signs and physical examination are
normal Which of the following hemoglobin compositions is most likely predominant
in this infant?
u
C A 02p2 [17% J
* <*
B o 2 V 2 [75% ]
O C o 262 [6%]
O D . p 4 [0%I
O E . y 4 [1%]
O F. £ t 2 [0%I
O G . (2 y2 [0%]
,
Explanation:
31
32
33
34
Hemoglobin
Type
Name
Components
Gowt? r 1
Qt?
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*
t alculdtor
Notes
4
&
6
7
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Explanation:
a
Hemoglobin
n
Name
Type
13
U
is
is
17
Components
b
Embryonic
13
Gower 1
C2c2
Portland
<2V2
Gower 2
2E 2
P
2y2
A
2fJ2
A2*
262
H
20
21
22
Fetal
2\
24
25
25
77
28
29
Adult
30
31
32
33
34
a thaiassemia intermedia
H
p4
a - thalassemia major
Barts
y4
*
.
* Piedorntri«ifrt kn p Bia *i,_ smsa.
© UWuHd
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Hemoglobin is a tetramer that consists of 2 pairs of globin chains total of 4 chains
In V
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*
Notes
t alculator
4
&
6
7
s
10
H
IS
13
17
13
19
20
21
n
23
24
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29
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31
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A
Hemoglobin is a tetramer that consists of 2 pairs of globin chains { total of 4 chains
per molecule ) . During the first few weeks of embryogenesis hemoglobin is
synthesized by the yolk sac and contains zeta or epsilon globin chains (Choices C
and F ) Thereafter , one pair of the globin charns should always be alpha and the
other should be non-alpha.
Fetal hemoglobin (Hb F) production begins around 8 weeks gestation and replaces
all embryonic hemoglobin by 14 weeks gestation , when erythropoiesis in the fetal
liver and spleen is established. Hb F consists of 2 alpha and 2 gamma protein
subunits (a 2y2 ). Production declines at birth , and Hb F comprises 60- 80% of all
hemoglobin in a term newborn. Hb F is gradually replaced by adult hemoglobin (Hb
A a2(32) ( Choice A) during the first 6 months of life after which Hb A composes the
vast majority of adult hemoglobin.
-
Compared to red blood cells with Hb A those with Hb F have a high oxygen
affinity as Hb F binds to 2.3 *bisphosphoglycerate poorly . The greater affinity of
Hb F facilitates transplacental oxygen delivery from the maternal circulation to that of
the fetus
(Choice C) Hemoglobin A2 ( a262 ) is a normal hemoglobin variant that makes up 2%
-3% of hemoglobin in a healthy adult and is functionally similar to Hb A Patients with
beta-thalassemia major have impaired beta globin production , resulting in an excess
of alpha globin chains (eg, Hb A2 Hb F ) and no Hb A.
.
(Choices D and E) Alpha -thalassemia results from a shortage of alpha globin
chains . Hemoglobin H ((34 ) and hemoglobin Barts ( y4 ) have a very high oxygen
affinity and cannot release oxygen , resulting in tissue hypoxia Hemoglobin H
disease manifests as chronic hemolytic anemia Hemoglobin Barts is incompatible
with life ( eg , hydrops fetaJis ) as normal fetal and adult hemoglobin cannot be
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lo k u f i i t o r
Hb F facilitates transplacental oxygen delivery from the maternal circulation to that of
the fetus
(C hoice C) Hemoglobin A2 ( a 262 ) is a normal hemoglobin variant that makes up 2%
-3% of hemoglobin in a healthy adult and is functionally similar to Hb A, Patients with
beta-thalassemia major have impaired beta globm production resulting in an excess
of alpha globin chains (eg Hb A2 Hb F) and no Hb A .
P
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(Choices D and E) Alpha -thalassemia results from a shortage of alpha globin
chains . Hemoglobin H ( £4 ) and hemoglobin Barts ( y 4 ) have a very high oxygen
affinity and cannot release oxygen, resulting in tissue hypoxia. Hemoglobin H
disease manifests as chronic hemolytic anemia Hemoglobin Barts is incompatible
with life (eg , hydrops fetalis } as normal fetal and adult hemoglobin cannot be
produced .
Educational objective:
Hemoglobin F (Hb F ) is the predominant hemoglobin type in the second and third
tnmesters of pregnancy and dunng the first few months after birth . Hb F consists of
2 alpha and 2 gamma protein subunits ( o 2 y2 } and has a high affinity for oxygen
which facilitates oxygen transport across the placenta to the fetus Hb A ia2(J2 } is
the major hemoglobin in adults .
,
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References:
1 . Human embryonic, fetal , and adult hemoglobins have different
subunit interface strengths, correlation with lifespan in the red cell
2 . Advances in the understanding of haemoglobin switching.
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A 21- year-old laboratory worker experiences rapid-onset breathing difficulty,
palpitations , and flushed skin . He has no significant past medical history and takes
only loratadine for seasonal allergies . The patient is suspected to have accidental
poisoning . Amyl nitrite from a laboratory safety kit is immediately administered via
inhalation Amyl nitrite affects the affinity of hemoglobin for which of the following9
O A . Carbon dioxide
B . 2.3 - biphosphoglycerate
13
C C . Carbon monoxide
20
O D . Cyanide
O E. Iron
O R Lead
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A
A 21-year-old laboratory worker experiences rapid-onset breathing difficulty,
palpitations , and flushed skin . He has no significant past medical history and takes
oniy loratadine for seasonal allergies . The patient is suspected to have accidental
poisoning. Amyi mtnte from a laboratory safety kit is immediately administered via
inhalation. Amyl nrtrite affects the affinity of hemoglobin for which of the following9
O A . Carbon dioxide [2%]
O B . 2 ,3 -biphosphoglycerate [5%]
O C Carbon monoxide [13%]
* D. Cyanide [75%]
•
O E, Iron [3%]
O F. Lead [2%J
Explanation:
Cyanide binds to a variety of iron-containing enzymes , the most important of which is
the cytochrome a - a3 complex . This complex is critical for electron transport dunng
oxidative phosphorylation By binding to this molecule , minute amounts of cyanide
can inhibit aerobic metabolism and rapidly result in death .
The typical clinical syndrome present in cyanide poisoning is rapidly -developing
cutaneous flushing , tachypnea , headache , and tachycardia , often accompanied by
nauseaVvomiting confusion, and weakness Respiratory distress and cardiac
dysfunction may follow. Laboratory studies indicate severe lactic acidosis in
conjunction with a lessened difference between arterial and venous O , content (i. e
tho
uonr>ijc
hlnoH ic elilLhinhKf
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NPK t
Notes
*
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A
Explanation:
Cyanide binds to a variety of iron-containing enzymes the most important of which is
the cytochrome a - a 3 complex . This complex is critical for electron transport during
oxidative phosphorylation. By binding to this molecule minute amounts of cyanide
can inhibit aerobic metabolism and rapidly result in death .
b
The typical clinical syndrome present in cyanide poisoning is rapidly -deveioping
cutaneous flushing , tachypnea headache and tachycardia , often accompanied by
nausea/ vomiting , confusion, and weakness. Respiratory distress and cardiac
dysfunction may follow Laboratory studies indicate severe lactic acidosis in
conjunction with a lessened difference between artena! and venous O, content (i. e
the venous blood is still highly oxygenated )
.
Drugs
(Norites )
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MI
Fe?*
( Hemoglobin)
.
FeJ*
(Met hemoglobin)
v
* tightly to cyanide
Binds
Causes dusky discoloration to skin
Cannot carry oxygen
The antidotal effect of nitrites for cyanide poisoning has been recognized since the
late nineteenth century Nitrites are oxidizers and act primarily in cyanide poisoning
,
by inducing the formation of methemoglobin . This occurs when ferrous iron in
hemoglobin is oxidized to ferric iron . Methemoglobin cannot carry oxygen , but it
does have a high affinity for cyanide . Methemoglobin can bind and sequester
cyanide in the blood, thereby keeping the poison away from mitochondrial (and
other ) enzymes where cyanide exerts its toxic effects Sodium thiosulfate is also
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A
The antidotal effect of nitrites for cyanide poisoning has been recognized since the
late nineteenth century . Nitrites are oxidizers , and act primarily in cyanide poisoning
by inducing the formation of methemoglobin This occurs when ferrous iron in
hemoglobin is oxidized to feme iron. Methemoglobin cannot cany oxygen, but it
does have a high affinity for cyanide Methemoglobin can bind and sequester
cyanide in the blood, thereby keeping the poison away from mitochondrial (and
other ) enzymes where cyanide exerts its toxic effects Sodium thiosulfate is also
used for cyanide poisoning ; it combines with cyanide to form the less -toxic
thiocyanate , which is excreted in the urine .
H
(Choices A, B and C ) Methemoglobin . which is formed when amyl nitrite is
administered, does not have a high affinity for molecules such as carbon monoxide ,
20
carbon dioxide , and 2,3 bisphosphoglycerate
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(Choice E) The affinity of hemoglobin for iron is not affected by nitrite administration,
though mtntes do oxidize the heme iron to its Fe( HI ) state .
(Choice F) Lead poisoning causes defective heme synthesis . Lead poisoning is
treated first and foremost by the avoidance of lead ingestion If acutely ingested,
chelation therapy , such as with dimercaprol or CaNaEDTA , should be initialed
Educational Objective;
Nitrites are oxidizing agents that are effective in the treatment of cyanide poisoning
due to their ability to cause methemoglobinemia , f / ethemoglobin contains ferric
rather than ferrous iron Cyanide binds to ferric iron more avidly than to mitochondrial
cytochrome enzymes, which saves these mitochondrial enzymes from cyanide's
toxic effect.
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The following vignette applies to the next 2 items The items m the set must be
answered in sequential order Once you click Proceed to Next Item , you will not be
able to add or change an answer .,
The genetic evaluation of a family with hereditary anemia reveals a point mutation in
the (J-globin gene , which results in the replacement of guanine (G) by cytosine ( C ) in
the {S-globin mRNA molecule three bases upstream from the AUG codon (position 6),
Item 1 of 2
Which of the following is most likely impaired in these patients?
O A . mRNA binding to ribosomes
C B . Translocation during translation
O C. Peptide bond formation
C D . Termination of polypeptide synthesis
O £ . Protein targeting
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A
The following vignette applies to the next 2 items . The items in the set must be
answered in sequential order Once you click Proceed to Next Item , you will not be
able to add or change an answer .
The genetic evaluation of a family with hereditary anemia reveals a point mutation in
the p-globin gene , which results in the replacement of guanine (G) by cytosine ( C ) in
the (S-globin mRNA molecule three bases upstream from the AUG codon (position 6),
Item 1 of 2
Which of the following is most likely impaired in these patients?
1B
H
2Q
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*
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*
A . mRNA binding to ribosomes [46%]
C 8 . Translocation during translation [11%]
O C Peptide bond formation [11% ]
D . Termination of polypeptide synthesis [13%]
O E. Protein targeting [18%]
**
Explanation:
The Kozak consensus sequence occurs on eukaryotic mRNA and is defined by the
following sequence : ( gcc )gccRccAUGG . where R is either adenine or
guanine . When the methionine codon (AUG } is positioned near the beginning of a
mRNA molecule and is surrounded by the Kozak sequence , it serves as the initiator
for translation (i e mRNA binding to ribosomes ) Among other factors a purine (G
or A ) positioned three bases upstream from the AUG appears to be a key factor in
this initiation process . A mutation in which guanine (G) is replaced by cytosine ( C ) in
Block Time Remaining :
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t alculdtor
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ft
&
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Explanation:
7
The Kozak consensus sequence occurs on eukaryotic mRNA and is defined by the
following sequence : ( gcc )gccRccAUGG , where R is either adenine or
guanine . When the methionine codon ( AUG ) is positioned near the beginning of a
mRNA molecule and is surrounded by the Kozak sequence , it serves as the initiator
for translation (i. e mRNA binding to ribosomes ) Among other factors , a purine (G
or A ) positioned three bases upstream from the AUG appears to be a key factor in
this initiation process . A mutation in which guanine ( G ) is replaced by cytosine ( C ) in
this particular position of the (3 -globtn gene has been associated with thalassemia
intermedia.
a
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(Choice B) Translocation is catalyzed by the elongation factor eEF2 and requires
GTP hydrolysis
:
(Choice C) Peptide bond formation is catalyzed by peptidyl transferase on
eukaryotic ribosomes. A defect in the mRNA coding for ribosomes would be
required to interfere with this function .
(Choice D) Termination of polypeptide synthesis occurs at stop codons. A mutation
producing a premature stop codon would have this effect .
30
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(Choice E) Protein targeting is achieved by the amino acid sequence of the Nterminal section of a formed protein. This sequence is often removed from the final
protein product once it reaches its destination .
Educational objective:
The Kozak sequence plays a role in the initiation of translation A mutation three
bases upstream from the start codon ( AUG) in this sequence is associated with
thalassemia intermedia.
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Item 2 of 2
The mutation results in a 30% decrease in the efficiency of protein synthesis.
Homozygous patients with the mutation are most likely to have which of the following ?
A Red blood cell sickling
n
u
O B . Spherocytosis
IS
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O C Microcytosis
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1B
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O D . Hyperchromia
O E. Iron deficiency
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Item 2 of 2
The mutation results in a 30% decrease in the efficiency of protein synthesis .
Homozygous patients with the mutation are most likely to have which of the following ?
( A. Red blood cell sickling [20%]
u
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17
13
H
O B. Spherocytosis [13%]
* # C. Microcytosis [57%]
O D. Hyperchromia [4%]
O E. Iron deficiency [5%I
20
21
Explanation:
23
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29
This particular genetic defect is a known cause of thalassemia intermedia , which is a
form of beta-thalassemia that is clinically less severe than beta -thalassemia
major. Hypochromic, microcytic anemia is the classic laboratory finding in patients
with thalassemia Red blood cell morphology is quite vanable depending on the type
of thalassemia and can include marked anisopoikilocytosis target cell formation
tear drop cells, and/ or Heinz bodies.
n
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32
33
34
(Choice A) Red blood cell sickling occurs in sickle cell anemia , which is caused by
an A to T mutation in the p-giobin gene that results in glutamate being substituted by
valine at position 6.
(Choice B) Spherocytosis is caused by a variety of molecular defects in the genes
that encode for certain red blood cell membrane proteins such as spectrin , ankyrin
band 3 protein, protein 4.1, and others .
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A
Explanation:
This particular genetic defect is a known cause of thalassemia intermedia , which is a
form of beta-thalassemia that is clinically less severe than beta -thalassemia
major. Hypochromic , microcytic anemia is the classic laboratory finding in patients
with thalassemia Red blood cell morphology is quite variable depending on the type
of thalassemia and can include marked anisopoikilocytosis, target cell formation
tear drop cells, and/ or Heinz bodies.
(Choice A) Red blood cell sickling occurs in sickle cell anemia which is caused by
an A to T mutation in the (3-globin gene that results in glutamate being substituted by
vatine at positron 6.
b
(Choice B) Spherocytosis is caused by a variety of molecular defects in the genes
that encode for certain red blood cell membrane proteins such as spectrin ankyrin
band 3 protein , protein 4.1 , and others .
,
(Choice D) Hypochromia, not hyperchromia, is a feature of thalassemia .
(Choice E ] Iron deficiency anemia is most commonly caused by chronic,
asymptomatic bleeding such as a slow Gl bleed or abnormal menstruation
Educational objective:
A mutation in the Kozak sequence of the beta - globin gene is associated with
thalassemia intermedia, which results in hypochromic, microcytic anemia.
References:
1 . Beta+ 45 G -> C: a novel silent beta -thalassaemia mutation, the first
in the Kozak sequence.
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alculdlor
r%
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H
A 6- year-old African American boy is brought to the physician because of easy
fatigability . Physical examination reveals splenomegaly and his complete blood
count shows mild anemia . Hemoglobin electrophoresis is performed at alkaline pH
on a cellulose acetate strip Findings for the patient are shown below compared to
individuals with normal hemoglobin and known sickle cell disease .
Starting point of
electrophoresis
J
20
21
n
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2X
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Normal
Individual
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I
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Sickle cell
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Patient
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Normal
Individual
A
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Individual
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u
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ir
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Patient
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Anode
Cathode
© UWorld
Which of the following is the most likely cause of this patient ' s condition'?
C A . Frameshift mutation
C B . Missense mutation
O C . Nonsense mutation
O D . Silent mutation
E . Trinucleotide expansion
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/%
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Sickle cell
Individual
Patient
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Anode
Cathode
© UWofid
Which of the following is the most likely cause of this patient ' s condition'?
C A . Frameshtft mutation [18%]
^
• B . Missense mutation [50%]
C C . Nonsense mutation [22%]
O D . Silent mutation [1%]
E . Trinucleotide expansion [8%]
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t dlculdior
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G
Hemoglobin electrophoresis
7
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Starling point of
electrophoresis
Hemoglobins are negatively charged
& move towards the anode
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Hemoglobin A
b
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Hemoglobin C
Anode
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Hemoglobin S
V
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Lalculdtor
A
Hemoglobin electrophoresis is used to analyze the different forms of hemoglobin in
patients with suspected hemoglobinopathy. Normal hemoglobin consists primarily of
hemoglobin A (HbA } which migrates rapidly toward the positive electrode (anode )
because of its negative charge Hemoglobin S (HbS ) is an abnormal type of
hemoglobin in which a nonpolar amino acid (vaiine ) replaces a negatively charged
amino acid ( glutamate ) in the beta globin chain . This amino acid replacement
decreases the negative charge on the HbS molecule , which causes HbS to move
more slowly toward the anode . Similarly, hemoglobin C (HbC) has a glutamate
residue replaced by lysine in the beta globin chain. Because lysine is a positively
charged amino acid HbC has even less total negative charge than HbS and moves
even more slowly toward the anode. Both HbC and HbS result from missense
mutations , a type of mutation in which a single base substitution results in a codon
that codes for a different ammo acid
f
b
Hemoglobinopathy electrophoresis patterns
Starting point of
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NormJ I
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Sickle cell
Trail
Sickle -cetl
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© uwono
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(Choice C) Nonsense mutations introduce a stop codon within gene sequences,
resulting in the formation of truncated proteins As a result of their decreased size,
these proteins tend to move further during electrophoresis .
31
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34
Note
1 4iI c u ! At or
*
A
Patients with sickle cell disease have HbS mutations in both beta chains those with
HbC disease have HbC mutations involving both beta chains Patients with
hemoglobin SC disease have 1 HbS allele and 1 HbC allele and will have 2
hemoglobin bands on electrophoresis This patient ' s electrophoresis results show a
single band that migrates less than the HbA and HbS bands, meaning that he has
HbC disease. Patients with HbC disease are typically asymptomatic and often have
mild hemolytic anemia and splenomegaly.
*
21
2
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(Choice A) Frameshift mutahons occur with the deletion or insertion of base pairs
that are not a multiple of 3 . They alter the reading frame of the genetic code
resulting in the formation of nonfunctional proteins , Frameshift mutations involving
the alpha globin genes can cause alpha thalassemia , which results in the production
of beta tetramers {hemoglobin H ) that migrate further than HbA during
electrophoresis.
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(Choice D ) Silent mutations are point mutations that have no effect on the protein
formed , A mutation from UCA to UCC does not result in any change in protein
structure as both codons result in the placement of serine into the growing
polypeptide chain .
(Choice E) Trinucleotide expansions increase the number of trinucleotide repeats
within a gene , resulting in large , unstable proteins or alterations in the epigenetic
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t aiculator
(Choice A ] f- rameshitt mutations occur with the deletion or insertion of base pairs
that are not a multiple of 3 . They alter the reading frame of the genetic code
resulting in the formation of nonfunctional proteins . Frameshift mutations involving
the alpha globin genes can cause alpha thalassemia , which results in the production
of beta tetramers (hemoglobin H ) that migrate further than HbA during
electrophoresis.
A
(Choice C) Nonsense mutations introduce a stop codon within gene sequences,
resulting in the formation of truncated proteins As a result of their decreased size
these proteins tend to move further during electrophoresis .
IS
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2\
(Choice D) Silent mutations are point mutations that have no effect on the protein
formed . A mutation from UCA to UCC does not result in any change in protein
structure as both codons result in the placement of serine into the growing
polypeptide chain .
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(Choice E) Trinucleotide expansions increase the number of trinucleotide repeats
within a gene , resulting in large , unstable proteins or alterations in the epigenetic
effects on a particular gene These proteins tend to move less dunng
electrophoresis due to their increased size. However , none of the
hemoglobinopathies are characterized by trinucleotide expansions
30
Educational objective:
Hemoglobin C is caused by a missense mutation that results in a glutamate residue
being substituted by lysine in the beta globin chain . This results in an overall
decrease in negative charge for the hemoglobin molecule The speed of
hemoglobin movement dunng gel electrophoresis is hemoglobin A > hemoglobin
S > hemoglobin C .
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A 6-year -old African American male is brought to your office for a routine check -up .
His mother remarks that he often seems uninterested in playing with his peers and
appears to "run out of breath quickly, ” His medical records reveal that he has
missed several pediatric vaccinations and has been hospitalized twice , once with a
“ chest infection” and once with abdominal pain , The patient mentions to you that
occasionally his "bones hurt /' Which of the following protein changes most likely
accounts for this patient's condition?
is
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O A Phenylalanine deletion
B . Valine substitution for glutamic acid
C C . Phenylalanine substitution for proline
O D. Valine substitution for lysine
< E . Early termination of polypeptide synthesis
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Notes
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A
-
A 6 year -old African American male is brought to your office for a routine check -up.
His mother remarks that he often seems uninterested in playing with his peers and
appears to "run out of breath quickly," His medical records reveal that he has
missed several pediatric vaccinations and has been hospitalized twice , once with a
"chest infection" and once with abdominal
pain , The patient mentions to you that
occasionally his "bones hurt," Which of the following protein changes most likely
accounts for this patient's condition?
IB
1G
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Lr
A, Phenylalanine deletion [3%]
* ® B, Valine substitution for glutamic acid [81%]
C Phenylalanine substitution for proline [3%J
O D , Valine substitution for lysine [10%]
E, Early termination of polypeptide synthesis [2%]
,
Explanation:
This patient is exhibiting signs and symptoms of sickle cell anemia Sickle cell
anemia is a hemoglobinopathy that typically affects patients of African ancestry A
point mutation in the 6 codon of the beta -globin gene, which causes the substitution
of valine (hydrophobic } for glutamic acid (hydrophilic ) , is responsible . The
incorporation of this abnormal beta -globin protein into hemoglobin results in the
formation of hemoglobin S (HbS ). HbS polymerizes at low oxygen tension causing
sickling and hemolysis of erythrocytes and resultant vascular occlusion This
patient' s poor exercise tolerance and exertional dyspnea are due to anemia His
history of acute chest syndrome abdominal pain , and bone pain are due to vaso HNCHJQ auQnfc
in
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(
akul -i t o r
A
D Valine substitution for lysine [10%]
E . Early termination of polypeptide synthesis [2%]
Explanation:
&
This patient is exhibiting signs and symptoms of sickle cell anemia Sickle cell
anemia is a hemoglobinopathy that typically affects patients of African ancestry A
point mutation m the 6: codon of the beta -globin gene, which causes the substitution
of valine (hydrophobic ) for glutamic acid (hydrophilic ) , is responsible The
incorporation of this abnormal beta - globin protein into hemoglobin results in the
formation of hemoglobin S (HbS ). HbS polymerizes at low oxygen tension causing
sickling and hemolysis of erythrocytes and resultant vascular occlusion This
patient' s poor exercise tolerance and exertional dyspnea are due to anemia His
history of acute chest syndrome abdominal pain , and bone pain are due to vaso occlusive events in the lungs, spleen and bone , respectively .
r
(Choice A ) A phenylalanine deletion ( AF508 ) is the most common cause of cystic
fibrosis, the most common fatal genetic disease of Caucasians.
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(Choice E } Early termination of polypeptide synthesis (nonsense mutation ) will
produce a truncated protein .
Educational Objective;
Exertional dyspnea , pneumonia resulting in life -threatening acute chest syndrome
and recurrent abdominal and bone pain are clinical features of sickle cell
anemia . Sickle cell anemia results from a point mutation that causes valine to
substitute for glutamic acid in the sixth position of the b-globin chain of hemoglobin.
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A 67-year -old man comes to the physician complaining of pallor , early satiety , and
severe fatigue . He has also lost 20 pounds (9.07 kg } over the past 6 months
Physical examination reveals hepatomegaly and massive splenomegaly A cytosolic
protein recovered from his white blood cells is found to have constitutive tyrosine
phosphorylation actrvity . Consequently there is persistent activation of STAT ( signal
transducers and activator of transcription ) proteins The patient is most likely
suffering from which of the following disorders?
.
A . Acute lymphocytic leukemia
O B . Acute promyelocytic leukemia
C C Chronic lymphocytic leukemia
D High-grade non-Hodgkin's lymphoma
O E . Myelofibrosis
2b
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ft
A 67-year-old man comes to the physician complaining of pallor , early satiety , and
severe fatigue . He has also lost 20 pounds (9.07 kg) over the past 6 months
Physical examination reveals hepatomegaly and massive splenomegaly. A cytosolic
protein recovered from his white blood cells is found to have constitutive tyrosine
phosphorylation actrvity . Consequently there is persistent activation of STAT ( signal
transducers and activator of transcription ) proteins The patient is most likely
suffering from which of the following disorders?
C A . Acute lymphocytic leukemia [8%]
B . Acute promyelocytic leukemia [15%]
C Chronic lymphocytic leukemia [36%]
D . High-grade non-Hodgkin’s lymphoma [1Q%]
v # E. Myelofibrosis [30%J
Explanation:
Chronic myeloproliferative disorders
30
31
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34
Diagnostic features
Disorder
Chronic
myelogenous
leukemia
Constitutional symptoms (eg, fatigue, weight loss ,
sweating ] , splenomegaly &
leukocytosis with marked left shift ( eg,
myelocytes, metamyelocytes, band forms )
excessive
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Mutation
Philadelphia
chromosome
t( 9 22 ) BCft - ABL
fusion protein
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Explanation:
8
9
10
Chronic myeloproliferative disorders
Diagnostic features
Disorder
Mutation
H
15
17
1 j
Chronic
myelogenous
H
leukemia
Constitutional symptoms (eg fatigue, weight loss,
^
excessive sweating l, splenomegaly &
leukocytosis with marked left shift ( eg
myelocytes, metamyelocytes, band Forms)
.
20
Philadelphia
chromosome
T( ?:22 } eCft 'ABL
fusion protein
2%
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Hemorrhagic & thrombotic symptoms ( eg, easy
bruising, microangiopathic occlusion),
Essential
thrombocytosis
thrombocytosis & megakaryocytic hyperplasia
Pruritus, erythromeialgia, splenomegaly,
thrombotic complications erythrocytosis &
thrombocytosis
Polycythemia
vera
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JAK2
Severe fatigue, splenomegaly ( often causing early
satiety /abdominal discomfort ), hepatomegaly,
anemia & bone marrow fibrosis
Primary
myelofibrosis
© UWorld
The chronic myeloproliferative disorders are a group of bone marrow diseases
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A
The chronic myeloproliferative disorders are a group of bone marrow diseases
characterized by the overproduction of myeloid cells. Primary myelofibrosis is
caused by atypical megakaryocytic hyperplasia . The clonally expanded
megakaryocytes activate fibroblast proliferation , resulting in progressive replacement
of the marrow space by extensive collagen deposition . In the early stages there is
marrow hypercellularity with minimal fibrosis . As the disease progress , pancytopenia
can result . Hepatomegaly and massive splenomegaly occur in myelofibrosis
because the loss of bone marrow hematopoiesis is compensated for by
extramedullary hematopoiesis . The peripheral smear characteristically shows
teardrop - shaped red blood cells ( daerocytes ) and nucleated red blood cells
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With the exception of chronic myelogenous leukemia , the chronic myeloproliferative
disorders { especially polycythemia vera ) frequently harbor a mutation in the
cytoplasmic tyrosine kinase Janus kinase 2 ( JAK2 ). This mutation ( V617F )
substitutes a bulky phenylalanine for a conserved valine at position 617, resulting in
constitutive tyrosine phosphorylation activity , and consequently, cytokineindependent activation of the JAK-STAT pathway , A JAK 2 inhibitor (mxolrtinib ) has
been approved for the treatment of primary myelofibrosis.
(Choice A) Acute lymphocytic leukemia predominantly affects children. Clinical
manifestations are nonspecific and include fever, fatigue , pallor, petechiae and
bleeding Leukemic spread can cause lymphadenopathy, hepatosplenomegaly . and
bone pain .
>
(Choice B) In acute promyelocytic leukemia the translocation t(15 ; 17 leads to the
formation of a fusion gene between the promyelocytic leukemia (PML) and the
retinoic acid receptor alpha (RARa ) genes This abnormal PML/RARo fusion protein
blocks the differentiation of myeloid precursors .
v
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bleeding l_ eukemic spread can cause lymphadenopathy hepatosplenomegaiy and
bone pain .
Notes
I alcufdtor
A
(Choice B) In acute promyelocytic leukemia the translocation t(l5;l7 ) leads to the
,
formation of a fusion gene between the promyelocytic leukemia (PML ) and the
retinoic acid receptor alpha ( RARo ) genes This abnormal PMLRARa fusion protein
blocks the differentiation of myeloid precursors .
(Choice C) Chronic lymphocytic Jeukemia is a lymphoproliferative disorder involving
B-lymphocytes . The most significant laboratory finding is marked lymphocytosis
with smudge cells ' seen on peripheral blood smear . The majority of cases exhibit
increased expression of the proto- oncogene BCL-2 , similar to follicular lymphomas.
"
'
(Choice D) Several high-grade non-Hodgkin lymphomas are associated with
cytogenetic abnormalities The t( 8.14 ) translocation is the most common in Burkitt
lymphoma and involves the c-myc oncogene . Burkitt lymphoma is associated with
Epstein- Barr virus infection and classically has a "starry sky" histologic appearance .
Educational objective:
The chronic myeloproliferative disorders (polycythemia vera , essential
thrombocytosis and primary myelofibrosis) often have a mutation (V617 F ) in the
cytoplasmic tyrosine kinase Janus kinase 2 ( JAK2 ). This results in constitutive
tyrosine kinase activity , and consequently , cytokine-independent activation of STAT
transcription factors .
References:
1. Myeloproliferative disorders.
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A
A group of investigators is researching the changes in oxygen-hemoglobin binding
that occur under various clinical conditions . They are especially interested in
situations that alter the shape and position of the oxygen-hemoglobin dissociation
curve . Which of the following processes would most likely cause a shift from the
blue curve to the red curve in the graph below?
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100% -
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1
100
pO, ( mm Hg)
A . Chronic high- altitude adaptation
O B . Hypothermia
O C . Hypoventilation
O D . Severe anemia
E . Strenuous exercise
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pOj ( mm Hg)
OL/Wl t Wortdj 111
O A . Chronic high-altitude adaptation [19%]
v # B . Hypothermia [64 % J
O C . Hypoventilation I5 %]
C D . Severe anemia [5 %]
E . Strenuous exercise [7% ]
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l alculalor
Explanation:
Oxygen hemoglobin dissociation curve
Left shift caused by:
+
1 Decreased H (increased pH )
.
2. Decreased 2, 3 DPG
A
«
3 - Decreased temperature
is
Think LUNGS = Left shift
13
H
100% -
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O
Right shift caused by:
1 . Increased H‘( decreased pH)
2, Increased 2> 3 DPG
l , Increased temperature
<0
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OU^MUMorid. LUT
po, { mm Hg )
The oxygen-hemoglobin dissociation curve describes the relationship between the
partial pressure of oxygen ( x -axis) and the hemoglobin oxygen saturation ( y - axis ) .
Oxygen saturation increases in a sigmoidal fashion as the pO increases because of
the increase in oxygen-binding affinity that occurs after the first oxygen molecule
binds to hemoglobin As more oxygen molecules bind to hemoglobin , the number of
available binding sites decreases and the curve eventually flattens out .
The partial pressure of oxygen in the blood at which hemoglobin is 50% saturated is
known as the PH ( dotted black line in diagram above ): this value is a standard
measure of hemoglobin's affinity for oxygen and is about 26 mm Hg in normal
individuals A leftward shift of the oxygen-hemoglobin dissociation curve occurs
when hemoglobin has increased affinity for oxygen (ie , a lower PK ). Because
decreased temperatures help to stabilize the bonds between oxygen and
hemoglobin hypothermia increases hemoglobin’s oxygen affinity and shifts the
dissociation curve to the left
27
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k
-
(Choice A) 2 , 3-diphosphoglycerate ( 2 , 3 DPG ) is an organophosphate created in
erythrocytes during glycolysis. The production of 2.3 -DPG is increased when
oxygen availability is reduced as occurs in chronic lung disease heart failure , and
chronic exposure to high altitudes Elevated levels of 2, 3- QPG decrease
hemoglobin O . affinity allowing the release of more 0._ in the peripheral tissues
(Choices C and D) Anemia severe enough to cause lactic acidosis wilt result in
lower blood pH, shifting the hemoglobin curve to the right . Similarly , hypoventilation
causes increased CO retention and respiratory acidosis that shifts the curve to the
right.
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measure of hemoglobin s affinity for oxygen and is about 26 mm Hg in normal
individuals. A leftward shift of the oxygen-hemoglobin dissocialion curve occurs
when hemoglobin has increased affinity for oxygen (ie, a lower P ), Because
^
decreased temperatures help to stabilize the bonds between oxygen and
hemoglobin hypothermia increases hemoglobin’s oxygen affinity and shifts the
dissociation curve to the left.
(Choice A) 2 , 3-diphosphoglycerate ( 2 , 3 -DPG ) is an organophosphate created in
erythrocytes during glycolysis. The production of 2,3 -DPG is increased when
oxygen availability is reduced as occurs in chronic lung disease heart failure , and
chronic exposure to high altitudes , Elevated levels of 2, 3-DPG decrease
hemoglobin O affinity allowing the release of more O in the peripheral tissues
u
(Choices C and D } Anemia severe enough to cause lactic acidosis will result in
lower blood pHf shifting the hemoglobin curve to the right Similarly , hypoventilation
causes increased CO retention and respiratory acidosis that shifts the curve to the
right,
(Choice E) Strenuous exercise will cause increased tissue oxidative
phosphorylation, increased tissue CO. levels , and decreased tissue pH. This results
in a shift of the dissociation curve to the right and decreased hemoglobin O affinity
Educational objective:
A left shift of the hemoglobin oxygen dissociation curve indicates increased
hemoglobin O. affinity and can be caused by increased pH. decreased
2 ,3 -diphosphoglycerate and decreased temperature A leftward shift of the
oxygen-dissociation curve means that O . is relatively less available to tissues
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21
Some cells have an alternative glycolytic pathway that produces no net ATP These
cells sometimes divert glycolytic intermediates into a reaction catalyzed by
bisphosphoglycerate mutase rather than using those intermediates to produce
energy . Which of the following cells are most likely to utilize this alternative pathway ?
,
A Hepatocytes
,
O B . Erythrocytes
O C Skeletal muscle cells
O D Adipocytes
C E . Neurons
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I dlculalor
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a
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15
IS
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20
A
Some cells have an alternative glycolytic pathway that produces no net ATP These
cells sometimes divert glycolytic intermediates into a reaction catalyzed by
bisphosphoglycerate mutase rather than using those intermediates to produce
energy . Which of the following cells are most likely to utilize this alternative pathway ?
^
•
(
A . Hepatocytes [13%|
B. Erythrocytes [64%]
C Skeletal muscle cells [6%]
O 0. Adipocytes [13%J
O E. Neurons [4%]
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Explanation:
Glucose
i
Glucose-6Phosphate
:
Fructose Phospnate
^
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Fructose 1,6Bisphosphdls
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g
Glucose-6Phosphate
20
21
22
23
24
25
2S
27
23
29
30
31
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Lab value
Ne t
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Notes
*
I alculdior
Glucose
9
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Explanation:
a
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Fructose-6Phosp^ate
i
Fruclose 1.6Bisphosphate
:
Glycer aJdehydo- 3P osphate
^
I
1.3 Bisphosoftoglyce 'ate
Phas.pttogtycerate ktnase
Mulasv
2,3 Bisphosphoglycerate
ATP
3-Phosphogtycerate
4
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Notes
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7
8
9
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GlyceraldehytJe- 3Pnosphate
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13
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1.3 Bisphosphoglycefaie
Phospfrogtyc&rate ktnase
Mttiase
2,3 Bisphosphoglycerate
ATP
20
21
S- Phosphoglyoerate
22
4
JPhosphatase
l
23
24
25
26
2 Pho&phogtycerate
I
27
28
29
PhosphoerwJpyruvate
30
31
32
33
34
Pyruvate
Glycolysis can occur in aerobic or anaerobic conditions . In aerobic environments
the NADH created dunng the conversion of glyceraldehyde 3- phosphate to
1 3-bisphosphoglycerate is regenerated to NAD through oxidation within the
mitochondrial electron transport chain The amount of NAD in cells is limited :
09 : 08
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23
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30
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Npxt
—
*
Note
*
t ale ultftlor
A
Glycolysis can occur in aerobic or anaerobic conditions . In aerobic environments
the NADH created during the conversion of glyceraidehyde 3-phosphate to
1.3-bisphosphoglycerate is regenerated to NAD through oxidation within the
mitochondrial electron transport chain . The amount of NAD in cells is limited :
therefore regeneration of NAD from NADH is essential. In anaerobic conditions
( and in erythrocytes under aerobic conditions ) , pyruvate cannot be oxidatively
decarboxylated to acetyl CoA . Instead pyruvate is converted to lactate by the
enzyme lactate dehydrogenase . The conversion of pyruvate to lactate also serves
to re-oxidize NADH to NAD in the absence of oxygen.
Erythrocytes are unique cells because they do not have mitochondria and cannot
generate energy from the citric acid cycle . Glycolysis is the major pathway used by
RBCs to produce energy . 2 , 3-bisphosphoglycerate ( BPG ) is generated as a
byproduct of glycolysis from 1, 3- BPG by the enzyme bisphosphoglycerate mutase
(producing no ATP ). It is catabolized to 3-phosphoglycerate by
bisphosphoglycerate phosphatase ( also producing no ATP). During normal
glycolysis, 1 , 3 BPG is converted to 3-phosphoglycerate by the enzyme
phosphoglycerate kinase which does produce ATP in the process , By generating
2.3 - BPG rather than proceeding with regular glycolysis , RBCs sacrifice the net ATP
-
gain
achieved in normal glycolysis .
The major function of RBCs is to carry hemoglobin-bound oxygen from the lungs to
the peripheral tissues , and 2 , 3-BPG is a very important regulator of oxygen-binding
to hemoglobin Increased 2.3-BPG concentrations within erythrocytes enable
increased oxygen delivery in the peripheral tissues in the presence of lower blood
oxygen concentrations because 2,3 - BPG allostencally decreases the affinity of
hemoglobin for oxygen The conversion of 1, 3-BPG to 2 , 3-BPG is increased in
hypoxia and chronic anemia.
Block Time Remaining :
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23
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I I I C U I I4 J
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Notes
(
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A
achieved in normal glycolysis.
The major function of RBCs is to carry hemoglobin-bound oxygen from the lungs to
the peripheral tissues and 2 , 3-BPG is a very important regulator of oxygen-binding
to hemoglobin. Increased 2 , 3- BPG concentrations within erythrocytes enable
increased oxygen delivery in the peripheral tissues in the presence of lower blood
oxygen concentrations because 2, 3-BPG allostencally decreases the affinity of
hemoglobin for oxygen The conversion of 1, 3 BPG to 2 , 3-BPG is increased in
hypoxia and chronic anemia
-
.
22
23
24
2&
26
I
i
Ldb Value
Newt
(producing no ATP ) . It is catabolized to 3-phosphoglycerate by
bisphosphoglycerate phosphatase ( also producing no ATP), During normal
glycolysis, 1.3-BPG is converted to 3-phosphoglycerate by the enzyme
phosphogiycerate kinase , which does produce ATP in the process . By generating
2 , 3 - BPG rather than proceeding with regular glycolysis , RBCs sacrifice the net ATP
gain
i
is
16
<
Hem: 17 of 34
o
(Choices A , C, D, and E) The enzyme bisphosphoglycerate mutase is present in
large amounts in RBCs, while in the other cel! types mentioned above, it is present in
insignificant quantities . Therefore the production of 2,3 -BPG in hepatocytes ,
adipocytes , myocytes , and neurons is virtually negligible.
Educational Objective;
Increased 2.3-BPG concentrations within erythrocytes enable increased oxygen
delivery in the peripheral tissues in the presence of lower blood oxygen
concentration because 2 , 3 -BPG decreases the affinity of hemoglobin for oxygen.
2 , 3 - BPG is produced from 1,3 -BPG by the enzyme bisphosphoglycerate mutase
This reaction consumes the energy that would have been otherwise used by the
erythrocyte to produce energy in the form of ATP .
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09
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(
ulrufdtor
4
&
6
7
a
9
n
H
IB
15
17
19
20
21
22
21
24
2&
26
27
A 52-year-old woman comes to the physician complaining of a rapidly enlarging neck
mass and persistent nighttime sweating After the appropriate workup she is
diagnosed with diffuse large B-cell lymphoma and admitted to the hospital She is
started on combination chemotherapy without incident . On the third day of treatment
she is noted to have decreased unne output . Laboratory analysis shows increased
levels of blood urea nitrogen and creatinine . Electrocardiography shows peaked T
waves. Administration of which of the following agents would have most likely
prevented this patients renal impairment?
L
A. Denosumab
O B . Folinic acid
C N-acetylcysteme
C D . Prednisone
O E. Probenecid
F . Rasburicase
28
29
30
31
32
33
34
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Lab Value
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(Notes
I iilcufdtDr
4
A
5
6
7
8
9
10
u
15
115
A 52-year -old woman comes to the physician complaining of a rapidly enlarging neck
mass and persistent nighttime sweating . After the appropriate workup she is
diagnosed with diffuse large B-ceil lymphoma and admitted to the hospital. She is
started on combination chemotherapy without incident . On the third day of treatment,
she is noted to have decreased urine output . Laboratory analysis shows increased
levels of blood urea nitrogen and creatinine . Electrocardiography shows peaked T
waves. Administration of which of the following agents would have most likely
prevented this patient's renal impairment?
17
19
20
21
n
23
24
25
26
27
23
29
30
31
32
33
34
O A. Denosumab [5%]
G B. Folinic acid [16%]
O C . N-acetylcysteine [16%]
C D Prednisone [8%]
O E, Probenecid [32%]
* ft F. Rasburicase [24%]
,
Explanation:
Purine catabolism
I
Hypoxanthine
Xanthine oxidase
r
Block Time Remaining :
09 : 3S
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(Notes
I alculalor
ft
Explanation:
e
7
Purine catabolism
8
9
10
L
b
Hypoxanthme
Xanthine oxidase
U
15
is
17
Allopunnol
Xanthine
Xanthine oxtdase
19
20
21
r
Uric add
(normal endpoint of purine
metabolism in humans )
22
23
24
25
26
27
23
29
Urate oxidase
( RasbuncaseJ
30
31
32
33
34
i
Allantoid
(excreted in urine
CUWtrtt
Tumor lysis syndrome is an oncologic emergency that can develop during
chemotherapy for high-grade lymphomas , leukemias and other cancers with rapid
cell turnover substantial tumor burden or high sensitivity to chemotherapy When
Iflma
pi
imhorc
p\
f li i m n r rollc aro HoctrnvoH Hurin/i s» chnrf
Block Time Remaining :
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Note
*
I iilculdtor
euworti
A
Tumor lysis syndrome is an oncologic emergency that can develop durmg
chemotherapy for high-grade lymphomas leukemias , and other cancers with rapid
cell turnover substantial tumor burden, or high sensitivity to chemotherapy . When
large numbers of tumor cells are destroyed during a short interval, intracellular ions
such as potassium and phosphorous are released into the serum along with nucleic
acids { which are catabolized into uric acid ) Elevated potassium levels can result in
arrhythmias that may be fatal Uric acid can precipitate in the kidney causing
obstructive uropathy and acute renal failure.
&
The nsk of tumor lysis syndrome can be reduced by aggressive fluid hydration.
Allopunnol ( a xanthine oxidase inhibitor ) or rasburicase can also be used to reduce
uric acid levels during the breakdown of tumor cells Rasburicase is a recombinant
version of urate oxidase an enzyme present in many mammals but not in humans
It catalyzes the conversion of uric acid to allantoin, which is 5-10 times more soluble
than uric acid. Rasburicase is effective in preventing and treating hyperuricemia and
the resulting renal manifestations of tumor lysis syndrome.
(Choice A) Denosumab is a humanized monoclonal antibody that prevents
osteoclast activation by binding to RANKL and preventing its interaction with RANK
{ similar to the function of osteoprotegerin ) . Osteoclast inhibiting agents decrease
the risk of bone loss and fracture in patients with bone metastasis .
(Choice B) Folinic acid (leucovorin ) is used to reverse the toxicity of methotrexate
on gastrointestinal mucosa and bone marrow,
(Choice C) N- acetylcysteine works in acetaminophen overdose by enhancing
glutathione production and conjugation of the toxic NAPQI metabolite
N- acetylcysteine is also used as a mucolytic agent in patients with influenza .
V
ai
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Sent
(Notes
*
I alculator
[ similar to the function of osteoprotegerin ) . Osteoclast inhibiting agents decrease
A
the risk of bone loss and fracture in patients with bone metastasis .
(Choice B) Foiinic acid (leucovorin ) is used to reverse the toxicity of methotrexate
on gastrointestinal mucosa and bone marrow ,
(Choice C) N- acetylcysteine works in acetaminophen overdose by enhancing
glutathione production and conjugation of the toxic NAPQI metabolite
N-acetylcystelne is also used as a mucolytic agent in patients with influenza ,
bronchitis , and cystic fibrosis and for the prevention of radiocontrast - induced
nephropathy in patients with renal insufficiency.
b
(Choice D) Prednisone is part of the CHOP therapy used in the treatment of
non-Hodgkin lymphoma , including diffuse large B -cell lymphoma . It induces
lymphocyte apoptosis , contributing to tumor lysis syndrome .
(Choice E) Probenecid is a uricosuric agent a substance that increases the
excretion of uric acid in the urine . Uncosuric agents are used for treating gout and
hyperuricemia and are effective only in patients with good renal function . They
should not be used in patients at nsk for nephrolithiasis or uric acid nephropathy
,
Educational objective:
Tumor lysis syndrome can develop during chemotherapy for cancers with rapid cell
turnover ( eg , poorly differentiated lymphomas and leukemias ), substantial tumor
burden , or high sensitivity to chemotherapy It is characterized by
hyperphosphatemia hypocalcemia , hyperkalemia and hyperuricemia Prevention of
tumor lysis syndrome often involves hydration and the use of hypouricemic agents
such as allopurinol or rasburicase .
,
,
References:
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Note
*
Ldlculdtor
4
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&
6
A 72-year -old Caucasian female presents to your office complaining of difficulty in
swallowing. She also complains of severe fatigue and progressive exertional
dyspnea Physical examination reveals pale conjuctiva . A photograph of her finger
JS shown below .
7
a
9
to
&
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19
115
17
n
20
21
n
21
2
29
26
*
27
28
29
30
31
32
33
34
Which of the following is the best treatment for this patient?
A Iron preparations
C B . Vitamin B12
V
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(N otes
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6
7
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9
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19
19
17
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20
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23
Which of the following is the best treatment for this patient?
29
30
A. Iron preparations
O B Vitamin B12
O C . Pyridoxine
O D. Vitamin C
O E. Folic acid
C F. Erythropoietin
O G. Filgrastim
.
31
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34
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Motes
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77
23
29
30
31
32
33
3
*
Which of the following is the best treatment for this patient?
* ® A. Iron preparations [73%]
O B . Vitamin B12 [8%1
O C . Pyridoxine [5%]
O D. Vitamin C [5%]
O E Folic acid [3%]
,
O F. Erythropoietin [4 %|
G . Filgrastim [2%l
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Notes
t alcufator
G Filgrastim [2%]
/%
6
7
a
9
ID
1
*
IB
IS
17
ia
20
21
n
23
24
2&
25
27
28
29
30
31
32
33
34
Explanation:
Man / clinical features that occur due to decreased hemoglobin are common for all
types of anemia These features include malaise increased fatigability pallor,
decreased exercise capacity and congestive heart failure. The symptoms of
difficulty in swallowing ( dysphagia ) and disfigured fingernails ( spoon nails or
koilonychia ), as shown in the above photograph , are more specific for iron deficiency
anemia . Dysphagia in a patient with iron deficiency anemia is often caused by the
formation of esophageal webs and is described as Plummer -Vinson or Patterson
Kelly syndrome .
b
Red cells are microcytic and hypochromic in iron deficiency . Iron deficiency anemia
is typically treated with an oral iron preparation. Vitamin C is sometimes added to
improve oral iron absorption,
(Choices B and E) Deficiency of vitamin B12 or folic acid causes megaloblastic
anemia.
(Choice C) Pyridoxine is required as a cofactor for the first step in heme synthesis .
Therefore , pyridoxine deficiency causes decreased heme synthesis and microcytic ,
hypochromic pyridoxine -responsive anemia (ie sideroblastic anemia) anemia,
(Choice F) Erythropoietin is commonly used to treat anemia associated with chronic
renal failure
Educational Objective:
The symptoms of difficulty in swallowing ( dysphagia ) and disfigured fingernails
( spoon nails or koilonychia ) are specific for iron deficiency anemia ,
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Notes
(
alcufator
Scientists studying the principles behind oxygen-hemoglobin dissociation have
discovered a way to successfully separate hemoglobin tetramers into individual
alpha and beta subunits . During an expenment , a solution is created that contains
only monomeric beta-hemoglobin subunits under physiologic conditions If
measured the oxygen dissociation curve of the dissolved beta subunits will most
likely resemble which of the following lines?
,
Li
1
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*
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19
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25
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23
29
-
Oxygen partial pressure (mm Hg)
Ol/' ViUttwIdALC
30
31
32
33
34
O A . Line A
O B . Line B
O C Line C
C D . LineD
O E . Line E
I
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9
6
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16
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21
22
23
24
29
25
27
23
29
Oxygen partial pressure (mm Kg)
Oi^witwtofUAK
30
31
32
33
34
* @> A. Line A [53%]
O B . Line B [9%]
O C . Line C [8%]
O D . Line D [9% )
O E . Line E [20% ]
I
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Explanation:
7
8
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ta
Oxygen- dissociation curves of hemoglobin & myoglobin
b
i
15
1G
17
*
100%
Myoglobin
-
Ij
Hemoglobin
19
• 20
21
o
22
23
24
25
26
27
23
29
r
*
o
50%
The P„of hemoglobin is 26 mm Hg
while the Pw of myoglobin is
1 mmHg This indicates that
myoglobin has a much higher affinity
for oxygen than hemoglobin.
30
31
32
33
34
T
25
CuSMUWarMLlUC
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50
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75
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100
pOi ( mm Hg )
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*
Notes
I dkuldtor
ft
pOj ( mm Hg )
OlftWLlWwULlC
Hemeproteins such as hemoglobin and myoglobin contain heme groups that are
used to reversibly bind oxygen for transportation and storage Hemoglobin A (the
major form of hemoglobin in adults ) is a tetramer consisting of 2 alpha and 2 beta
chains Each hemoglobin subunit is associated with a heme moiety so each
hemoglobin molecule has 4 heme groups . After binding to 1 oxygen molecule , the
oxygen affinity of other heme molecules increases ; this heme -heme interaction is
responsible for the characteristic sigmoid shape of the oxygen-hemoglobin
dissociation curve .
In contrast to hemoglobin, myoglobin is a monomeric protein and the primary oxygenstoring protein in skeletal and cardiac muscle tissue ; it is only found in the
bloodstream after muscle injury. The partial pressure of oxygen at which 50% of
myoglobin molecules are oxygen saturated ( Pw ) is only 1 mm Hg. which is much
lower than the PK of hemoglobin ( 26 mm Hg). Myoglobin also has only a single
heme group and so does not experience heme-heme interactions ; therefore , its
oxygen-dissociation curve is hyperbolic.
The secondary and tertiary structures of myoglobin and the hemoglobin beta subunit
are almost identical (the o - subunits are also very similar to myoglobin ) . Because
individual hemoglobin subunits are structurally similar to myoglobin, their oxygenbinding behavior is also similar . That is , if a hemoglobin molecule is dissociated the
individual subunits will have a hyperbolic oxygen- dissociation curve (Choice A )
(Choice B) Line B is left- shifted compared to normal hemoglobin A, which could
indicate a high oxygen affinity hemoglobin ( eg fetal hemoglobin ). Hemoglobin A will
also undergo a left shift in the presence of factors such as increased pH, decreased
temperature , and decreased 2,3-diphosphoglycerate .
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Lab Valuer
IWWII H
!
Notes
l alculaior
“
A
individual hemoglobin subunits are structurally similar to myoglobin their oxygenbinding behavior is also similar . That is , if a hemoglobin molecule is dissociated the
individual subunits will have a hyperbolic oxygen-dissociation curve (Choice A )
(Choice B) Line B is left- shifted compared to normal hemoglobin A which could
indicate a high oxygen affinity hemoglobin ( eg fetal hemoglobin ). Hemoglobin A will
also undergo a left shift in the presence of factors such as increased pH . decreased
temperature and decreased 2 , 3-diphosphoglycerate .
.
,
1
19
15
*
17
(Choice C) Line C depicts the normal oxygen-hemoglobin dissociation curve with
the typical sigmoidal relationship of pO. to hemoglobin O saturation.
,
I
,
13
14
V
21
n
23
24
2S
25
27
28
29
30
31
32
33
34
(Choices D and E) Lines D and E are shifted to the right , indicating a low oxygen
affinity hemoglobin Hemoglobin A has low affinity for oxygen under conditions such
as decreased pH , increased temperature and increased 2.3-diphosphoglycerate
Educational objective:
The individual subunits of the hemoglobin molecule are structurally analogous to
myoglobin If separated the monomeric subunits will demonstrate a hyperbolic
oxygen-dissociation curve similar to that of myoglobin.
References:
1. Myoglobin oxygen dissociation by multiwavelength spectroscopy.
2 . Configuration of the hemoglobin oxygen dissociation curve
demystified: a basic mathematical proof for medical and biological
sciences undergraduates.
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A 42-year -old obese man consumed a varied , calorie -rich diet throughout most of his
life . After he suffers a mild myocardial infarction he vows to lose weight by changing
his diet and exercising regularly. He strictly adheres to his new diet which has
minimal variation and is entirely devoid of a certain vitamin. After four years ,
symptoms of a specific vitamin deficiency develop Which of the following vitamins
is missing from his diet?
b
O A. Vitamin A
B . Vitamin D
C C Folic acid
C D Vitamin K
C E. Thiamin
F . Riboflavin
G Cobalamin
23
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34
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I olc uhHor
A
&
6
7
s
9
10
/%
A 42-year -old obese man consumed a varied , calorie -rich diet throughout most of his
life . After he suffers a mild myocardial infarction he vows to lose weight by changing
his diet and exercising regularly. He strictly adheres to his new diet, which has
minimal variation and is entirely devoid of a certain vitamin. After four years ,
symptoms of a specific vitamin deficiency develop , Which of the following vitamins
is missing from his diet?
1
IB
is
*
I
O A . Vitamin A [5%]
O B , Vitamin D [4%]
17
13
OC
14
20
,
Folic acid [7%]
O D Vitamin K [4%]
,
O E . Thiamin [5%]
O F. Riboflavin [3%]
22
23
n
25
25
v
*
ilamm [72%]
G.
27
23
29
30
31
32
33
34
Explanation:
The individual described in this question demonstrated symptoms of vitamin
deficiency only after four years of complete dietary absence This long of a delay
between cessation of dietary vitamin intake and symptom development can only
occur with cobalamin ( vitamin B , ). Previously well-nourished individuals have hepatic
vitamin B ; reserves sufficient to last for up to several years of complete dietary
deprivation
1
Vitamin B,: deficiency presents as megaloblastic anemia . Neurologic symptoms can
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G . C abalamin [72%]
7
8
9
10
H
18
IS
17
Explanation:
The individual described in this question demonstrated symptoms of vitamin
deficiency only after four years of complete dietary absence. This long of a delay
between cessation of dietary vitamin intake and symptom development can only
occur with cobalamin ( vitamin B ..), Previously well-nourished individuals have hepatic
vitamin BI: reserves sufficient to last for up to several years of complete dietary
deprivation.
18
2Q
n
23
24
28
25
27
23
29
30
31
32
33
34
Vitamin B . . deficiency presents as megaloblastic anemia Neurologic symptoms can
also develop with characteristic subacute combined degeneration of the dorsal and
lateral spinal columns . Vitamin B,.-related neurologic degeneration may be
irreversible particularly when the neurological symptoms have been longstanding
(Choice A) More than 90% of the body ' s vitaminA reserves are stored in the liver ,
mainly in the perisinusoidal stellate (Ito ) cells . These stores are often sufficient
enough to last around 0 months
.
(Choice B) Vitamin D is stored in adipose tissue but vitamin D deficiency can
develop over the course of several months in the face of inadequate dietary intake
and minimal sunlight exposure
.
(Choice C) The liver contains about half of the total body stores of folate . Because
of this supplemental liver storage, folate levels tend not to drop as quickly as some
other water -soluble vitamins , and deficiency presents after 3 or 4 months of dietary
deficit.
V
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(Choice A) More than 90% of the body ' s vitaminA reserves are stored in the liver ,
mainly in the perisinusoida! stellate ( Ito ) cells . These stores are often sufficient
enough to last around 6 months.
*
Notes
(
alculator
A
b
(Choice B) Vitamin D is stored in adipose tissue but vitamin D deficiency can
develop over the course of several months in the face of inadequate dietary intake
and minimal sunlight exposure
(Choice C) The liver contains about half of the total body stores of folate . Because
of this supplemental liver storage , folate levels tend not to drop as quickly as some
other water -soluble vitamins, and deficiency presents after 3 or 4 months of dietary
deficit .
(Choice D) The small quantity of vitamin K normally stored in the liver is only
sufficient enough to meet the body 's biochemical requirements for 1-3 weeks .
However, severe vitamin K deficiency generally does not develop from dietary
deprivation alone , since bacteria in the large intestine normally produce functional
forms of vitamin K
(Choices E and F) With the exception of vitamin B ( cobalamin ) and folate , ail
water -soluble vitamins (including thiamine and riboflavin ) are flushed from the body
relatively quickly .
Educational objective:
With the exception of vitamin B ,, the body' s stores of most water - soluble vitamins
are rapidly depleted without dietary intake . In contrast, hepatic stores of vitamin B,:
may last up to several years Severe vitamin K deficiency rarely results from poor
dietary intake because colonic bacteria produce functional forms of vitamin K.
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io
Molecular biologists studying the properties of hemoglobin are investigating the
structural changes associated with oxygen loading and unloading . During the
transition from point 1 to point 2 on the graph shown below, hemoglobin molecules
are most likely to release which of the following?
i
15
15
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17
100%
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20
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23
24
25
25
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pO, ( mm fig)
O . "vV, FW
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C k f.w n n r
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34
,
pO ( mm Hg)
O A . Chloride
O B . Heme
C C . Oxygen
.
I
D . Phosphate
E . Protons
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9
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100%
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pOjImmHg)
30
MC
31
32
33
34
O A. Chloride [15%]
O B Heme [2%J
O C . Oxygen [2G%]
.
v
*
D . Phosphate [6% ]
E . Protons [57%]
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• E ^rotons [57%]
*
N o t e*
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A
Explanation:
The transition from point 1 to point 2 on the graph above represents the loading of O
onto partially deoxygenated hemoglobin. At very low pO . the hemoglobin molecule
is fully deoxygenated and binding of the first O molecule is relatively difficult ( as
indicated by the early flatness of the curve )L As pO. increases O, binds to 1 of the 4
heme moieties on the hemoglobin molecule causing the oxygen-binding affinity of
the other hemoglobin subunits to increase ( steepening of the curve ). Additional O
molecules bind as the oxygen partial pressure increases . As hemoglobin becomes
saturated with oxygen, very little additional binding occurs, and the curve levels out .
b
In the peripheral tissues, the release of 0? from hemoglobin is enhanced by
,
increased pCO and the resultant decrease in pH ( Bohr effect ) . This effect occurs
^
due to the histidine side chains found on the alpha and beta hemoglobin subunits.
As the tissues release CO , the majonty is converted by erythrocyte carbonic
^
.
anhydrase to bicarbonate and H While bicarbonate is shifted out of the
erythrocytes in exchange for chloride ions found in the plasma the hydrogen ions
remain within the erythrocytes . These hydrogen ions are buffered by the histidine
residues on hemoglobin and in the process, help stabilize the deoxygenated form of
hemoglobin and decrease its affinity for oxygen
.
*
The Bohr -Haldarte effect
,
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-
wtiirwt bi ILK
ti
When deoxygenated blood enters the alveolar capillaries of the lung, the rise in pO.
increases the binding of oxygen to hemoglobin and causes the release of H and
CO. from hemoglobin iHaldane effect } . As bicarbonate shifts back into the
erythrocytes in exchange for chloride carbonic anhydrase converts H and
bicarbonate ions back into CO and water The CO is then excreted through the
lungs.
-
*
(Choices A and D) Erythrocytes contain a higher concentration of chloride ions in
venous blood than in arterial blood due to the chloride shift However, neither
chloride nor phosphate is transported in any important manner by hemoglobin .
(Choice B) Heme is not released from hemoglobin in the context of oxygen loading
or dissociation However , it is released dunng the normal destruction of aged red
blood celts by the spleen .
(Choice C) En the lungs higher pH, lower temperatures , and decreased
concentration of 2 , 3-DPG shift the oxygen- dissociation curve to the left increasing
the affinity of hemoglobin for oxygen.
Educational objective:
In the lungs, the binding of oxygen to hemoglobin drives the release of H* and CO,
from hemoglobin (Haldane effect ). In the peripheral tissues , high concentrations of
CO. and H' facilitate oxygen unloading from hemoglobin (Bohr effect ).
v
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n
A 20-year-old African American student presents to the ER with malaise and dark
urine He was diagnosed with a mild urinary tract infection several days ago Anemia
is evident on the complete blood count and erythrocyte fragments are seen on a
peripheral smear . Which of the following substrate flow pathways is most likely
deficient in this patient?
u
k
Glucose - 1 -Ptiosphaie
IS
IS
17
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H
6-PtiosphoghJCoriaife
20
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Glucose
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-P h o S C h a t e
O A. A
OB B
O C. C
ODD
O E. E
O F. F
C G. G
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I dlculdlor
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7
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9
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A
A 20-year-old African American student presents to the ER with malaise and dark
urine He was diagnosed with a mild urinary tract infection several days ago Anemia
is evident on the complete blood count and erythrocyte fragments are seen on a
peripheral smear. Which of the following substrate flow pathways is most likely
deficient in this patient?
u
Glucose - 1
15
16
17
18
6-Ph < Kf ) N .
1 -1
20
:
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tos*^-b - Pliosclr rfl ^
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O A . A [8%]
23
29
O B. B [6%]
30
31
32
33
34
©
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[ 58%)
C D. D [ 11%)
O E. E [5%]
O F. F [10%]
O G. G [2%]
*
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Explanation:
7
--
Glucose 5 phosphate
8
9
n
G/ucose -6 -PhospAj<e Dehydrogenase
(Ra ® //mring step )
IWDPH4H*
13
U
15
16
OXIDATIVE
( IRREVERSIBLE )
6- phosphogluconate
UfiDP*
6 - pftospftogfuconate Defr/ drogenaee
17
MADPH*H*
13
RiDuiose - 5-pho phate
14
^
20
2\
n
24
25
26
27
Ribtiose -5-pl >$phate
NONOXIDATIVE
/
(REVERSIBLE )
A
"ibose
5 ph &sphate
*
-
Xylul ose - 5-phosphate
23
29
30
31
32
33
34
TtsnshetbJase
i
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-
SeaoheptsJose - 7 - phosphate
phgsphate
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5-phcsphate
NONOXIDATIVE
(REVERSIBLE )
Ribose -5- phosphaie
1
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15
Lab Value
Xylul ose - 5-p ho sp h ate
0
Lf .u
17
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20
SedohepU ose - 7- phosphate
GlyceralEtehyde - 3 - phosphate
21
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Fructose -(> phosphate
E:v
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31
32
33
34
Glycolysis
Metabolism of glucose through the hexose monophosphate (HMP ) shunt serves two
major functions 1 production of NADPH as a reducing equivalent and 2 . synthesis
of ribose 5 -phosphate for nucleotide synthesis The HMP shunt consists of two
different types of reactions , oxidative (irreversible ) and non- oxidative (reversible )
reactions . All reactions of HMP shunt occur exclusively in the cytoplasm . In the
la
r* f
fit
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ivieidDuiibrn ui yiu
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muiiupilu^
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Lab Value
MIUIIL s>t?i vw: LWU
*
Note
*
(
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*
major functions 1 . production of NADPH as a reducing equivalent and 2 . synthesis
of ribose 5 -phosphate for nucleotide synthesis . The HMP shunt consists of two
different types of reactions oxidative (irreversible } and non-oxidative i reversible }
reactions All reactions of HMP shunt occur exclusively in the cytoplasm . In the
oxidative portion of HMP shunt , glucose 6-phosphate is first converted to
6 -phosphogluconolactone producing one molecule of NADPH This reaction is
catalyzed by glucose 6 -phosphate dehydrogenase , the rate limiting enzyme of the
HMP shunt . In the second reaction of the oxidative portion of HMP shunt . 6
phosphogiuconolactone is hydrolyzed to ribulose 5-phosphate by
6 -phosphogluconate dehydrogenase producing a second molecule of NADPH The
non-oxidative reactions of the HMP are primarily designed to generate ribose
5-phosphate from intermediates of glycolysis
A
.
Erythrocytes utilize the reactions of the HMP shunt to generate large amounts of
NADPH to maintain glutathione in a reduced state by the action of glutathione
reductase Reduced glutathione is important in protecting erythrocytes from
oxidative damage resulting from oxidant drugs and oxidizing environmental toxins In
erythrocytes , the HMP shunt is the only major pathway that generates NADPH. Thus ,
defects in the oxidative portion of the HMP shunt result in poor protection of these
ceils against free radicals, hydrogen peroxide and other forms of oxidant stress .
30
31
32
33
34
Oxidative damage to red cells causes denatured hemoglobin to form insoluble Heinz
bodies resulting in erythrocyte destruction in the spJeen Additionally oxidative
stress results in stiffening of the erythrocyte membrane and hemolysis in the
microvasculature due to an inability of the erythrocyte to deform and fit through
capillary beds. The patient described in the vignette most likely has glucose
6 -phosphate dehydrogenase deficiency ( G6PD ). G6PD is an X -linked disorder that
results in episodes of hemolysis during oxidative and infective stress The patient in
thisvionette was likeiv prescribed trimethoorim-sulfamethoxazoie for UTf this druo
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Oxidative damage to red cells causes denatured hemoglobin to form insoluble Heinz
bodies resulting in erythrocyte destruction in the spleen Additionally , oxidative
stress results in stiffening of the erythrocyte membrane and hemolysis in the
microvasculature due to an inability of the erythrocyte to deform and fit through
capillary beds. The patient described in the vignette most likely has glucose
6 -phosphate dehydrogenase deficiency ( G6PD ) G6PD is an X -linked disorder that
results in episodes of hemolysis during oxidative and infective stress The patient in
this vignette was likely prescribed trimethoprim-sulfamethoxazole for UTI this drug
has oxidant properties and can precipitate hemolysis in patients with this disease
*
(Note
*
(
alculdior
b
.
(Choices A and B] Phosphoglucomutase interconverts glucose 6-phosphate and
glucose 1-phosphate This enzyme provides a link between glycogenolysis ,
glycolysis, galactose metabolism and glucoronate synthesis . Deficiency of this
enzyme is rare .
(Choices D and E) Reaction E is the first step in glycolysis catalyzed by hexokinase
or gfucokinase Reaction D is catalyzed by glucose 6-phosphatase This last step in
gluconeogenesis and glycogenolysis allows glucose to be liberated into the
circulation Deficiency of glucose 6-phosphatase causes glycogen storage disease
type 1.
(Choices F and G) Interconversion of glucose 6 -phosphate and fructose
6 -phosphate is catalyzed by the bidirectional enzyme phosphoglucoisomerase
Educational Objective:
Glucose 6-phosphate dehydrogenase deficiency is a common X -linked disorder of
the hexose monophosphate pathway that results in episodes of hemolytic anemia
due to oxidative stress
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6
7
a
9
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13
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20
A group of investigators is studying the mechanisms involved in cancer
pathogenesis Their research focuses on the intracellular signaling cascades that
begin after receptor tyrosine kinases are activated by their respective ligands They
find that interaction of a certain growth factor with its receptor leads to the following
sequence of events:
Binding of growth factor
i
Autophosphorylation of tyrosine residues
I
Activation of phosphomositide 3-kinase
i
21
Activation of protein kinase B ( Akt )
22
1
23
2&
25
0
Activation of X
Which of the following is the most likely direct effect of X upon activation?
27
28
29
30
31
32
33
34
C A . Ca - efflux from endoplasmic reticulum
O B cAMP accumulation
O C Dimerization of STAT proteins
O D . Rapid decrease in cGMP levels
. E. Translocation to the nucleus and gene transcription
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Which of the following is the most likely direct effect of X upon activation?
O A Ca:* efflux from endoplasmic reticulum [17%]
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O B. cAMP accumulation [9%]
C C . Dimerization of STAT proteins [16%]
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Activation of X
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Activation of protein kinase B ( Akt )
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O D . Rapid decrease in cGMP levels [4%]
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eus and gene transcription [54%]
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Explanation:
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mTOR pathway
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Growth factor
receptor
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Growth factor
receptor
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mTOR
*
Cell growth, proliferation
& survival ( anti- apoptosis )
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Growth factors can stimulate cell proliferation by altering the expression of certain
genes in the nucleus After a growth factor binds to its cell membrane receptor,
signal transduction systems transfer the signal to the nucleus Examples of signal
transduction systems include;
1. MAP-kinase pathway
2. PI 3K/Akt/mTOR pathway
3. Inositol phospholipid pathway
4. cAMP pathway
5. JAK/STAT pathway
b
The PI3K/ Akt /mTOR pathway is an intracellular signaling pathway that is important for
cellular proliferation This pathway is typically activated when a growth factor binds to
its receptor tyrosine kinase, causing auto-phosphorylation of specific tyrosine
residues within the receptor These phosphotyrosine residues activate
phosphoinositide 3-kmase ( PI 3 K }. which then phosphorylates PIP3 found in the
plasma membrane to PIP . This leads to activation of a protein called Akt ( or protein
kinase Bj a sermetthreonine - specific protein kinase . Subsequently . Akt activates
mTOR (mammalian target of rapamycin ) , which translocates to the nucleus to induce
genes involved in cell survival , anti- apoptosis , and angiogenesis . mTOR activation is
inhibited by PTEN ( phosphatase and tensin homolog } a tumor suppressor protein
that removes the phosphate group from PIP,.
The PI3K/Akt/mTOR pathway is highly active in many cancer cells as a result of
mutations causing increased activity of PI 3 K or Akt or loss of function of
PTEN , Mutations involving certain growth factor receptors ( eg , epidermal growth
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I alculaior
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residues within the receptor . These phosphotyrosine residues activate
phosphoinositide 3-kmase ( PI3 K ) . which then phosphorylates PIP. found in the
plasma membrane to PIP . This leads to activation of a protein called Akt ( or protein
kinase B), a serine / threonine - specific protein kinase Subsequently Akt activates
mTOR (mammalian target of rapamycin ) , which translocates to the nucleus to induce
genes involved in cell survival, anti apoptosis , and angiogenesis mTOR activation is
inhibited by PTEN (phosphatase and tensin homolog ) a tumor suppressor protein
that removes the phosphate group from PIP,.
-
1
IB
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*
17
n
H
20
21
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as
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The PI3K/Akt/mTOR pathway is highly active in many cancer cells as a result of
mutations causing increased activity of PI 3 K or Akt or loss of function of
PTEN , Mutations involving certain growth factor receptors (eg , epidermal growth
factor ) can also enhance activity . Several drugs targeting this pathway ( eg , mTOR
inhibitors including rapamycin [ sirolimus ] ) have shown benefit in treating certain
cancers .
(Choice A) The inositol phospholipid pathway utilizes G. proteins that stimulate
hydrolysis of membrane -bound phospholipids via phospholipase C . This pathway
increases cytoplasmic Ca:* levels through IP,- mediated Ca; efflux from the
endoplasmic reticulum .
(Choice B) Some G protein-coupled receptors utilize cAMP as a second
messenger . For example . (S- adrenergic receptors activate Gt, which in turn activates
adenylate cyclase and increases intracellular cAMP levels
(Choice C) Most cytokine receptors lack intrinsic kinase activity and instead
transduce their signals through associated intracellular tyrosine kinases known as
Janus kinases ( JAK ). These m turn activate cytoplasmic STAT ( signal transducer
and activator of transcription) proteins , which dimerize and translocate to the nucleus .
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6
Several drugs targeting this pathway ( eg , mTOR
inhibitors including rapamycin [ sirolimus ] ) have shown benefit in treating certain
cancers.
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29
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factor )
MULOLiuiia nnoivn
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can also enhance activity
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.
yniniiii
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Notes
( <ilc
ufdlor
A
(Choice A) The inositol phospholipid pathway utilizes G proteins that stimulate
hydrolysis of membrane-bound phospholipids via phospholipase C, This pathway
increases cytoplasmic Ca - levels through IP,- mediated Ca - efflux from the
endoplasmic reticulum .
k
-
(Choice B) Some G protein coupled receptors utilize cAMP as a second
messenger . For example (S- adrenergic receptors activate G . which in turn activates
adenylate cyclase and increases intracellular cAMP levels
:
(Choice CJ Most cytokine receptors lack intrinsic kinase activity and instead
transduce their signals through associated intracellular tyrosine kinases known as
Janus kinases ( JAK ), These in turn activate cytoplasmic STAT ( signal transducer
and activator of transcription ) proteins, which dimerize and translocate to the nucleus,
(Choice D) Phosphodiesterases ( POE ) degrade cGMP by hydrolyzing it into
GMP . PDE inhibitors prevent the degradation of cGMP . thereby enhancing and
prolonging its effects For example sildenafil enhances the vasodilatory effects of
cGMP within the corpus cavernosum by inhibiting PDE 5 .
Educational objective;
The PI3K/ Akt/mTOR pathway is an intracellular signaling pathway important for antiapoptosis, cellular proliferation , and angiogenesis . Mutations in growth factor
receptors , Akt mTOR . or PTEN that enhance the activity of this pathway contribute
to cancer pathogenesis .
-
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Notes
(
alculdtor
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H
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Hemoglobin C ( HbC ) disease is caused by a single amino acid substitution ( glutamic
acid > lysine ) at position 6 in the (J-globin chain of the Hb molecule. Patients
homozygous for HbC have mild chronic hemolytic anemia , whereas those with
hemoglobin (HbS ) generally have a more severe condition . Which of the following
properties of HbS best explains why HbS disease is associated with more
pronounced clinical manifestations than HbC disease?
—
C A HbS allows hydrophobic interaction among hemoglobin molecules
B . HbS decreases (3 -globin interaction with 2,3-diphosphoglycerate
O C HbS impairs oxygen binding to the heme moiety
C D HbS impairs proper folding of the o -helix in the f3 -globin chain
E . HbS stabilizes the iron moiety at the ferric ( Fe*) state
26
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Hemoglobin C ( HbC ) disease is caused by a single amino acid substitution ( glutamic
acid * lysine ) at position 6 in the p-globin chain of the Hb molecule . Patients
homozygous for HbC have mild chronic hemolytic anemia , whereas those with
hemoglobin (HbS ) generally have a more severe condition. Which of the following
properties of HbS best explains why HbS disease is associated with more
pronounced clinical manifestations than HbC disease'?
—
^ # A . HbS allows hydrophobic interaction among hemoglobin molecules [41%]
B . HbS decreases p-globin interaction with 2 , 3-diphosphoglycerate [12%]
C . HbS impairs oxygen binding to the heme moiety [18%]
D . HbS impairs proper folding of the a -helix in the p-globin chain [25%]
C E . HbS stabilizes the iron moiety at the ferric (Fe '} state [3%]
(
Explanation:
Globm chains of the hemoglobin (Hb ) tetramer are compactly folded due to nonpolar
hydrophobic residues in the interior and charged polar residues on the surface In
sickle cell disease (SCD ) with HbS , the usual acidic (negatively charged) glutamic
acid ( glut residue at the sixth position on the p- globin chain is replaced by a
nonpolar (neutral charge ) valine ( vai ) residue. This single glu ^val substitution
leads to the alteration of a region on the p-globm surface that interacts with a
complementary 9ite on another Hb molecule. As a result of the charge difference,
the hydrophobic interactions that occur cause aggregation of Hb molecules
(under anoxic conditions ) and subsequent erythrocyte sickling ( distortion and
inflexibility ) , which is promoted by low oxygen levels , increased acidity , or
v
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I dlculdior
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8
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1
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13
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A
Explanation:
Globin chains of the hemoglobin (Hb ) tetramer are compactly folded due to nonpolar
hydrophobic residues in the interior and charged polar residues on the surface. In
sickle cell disease (SCD ) with HbS , the usual acidic ( negatively charged ) glutamic
acid ( glu) residue at the sixth position on the p - globm chain is replaced by a
nonpolar (neutral charge ) valine (vai ) residue This single glu^val substitution
leads to the alteration of a region on the p-globin surface that interacts with a
complementary site on another Hb molecule . As a result of the charge difference ,
the hydrophobic interactions that occur cause aggregation of Hb molecules
(under anoxic conditions ) and subsequent erythrocyte sickling ( distortion and
inflexibility ) , which is promoted by low oxygen levels increased acidity, or
dehydration,
in patients with HbC . another common Hb variant, glu is replaced by a basic polar
(positively charged ) lysine ( lys ) residue Because lys is charged ( although it has
opposite polarity to glu ), hydrophobic interactions between Hb molecules do not
occur. The presence of lys causes HbC to have decreased mobility on
.
electrophoresis
(Choices B and C) HbS and HbC do not differ significantly from normal Hb in
binding or affinity to oxygen or 2.3- diphosphoglycerate ( 2,3-DPG ), If HbS (S-globin
had weaker interactions with 2 , 3 - DPG this would likely prevent sickling as it would
lead to increased oxygen affinity (left shift on oxygen-hemoglobin dissociation
curve ) , thereby limiting oxygen unloading in tissues .
.
(Choice D) A glu ^val substitution affects the 3 -dimensional (tertiary ) structure of
Hb, but it does not result in a significant change in the a -helical ( secondary )
structure . Introducing proline into the primary structure could distort the q-helix due
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Note
*
(
*slc ulcitor
dehydration .
A
In patients with HbC another common Hb variant glu is replaced by a basic polar
(positively charged ) lysine ( lys ) residue Because lys is charged ( although it has
opposite polarity to glu ), hydrophobic interactions between Hb molecules do not
occur . The presence of lys causes HbC to have decreased mobility or
electrophoresis.
,
(Choices B and C) HbS and HbC do not differ significantly from normal Hb in
binding or affinity to oxygen or 2.3-diphosphoglycerate ( 2.3- DPGi If HbS j3-globin
had weaker interactions with 2.3- DPG . this would likely prevent sickling as it would
lead to increased oxygen affinity (left shift on oxygen -hemogiobin dissociation
curve ) , thereby limiting oxygen unloading in tissues .
b
20
21
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(Choice D) A glu-+vai substitution affects the 3-dimensional (tertiary ) structure of
Hb . but it does not result in a significant change in the a- helical ( secondary )
structure Introducing proline into the primary structure could distort the o- helix due
to proline ’ s ngid cyclic structure ,
(Choice E) HbS and HbC mutations do not involve amino acids lining the heme
pocket Heme iron oxidation ( ferrous [Fe- *] — * ferric [Fe *] ) results in methemoglobin
formation HbM disease is a congenital cause of methemoglobinemia due to a
mutation in the heme-binding pocket
.
Educational objective:
Hemoglobin S (HbS) contains valine in place of glutamic acid at the sixth amino acid
position of the p-globin chain . This promotes hydrophobic interaction among Hb
molecules and results in HbS polymerization and erythrocyte sickling.
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A 22-year-old Caucasian female with recurrent abdominal pain and anxiety has had
marked improvement in her symptoms after intravenous administration of a heme
preparation . Rapid improvement of symptoms in this patient is most likely due to
significant repression of:
(
A 5 -Aminolevulinate synthase
v
B 6 -Aminolevuliinate dehydratase
h
C C Hydroxymethylbilane synthase
O D . Ferrochelatase
C E . Bilirubin glucuronyl transferase
n
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A
A 22-year-old Caucasian female with recurrent abdominal pain and anxiety has had
marked improvement in her symptoms after intravenous administration of a heme
preparation . Rapid improvement of symptoms in this patient is most likely due to
significant repression of:
* ® A. 5 -Aminolevulinate synthase [58%1
B 5 -Aminolevulinate dehydratase [17%]
O C Hydroxymethylbilane synthase [7%]
O D, Ferrochelatase J14%]
- E. Bilirubin glucuronyl transferase [5%]
21
22
21
24
Explanation:
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( Uroporphynnogen I
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A deficiency of any of the enzymes responsible for porphyrin synthesis can result in
porphyria , which can be broadly classified as either hepatic or erythropoietic,
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A deficiency of any of the enzymes responsible for porphyrin synthesis can result in
porphyria , which can be broadly classified as either hepatic or erythropoietic,
depending on the site of the enzymatic deficiency.
Heme is synthesized in the liver for use in the cytochrome p 450 enzyme system ,
while heme in the bone marrow is generated for hemoglobin use The synthesis
pathways in the liver and bone marrow are regulated differently because the heme
generated by these two separate tissues serves different functions. The clinical
manifestations of porphyria result from the accumulation of precursors of porphyrins
in the blood , tissues, and urine .
Acute attacks of intermittent hepatic porphyria can be precipitated by administration
of drugs such as phenobarbital griseofulvin , and phenytoin Additionally alcohol and
a low caloric diet can induce an acute attack of porphyria. All of these factors
precipitate porphyria symptoms by decreasing the hepatic concentration of heme
which causes an increase in hepatic ALA synthase activity and leads to increased
formation of 5-aminolevulinic acid and porphobilinogen . Since heme serves to
negatively feedback inhibit the synthesis of ALA synthase , a reduction in heme
synthesis subsequently leads to increased 6-aminolevulinic acid and
porphobilinogen . In patients with acute intermittent porphyrias , 5- aminolevulinic acid
and porphobilinogen accumulate because of a congenital blockage , and high
concentrations of these intermediates are responsible for the acute abdominal pain
and neurologic symptoms observed in this disease The diagnosis of acute
intermittent porphyria is made by demonstrating elevated 5 - aminolevulinic acid and
porphobilinogen during acute attacks.
(Choices B and D) ALA dehydratase and ferrochelatase are inhibited by lead
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I alculdtor
A
&
6
7
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1
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A
Acute attacks of intermittent hepatic porphyria can be precipitated by administration
of drugs such as phenobarbital . griseofulvin , and phenytoin . Additionally , alcohol and
a low caloric diet can induce an acute attack of porphyria . All of these factors
precipitate porphyria symptoms by decreasing the hepatic concentration of heme
which causes an increase in hepatic ALA synthase activity and leads to increased
formation of 5-aminolevulinic acid and porphobilinogen Since heme serves to
negatively feedback inhibit the synthesis of ALA synthase, a reduction in heme
synthesis subsequently leads to increased 5-aminolevulinic acid and
porphobilinogen In patients with acute intermittent porphyrias , 6- aminolevulinic acid
and porphobilinogen accumulate because of a congenital blockage , and high
concentrations of these intermediates are responsible for the acute abdominal pain
and neurologic symptoms observed in this disease . The diagnosis of acute
intermittent porphyria is made by demonstrating elevated 5- aminolevulinic acid and
porphobilinogen during acute attacks .
(Choices B and D ) ALA dehydratase and ferrochelatase are inhibited by lead .
(Choice E) Bilirubin glucuronyltransferase is a hepatic enzyme that is responsible for
the conjugation of bilirubin ( a byproduct of heme catabolism ) with glucuronide , a
polar molecule that improves bilirubin solubility for subsequent biliary excretion A
decrease in giucuronyl transferase activity results in unconjugated hyperbilirubinemia
Educational Objective:
Decreased heme concentration results in an increase in hepatic ALA synthase
activity , which in turn leads to increased formation of 5- aminolevulinic acid and
porphobilinogen Increased formation of 6- aminolevulinic acid and porphobilinogen
occurs because heme normally serves to inhibit the synthesis of ALA synthase
,
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15 : 32
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I dlculdlor
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13
A 17-year -old girl comes to the office for follow-up on anemia She has taken iron
supplements regularly since being diagnosed with iron- deficiency anemia 3 months
ago. However , the patient still feels fatigued and does not think that the supplements
have improved her symptoms. On review of systems , she has occasional gingival
bleeding when brushing her teeth . Menses occur every 27- 28 days and last 7- 8
days with heavy flow , sometimes requiring her to change her pad every hour
Platelet count is normal Further evaluation reveals that the patient' s platelets do not
aggregate appropriately in response to ristocetin . When normal plasma is added to
the solution of patient platelets and ristocetin , appropriate platelet aggregation
occurs . Which of the following is most likely deficient in this patient?
14
20
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25
*
-
*
-
-
-*
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28
A . Glycoprotein lb receptors
C B . Glycoprotein llb-llla receptors
O C . Hageman factor
O 0 Thromboxane A2
E . von Willebrand factor
29
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Iolcuhitor
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15
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13
A
A 17-year -old girl comes to the office for follow-up on anemia She has taken iron
supplements regularly since being diagnosed with iron-deficiency anemia 3 months
ago. However , the patient still feels fatigued and does not think that the supplements
have improved her symptoms. On review of systems , she has occasional gingival
bleeding when brushing her teeth . Menses occur every 27- 28 days and last 7- 8
days with heavy flow , sometimes requiring her to change her pad every hour
Platelet count is normal Further evaluation reveals that the patient’s platelets do not
aggregate appropriately in response to ristocetin . When normal plasma is added to
the solution of patient platelets and ristocetin , appropriate platelet aggregation
occurs . Which of the following is most likely deficient in this patient?
H
20
21
22
21
24
2&
25
-
-
-
-
23
O A Glycoprotein lb receptors [6%]
B Glycoprotein llb- illa receptors [16%]
. C . Hageman factor [5%]
O D, Thromboxane A2 [3%]
V @ E von WHlebrand factor [70%)
.
29
30
31
32
33
34
Explanation:
Platelet adhesion & activation via vWF
Collagen
En dothelium
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Notes
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A
* ® E. von Willebrand factor [70%]
7
a
9
Explanation:
n
&
Platelet adhesion & activation via vWF
i
IS
15
*
CoHagen
>
Endothelium
17
13
H
20
Coagulation
cascade
21
22
21
24
2S
25
Circulating
vWF
Fibrinogen ^
GpllttllJa
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Gplb
Plate !
23
-
-*
-
29
30
«
Endothelial injury
31
32
33
34
Subendothelial vVYF
© UWortd
von Willebrand factor ( vWF ) is an important hemostatic glycoprotein synthesized
by endothelial cells and megakaryocytes After endothelial damage , vWF binds
glycoprotein (GP) !b receptors on the platelet membrane and mediates platelet
4
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t dlculdtor
CUWorkJ
5
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7
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15
15
1?
13
14
20
21
22
23
24
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23
*
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29
31
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33
von Willebrand factor ( vWF ) is an important hemostatic glycoprotein synthesized
by endothelial cells and megakaryocytes After endothelial damage , vWF binds
glycoprotein (GP) lb receptors on the platelet membrane and mediates platelet
aggregation and adhesion to subendothehal collagen. Subsequently, vWF serves as
a carrier for factor VIII and prolongs its half -life .
Deficiency of vWF impairs platelet function Decreased platelet adhesion and
aggregation cause easy bruising and mucocutaneous bleeding (eg , gingival
bleeding heavy menses ). Laboratory workup reveals a norma! platelet count The
ristocetin aggregation test measures in vitro vWF -dependent platelet aggregation.
Ristocetin activates GP lb receptors on platelets and makes them available for vWF
binding . When the vWF level is decreased there is poor platelet aggregation in
the presence of ristocetin . When normal plasma that contains vWF is added»
appropriate platelet aggregation occurs .
k
vWF deficiency also leads to functional deficiency of factor VIII, This results in
prolonged bleeding after tooth extraction and other minor surgeries Partial
thromboplastin time ( PTT ) may be normal or prolonged depending on the level of
factor VIII deficiency.
Combined oral contraceptives are the first-line therapy for menorrhagia due to von
Willebrand disease. Patients can also be treated with desmopressin , which
stimulates vWF release from endothelium .
34
(Choice A ) Bernard-Soulier syndrome ihereditary deficiency of GP lb receptors ) is
characterized by thrombocytopenia , enlarged platelets and mucocutaneous
bleeding Platelet aggregation to ristocetin will be abnormal However, because
vWF levels are normal, addition of normal plasma will not correct aggregation.
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26
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Notes
t dkuldtor
/s
stimulates vWF release from endothelium ,
(Choice A) Bernard-Soulier syndrome [ hereditary deficiency of GP lb receptors) is
characterized by thrombocytopenia , enlarged platelets and mucocutaneous
bleeding Platelet aggregation to ristocetin will be abnormal. However , because
vWF levels are normal, addition of normal plasma will not correct aggregation
(Choice B) Hereditary deficiency of GP llb- llla receptors occurs in Glanzmann
thrombasthenia which manifests with mucocutaneous bleeding Platelet
aggregation in response to ristocetin is normal as levels of vWF and GP lb
receptors are normal .
(Choice C) Congenital deficiency of factor XII (Hageman) causes marked PTT
prolongation without bleeding diathesis Instead patients may have a tendency for
thromboembolic complications .
(Choice D) Thromboxane A2 deficiency is associated with aspirin treatment due to
irreversible inactivation of cyclooxygenase (COX ) in platelets . The result is
decreased platelet adhesion and aggregation The ristocetin aggregation test is
normal
Educational objective :
von Willebrand factor (vWFj binds glycoprotein lb receptors on platelets after
endothelial damage and carries factor VIII. vWF deficiency results in
mucocutaneous bleeding due to defects in platelet aggregation and coagulation
pathway abnormalities Ristocetin aggregation test will show decreased aggregation
of platelets
References:
1. Diagnosis and management of von Willebrand disease: guidelines
16 : 17
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dlculdtor
4
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to
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IB
15
17
13
19
20
21
A 13- year-old girl undergoing hematologic evaluation is found to have a hemoglobin
abnormality that decreases the partial pressure of oxygen at which hemoglobin is
50% saturated from 26 to 20 mm Hg . Which of the following sequelae is this patient
most likely to develop?
O A . Erythrocytosis
O B . Hypoxia-induced hemolysis
C Increased erythrocyte osmotic fragility
C D Megaloblastic erythrocyte changes
O E . Oxidant -induced hemolysis
n
23
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27
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34
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I olcuhHor
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A
5
6
7
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14
IS
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H
20
A 1 3 - year -oid girl undergoing hematologic evaluation is found to have a hemoglobin
abnormality that decreases the partial pressure of oxygen at which hemoglobin is
50% saturated from 26 to 20 mm Hg . Which of the foilowing sequelae is this patient
most likely to develop?
*
A . Erythrocytosis [49%]
O 6. Hypoxia-mduced hemolysis [25%]
C Increased erythrocyte osmotic fragility [10%]
O D Megaloblastic erythrocyte changes [4%]
O E. Oxidant-induced hemolysis [13%]
21
n
21
24
2&
25
27
*
Oxygen - hemoglobin dissociation curve
29
30
-
31
*
32
-
33
34
-
Explanation:
Left shift caused by;
1. Decreased H * (increased pH )
2 . Decreased 2 3 DPG
3. Decreased temperature
t
A
Think LUNGS Left shift
-
100% -
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t nkuldlor
Explanation:
/%
Oxygen hemoglobin dissociation curve
7
-
a
9
10
Left shift caused by:
1 . Decreased H * (increased pHj
2. Decreased 2 , 3 DPG
3. Decreased temperature
+
i
IS
IS
17
13
*
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Think LUNGS Left shift
100% -
H
20
21
22
21
n
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Right shift caused by:
1 , Increased H * ( decreased pH)
2 Increased 2, 3 DPG
3 , Increased temperature
M
2&
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27
3
,
rv
-
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*
-
29
30
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34
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50%
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(Notes
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/%
pOj ( mm Hg }
© U^U W o f k l L U
The oxygen-hemoglobin dissociation curve describes the relationship between the
partial pressure of oxygen in the blood and the oxygen saturation of hemoglobin .
The partial pressure of oxygen at which hemoglobin is 50% saturated is known as
the PK; this value is used as a conventional measure of hemoglobin' s affinity for
oxygen . The PM is about 26 mm Hg in normal individuals . A Ps shift from 26 to 20
mm Hg indicates that the affinity of hemoglobin for oxygen is increased (left shift of
the O, dissociation curve )
.
There are numerous high-oxygen- affinity hemoglobin mutations that reduce the
ability of hemoglobin to reiease oxygen in the tissues (eg hemoglobins Chesapeake
and Kempsey ). However, the low tissue oxygen levels stimulate the kidneys to
increase erythropoietin synthesis, which results in a compensatory erythrocytosis that
helps maintain normal oxygen delivery Thus patients with high-oxygen- affinity
hemoglobins are typically asymptomatic.
.
b
(Choice B) Sickle cell disease can result in hypoxia-induced hemolysis due to the
ability of deoxygenated hemoglobin S to polymerize and cause excessive
erythrocyte sickling and irreversible cell membrane damage The
oxygen-dissociation curve for hemoglobin S is shifted to the right not to the left .
(Choice C) Hereditary spherocytosis results from a variety of molecular defects
involving erythrocyte structural proteins responsible for linking the plasma membrane
to the cytoskeleton ( eg , spectnn. ankynn ). The resulting membrane instability and
loss leads to the formation of sphenocytes . small rounded erythrocytes with
increased susceptibility to lysis in hypotonic solutions .
^
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Note
*
*
I alculaior
A
&
6
7
8
9
n
1
15
15
*
ft
(Choice B) Sickle cell disease can result in hypoxia- induced hemolysis due to the
ability of deoxygenated hemoglobin S to polymerize and cause excessive
erythrocyte sickling and irreversible cell membrane damage. The
oxygen-dissociation curve for hemoglobin S is shifted to the right not to the left .
(Choice C) Hereditary spherocytosis results from a variety of molecular defects
involving erythrocyte structural proteins responsible for linking the plasma membrane
to the cytoskeleton ( eg , spectrin ankyrm ). The resulting membrane instability and
loss leads to the formation of spherocytes small rounded erythrocytes with
increased susceptibility to lysis in hypotonic solutions .
b
,
17
1j
20
21
22
23
24
25
25
27
-
29
* 30
*
31
32
33
(Choice D) Megaloblastic erythrocyte changes are characteristically seen in vitamin
B and folic add deficiency due to impaired DNA synthesis and mitosis
.
(Choice E) Individuals with G6 PD deficiency develop rapid hemolysis under oxidant
stress , such as when taking antimalanals or sulfonamides or after ingestion of fava
beans
Educational objective:
Ps refers to the partial pressure of oxygen at which hemoglobin is 50% saturated .
High-oxygen-affinity hemoglobins have a decreased P, that is represented by a
leftward shift of the oxygen-dissociation curve High- oxygen- affinity hemoglobins
have reduced ability to release oxygen within the peripheral tissues leading to renal
hypoxia increased erythropoietin synthesis and compensatory erythrocytosis
,
,
,
34
References:
1 . OXYGEN DISSOCIATION CURVES IN SICKLE CELL ANEMiA AND IN
SUBJECTS WITH THE SICKLE CELL TRAIT
V
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Notes
t dlculdtor
A
&
6
7
8
9
A 14- year -old male is being evaluated for splenomegaly. Enzyme assays performed
on circulating blood cells demonstrate low pyruvate kinase activity Which of the
following is the most likely cause of this patient’s splenomegaly?
10
A Intracellular substance accumulation
i
is
115
*
17
13
14
20
O B . Passive congestion
O C. Inflammatory infiltration
b
O D . Work hypertrophy
E Neoplastic lesion
21
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25
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26
*
*
*
-
-
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34
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Notes
I olcuhilor
A
&
G
7
£
9
A
A 14- year-old male is being evaluated for splenomegaly. Enzyme assays performed
on circulating blood cells demonstrate low pyruvate kinase activity Which of the
following is the most likely cause of this patient' s splenomegaly?
10
C A . Intracellular substance accumulation [ 33% ]
u
15
IS
17
13
H
_ B. Passive congestion [15%]
O C . Inflammatory infiltration [6 %]
* # D. Work hypertrophy [43%]
O E . Neoplastic lesion [2%]
20
21
22
23
2i
2S
25
27
23
-
--
30
31
32
33
Explanation:
The spleen is an organ of the reticuloendothelial system that contains approximately
25% of body’s lymphoid tissue One of the main functions of the spleen in adult
humans is maintenance of erythrocyte quality in the red pulp by removal of
senescent and defective red blood cells The spleen accomplishes this function
through the unique organization of its parenchyma and vasculature. Antibody
production and B cel! affinity maturation occur in the white pulp of the spleen and the
spleen also serves to remove antibody-coated bacteria and other opsonized material
and cells from the circulation An increase in any of these normal functions may
result in splenomegaly
Pyruvate kinase is the enzyme in the glycolytic pathway that converts
phosphoenolpyruvate to pyruvate resulting in the generation of a molecule of ATP .
Pyruvate kinase is allosteric ally stimulated by fructose 1.6- bisphosphate . which is
produced from fructose -6- phosphate by the enzyme phosphofructokinase .
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20
21
n
23
n
2&
26
27
28
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*
4*
30
31
32
33
H
rt
'
Mil i
Explanation:
<
Previous
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Lab Value
r
*
Notes
(
okuldtor
A
The spleen is an organ of the reticuloendothelial system that contains approximately
25% of body's lymphoid tissue . One of the main functions of the spleen in adult
humans is maintenance of erythrocyte quality in the red pulp by removal of
senescent and defective red blood cells The spleen accomplishes this function
through the unique organization of its parenchyma and vasculature . Antibody
production and B cell affinity maturation occur in the white pulp of the spleen , and the
spleen also serves to remove antibody-coated bacteria and other opsonized material
and cells from the circulation . An increase in any of these normal functions may
result in splenomegaly.
u
Pyruvate kinase is the enzyme in the glycolytic pathway that converts
phosphoenolpyruvate to pyruvate resulting in the generation of a molecule of ATP .
Pyruvate kinase is aliostencally stimulated by fructose 1,6-bisphosphate , which is
produced from fructose -6-phosphate by the enzyme phosphofructokinase .
Allosteric stimulation of pyruvate kinase by fructose 1 6-bisphosphate results in
stimulation of glycolysis . Red blood cells do not contain mitochondria , so the main
metabolite of glycolysis is lactate Any deficiency of glycolysis in red blood cells
leads to hemolysis because of insufficient production of ATP and defective
maintenance of red blood cell architecture Excessive erythrocyte destruction by the
spleen causes splenomegaly due to work hypertrophy (choice DJ . Work
hypertrophy results from hypertrophy of the reticuloendothelial cells of the splenic
parenchyma as these cells are involved in the removal of damaged RBCs
(Choices A, B, C and E ) The other choices listed will also result in splenomegaly .
Accumulation of sphingomyelin and glucocerebrosides are responsible for
splenomegaly in Niemann- Pick disease and Gaucher disease , respectively ( A ) .
Passive congestion of spleen occurs with portal hypertension, splenic vein
thrombosis and ccncest ve rteart fa : ve ( B) ^
d atation o - W e
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Notes
(
akulator
produced from fructose -6-phosphate by the enzyme phosphofructokinase
Allosteric stimulation of pyruvate kinase by fructose 1 , 6-bisphosphate results in
stimulation of glycolysis Red blood cells do not contain mitochondria , so the main
metabolite of glycolysis is lactate. Any deficiency of glycolysis in red blood cells
leads to hemolysis because of insufficient production of ATP and defective
maintenance of red blood cell architecture Excessive erythrocyte destruction by the
spleen causes splenomegaly due to work hypertrophy ( choice 0) Work
hypertrophy results from hypertrophy of the reticuloendothelial cells of the splenic
parenchyma as these cells are involved in the removal of damaged RBCs .
A
(Choices A . B, C and E ) The other choices listed will also result in splenomegaly .
Accumulation of sphingomyelin and glucocerebrosides are responsible for
splenomegaly in Niemann- Pick disease and Gaucher disease , respectively ( A )
Passive congestion of spleen occurs with porta! hypertension, splenic vein
thrombosis and congestive heart failure ( B) In these conditions , dilatation of the
splenic sinusoids leads to congestion of the spleen with blood resulting in
splenomegaly. The spleen is very important for fighting infectious pathogens in the
body ; therefore many acute and chronic infections lead to enlargement of the
spleen due to proliferation of lymphoid tissue (C) Disorders such as leukemia and
lymphoma also result in splenomegaly from neoplastic proliferation of lymphoid
tissue within the spleen ( E)
u
Educational Objective;
Pyruvate kinase deficiency causes hemolytic anemia due to failure of glycolysis and
resultant failure to generate sufficient ATP to maintain erythrocyte structure In this
case , splenic hypertrophy results from increased work of the splenic parenchyma
which must remove these deformed erythrocytes from the circulation.
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Last updated [8/19 ^20151
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ulculator
A
&
6
7
a
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H
IS
1&
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13
H
20
A 34-year-old man comes to the emergency department due to a facial injury . He
reports getting hit on the face during a fistfight at a bar . Examination shows dark blue
periorbital ecchymosis on the right side . Ophthalmic and neurologic examinations
are otherwise normal. After appropriate evaluation , the patient is discharged home.
Several days later the bruise becomes greenish in color . This change in color is
best explained by the activity of which of the following enzymes?
A . Bilirubin glucuronyl transferase
O B . Ferrocheiatase
O C Heme oxygenase
21
D . Porphobilinogen deaminase
23
2t
2&
25
E Uroporphynnogen decarboxylase
n
27
28
29
*
i
31
32
33
M
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*
Notes
I olc uhHor
A
A
&
6
-
A 34 year-old man comes to the emergency department due to a facial injury . He
reports getting hit on the face during a fistfight at a bar . Examination shows dark blue
periorbital ecchymosis on the right side . Ophthalmic and neurologic examinations
are otherwise normal . After appropriate evaluation , the patient is discharged home.
Several days later , the bruise becomes greenish in color . This change in color is
best explained by the activity of which of the following enzymes?
7
8
9
n
1
IB
is
*
.
lr
O A. Bilirubin glucuronyl transferase [17%]
O B . Ferrochelatase [8%]
17
n
v @ C . Heme oxygenase [56%)
H
20
;
21
22
(
21
D Porphobilinogen deaminase [12%]
E . Uroporphyrinogen decarboxylase [7%]
n
Explanation:
2S
25
27
Bilirubin metabolism
23
29
Heme
Hem
*
31
• 32
• 33
• 34
•
* Biliverdin
BrtivQftim
reductase
Unconjugated bilirubin
Glucvrwtyt In
Conjugate
bilirubin
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Iolculator
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Bilirubin metabolism
7
Heme
* B'lpverdin
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8
9
* UrKonjugaied bilirubin
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13
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Conjugated
is
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17
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20
Bacterial
21
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22
23
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27
Urobiftaogen
23
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*
31
‘ 32
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31
Brown color to feces
Ouworu
This patient has a resolving hematoma after a traumatic injury . Following the injury ,
I
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Notes
t
alculalor
A
©UWortd
This patient has a resolving hematoma after a traumatic injury . Following the injury ,
hemoglobin-containing erythrocytes escape into the periorbital tissues , giving the
bruise its initial purple or bluish color Erythrocyte destruction causes the release of
iron-containing heme molecules Heme oxygenase ( contained in macrophages
among other cells ) degrades heme into biliverdin , carbon monoxide and ferric iron
while consuming oxygen and electrons provided by NADH and NADPH-cytochrome
P 450 reductase Biliverdin is green in color and is further reduced (by the enzyme
biliverdin reductase ) to the yellow pigment bilirubin, which is then transported to the
liver bound to albumin
(Choice A) Bilirubin glucuronyltransferase . or uridine 5'- diphospho -glucuronyl
transferase (UGT) is the enzyme necessary for bilirubin conjugation to glucuronic
acid Lack of UGT or the use of medications that interfere with its activity impairs the
liver's ability to conjugate bilirubin.
b
(Choices B, D, and E) Ferrochelatase ( inhibited by lead) is the final enzyme in the
heme synthetic pathway Porphobilinogen deaminase (PBG deaminase ) and
uroporphyrinogen decarboxylase (UROO ) are also involved in heme production , not
degradation: UROD deficiency can be seen in porphyria cutanea tarda , the most
common porphyria , and PBG deaminase deficiency can be seen in acute
intermittent porphyria.
Educational objective:
Heme oxygenase converts heme to biliverdin, a pigment that causes the greenish
color to develop in bruises several days after an injury.
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H
An infant born to a Greek immigrant appears healthy at birth but develops
transfusion-dependent hemolytic anemia by the age of 6 months His erythrocytes
contain insoluble aggregates of hemoglobin subunits . The child developed normally
in utero because at that time he produced high quantities of:
C A . a -globin
O B . p-globin
C . y - globin
D 6 -globin
b
O E . (-globin
20
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Notes
I alculator
4
5
G
7
£
9
10
*v
<
An infant born to a Greek immigrant appears healthy at birth but develops
transfusion-dependent hemolytic anemia by the age of 6 months His erythrocytes
contain insoluble aggregates of hemoglobin subunits . The child developed normally
in utero because at that time he produced high quantities of:
O A, a -globin [3%J
C B p -globin [4%J
,
i
15
IS
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17
13
19
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-
-
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34
* § C.
.
globin [83% ]
O D . 5 -globm [6%I
O E. -globin [3%]
^
Explanation:
In adults , hemoglobin A is the predominant form of hemoglobin. It is a tetramer
consisting of two alpha and two beta chains . Normally the synthesis of alpha and
beta chains is tightly regulated such that one a -chain is synthesized for every (3 chain Hemoglobin formation begins within a few weeks of conception . The initial
hemoglobin formed by a fetus in utero is called embryonic hemoglobin
(Gower ) . This hemoglobin is composed of two zeta (£ } and two epsilon ( e J chains
( £2t 2 ) and is produced in the embryonic yolk sac. Within a few weeks the fetal liver
starts synthesizing hemoglobin F ( fetal hemoglobin ). This form of hemoglobin is
composed of two alpha and two gamma chains ( a 2 y 2 ). HbF is the major
hemoglobin in the fetus during last few months of gestation and in infants during first
few weeks of postnatal life HbA synthesis starts during the final month of gestation
and gradually replaces HbF during postnatal life Knowing the chronology of feta!
hemoglobin formation and the gradual transition to adult hemoglobin (HbA ) is
important in understanding the relationship between clinical manifestations and
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17
ia
20
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2&
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ft
Explanation:
In adults hemoglobin A is the predominant form of hemoglobin. It is a tetramer
consisting of two alpha and two beta chains . Normally the synthesis of alpha and
beta chains is tightly regulated such that one a -chain is synthesized for every p chain Hemoglobin formation begins within a few weeks of conception . The initial
hemoglobin formed by a fetus in utero is called embryonic hemoglobin
(Gower ) This hemoglobin is composed of two zeta (£ ) and two epsilon ( E ) chains
( £2t 2 ) and is produced in the embryonic yolk sac. Within a few weeks the fetal liver
starts synthesizing hemoglobin F ( fetal hemoglobin ). This form of hemoglobin is
composed of two alpha and two gamma chains (a2y 2). HbF is the major
hemoglobin in the fetus during last few months of gestation and in infants during first
few weeks of postnatal life HbA synthesis starts dunng the final month of gestation
and gradually replaces HbF during postnatal life Knowing the chronology of fetal
hemoglobin formation and the gradual transition to adult hemoglobin (HbA ) is
important in understanding the relationship between clinical manifestations and
postnatal age in beta thalassemia .
Thalassemias are hereditary hemolytic
resulting from defective synthesis of
globin chains As described above the synthesis of alpha and beta globin chains is
very coordinated . In patients with thalassemia, the synthesis of either alpha or beta
chains is defective . Beta thalassemia is caused by defective synthesis of beta
chains. There are two copies of the beta globin gene ( one from each parent ). If only
one gene is defective the patient will have beta thalassemia trait (beta thalassemia
minor ) and lack significant anemia A defect in both beta globin genes results in
severe hemolytic anemia known as beta thalassemia major !n this disease alpha
chains are produced normally but they cannot form stable tetramers due to the lack
of beta globin chains This failure to form stable hemoglobin leads to precipitation of
al^h a fllnhin rkaint snr| nromiitMffl IMCIC Q # raH
c o r r a i a n n^t
rftHc B Q f r g i 4h
IS : 01
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I alculaior
one gene is defective the patient will have beta thalassemia trait (beta thalassemia
minor ) and lack significant anemia A defect in both beta globin genes results in
severe hemolytic anemia known as beta thalassemia major In this disease alpha
chains are produced normally but they cannot form stable tetramers due to the lack
of beta globin chains This failure to form stable hemoglobin leads to precipitation of
alpha globin chains and premature lysis of red blood cells . Beta thalassemia cannot
become symptomatic as long as there are significant amounts of gamma chains
present because gamma chains make up for the absence of HbA beta chains in
forming tetramers . Thus , in late gestation and early postnatal life the expression of
hemoglobin A is offset by gamma chain production (Choice C ) As gamma chain
production wanes, patients will become symptomatic.
(Choice A) o -globin is a normal component of both HbA and HbF a -globin
synthesis is defective in a - thalassemia but it is normal in (3 -thalassemia a-globin is
able to combine with y -globin to form HbF. This allows patients with (3 -thalassemia
major to be asymptomatic in utero and in the first few months following birth.
(Choice B) (3-globin synthesis is defective in patients with b -thalassemia
(Choice D) 5-globin is a minor globin gene that is expressed at very low levels in
normal adults Two a -globins and two 5-globins combine to form hemoglobin A2 .
- globin a component of hemoglobin Gower , the
^
formed by the embryo very early in embryogenesis .
(Choice E)
is
initial hemoglobin
Educational Objective:
HbF contains y -globin instead of (3 -globin Patients with homozygotic (3-thalassemia
(P-thalassemia major ) are asymptomatic at birth due to the presence of y -globins
and HbF . Switching to HbA production and the cessation of y -globin synthesis
precipitates the symptoms of (3-thalassemia .
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1 dkufdior
4
b
6
7
8
9
1
18
16
*
17
13
19
20
21
22
21
-
A 6 year - old male with hemolytic anemia is found to have an abnormality due to an
inactive erythrocyte enzyme The defective enzyme contains 166 amino acid
residues instead of the normal 190 residues . A point mutation in exon 2 of the
enzyme gene is identified as the cause for this patient's disease . Which of the
following mRNA code changes is most likely in this case?
b
O A. UAA - UAG
O B . UUU ^ UUA
O C . CUU ^ AUU
O D . UCA UGA
O E. UAC CAC
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I alculaior
A
&
6
7
a
9
10
r\
-
A 6 year - old male with hemolytic anemia is found to have an abnormality due to an
inactive erythrocyte enzyme The defective enzyme contains 156 amino acid
residues instead of the normal 190 residues . A point mutation in exon 2 of the
enzyme gene is identified as the cause for this patient's disease . Which of the
following mRNA code changes is most likely in this case?
H
is
is
O A UAA
O B . UUU
17
13
19
20
21
*
—
—
UUA [3%J
O C . CUU ^ AUU [4%]
# D . UCA UGA [74%J
O E. UAC
22
21
n
UAG [16%]
-
—
CAC [ 2% ]
Explanation:
2&
25
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29
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31
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-
33
3J
Messenger RNA (mRNA ) is produced from DNA by RNA polymerase II . After
processing which includes removal of noncoding regions of mRNA (introns ) , the
mRNA molecule is transported to the cytoplasm for translation Remember that the
mRNA molecule is composed of groups of three sequential bases known as
codons . Since there are 64 possible combinations of the four bases found in DNA .
there are 64 possible codons . Because there are only twenty amino acids , most
amino acids have more than one potential codon . For instance UUU and UUA both
code for the amino acid phenylalanine Some codons call for the termination of
synthesis of the polypeptide chain and are referred to as stop codons , including
UAA. UAG, and UGA .
Changes in the DNA sequence (mutations ) can result in alterations of protein
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Notes
I tilculdior
Explanation:
A
Messenger RNA (mRNA ) is produced from DNA by RNA polymerase II . After
processing which includes removal of noncoding regions of mRNA (introns ), the
mRNA molecule is transported to the cytoplasm for translation Remember that the
mRNA molecule is composed of groups of three sequential bases known as
codons Since there are 64 possible combinations of the four bases found in DNA
there are 64 possible codons . Because there are only twenty amino acids most
amino acids have more than one potential codon . For instance . UUU and UUA both
code for the amino acid phenylalanine Some codons call for the termination of
synthesis of the polypeptide chain and are referred to as stop codons , including
UAAt UAG , and UGA .
,
H
IB
15
17
18
14
20
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2t
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Changes in the DNA sequence ( mutations ) can result in alterations of protein
structure affecting enzyme function . In the case descnbed in the question stem the
structure of the protein is significantly shortened secondary to an alteration of the
genetic code through a point mutation , which most likely resulted in the introduction
of a premature stop codon Of the choices listed , the single base change of UCA
( serine ) to UGA results in premature termination of protein synthesis and the
formation of a truncated protein molecule . The introduction of a stop codon in the
middle of a protein sequence is called a nonsense mutation (Choice D)
30
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*
31
(Choices A and BJ Changing UAA to UAG would not modify the structure of the
protein because both of these are stop codons This type of mutation is called a
silent mutation . Similarly , UUU and UUA both code for phenylalanine , and this
mutation will not alter the protein structure either as it is also a silent mutation.
,
(Choices C and E) Changing CUU ( leucine ) to AUU (ISOLEUCINEI will result in an
amino acid change at one position. The function of this protein may be altered
depending on a variety of factors , but the ultimate size of the protein will remain the
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there are 64 possible codons Because there are only twenty amino acids most
amino acids have more than one potential codon . For instance UUU and UUA both
code for the amino acid phenylalanine Some codons call for the termination of
synthesis of the polypeptide chain and are referred to as stop codons , including
Previou
*
Note
*
I ulculdtor
A
UAAT UAG, and UGA.
Changes in the DNA sequence ( mutations ) can result in alterations of protein
structure affecting enzyme function In the case described in the question stem the
structure of the protein is significantly shortened secondary to an alteration of the
genetic code through a point mutation , which most likely resulted in the introduction
of a premature stop codon. Of the choices listed , the single base change of UCA
( serine ) to UGA results in premature termination of protein synthesis and the
formation of a truncated protein molecule . The introduction of a stop codon in the
middle of a protein sequence is called a nonsense mutation (Choice D)
(Choices A and B) Changing UAAto UAG would not modify the structure of the
protein because both of these are stop codons . This type of mutation is called a
silent mutation. Similarly , UUU and UUA both code for phenylalanine , and this
mutation will not alter the protein structure either as it is also a silent mutation .
21
23
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30
31
*
33
3
*
(Choices C and E ) Changing CUU (leucine ) to AUU (isoleucinei will result in an
amino acid change at one position The function of this protein may be altered
depending on a variety of factors , but the ultimate size of the protein will remain the
same. This type of mutation is called a missense mutation.
Educational Objective:
UGA , UAG and UAA are stop codons , and mutations producing abnormally placed
stop codons are called nonsense mutations.
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(Note
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6
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n
A 46-year -old male is brought to the emergency department by EMS. He has
attempted suicide twice in the past once by cutting his wrists and another time by
taking an overdose of amitriptyline. His current medications include quetiapine and
fluoxetine . He has no known drug allergies . This evening a neighbor found him in a
closed garage with the car running. As you examine him he loses consciousness
and begins to seize The toxic substance causing this patient s condition affects
hemoglobin by:
O A Oxidation of the iron moiety
O B . Oxidation of the porphyrin ring
O C. Covalent linking to heme
!
D Competitive binding to heme
E . Altering the partial pressure of oxygen
2b
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Notes
I dlculdtor
4
&
6
7
a
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u
A
A 46- year-old male is brought to the emergency department by EMS He has
attempted suicide twice in the past once by cutting his wrists and another time by
taking an overdose of amitriptyline. His current medications include quetiapine and
fluoxetine He has no known drug allergies This evening a neighbor found him in a
closed garage with the car running. As you examine him he loses consciousness
and begins to seize The toxic substance causing this patient s condition affects
hemoglobin by:
b
IS
IS
17
13
14
20
21
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A Oxidation of the iron moiety [3%1
O B Oxidation of the porphyrin ring [2%]
C Covalent linking to heme [10%]
* * D . Compe t e bin :: i ng to heme [81%]
O E . Altering the partial pressure of oxygen [2%]
'
'
Explanation:
Carbon monoxide (CO ) is a colorless , odorless nonirritant gas that is generated as a
byproduct of incomplete hydrocarbon combustion Carbon monoxide emission from
automobiles can result in carbon monoxide poisoning in poorly ventilated spaces
The scenano described above is ty pical for carbon monoxide poisoning Another
classic source of carbon monoxide poisoning is a faulty home heater
,
Carbon monoxide has 220 times more affinity for hemoglobin than does oxygen
Inhaled carbon monoxide rapidly diffuses across the alveolar membrane and binds
tightly with heme- bound iron in hemoglobin and other hemeproteins. Carbon
monoxide binding to hemoglobin results in formation of carboxyhemoglobin . Carbon
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Notes
Lalcufdtor
E Altering the partial pressure of oxygen [2% ]
A
6
7
8
9
10
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17
13
*
H
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n
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24
2S
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-
34
Explanation:
Carbon monoxide (CO ) is a colorless odorless , nonirritant gas that fs generated as a
byproduct of incomplete hydrocarbon combustion. Carbon monoxide emission from
automobiles can result in carbon monoxide poisoning in poorly ventilated spaces.
The scenano described above is typical for carbon monoxide poisoning Another
classic source of carbon monoxide poisoning is a faulty home heater.
,
Carbon monoxide has 220 times more affinity for hemoglobin than does oxygen .
Inhaled carbon monoxide rapidly diffuses across the alveolar membrane and binds
tightly with heme-bound iron in hemoglobin and other hemeprotems, Carbon
monoxide binding to hemoglobin results in formation of carboxyhemoglobin . Carbon
monoxide decreases the oxygen content of the blood by occupying oxygen binding
sites . Carbon monoxide also inhibits the release of oxygen from hemoglobin in
tissues by altering hemoglobin conformation into the relaxed form that has a very
high affinity for oxygen . This results in a leftward shift of the oxygen dissociation
curve and tissue hypoxia via deficient unloading of oxygen. Treatment of carbon
monoxide toxicity is with 100% or hyperbaric oxygen,
(Choice A ) Iron bound to heme is in the reduced ferrous (Fe“ ) state Oxidation of
ferrous iron to ferric iron (Fe-’ ) leads to the formation of methemoglobin .
Hemoglobin containing ferric iron (methemoglobin ) is unable to bind to oxygen.
(Choices B and C) Heme is an iron containing protoporphyrin IX . where ferrous iron
is held in the center of heme molecule by four nitrogens in the porphyrin ring
Carbon monoxide binds to iron in metalloproteins , but it does not alter porphyrin
oxidation in cytochrome oxidase or covalent linkage in cytochrome C.
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Inhaled carbon monoxide rapidly diffuses across the alveolar membrane and binds
tightly with heme-bound iron in hemoglobin and other hemeproteins. Carbon
monoxide binding to hemoglobin results in formation of carboxyhemoglobin. Carbon
monoxide decreases the oxygen content of the blood by occupying oxygen binding
sites . Carbon monoxide also inhibits the release of oxygen from hemoglobin in
tissues by altering hemoglobin conformation into the relaxed form that has a very
high affinity for oxygen This results in a leftward shift of the oxygen dissociation
curve and tissue hypoxia via deficient unloading of oxygen Treatment of carbon
monoxide toxicity is with 100% or hyperbaric oxygen
(Notes
*
t alcutator
A
U
.
-
(Choice A) Iron bound to heme is in the reduced ferrous (Fe ) state . Oxidation of
ferrous iron to feme iron ( Fe * - ) leads to the formation of methemoglobin .
Hemoglobin containing ferric iron ( methemoglobin ) is unable to bind to oxygen.
1
20
21
22
23
n
25
(Choices B and C) Heme is an iron containing protoporphyrin IX. where ferrous iron
is held in the center of heme molecule by four nitrogens in the porphyrin ring
Carbon monoxide binds to iron in metalloproteins , but it does not alter porphyrin
oxidation in cytochrome oxidase or covalent linkage in cytochrome C.
25
27
23
29
30
31
32
• 3
*
(Choice E) The partial pressure of oxygen ( pO . ) is dependent on the concentration
of oxygen dissolved in the plasma and not on actual concentration or content of
oxygen in hemoglobin . The amount of oxygen dissolved in plasma does not change
in CO poisoning, so the pO. is not decreased.
Educational Objective:
CO binds to hemoglobin with an affinity that is 220 times that of oxygen for
hemoglobin. The binding of CO and O. to hemoglobin are reversible . CO.
therefore , competes with O . for binding on the heme iron of hemoglobin
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b
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IS
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17
A 7-year - old African American boy is brought to the office by his parents . The
parents state that the boy has been hospitalized several times for severe pains in his
back and extremities The patient is very active when he is not in pain but gets quite
tired by the end of the day . He has no other medical problems and takes no
medications except acetaminophen for pain control. On examination , the
conjunctivae are pale Blood count reveals a hemoglobin level of 7.8 mgfcIL and a
reticulocyte count of 15% . A valine for glutamic acid substitution at position 6 of the
P globin chain of the hemoglobin molecule is suspected This patient 's hemoglobin
would most likely aggregate upon which of the following?
n
H
20
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n
2b
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27
C A. 2 ,3 -bisphosphoglycerate depletion
O B p globin chain folding
b
O C Capillary pH values >7.4
C D. Interaction with fetal hemoglobin
O £ Oxygen unloading
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*
Note
*
t alt ufalor
A
A 7*year - old African Amencan boy is brought to the office by his parents. The
parents state that the boy has been hospitalized several times for severe pains in his
back and extremities The patient is very active when he is not in pain but gets quite
tired by the end of the day. He has no other medical problems and takes no
medications except acetaminophen for pain control. On examination, the
conjunctivae are pale Blood count reveals a hemoglobin level of 7 8 mg/ dL and a
reticulocyte count of 15% . A valine for glutamic acid substitution at position 6 of the
P globin chain of the hemoglobin molecule is suspected This patient 's hemoglobin
would most likety aggregate upon which of the following?
1B
14
20
2\
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23
2i
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25
21
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29
C A. 2.3 -bisphosphogiycerate depletion [14%]
B. p globin chain folding [12%J
O C Capillary pH values >7.4 [5%]
O D . Interaction with fetal hemoglobin [5%]
* # E. Oxygen unloading [63%]
Explanation:
30
31
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33
In sickle cell ( hemoglobin S {HbS ]) anemia the nonpolar amino acid valine
replaces the charged ammo acid glutamate at position 6 of the p globin chain This
results in the alteration of a hydrophobic portion of the p globin chain that fits into a
complementary site on the a globin chain of another hemoglobin molecule As a
result , hemoglobin molecules aggregate under anoxic conditions After
polymerization , HbS initially forms a gel and then a meshwork of fibrous polymers
causing the red blood cells to distort into an abnormal sickle shape .
V
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(Notes
(
alcufdtor
jpovoj
A
Explanation:
In sickle ceil ( hemoglobin S [HbS]) anemia the nonpolar amino acid valine
replaces the charged amino acid glutamate at position 6 of the p globin chain This
results in the alteration of a hydrophobic portion of the p globin chain that fits into a
complementary site on the a globin chain of another hemoglobin molecule . As a
result , hemoglobin molecules aggregate under anoxic conditions After
polymerization , HbS initially forms a gel and then a meshwork of fibrous polymers
causing the red blood cells to distort into an abnormal sickle shape
b
Sickling is promoted by conditions associated with low oxygen levels, increased
acidity, or low blood volume ( dehydration). Sickfed cells are not flexible enough to
pass through microvasculature , As a result they impede blood flow and cause
microinfarcts in tissues and painful vasoocclusive crises. Organs in which blood
moves slowly ( eg , spleen, liver ) are predisposed to lower oxygen levels or
acidity. Organs with particularly high metabolic demands ( eg brain , muscles
placenta ) promote sickling by extracting more oxygen from the blood ( oxygen
unloading ) . The sickling process is complex and incompletely understood.
(Choices A and C) The molecule 2 , 3-bisphosphoglycerate ( 2 , 3-BPG ) binds the
globin chains ionically and stabilizes the taut ( T ) deoxyhemoglobin This binding
decreases hemoglobin' s oxygen affinity facilitating oxygen release at the tissue
2p
level . With 2.3-BPG depletion, hemoglobin affinity for oxygen increases (left shift on
oxygen hemoglobin dissociation curve ), and this results in oxygen uptake by
hemoglobin: therefore , erythrocyte sickling will decrease . Similarly, increased acidity
or low pH is associated with sickling , so decreased acidity with elevated capillary pH
values >7.4 would not promote sickling
-
.
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7
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A
(Choices A and C) The molecule 2 , 3- bisphosphoglycerate ( 2 , 3- BPG ) binds the 2 p
globin chains ionically and stabilizes the taut ( T ) deoxyhemoglobin . This binding
decreases hemoglobin' s oxygen affinity , facilitating oxygen release at the tissue
level With 2 3-BPG depletion , hemoglobin affinity for oxygen increases (left shift on
oxygen- hemoglobin dissociation curve ) and this results in oxygen uptake by
hemoglobin therefore erythrocyte sickling will decrease. Similarly , increased acidity
or low pH is associated with sickling , so decreased acidity with elevated capillary pH
values >7.4 would not promote sickling,
t
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19
20
21
22
23
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29
30
31
32
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(Choice B) The globin chains in the hemoglobin tetramer are folded compactly , with
nonpolar hydrophobic residues in the interior and charged polar residues on the
surface A valine for glutamic acid substitution does not result in a significant change
in p globin folding It is the mature hemoglobin tetramer that undergoes
polymerization, not Individual globin chains during folding.
b
(Choice D) HbS does not polymerize when fetal hemoglobin (HbF ) is present so
patients with sickle cel! anemia often do not have symptoms until the HbF fraction
decreases a few months after delivery . Some patients with HbS may have fewer
clinical manifestations because they produce larger amounts of HbF as adults
Educational objective:
Hemoglobin S ( HbS) aggregates in the deoxygenated state HbS polymers form
fibrous strands that reduce red blood cell membrane flexibility and promote
sickling Sickling occurs under conditions associated with anoxia including low pH
and high levels of 2 ,3-bisphosphoglycerate These inflexible erythrocytes
predispose to microvascular occlusion and microinfarcts.
|
References:
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A
5
6
7
8
9
10
A mutation m a non-coding sequence is believed to affect expression of the gene
coding for a specific fetal enzyme Liver and bone marrow cells from the fetus and
his parents are obtained Which of the following is the best method to determine if
this gene is being transcribed in cultures of isolated cells?
12
13
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15
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20
21
O A. Northern blot
< B . Western blot
k
O C Southern blot
D . Southwestern blot
O E. ELISA
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A
A mutation in a non-coding sequence is believed to affect expression of the gene
coding for a specific fetal enzyme Liver and bone marrow cells from the fetus and
his parents are obtained Which of the following is the best method to determine if
this gene is being transcribed in cultures of isolated cells?
b
12
13
15
15
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A . Northern blot [48%]
B . Western blot [17%]
O
O C . Southern blot [21%)
D . Southwestern blot [7%]
O E. ELISA [6 %J
21
22
23
n
25
25
27
23
29
Explanation:
The Southern Western , Northern and Southwestern blot procedures are techniques
used to analyze and identify DNA fragments, proteins , mRNA , and DNA-bound
proteins , respectively The best method for determining whether a gene is being
expressed is to analyze for the presence of its mRNA using a Northern blot ( Choice
A) In the case described above Northern blot analysis of each of the cell culture
samples can determine if mRNA corresponding to the gene of interest is being
transcribed
,
All of the "blot" tests rely on the same basic techniques First the components of
the unknown sample - DNA for Southern blots (Choice C) . mRNA for Northern blots
(Choice A) protein for Western blots (Choice B ) , and DNA -bound protein for
Southwestern blots (Choice D } - are separated by size and charge via gel
electrophoresis. The resultant bands are then blotted onto a nitrocellulose
,
,
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E. ELISA [6%J
&
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Explanation:
The Southern Western, Northern and Southwestern blot procedures are techniques
used to analyze and identify DNA fragments, proteins , mRNA , and DNA -bound
proteins , respectively . The best method for determining whether a gene is being
expressed is to analyze for the presence of its mRNA using a Northern blot ( Choice
A) In the case described above Northern blot analysis of each of the cell culture
samples can determine if mRNA corresponding to the gene of interest is being
transcribed
b
AH of the 'blot" tests rely on the same basic techniques . First the components of
the unknown sample - DNA for Southern blots (Choice C), mRNA for Northern blots
(Choice A ) , protein for Western blots (Choice B ) , and DNA -bound protein for
Southwestern blots (Choice D ) - are separated by size and charge via gel
electrophoresis . The resultant bands are then blotted onto a nitrocellulose
membrane and incubated with a labeled hybridization probe or antibody to identify
the specific DNA fragment , mRNA molecule or protein of interest
(Choice B) A complete failure of gene expression would lead to a failure to produce
protein while proper gene expression would presumably result in translation into
protein thus one could argue that Western blotting could also be used in this
case . The question , however, asks howto determine if the gene is being
transcribed As transcription refers to the production of mRNA from a DNA
template , a test that detects specific mRNA sequences (Northern blot ) would be
most desirable
(Choice E) The ELISA ( enzyme -linked immunosorbent assay ) is a test commonly
,
.
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expressed is to analyze for the presence of its mRNA using a Northern blot ( Choice
A) In the case described above Northern blot analysis of each of the cell culture
samples can determine if mRNA corresponding to the gene of interest is being
transcnbed
A
AH of the ''blot ' tests rely on the same basic techniques . First, the components of
the unknown sample - DNA for Southern blots (Choice C), mRNA for Northern blots
(Choice A) , protein for Western blots (Choice B) , and DNA -bound protein for
Southwestern blots (Choice D) - are separated by size and charge via gel
electrophoresis. The resultant bands are then blotted onto a nitrocellulose
membrane and incubated with a labeled hybridization probe or antibody to identify
the specific DNA fragment , mRNA molecule or protein of interest
(Choice B) A complete failure of gene expression would lead to a failure to produce
protein while proper gene expression would presumably result in translation into
protein , thus one could argue that Western blotting could also be used in this
case. The question, however, asks howto determine if the gene is being
transcnbed As transcription refers to the production of mRNA from a DNA
template a test that detects specific mRNA sequences (Northern blot) would be
most desirable
(Choice E) The ELISA ( enzyme -linked immunosorbent assay ) is a test commonly
employed to measure the amount of a protein in body fluids It can be quantitative
for example , to measure plasma insulin levels .
Educational Objective:Northern blots detect target mRNA in a sample to assess
gene expression.
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A 21- year -old male with vision impairment is found to have bilateral lens opacities .
He is otherwise asymptomatic and does not follow any specific diet . Laboratory
evaluation reveals urinary excretion of large amounts of galactose Activity of which
of the following enzymes contributes most to this patient's eye condition?
b
C A UDP-Hexose 4- epimerase
O B p-Galactosidase
O C . Aldolase B
O D Aldose reductase
C E . Glucose-6-phosphatase
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A
A 21-year-old male with vision impairment is found to have bilateral lens opacities .
He is otherwise asymptomatic and does not follow any specific diet . Laboratory
evaluation reveals urinary excretion of large amounts of galactose Activity of which
of the following enzymes contributes most to this patient's eye condition?
&
A UDP-Hexose 4- epimerase [ 7%]
C ) B. 3-Galactosidase [21%]
'
O C . Aldolase B [16%]
* ® 0. Aldose reductase [52%]
C E. Glucose - 6-phosphatase [3%]
21
n
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25
Explanation:
GHJiWTftJAJfJdTSe
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Gcilacbtol
(Causes cataract)
-—
Galactose
* Galactose IP
UDP-giucose
27
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ufXfyt tfftfTSfcwnse
0 - Gsuictostdaso
Lactose
synthase
Lactose *(Galactosyl P ~ 1 4 -glucose )
<
UDP-galactose Glucose IP
Glucose
Glucose 6pbosphate
. 1
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E. Glucose - 6-phosphatase [ 3%]
Explanation:
Gafnctokmas®
Afdoso rtKiuaaim
Ga lacLtd
( Cautes cataract )
—
Galaciosc
I
Galactose IP
UDP-glucase
12
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ufXJyi fran$f f9$P
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Lactose
(Galactosyl fH 4 -glucose )
UOP - qaiactoso Grucose IP
Glucose 6phosphale
21
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t
Pyruvate
The patient described in this question stem has a clinical presentation of bilateral
cataracts , but he is otherwise asymptomatic . This presentation is most consistent
with galactokinase deficiency , a form of galactosemia that causes a benign disorder
characterized by cataracts without hepatocellular manifestations. Classic
galactosemia , by comparison, results from ga!actose -1-phosphate uridyl transferase
(GALT) deficiency: this is the most common form of galactosemia. Patients with
GALT deficiency present with vomiting lethargy and failure to thrive soon after
feeding is begun. Other clinical findings of this disorder include impaired liver
function , hyperchloremic metabolic acidosis , and aminoaciduria. This disorder
results in severe symptoms after initiation of breast feeding . A normal newborn
obtains a large amount of their daily calories from lactose present in breastmilk .
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*
Motes
The patient described in this question stem has a clinical presentation of bilateral
cataracts , but he is otherwise asymptomatic. This presentation is most consistent
with galactokmase deficiency , a form of galactosemia that causes a benign disorder
characterized by cataracts without hepatocellular manifestations Classic
galactosemia , by comparison , results from ga = actose -1-phosphate undyl transferase
(GALT ) deficiency ; this is the most common form of galactosemia Patients with
GALT deficiency present with vomiting lethargy and failure to thrive soon after
feeding is begun. Other clinical findings of this disorder include impaired liver
function , hyperchloremic metabolic acidosis , and aminoaciduria. This disorder
results in severe symptoms after initiation of breast feeding A normal newborn
obtains a large amount of their daily calories from lactose present in breastmilk .
Following degradation of lactose and absorption of galactose and glucose
galactose is phosphorylated to galactose -1-phosphate by the enzyme
galactoklrase . A deficiency of galectokinase results in elevation of galactose levels
Excess circulating galactose is converted to galactrtol by aldose reductase and to
galactonic acid by galactose oxidase While galactonic acid can be metabolized by
the HMP shunt galactitol accumulates in cells Excess galactitol is responsible for
the formation of cataracts in patients with galactokinase deficiency. Dietary
restriction of lactose results in improvement in symptoms in all forms of
galactosemia.
t alculator
&
.
(Choice A) UDP galactose-4-epimerase converts UDP-galactose to UDP-glucose .
Deficiency of UDP galactose 4 - epimerase is an uncommon cause of disorders of
galactose metabolism .
-
(Choice B) Beta-galactosidase is one of the lysosomal enzymes responsible for
breakdown of glycosaminoglycans Beta -galactosidase deficiency results in
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(Choice B) Beta-galactosidase is one of the lysosomal enzymes responsible for
breakdown of glycosaminoglycans Beta-galactosidase deficiency results in
accumulation of keratin sulfate within lysosomes and manifests with short stature ,
normal intelligence atlantoaxial instability and valvular heart disease This enzyme is
also responsible for breakdown of lactose .
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(Notes
I alculalor
the formation of cataracts in patients with galactokinase deficiency, Dietary
restriction of lactose results in improvement in symptoms in all forms of
galactosemia.
(Choice A) UDP galactose-4 - epimerase converts UDP - galactose to UDP-glucose.
Deficiency of UDP galactose-4-epimerase is an uncommon cause of disorders of
galactose metabolism .
20
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Newt
A
&
(Choice C) Deficiency of aldolase B results in hereditary fructose intolerance
Patients with aldolase B deficiency become symptomatic after ingesting sucrose
and fructose -containing foods for the first time , typically during infancy.
(Choice E) Glucose - 6-phosphatase converts glucose - 6 -phosphate to glucose This
is the last step in production of glucose from gluconeogenesis and glycogenolysis.
Deficiency of glucose -6-phosphatase causes glycogen storage disease type 1, von
Gierke’s disease The main clinical manifestations are hypoglycemia lactic acidosis
hepatomegaly and hypertriglyceridemia .
Educational Objective:
Gaiactitol accumulates in lens of patients with galactosemia and causes osmotic
damage leading to cataract formation Gaiactitol is formed from excess circulating
galactose in galactosemia by aldose reductase
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A 47-year -ofd woman comes to the physician with progressive joint pain and swelling
in her hands for the past several months . She also complains of easy fatigability that
has gradually worsened . Morning activities are especially difficult for her due to
prolonged stiffness . Examination of her joints reveals warmth, swelling , and
tenderness involving the proximal interphaiangeal joints metacarpophalangeal joints
and wrists bilaterally Serum autoantibodies with high specificity for this patient's
condition are most likely to react with which of the following?
.
A Centromeres
O B . Crtrullinated peptides
C C Double - stranded DNA
D Fc portion of human IgG
C E . Nuclear basic proteins
O F . Phospholipids
27
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A 47- year-ofd woman comes to the physician with progressive joint pain and swelling
in her hands for the past several months . She also complains of easy fatigability that
has gradually worsened . Morning activities are especially difficult for her due to
prolonged stiffness . Examination of her joints reveals warmth swelling , and
tenderness involving the proximal interphaiangeal joints metacarpophalangeal joints
and wrists bilaterally Serum autoantibodies with high specificity for this patient' s
condition are most likely to react with which of the following?
,
b
O A . Centromeres [4 %]
*
20
21
•B
Citrullinated peptides (41%]
O C Double- stranded DNA [8%J
,
D . Fc portion of human IgG [42%|
22
23
24
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25
C E . Nuclear basic proteins [3%]
O F. Phospholipids [2% ]
27
23
29
Explanation:
This patient has symmetric polyarthritis (involving the MCP and PIP joints )
associated with morning stiffness lasting for >30 minutes her symptoms have been
present for >6 weeks These findings are suggestive of rheumatoid arthritis (RA )
The diagnosis is made clinically , but the presence of anti-cyclic citrullinated peptide
( anti-CCP) antibodies is helpful for confirmation.
,
Tissue inflammation causes arginine residues in proteins such as vimentin to be
enzymatically converted into citrulline through a process called citrullination This can
Ql
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F. Phospholipids [2%]
&
6
7
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14
IS
16
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14
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26
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29
*
Explanation:
This patient has symmetric polyarthritis (involving the MCP and PIP joints )
associated with morning stiffness lasting for > 30 minutes : her symptoms have been
present for >6 weeks . These findings are suggestive of rheumatoid arthritis (RA ),
The diagnosis is made clinically but the presence of anti-cyclic citrullinated peptide
( anti CCP) antibodies is helpful for confirmation .
b
-
Tissue inflammation causes arginine residues in proteins such as vimentin to be
enzymatically converted into citrulhne through a process called citrullination, This can
significantly alter the shape of these proteins , which can then serve as antigens and
generate an immune response In RA , this immune response is exaggerated ,
resulting in high titers of anthCCP antibodies that are not usually present in other
inflammatory conditions Thus , anti-CCP antibodies have a high specificity
( 95%-98% ) for RA . Antibodies to citrullinated peptides proteins are usually
measured by enzyme -iinked immunosorbent assay ( ELISA ) using a mixture of cyclic
citrullinated peptides as the antigen,
(Choice A) Anti-centromere antibodies are found in the majority of patients with
CREST syndrome .
(Choice C) Antibodies to double - stranded DNA lanti- dsDNA ) are specific for
systemic lupus erythematosus .
(Choice D) Rheumatoid factors (RF ) are autoantibodies targeting the Fc portion of
human IgG that occur in approximately 80% of patients with RA Their diagnostic
utility is limited by their poor specificity as they are found in approximately 10% of
lit
.1
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t dlculdlor
inTiarnrnaioryconamons . i nus , anu-uv,K aniioooies nave a mgn specmcuy
(95%-98% ) for RA . Antibodies To citrullinated peptides proteins are usually
measured by enzyme -linked immunosorbent assay ( ELISA ) using a mixture of cyclic
citrullinated peptides as the antigen
A
.
(Choice A) Anti-centromere antibodies are found in the majority of patients with
CREST syndrome .
&
(Choice C) Antibodies to double - stranded DNA lanti- dsDNAj are specific for
systemic lupus erythematosus .
(Choice D) Rheumatoid factors ( RF ) are autoantibodies targeting the Fc portion of
human IgG that occur in approximately 80% of patients with RA . Their diagnostic
utility is limited by their poor specificity as they are found in approximately 10% of
healthy individuals , in 30% of patients with systemic lupus erythematosus in nearly
all patients with mixed cryoglobulinemia , and in many other inflammatory conditions .
(Choice E) The presence of antinuclear antibodies ( ANA ) is a nonspecific finding in
many connective tissue disorders . Antinuclear antibodies characteristically occur in
IgM form in patients with RAt but they are found less frequently than rheumatoid
factors.
(Choice F) Anti-phospholipid antibodies are found in patients with systemic lupus
erythematosus and antiphospholipid antibody syndrome Antiphosphohpid antibody
syndrome causes hypercoagulability , paradoxical partial thromboplastin time ( PTT )
prolongation , and recurrent miscarriages ( spontaneous abortions ).
Educational objective:
Antibodies to citrullinated peptides /proteins have a high specificity for rheumatoid
arthritis.
v
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( <aicu ( dtor
A 24-year -old female presents to her primary care physician with complaints of a
rash that will not go away . The rash is present on the patient's face in a butterfly
distribution sparing the nasolabial folds She easily gets sunburned and thus avoids
going to beach with her family . Additionally, she states that she is getting tired more
and more easily. Laboratory studies show elevated antinuclear antibodies and
anti- dsDNA levels Further testing also reveals elevated anti-U1-snRNP titers .
These reflect increased levels of antibodies to U1- snRNP molecules snRNPs are
involved in what critical cellular function?
17
U
H
20
O A . Charging tRNA with amino acids
O B . Synthesizing Okazaki fragments
21
22
23
24
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26
27
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29
.
C Removal of introns from RNA transcripts
O D . Potyadenylation of RNA transcripts
O E . Aiding in mRNA to exit the nucleus
F . Allowing proper functioning of DNA ligase
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(Notes
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IS
115
A 24- year -old female presents to her primary care physician with complaints of a
rash that will not go away . The rash is present on the patient 's face in a butterfly
distribution sparing the nasolabial folds . She easily gets sunburned and thus avoids
going to beach with her family . Additionally, she states that she is getting tired more
and more easily . Laboratory studies show elevated antinuclear antibodies and
anti-dsDNA levels Further testing also reveals elevated anti-U1-snRNP titers .
These reflect increased levels of antibodies to U1- snRNP molecules . snRNPs are
involved in what critical cellular function ?
17
O A. Charging tRNA with amino acids [3%l
O B . Synthesizing Okazaki fragments [3%]
Ij
n
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*
* C. Removal of introns from RNA tnanscnpts [81%]
D. Polyadenylation of RNA transcripts [4%]
O E. Aiding in mRNA to exit the nucleus [7%]
F. Allowing proper functioning of DNA ligase [2%1
.
Explanation:
Synthesis of RNA from DNA (transcription ) occurs in the nucleus and is catalyzed by
three types of RNA polymerases. Transcription leads to the formation of messenger
RNA (mRNA ), ribosomal RNA (rRNA ) , transport RNA (tRNA ) and small nuclear
ribosomal proteins ( snRNPs).
Messenger RNA transfers the genetic code to the cytoplasm and serves as a
template for protein synthesis . Synthesis of mRNA is catalyzed by RNA polymerase
II smrl
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t alruhitor
Explanation:
/s
Synthesis of RNA from DNA (transcription ) occurs in the nucleus and is catalyzed by
three types of RNA polymerases Transcription leads to the formation of messenger
RNA (mRNA ), ribosomal RNA (rRNA ) . transport RNA (tRNA ) and small nuclear
ribosomal proteins ( snRNPs),
,
b
Messenger RNA transfers the genetic code to the cytoplasm and serves as a
template for protein synthesis . Synthesis of mRNA is catalyzed by RNA polymerase
II and occurs in two stages . During the first stage the RNA transcript ( exact copy of
the DNA template ) is produced . In the second stage the RNA transcript is
converted into mRNA . Processing of the RNA transcript involves:
,
,
1 -1
1 . RNA capping : addition of methylated guanine nucleotide to the 5 ' end
2 . RNA polyadenylation: addition of several adenine nucleotides to the 3 ' end
(poly-A tail).
3. RNA splicing removal of introns (non- coding regions ) Splicing is performed
by spliceosomes . which consist of snRNPs plus proteins . Synthesis of
snRNP also occurs in the nucleus catalyzed by RNA polymerase II.
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(Choice A) Ammoacyl-tRNA synthetases catalyze the linkage of tRNAs to the
corresponding ammo acids There is an enzyme specific for each amino acid
snRNPs are not involved in this reaction
(Choices B and F ) During DNA synthesis ( replication ) segmented fragments of
DNA are formed along the lagging strand of the template DNA These Okazaki
fragments are then joined by DNA ligase . Absence of snRNPs would not affect any
of these processes because snRNPs do not participate in replication
(Choices D and E) snRNPs are not responsible for RNA polyadenylation or mRNA
.
*
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(Notes
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2. RNA polyadenylation : addition of several adenine nucleotides to the 3‘end
(poly-A tail).
3. RNA splicing removal of introns (non- coding regions ) Splicing is performed
by spliceosomes which consist of snRNPs plus proteins . Synthesis of
snRNP also occurs in the nucleus catalyzed by RNA polymerase II.
(Choice A) Aminoacyl-tRNA synthetases catalyze the linkage of tRNAs to the
corresponding amino acids There is an enzyme specific for each amino acid
snRNPs are not involved in this reaction
(Choices B and F) During DNA synthesis (replication ) , segmented fragments of
DNA are formed along the lagging strand of the template DNA These Okazaki
fragments are then joined by DNA ligase Absence of snRNPs would not affect any
of these processes because snRNPs do not participate in replication
22
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(Choices D and E) snRNPs are not responsible for RNA polyadenylation or mRNA
ext from the nucleus .
Educational objective :
snRNPs ( small nuclear nbonucleoproteins ) are synthesized by RNA polymerase II in
the nucleus. They help to remove introns from the RNA transcript and are thus
necessary for synthesis of messenger RNA .
References:
t . The U1 - snRNP complex : structural properries relating to
autoimmune pathogenesis in rheumatic diseases.
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U
IS
115
17
13
14
20
A mutation in the TATA box of a eukaryotic gene that codes for a transmembrane
protein is most likely to affect which of the following functions?
O A. DNA methylation
O B . Transcription initiation
O C. Translation initiation
O D . RNA elongation
O E . Posttranscriptional RNA splicing
F . Polypeptide folding following translation
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15
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A
A mutation in the TATA box of a eukaryotic gene that codes for a transmembrane
protein is most likely to affect which of the following functions?
*
17
13
H
A . DNA methylation [2% ]
B . Transcnption initiation [80%J
C . Translation initiation [10%]
O 0 . RNA elongation [2%]
O E. Posttranscriptional RNA splicing [ 4%]
F, Polypeptide folding following translation [2%]
*
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Explanation:
Start
5*
transaction
Exon 1
CAAT
< 75bp >
-
TATA
t 25bp)
-
fc on 2
1
Irrtmn
*
V
{
'
Exon 3 {
T
ir.tron
5 upsiream promoter regions
Genetic information flows from DNA to RNA to proteins Most eukaryotic DNA
sequences consist of coding exons, non-coding introns, and two promoter regions
(the CAAT box and the TATA box ). The CAAT box is located 60 to 80 bases
upstream of the beginning (5' end } of the coding region and the TATA box is located
25 bases upstream from the beginning of the coding region . Gene transcription
begins when RNA polymerase II attaches at one of these promoter sites in a
V
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Explanation:
Start of transcription
\
5'
}
[
T
TATA
CAAT
i - 75bp
Ejtoffi 1
>
i - jrsbp )
irrtron
Eion 2
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3
IntrOfl
S' upstream promote regions.
*
Genetic information flows from DNAto RNA to proteins Most eukaryotic DNA
sequences consist of coding exons, non- coding introns, and two promoter regions
(the CAAT box and the TATA box ) . The CAAT box is located 60 to 80 bases
upstream of the beginning ( 5‘ end ) of the coding region and the TATA box is located
25 bases upstream from the beginning of the coding region Gene transcription
begins when RNA polymerase II attaches at one of these promoter sites in a
process facilitated by numerous general transcription factors i In eukaryotes . RNA
polymerase II alone is unable to recognize the TATA box ). Transcription factors and
DNA enhancer regions can associate with these promoter sites and increase the
affinity of RNA polymerase II for the promoter site thereby increasing gene
expression Though promoters are not directly translated into protein promoter
mutations can cause abnormal gene expression by altering the ability of RNA
polymerase M and transcription factors to bind.
(Choice A) DNA methylation is part of the epigenetic code It is earned out by DNA
methyitransferases and serves to silence the genes it affects .
(Choice C ) In eukaryotes translation initiation requires both ribosomal subunits (60S
illUKthair assneial
factors initial * tRNA char
SHUI 40S \ with
am*I rRNA inRNA initiati
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Notes
l olc uhitor
affinity of RNA polymerase II for the promoter site thereby increasing gene
expression Though promoters are not directly translated into protein , promoter
mutations can cause abnormal gene expression by altering the ability of RNA
polymerase II and transcription factors to bind .
A
(Choice A) DNA methyiation is part of the epigenetic code . It is earned out by DNA
methyttransferases and serves to silence the genes it affects.
(Choice C) In eukaryotes , translation initiation requires both ribosomal subunits ( 60S
and 40 Si with their associated rRNA mRNA initiation factors , initiator tRNA charged
with methionine (met-tRNAi) and GTP . The assembled ribosome then recognizes
the AUG start codon on mRNA to begin the process .
(Choice D) The TATA box only participates in the initiation of transcription .
Nucleotide addition to the forming RNA molecule continues until RNA polymerase
encounters a termination signal .
(Choice E) Posttranscoptional RNA splicing removes the introns from hnRNA .
Small nuclear ribonucleoprotein particles ( snRNPs ) facilitate this process.
(Choice F) The folding of a formed polypeptide into its secondary and tertiary
structures is entirely spontaneous and is determined completely by the amino acid
sequence in the protein's primary structure, Heat shock proteins assist in the
spontaneous refolding of proteins
.
Educational Objective:
The TATA box is a promoter region that binds transcription factors and RNA
polymerase II during the initiation of transcription It is located approximately 25
bases upstream from the beginning of the coding region .
fimo Qnanf R
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t akuldtor
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A 38-year -old man comes to the office due to pain in multiple joints . He has a 5 - year
history of lumbar pain and a 2- year history of bilateral knee pain . The patient works in
construction and his pain is worst after a long day on his feet . He has taken
ibuprofen intermittently , but the pain is no longer tolerable . The patient has a paternal
aunt with osteoarthritis . Physical examination shows blue -black spots on his sclera ©
and diffuse darkening of the auricular helices Which of the following is the most
likely cause of this patient's arthritis?
IS
1S
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A . Homogentisic acid dioxygenase deficiency
O B . Hyperuricemia
O C. Multifactorial articular cartilage failure
C D Recent infection with Salmonella
C E. Tyrosinase deficiency
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IS
115
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2X
A 38-year-old man comes to the office due to pain in multiple joints . He has a 5 *year
history of lumbar pain and a 2-year history of bilateral knee pain The patient works in
construction and his pain is worst after a long day on his feet . He has taken
tbuprofen intermittently, but the pain is no longer tolerable. The patient has a paternal
aunt with osteoarthritis . Physical examination shows blue -black spots on his sclerae
and diffuse darkening of the auricular helices Which of the following is the most
likely cause of this patient's arthritis?
@ A. Homogentisic acid dioxygenase deficiency [70%]
O B . Hyperuricemia [5%]
C Multifactonal articular cartilage failure [16%]
O D . Recent infection with Salmonella [1%]
O E . Tyrosinase deficiency [7%]
as
25
27
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Explanation:
Alkaptonuria
25
Dihydroptendme
reductase
BHi
BH2
Tyrosine
Phenylalanine
.. .
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Notes
I dlruldlor
Explanation:
/s
Alkaptonuria
7
a
5
10
D‘hydtoptf‘' rdine
reductase
12
13
Melanin
u
is
115
17
Phenylalanine
Tyrosine
13
Phenylalanine
H
hydroxylase
DORA
Catecholamines
20
21
n
Homogenhsate
23
2X
as
25
Homogent / sic ac -d
dioxygenase
?7
23
+
Maleylacetoacetate
Fumarylacetoacetate
Fumarate
V
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ft
ouWorld
Alkaptonuria is a relatively benign childhood disorder that is marked by severe
arthritis in adult life This autosomal-recessive disorder is caused by deficiency
of homogentisic acid dioxygenase which normally metabolizes homogentisic
acid into maleylacetoacetate Accumulated homogentisic acid causes pigment
deposits in connective tissue throughout the body . During adulthood these
blue-black deposits become apparent in the sclerae and ear cartilage . Deposits
also occur in the large joints and spine , causing ankylosis , motion restriction , and
significant pain A distinctive characteristic of alkaptonuria is that the urine of these
patients turns black when exposed to air due to oxidization of homogentisic acid .
,
(Choice B) Hyperuricemia can cause acute monoarticular gouty arthritis due to urate
crystal deposition in joints ( usually the great toe or knee ) These acute attacks
resolve in days to weeks and are not associated with connective tissue
hyperpigmentation.
(Choice C) Osteoarthritis is due to combined genetic , metabolic, and mechanical
factors that result in defects in articular cartilage Polyarticular joint involvement of
the fingers (including Heberden and Bouchard nodes), knees , hips , and spine
classically occur Joint pain typically peaks in the afternoon or evening after activity ,
but osteoarthritis is not associated with the blue - black deposits .
(Choice D) Reactive arthritis can occur following enteric or genitourinary infections
with organisms such as Salmonella , Shigella Campylobacter and Chlamydia ,
The typical pattern is asymmetric involvement of lower extremity joints accompanied
by enthesitis (inflammation at insertion of tendons), conjunctivitis , and urethritis .
Reactive arthritis has no associated skin findings in contrast to alkaptonuria
If * h /HPS Ci
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(Choice C) Osteoarthritis is due to combined genetic , metabolic , and mechanical
factors that result in defects in articular cartilage Polyarticular joint involvement of
the fingers (including Heberden and Bouchard nodes ) , knees , hips , and spine
classically occur Joint pain typically peaks in the afternoon or evening after activity ,
but osteoarthritis is not associated with the blue - black deposits .
it
(Choice D) Reactive arthritis can occur following enteric or genitourinary infections
with organisms such as Salmonella , Shigella Campylobacter and Chlamydia ,
The typical pattern is asymmetric involvement of lower extremity joints accompanied
by enthesitis (inflammation at insertion of tendons ), conjunctivitis , and urethritis .
Reactive arthntis has no associated skin findings in contrast to alkaptonuria,
H
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(Choice E) Albinism is caused by defects in biosynthesis and distribution of
melanin Melanocytes synthesize melanin from tyrosine via the enzyme tyrosinase
Educational objective:
Alkaptonuria is an autosomal-recessive disorder caused by a deficiency of
homogentisic acid dioxygenase , an enzyme involved in tyrosine metabolism.
Excess homogentisic acid causes diffuse blue-black deposits in connective tissues
Adults have sclerae and ear cartilage hyperpigmentation along with osteoarthropathy
of the spine and large joints .
References:
1. Alkaptonuria.
2 Alkaptonuria .
,
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t alculdtor
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n
The DNA replication process in eukaryotic cells closely mimics that in prokaryotic
cells , but the volume of genetic material to be replicated is typically much greater in
eukaryotic cells Which of the following ensures fast DNA replication in eukaryotic
cells?
C A . Energy-independent DNA unwinding
is
18
1?
13
O B Multiple origins of replication
2Q
C E . No proofreading by DNA polymerase (pol 5 )
21
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C . No RNA pnmers synthesized dunng replication
O D . Continuous synthesis of the lagging strand
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(Notes
4
&
6
a
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A
The DNA replication process in eukaryotic cells closely mimics that in prokaryotic
cells , but the volume of genetic material to be replicated is typically much greater in
eukaryotic cells Which of the following ensures fast DNA replication in eukaryotic
cells?
&
12
13
14
IS
15
17
C A Energy-independent DNA unwinding [3%]
* • EL Multip e ongins of replication [85%]
(
C No RNA pnmers synthesized during replication [3%]
D . Continuous synthesis of the lagging strand [6%]
n
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20
O E. No proofreading by DNA polymerase (pol 5) [2%]
21
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Explanation:
Pfokflfyolo replication
Eukaryotic replication
on
28
On iOrigin of replication )
29
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Notes
A
E. No proofreading by DNA polymerase (pol 5 } [2%]
6
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9
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Explanation:
Prokaryote replication
&
Eukaryotic replication
On
13
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The process of DNA replication is similar in eukaryotes and prokaryotes The key
steps involved in DNA replication are :
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The process of DMA replication is similar in eukaryotes and prokaryotes The key
steps involved in DNA replication are :
u
1. Unwinding of double stranded DNA ( dsDNA ) by helicase to produce single
stranded DNA ( ssDNA )
2. Formation of a replication fork
3. Formation of an RNA primer by the action of the enzyme primase
4. Synthesis and concurrent proofreading of daughter DNA strands by DNA
polymerases
5. Ligation of Okazaki fragments on lagging strands by hgase and removal and
replacement of RNA primers with DNA by DNA polymerase I
6. Reconstitution of chromatin and ligation of daughter strands
In £. co( i , a prokaryote the three major types of DNA polymerase are DNA
polymerase I, II and III . In eukaryotes , there are five major DNA polymerases :
alpha , beta, gamma delta and epsilon Though the eukaryotic genome is much
larger and more complex than the prokaryotic genome , interestingly the size of the
eukaryotic genome is not the source of its complexity Its complexity results from
the presence of a large number of non-coding DNA regions between coding
regions . Wtthm genes there are introns (Non-coding regions - Think "IN” between )
separating exons (Coding regions Think ‘'EX 'pressed) . Prokaryotes rarely have
introns within their genes .
-
In contrast to prokaryotes, which typically have a single origin of replication
eukaryotes have multiple origins of replication With multiple origins of replicationi
the genome can be copied much more quickly because multiple regions are being
replicated at once .
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A
polymerases
5. Ligation of Okazaki fragments on lagging strands by ligase and removal and
replacement of RNA primers with DNA by DNA polymerase I
6. Reconstitution of chromatin and ligation of daughter strands
b
In E . coli , a prokaryote , the three major types of DNA polymerase are DNA
polymerase I, II and III . In eukaryotes there are five major DNA polymerases
alpha , beta, gamma delta and epsilon. Though the eukaryotic genome is much
larger and more complex than the prokaryotic genome interestingly , the size of the
eukaryotic genome is not the source of its complexity . Its complexity results from
the presence of a large number of non-coding DNA regions between coding
regions . Within genes there are introns (Non-coding regions - Think "IN" between )
separating exons (Coding regions - Think "EX "pressed|. Prokaryotes rarely have
introns within their genes .
In contrast to prokaryotes, which typically have a single origin of replication ,
eukaryotes have multiple origins of replication . With multiple origins of replication ,
the genome can be copied much more quickly because multiple regions are being
replicated at once ,
.
(Choices A . C D and E) The processes of unwinding synthesis of an RNA primer,
synthesis of leading and lagging strands , and the proofreading activity of DNA
polymerases are similar in prokaryotes and eukaryotes .
Educational Objective:
Multiple origins of replication make eukaryotic DNA synthesis quick and effective
despite the large size of the genome compared to that of prokaryotic organisms .
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20
Several tissues use triglyceride breakdown products as intermediates in energy
generation and glucose synthesis Which of the following is a liver - specific enzyme
that facilitates such reactions?
O A. Acyl CoA synthetase
O B . Glycerol kinase
C Glucose-6-phosphate dehydrogenase
OD
Acetyl CoA carboxylase
O E ATP-citrate lyase
21
n
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t ulcuidtor
A
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7
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13
U
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15
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19
A
Several tissues use triglyceride breakdown products as intermediates in energy
generation and glucose synthesis Which of the following is a liver - specific enzyme
that facilitates such reactions?
I-
O A. Acyl CoA synthetase [22%]
v ® B. Glycerol kinase [36%]
. C . Glucose 6 -phosphate dehydrogenase [10%]
O D. Acetyl CoA carboxylase [28%]
O E. ATP-citrate lyase [ 4%]
20
21
n
Explanation:
23
21
Triglycende
2S
26
1
1 [
27
23
29
Lipase
Glycerol
Klfl
{lived
+
Fatly acids
—
Beta oxidalion/Kelogenesis
.
***
Glycerol 3-phosphate
i
DHAP
* ^/
G7ycoV o
Energy
\ GhJcon aQ
,
8
9nests
Glucose
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t alculator
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Glycerol
+
Fatly acids
> Beta oxidalion/Kelogenesis
.
GiycQrvi KmaSM
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Glycerol 3 - phosphate
I
DHAP
21
21
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71
23
29
Lab Value
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' -J
Triglyceride
20
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Explanation:
9
10
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115
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Gtycotysis
/
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Energy
\
( jfucon6og6nesis
Glucose
Triglyceride stored in adipose tissue is metabolized to free fatty acids and glycerol
by hormone sensitive lipase. Glycerol is then transported to the liver where it is
phosphorylated to glycerol-3 -phosphate by the iiver - specific enzyme glycerol hinase
(Choice B ). Subsequently , glycerol-3-phosphate is converted to dihydroxyacetone
phosphate ( DHAP ) by giycerol- 3 -phosphate dehydrogenase DHAP can be used to
produce ATP through glycolysis or glucose through gluconeogenesis . Glycerol in
the liver can also be utilized for triglyceride synthesis.
(Choice A) In order to undergo beta -oxidation in mitochondria , fatty acids must first
be activated to the acyl-CoA form in the cytoplasm . This reaction is catalyzed by
fatty acyl-CoA synthetase The fatty acyl- CoA must then be combined with carnitine
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A
phosphorylated to glycerol-3 phosphate by the liver -specific enzyme glycerol kinase
(Choice B) . Subsequently , glycerol-3-phosphate is converted to dihydroxyacetone
phosphate ( DHAP ) by glyceroi-3 -phosphate dehydrogenase DHAP can be used to
produce ATP through glycolysis or glucose through gluconeogenesis . Glycerol in
the liver can also be utilized for triglyceride synthesis.
(Choice A) In order to undergo beta -oxidation in mitochondria , fatty acids must first
be activated to the acyl-CoA form in the cytoplasm . This reaction is catalyzed by
fatty acyl-CoA synthetase . The fatty acyl-CoA must then be combined with carnitine
in order to be transported into the mitochondrion,
(Choice C) G!ucose-6-phosphate dehydrogenase is the first enzyme of the hexose
monophosphate (HMP ) shunt The HMP shunt produces NADPH and pentose
b
sugars for nucleotide synthesis .
(Choices D and E) Acetyl CoA carboxylase (ACC ) is a biotin-dependent enzyme
present in both liver and adipose tissues it catalyzes the first committed step in fatty
acid synthesis , the conversion of acetyl CoA to malonyl CoA, In contrast to fatty acid
oxidation , which occurs in mitochondria , fatty acid synthesis occurs in the
cytosol . Acetyl CoA generated in mitochondria by p-oxidation is transferred to the
cytosol as citrate , In the cytoplasm. ATP-crtrate lyase converts citrate back to
oxaloacetate and acetyl CoA .
Educational Objective:
Only the liver can utilize the glycerol produced by the degradation of triglycendes by
hormone sensitive lipase In the liver , glycerol is used for triacylglycerol synthesis,
gluconeogenesis and as an intermediate in glycolysis . ( Adipose tissue synthesizes
the glycerol phosphate needed for triacylglycerol synthesis from dihydroxyacetone
|
)
phosphate (DHAP
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(Notes
1 4ilc ufator
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29
A 6-month-old full-term boy is brought to the emergency department with lethargy
and vomiting . He was bom by uncomplicated spontaneous vaginal delivery and has
been growing and developing normally. The patient was breastfed exclusively until 2
days ago when homemade pureed food was added to his diet . He has had no fever
or diarrhea . His parents are healthy and he has had no sick contacts , Examination
shows a pale , diaphoretic , and ill- appearing infant . Serum glucose is 30 mg/ dL
Diagnostic testing confirms aldolase B deficiency Which of the following should be
removed from this patient's diet?
b
A Amylose
(
)
B Cellulose
C Galactose
O D Glucose
O E . Lactose
O F . Maltose
O G Sucrose
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I ulcutator
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a
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15
A
A 6-month-old full-term boy is brought to the emergency department with lethargy
and vomiting . He was born by uncomplicated spontaneous vaginal delivery and has
been growing and developing normally. The patient was breastfed exclusively until 2
days ago when homemade pureed food was added to his diet . He has had no fever
or diarrhea . His parents are healthy and he has had no sick contacts Examination
shows a pale , diaphoretic , and ill- appearing infant Serum glucose is 30 mg^ dL
Diagnostic testing confirms aldolase B deficiency . Which of the following should be
removed from this patient's diet ?
&
17
13
H
20
21
22
23
24
25
25
71
23
29
O A. Amylose [1%]
O B . Cellulose [1%]
O C . Galactose J16%]
O D . Glucose [2%I
O E. Lactose [5%]
C F. Maltose [4%]
' <• G. Sucrose [71%]
Explanation:
Disorders of fructose metabolism
Glucose
He
Glucose 6'phospbate
Sucrose
V
1
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I / 0J
*
(Notes
Ialculdlor
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Explanation:
7
a
b
Disorders of fructose metabolism
10
Glucose
12
HexOktn&se
13
14
is
IS
Glucose 6 phosphate
Sucrose
PtKiSphOgiuCO
17
H
Essential
fructosuris
20
( benign coodrtion )
13
isomerose
Fructose
Hexahmase
- Fmcrose 6 phosphate
-
21
22
-
Fructose - 1, 6
bisphospfratose
FructoksnAse
23
24
25
25
77
23
Fructose - 1. 6 -bisphosphate
Fructose 1 - phosphale
Akfoiase B
29
/
DHAP
Aldolase A & B
Glyceratdehyde
Hereditary fructose intolerance
Hypoglycemia 4 vomiting after
fmdose ingestion
Block Time Remaininq :
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Glyceraklehyde
3 -phosphate
Tnose
t
i
i
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(
1
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I
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18 : 12
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*
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( <i( culdior
4
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7
Carbohydrates are classified as monosaccharides , disaccharides and
polysaccharides Disaccharides and polysaccharides must be broken down to
their monosaccharide components for energy production and utilization.
£
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,
b
Aldolase B metabolizes fructose- 1-phosphate , a product of fructose metabolism, to
dihydroxyacetone phosphate (DHAP ) and glyceraldehyde, which can then enter the
glycolytic pathway Aldolase B deficiency can result in fructose-1-phosphate
accumulation , this toxic metabolite depletes intracellular phosphate and inhibits the
activation of hepatic phosphorylase and gluconeogenesis . The resulting condition
hereditary fructose intolerance ( eg . fructosemia ) , is an autosomal recessive
disorder .
Because gluconeogenesis is impaired fructosemia typically presents with
life- threatening hypoglycemia Consequences of hypoglycemia include lethargy ,
sweating vomiting and dehydration These symptoms manifest after intake of
fructose or sucrose such as from formula or fruit . Strict abstinence from dietary
fructose and sucrose can result in dramatic recovery.
,
(Choices A and D) Starch ( similar to glycogen in mammals ) is the major storage
form of carbohydrates in plants and contains only glucose molecules Starch
consists of an unbranched portion composed of amylose and a branched portion
called amylopectin . Patients with fructosemia have normal metabolism of glucose .
(Choice B) Cellulose is a linear polysaccharide of glucose that is mainly present in
the cell wall of plant cells . Cellulose is an insoluble , indigestible dietary fiber that is
responsible for the bulk of fecal matter
(Choices C , E. and F) Breast milk contains the disaccharides lactose ( composed
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yijy i MCIT
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(Notes
Because gluconeogenesis is impaired fructosemia typically presents with
life-threatening hypoglycemia Consequences of hypoglycemia include lethargy ,
sweating , vomiting , and dehydration . These symptoms manifest after intake of
fructose or sucrose such as from formula or fruit Strict abstinence from dietary
fructose and sucrose can result in dramatic recovery.
l
(Choices A and D) Starch ( similar to glycogen in mammals i is the major storage
form of carbohydrates in plants and contains only glucose molecules . Starch
consists of an unbranched portion composed of amylose and a branched portion
called amylopectin . Patients with fructosemia have normal metabolism of glucose.
(Choice B) Cellulose is a linear polysaccharide of glucose that is mainly present in
the cell wall of plant cells Cellulose is an insoluble indigestible dietary fiber that is
responsible for the bulk of fecal matter
(Choices C, E . and F) Breast milk contains the disaccharides lactose (composed
of galactose and glucose ) and maltose { composed of 2 glucose molecules ).
Patients with aidolase B deficiency can consume these disaccharides as their
breakdown will not produce fructose However , patients with galactosemia cannot
metabolize galactose in breast milk or cow's milk -based formula These patients
typically present in the first few days of Life with jaundice, vomiting, poor feeding , and
hepatomegaly.
Educational objective:
Aldolase B deficiency, or hereditary fructose intolerance , leads to accumulation of
the toxic metabolite fructose-1-phosphate . Patients have hypoglycemia and vomiting
when fructose or sucrose is consumed. Treatment involves strict removal of both
carbohydrates from the diet .
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*
( <ilr
uitilor
4
b
G
7
a
TT
*
12
13
U
IS
IS
17
The following vignette applies to the next 2 items The items in the set must be
answered in sequential order Once you click Proceed to Next Item , you will not
be able to add or change an answer .
b
As part of a long-term cohort study , members of a large extended family undergo
periodic analysis of multiple serum markers Many male participants are found to
have abnormal laboratory results despite no obvious signs of disease , Further
analysis shows that these men have an X -linked mutation affecting the
phosphoribosyl pyrophosphate ( PRPP ) synthetase gene resulting in greatly
increased substrate conversion.
13
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2S
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Item 1 of 2
Which of the following organs is most likely to develop pathology secondary to this
mutation?
O A. Aorta
O B . Heart
O C. Joints
O D . Liver
.
E. Pancreas
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I alculdtor
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1fi
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20
21
n
The following vignette applies to the next 2 items The items in the set must be
answered in sequential order Once you click Proceed to Next Item , you will not
be able to add or change an answer .
As part of a long-term cohort study , members of a large extended family undergo
periodic analysis of multiple serum markers Many male participants are found to
have abnormal laboratory results despite no obvious signs of disease , Further
analysis shows that these men have an X -linked mutation affecting the
phosphoribosyl pyrophosphate ( PRPP ) synthetase gene resulting in greatly
increased substrate conversion .
Item 1 of 2
Which of the following organs is most likely to develop pathology secondary to this
mutation?
24
2&
25
?7
23
29
O A. Aorta [3%]
O B . Heart [5%J
* » C . Joints [55%]
O D . Liver [33%]
O E. Pancreas [4%]
Explanation:
De novo purine synthesis
ffc ? lu «
^
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Ue novo purine synthesis
A
&
6
*
Notes
(
a l l ufalor
A
Ribose 5 phosphate
-
7
I
6
"r
®
12
13
U
IS
115
17
13
'
PRPP
synthetase
PRPP
AMP
IMP
o—
GMP
u
PRPP
amidotransferase
H
20
Phosphoribosylamine
5‘
21
22
21
24
2S
26
Nitrogen source
Carbon donor
let rahydrofolate
Glycine
Aspartate
Glutamine
27
23
29
Inoslne monophosphate
(IMP)
ATP
ADP
GTP
7V
-
GMP
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© UWwta
GMP
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Note
I alculalor
*
AMP
A
Gout is a disease caused by tissue deposition of monosodium urate crystals .
Elevated uric acid levels are a known nsk factor for gout and increased punne
metabolism is one possible cause of hyperuncemia Phosphoribosyl pyrophosphate
( PRPP ) synthetase is the enzyme responsible for the production of the activated
ribose necessary for de novo synthesis of purine and pyrimidine nucleotides The
mutation described in the question stem will cause increased production of punnes
due to feed-forward activation of the punne synthesis pathway . As a result , more
purine molecules will undergo degradation resulting in hyperuricemia and an
increased risk of gout .
(Choice A) The aorta can develop aneurysms in patients with Marfan syndrome ,
which results from defects in fibrillm-1.
b
(Choices B and D) The heart and liver car be affected by glycogen storage
diseases resulting from a variety of enzyme deficiencies such as
glucose-6 -phosphatase deficiency { von Gierke disease ) and acid maltase deficiency
(Pompe disease).
(Choice E) The pancreas is affected in patients with cystic fibrosis , which results
from a mutation in the cystic fibrosis transmembrane conductance regulator gene
Common sequelae include pancreatitis, pancreatic insufficiency, and destruction of
islet cells
Educational objective:
Gout occurs with increased frequency in patients with activating mutations involving
phosphoribosyl pyrophosphate synthetase due to increased production and
degradation of purines.
v
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Notes
(
alculdior
Item 2 of 2
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15
1?
ia
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2Q
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2i
Incidentally one of the male patients followed in the study is hospitalized with right
knee pain and swelling . A sample of his synovial fluid shows negatively birefringent
crystals under polarized light microscopy . To achieve rapid improvement in this
patient' s symptoms, therapy should be directed toward inhibiting which of the
following types of cells?
b
A . Eosinophils
C B. Lymphocytes
C . Neutrophils
D Synovial cells
0 E . Mast cells
25
25
21
23
29
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Item 2 of 2
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13
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IS
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1?
ia
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2X
2&
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27
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29
Incidentally one of the male patents followed in the study is hospitalized with right
knee pain and swelling A sample of his synovial fluid shows negatively birefringent
crystals under polarized light microscopy . To achieve rapid improvement in this
patient' s symptoms, therapy should be directed toward inhibiting which of the
following types of cells?
A Eosinophils [3%]
B Lymphocytes [15%]
v < C . Neutrophils [70%]
*
O D . Synovial cells [7%]
O E, Mast cells [6%]
f
Explanation:
Acute gouty arthritis
Signs &
symptoms
Diagnosis
* Usually involves first metatarsophalangeal joint or knee
* Swelling, erythema & exquisite tenderness
• Symptoms develop rapidly over 24 hours
* Joint aspiralion shows needle -shaped, negatively
birefringent crystals
V
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Notes
(
alculator
A
Explanation:
Acute gouty arthritis
12
13
H
IS
15
17
13
Signs &
symptoms
* Usually involves first metatarsophalangeal joint or knee
• Swelling
*
Diagnosis
nrylhema & exquisite tenderness
Symptoms develop rapidly over 24 hours
* Join I aspiration show needle - shaped , negatively
blrefringent crystals
H
2Q
• NSAIDs (eg
21
n
23
2X
2&
25
27
23
29
Treatment
,
naproxen, indomothacin) preferred if
no contraindications
* Colchicine used as second line therapy
GUjVLfcWo - ' . J. LLC
This patient' s synovial fluid analysis shows negatively blrefringent crystals (ie
monosodmm urate crystals ) under polarized light , which is diagnostic for gouty
arthritis Neutrophils are the primary cells responsible for the intense inflammatory
response seen in patients with gout Phagocytosis of urate crystals by neutrophils
causes the release of various cytokines and inflammatory mediators that lead to
further neutrophil activation and chemotaxis . resulting in a positive feedback loop that
amplifies the Inflammatory response .
Nonsteroidal anti-inflammatory drugs { NSAIDs ) are first- line therapy for treating acute
gouty arthritis , They inhibit prostanoid biosynthesis (eg , prostaglandins prostacyclin
thromboxanes ), exerting a broad anti-inflammatory effect that includes inhibition of
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i
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r
12
13
14
IS
16
1?
13
19
20
21
n
23
21
26
25
2T
26
26
Notes
I dlculdtor
d
gouty arthritis. They inhibit prostanoid biosynthesis Leg. prostaglandins prostacyclin ,
thromboxanes ), exerting a broad anti-inflammatory effect that includes inhibition of
neutrophils . Patients with contraindications to NSAIDs ( eg , peptic ulcer disease,
renal impairment ) are often treated with colchicine , which impairs neutrophil migration
and phagocytosis by interfering with microtubule formation . Colchicine also
decreases tyrosine phosphorylation in response to monosodium urate crystals
resulting in decreased neutrophil activation.
(Choice A ) Eosinophils function in defense against parasitic infections and are also
pathogenic in patients with asthma , allergy , hypereosinophilic syndromes and
vasculitides such as Churg-Strauss syndrome .
(Choice B) Lymphocytes produce delayed-type hypersensitivity reactions that do
not play a role in gout.
(Choice D) Synovial cells and macrophages play a role in initiating the inflammatory
response in gouty arthritis However targeting these cells would not eliminate the
inflammatory amplification caused by neutrophils , which is the central mechanism
involved in precipitating an acute gouty attack.
,
(Choice E) Mast cell degranulation can be inhibited by medications such as
cromolyn sodium, which is used in conditions such as asthma and allergic rhinitis.
Educational objective:
Nonsteroidal anti-inflammatory drugs ( NSAIDs ) are first - line therapy for treating acute
gouty arthritis . They inhibit cyclooxygenase and exert a broad anti-inflammatory
effect that includes inhibition of neutrophils. When NSAIDs are contraindicated ( eg
peptic ulcer disease , renal impairment ), colchicine is useful in the acute
management of gout as it inhibits neutrophil chemotaxis and phagocytosis by
preventing microtubule formation.
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(
alculaior
4
&
6
7
a
9
10
11
13
U
IS
115
RNA polymerase II is the enzyme responsible for eukaryotic gene transcnption
Gene-associated enhancer sequences can increase the rate at which transcription is
initiated . Most enhancers are located where, with respect to the transcription start
site?
O A. 25-30 base pairs upstream
O B At least 70 base pairs upstream
13
C C . Within the transcnbed portion of the gene
O 0 Downstream of the gene
20
C E. Variable locations
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4
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IS
115
17
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RNA polymerase II is the enzyme responsible for eukaryotic gene transcription
Gene-associated enhancer sequences can increase the rale at which transcription is
initiated Most enhancers are located where, with respect to the transcription start
site?
O A . 25-30 base pairs upstream [28%]
O B. At least 70 base pairs upstream [15%]
C Within the transcribed portion of the gene [6%]
O D . Downstream of the gene [5%]
* ® E. Vanable locations [46%]
Explanation:
In eukaryotic gene transcription , nuclear RNA polymerase II uses a DNA template to
generate complementary mRNA , which is then processed and translated into
protein Eukaryotic genes have associated promoter and enhancer sequences .
Promoter regions serve as binding sites for transcription factors and RNA
polymerase 11 . There are two types of eukaryotic promoter regions : 1 ) The TATA , or
Hogness , box is located approximately 25 nucleotides upstream from the gene
being transcnbed , and 2} The CAAT box is 70 to 80 bases upstream from the gene
In contrast to promoters , enhancers increase the rate of transcnption initiation
through protein binding and interactions with transcription factors bound to promoter
sequences Enhancers can be located upstream or downstream from the gene
being transcribed , and may be near the gene or thousands of base pairs away.
( Enhancers have been identified both within introns of the gene being transcribed as
luoll
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*
I ulculdtor
C B . At least 70 base pairs upstream [15%]
O
A
C Within the transcribed portion of the gene [6%]
D. Downstream of the gene [5%]
* # E. Variable locations [46%]
Explanation:
13
14
15
IS
17
13
2Q
21
n
23
24
25
25
?T
28
29
u
In eukaryotic gene transcription , nuclear RNA polymerase II uses a DNA template to
generate complementary mRNA , which is then processed and translated into
protein. Eukaryotic genes have associated promoter and enhancer sequences.
Promoter regions serve as binding sites for transcription factors and RNA
polymerase II . There are two types of eukaryotic promoter regions : 1) The TATA, or
Hogness , box is located approximately 25 nucleotides upstream from the gene
being transcribed , and 2 } The CAAT box is 70 to 80 bases upstream from the gene
In contrast to promoters , enhancers increase the rate of transcription initiation
through protein binding and interactions with transcripton factors bound to promoter
sequences Enhancers can be located upstream or downstream from the gene
being transcribed , and may be near the gene or thousands of base pairs away .
{ Enhancers have been identified both within introns of the gene being transcribed as
well as on separate chromosomes ) Repressor elements are similar to enhancer
elements , but they decrease instead of enhance transcription rates.
Educational Objective:
Enhancers/ repressors may be located anywhere upstream downstream or even
within the transcribed gene In contrast , promoter regions are typically located 25 or
70 bases upstream from their associated genes
Time Spent 6 seconds
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7
a
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n
w
A 5-year - old boy is brought to the physician by his parents because of easy
bruising. Physical examination reveals soft and loose skin as well as multiple
ecchymoses over the forearm and pretibial regions. Histologic evaluation with
electron microscopy shows type 1 collagen fibnlsthat are abnormally thin and
irregular . Biochemical evaluation reveals the presence of disulfide -rich globular
domains within a purified sample of mature collagen fibrils Which of the following
stages of collagen synthesis is most likely impaired in this patient?
c>
19
115
21
C A. Amino acid incorporation into proteins
O B . Extracellular protein cleavage
C C . Lysine residue hydroxylation
23
24
C D Signal peptide recognition
C E. Triple helix formation
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Notes
I dlculaior
4
&
6
i
8
9
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11
12
14
IS
15
1?
13
H
20
21
n
21
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25
25
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ft
-
A 5 year - old boy is brought to the physician by his parents because of easy
bruising Physical examination reveals soft and loose skin as well as multiple
ecchymoses over the forearm and pretibial regions. Histologic evaluation with
electron microscopy shows type 1 collagen fibrils that are abnormally thin and
irregular . Biochemical evaluation reveals the presence of disulfide -rich globular
domains within a purified sample of mature collagen fibrils Which of the following
stages of collagen synthesis is most likely impaired in this patient?
i-
A Amino acid incorporation into proteins [3%]
v ® B. Extracellular protein cleavage [34%]
;
C Lysine residue hydroxylation [26%]
O D . Signal peptide recognition [1%]
O E. Triple helix formation [36% ]
Explanation:
Collagen synthesis
Signal sequence directs
growing polypeptide chain
into endoplasmic reticulum
Preproa- chains
Signal sequence is cleaved
PrOd Chains
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Signal sequence directs
growing polypeptide chain
into endoplasmic reticulum
Preproa- chains
Signal sequence is cleaved
Pro O Chains
-
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23
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I alculdior
Collagen synthesis
20
21
Notes
*
/s
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Lab Value
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Explanation:
i
8
9
10
'
Hydroxylation of selected
proiine S lysine residues
{ vitamin C dependent )
OH
Glycosylation of
selected hydroxylysine
OH
OH
residues
i
I
-
t
OH
OH
f
OH
Galactose
Glucose
•
Assembly of pro -a - chams into
procollagen triple helix
Procollagen
V
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Notes
t <i l t utdlor
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procollagen triple helix
b
6
7
a
9
10
11
1?
Procollagen transferred to
Golgi apparatus & secreted
into extracellular matrix
H
IS
is
17
Terminal propeptides cleaved
by N & C procollagen
ij
i
-
H
peptidases
20
Tropocollagen
21
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21
2i
2b
25
27
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I
Collagen molecules
Collagen fibrils
spontaneously
assemble
Covalent cross links
formed by lysyl oxidase
I
a»j
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OUWortd
The process of type I collagen synthesis begins with collagen c -chain translation in
I
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(
aicutalor
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CUWorld
The process of type I collagen synthesis begins with collagen ct -chain translation in
the cytosol . Shortly after translation begins , a hydrophobic signal sequence at the Nterminus of the protein directs the ribosome to the rough endoplasmic reticulum
[ RER ), where it extrudes the growing polypeptide chain into the RER cisternae
(Choices A and D) . Within the RER . numerous posttranslahonal modifications are
made to the peptide, including cleavage of the signal sequence and hydroxylation
and glycosylation of certain residues Intrachain disulfide bonds then form at the Nand C -terminal globular regions, helping to stabilize them . Special interchain
disulfide bonds also form at the C -terminus between 3 pro -o - chains aligning the
helical domains into a conformation favorable for triple helix formation (Choice
E) The resulting procollagen molecule then undergoes exocytosis after which the
N- and C-terminal nonhelical regions are cleaved from the triple helix by procollagen
peptidases forming tropocollagen Tropocoilagen subunits then self -assemble into
collagen fibrils that are subsequently crosslinked by lysyl oxidase
Ehlers - Danlos syndrome (EDS ) is a group of rare hereditary disorders involving
connective tissues found in skin , tendons , ligaments , muscles , and
vasculature EDS usually results in hypermobile joints; fragile hyperelastic skin ; and
easy bruising due to decreased tissue strength and support. It is caused by
mutations affecting the collagen genes or the enzymes involved in collagen
synthesis such as lysyl -hydroxylase or procollagen peptidase . In the case of EDS
due to procollagen peptidase deficiency, impaired cleavage of the procollagen island C -termim causes the formation of more soluble collagen that does not properly
crosslink with other collagen molecules . This results in joint laxity , loose skin, and
easy bruisability .
(Choice C) Lysyl hydroxylase deficiency can cause a form of EDS characterized by
anH
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iwuiitiiiwui i
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Lab Valuer
Newt
iiipiw
ii t nA
Notes
I <i!c ulator
u j pi
peptidases forming tropocollagen Tropocollagen subunits then self -assemble into
collagen fibrils that are subsequently crosslinked by lysyl oxidase.
Ehlers - Danlos syndrome (EDS ) is a group of rare hereditary disorders involving
connective tissues found in skin , tendons , ligaments , muscles , and
vasculature EDS usually results in hypermobile joints; fragile, hyperelastic skin ; and
easy bruising due to decreased tissue strength and support It is caused by
mutations affecting the collagen genes or the enzymes involved in collagen
synthesis such as lysyl-hydroxyiase or procollagen peptidase . In the case of EDS
due to procollagen peptidase deficiency , impaired cleavage of the procollagen Nand C - termini causes the formation of more soluble collagen that does not properly
crosslink with other collagen molecules. This results in joint laxity, loose skin , and
easy bruisability .
k
(Choice C) Lysyl hydroxylase deficiency can cause a form of EDS characterized by
kyphoscoliosis and ocular fragility . However, biochemical analysis would show a
decreased amount of hydroxylysme within collagenous tissues. Collagen containing
disulfide - rich globular domains is more suggestive of procollagen peptidase
deficiency .
Educational objective:
Extracellular peptidases cleave disulfide -rich terminal extensions from the
procollagen molecule . This results in formation of water -insoluble thple helical
collagen subunits ( tropocollagen ) that self - assemble and undergo crosslinking by
lysyl oxidase to form mature collagen fibrils Impaired cleavage of procoilagen
causes the formation of more soluble collagen that does not properly crosslink with
other collagen molecules .
Time Spent : 6 seconds
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Notes
l akuldtor
4
&
6
7
a
9
10
11
1?
13
IS
1&
17
13
20
21
n
21
24
25
25
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29
A 2 - year - old boy is brought to the office by his parents . He is currently toilet-training
during the day and at nighttime After he urinated in the toilet last night his parents
forgot to flush the toilet and noticed that the boy s urine turned black overnight . The
child has no significant past medical history and takes no medications . He can say
2 -word sentences , follow 2- step directions , and jump with 2 feet off the ground
Examination shows a well nourished child with no swelling or tenderness of any
joints . Urinalysis results are as follows :
black
Color
1.022
Specific gravity
Protein
none
Blood
negative
&
'
-
Glucose
Ketones
Leukocyte esterase
negative
negative
negative
Which of the following conversion pathways is most likely deficient in this patient?
O A. Leucine to acetoacetate
B Phenylalanine to tyrosine
O C Serine to cysteine
C D . Tyrosine to fumarate
E Valine to glutamic acid
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Lab Values
Newt
Notes
I olc uhHor
4
&
6
7
a
9
10
11
1?
13
15
is
17
13
H
20
21
22
23
24
2&
25
27
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29
/%
A 2-year -old boy is brought to the office by his parents . He is currently toilet-training
during the day and at nighttime After he urinated in the toilet last night his parents
forgot to flush the toilet and noticed that the boy' s urine turned black overnight . The
child has no significant past medical history and takes no medications . He can say
2 - word sentences , follow 2- step directions , and jump with 2 feet off the ground
Examination shows a well-nourished child with no swelling or tenderness of any
joints . Urinalysis results are as follows :
black
Color
1.022
Specific gravity
Protein
none
Blood
negative
Glucose
Ketones
negative
negative
Leukocyte esterase negative
Which of the following conversion pathways is most likely deficient in this patient?
A Leucine to acetoacetate [6%]
B Phenylalanine to tyrosine [22%]
O C Serine to cysteine [13%]
* * D . T / ros ne to fumarate [54 %]
O E. Valine to glutamic acid [ 4%]
Explanation:
V
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Notes
(
alculdlor
4
A
&
6
Alkaptonuria
7
a
Diftydi'cpfefidfrre
9
10
11
12
Melanin
13
15
15
17
13
Phenylalanine
Tyrosine
b
DOPA
Pheti a/ani/w
^
TjydftMytese
Catecholamines
H
20
21
Homogentisate
22
23
24
2&
25
TVomogenta /c a& tf
d / oxygenase
27
+
Maleylacetoacetate
23
29
Fumarylacetoacetate
L
Fumarate
TCA
Cycle
© UWortd
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*
r
Notes
I <ilc u f d t o r
A
&
6
7
£
9
10
11
1?
13
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17
13
H
20
21
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2&
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A
Alkaptonuria is an autosomal recesstve disorder of tyrosine metabolism
Deficiency of homogentisic acid dioxygenase blocks homogentisic acid metabolism ,
preventing the conversion of tyrosine to fumarate. Homogentisic acid accumulates
in the body and is excreted in the urine , imparting a black color to the urine if
allowed to sit and undergo oxidation . In patients with alkaptonuna . the retained
homogentisic acid selectively binds to collagen in connective tissues tendons , and
cartilage This leads to ' ochronosis " a blue - black pigmentation most evident in the
ears, nose and cheeks , and ochronotic arthropathy , which typically manifests
during adulthood .
(Choice A) Leucine is a branched- chain amino acid that is elevated in maple syrup
urine disease Isoleucine and valine are also increased Impaired metabolism of
these amino acids leads to cerebral edema , seizures , and a sweet smell of the urine
(Choice B) Conversion of phenylalanine to tyrosine is defective in phenylketonuria
and usually occurs due to a defect in phenylalanine hydroxylase, Undiagnosed and
untreated phenylketonuria results in significant intellectual disability not seen in this
patient .
(Choice C) Impaired renal cystine ( a homodimer of cysteine ) transport leads to
cystinuria a disease characterized by flank pain hematuria , and renal stones in
childhood or adolescence
(Choice E) Sickle cell anemia results from the substitution of valine for glutamic
acid due to a single -nucleotide polymorphism This mutation leads to loss of red cell
elasticity polymerization of sickle hemoglobin and sickling of red blood cells which
results in vasoocclusive crises
Educational objective:
Alkaptonuria is an autosomal recessive disorder in which the lack of homogentisic
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i
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Notes
t alculdtor
anowea to sit ana unaergo oxiaation . in patients witn alkaptonuria tne retainea
homogentisic acid selectively binds to collagen in connective tissues , tendons , and
cartilage This leads to ochronosis " a blue - black pigmentation most evident in the
ears nose and cheeks, and ochronotic arthropathy , which typically manifests
during adulthood .
,
A
,
{ Choice A) Leucine is a branched- chain amino acid that is elevated in maple syrup
urine disease . Isoleucine and valine are also increased. Impaired metabolism of
these ammo acids leads to cerebral edema , seizures and a sweet smell of the urine
(Choice B) Conversion of phenylalanine to tyrosine is defective in phenylketonuria
and usually occurs due to a defect in phenylalanine hydroxylase . Undiagnosed and
untreated phenylketonuria results in significant intellectual disability not seen in this
patient .
I
(Choice C) Impaired renal cystine ( a homodimer of cysteine ) transport leads to
cystinuria a disease characterized by flank pain , hematuria , and renal stones in
childhood or adolescence.
(Choice E) Sickle cell anemia results from the substitution of valine for glutamic
acid due to a single-nucleotide polymorphism . This mutation leads to loss of red cell
elasticity , polymerization of sickle hemoglobin and sickling of red blood cells which
results in vasoocclusive crises.
Educational objective:
Alkaptonuria is an autosomal recessive disorder in which the lack of homogentisic
acid dioxygenase blocks the metabolism of tyrosine , leading to an accumulation of
homogentisic acid Clinical features include a black urine color when exposed to air
a blue-black pigmentation on the face , and ochronotic arthropathy.
References:
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*
Notes
I alculaior
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7
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16
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20
21
n
A 34-year-old man who died of internal hemorrhage inherited a defect of an
elastin-associated glycoprotein that is abundant in the zonular fibers of the tens
periosteum , and the aortic media. The patient most likely suffered from :
b
C A Osteopetrosis
B . Ankylosing spondylitis
C Osteogenesis imperfecta
C D Marfan's syndrome
E . Ehlers -Danlos syndrome
O F . Achondroplasia
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I dlculdtor
Notes
4
&
6
7
a
9
10
11
12
13
U
1S
17
13
19
20
ft
A 34- year-old man who died of internal hemorrhage inherited a defect of an
elastin-associated glycoprotein that is abundant in the zonular fibers of the lens
periosteum , and the aortic media. The patient most likely suffered from
O A . Osteopetrosis [1%]
b
B . Ankylosing spondylitis [1%]
C C . Osteogenesis imperfecta [3%]
* D . Marfan s s > rsdrome J81%J
E . Ehlers- Danlos syndrome [14 %]
C F . Achondroplasia [1%]
.. .
•
21
22
23
24
2&
25
27
23
29
Explanation:
Fibrillin-1 is a major component of the microfibrils that form a sheath around elastin
fibers . Microfibrils are abundantly present in blood vessels periosteum and the
suspensory ligaments of the lens . Fibrillin in the extracellular space acts as a
scaffold for deposition of elastin extruded from connective tissue ceils Defects in
the fibrillin- 1 genes cause classic Marfan' s syndrome Clinically , Marfan' s syndrome
has three prominent features :
1 . Long thin extremities: loose joints; and long fingers ( arachnodactyly )
2. Ocular abnormalities such as dislocation of lens ( ectopia lentis ).
3 . Cardiovascular abnormalities including ascending aortic aneurysm , aortic
dissection , and mitral valve prolapse The patient is this question likely
suffered from an aortic dissection.
,
V
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.
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IS
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13
14
20
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<1
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*
Note
*
(
akuidtor
ft
Explanation:
Fibrillin-1 is a major component of the microfibrils that form a sheath around elastin
fibers. Microfibnls are abundantly present in blood vessels , periosteum and the
suspensory ligaments of the lens. Fibrillin in the extracellular space acts as a
scaffold for deposition of elastin extruded from connective tissue ceils Defects in
the fibrillin-1 genes cause classic Marfan' s syndrome . Clinically , Marfan’s syndrome
has three prominent features:
b
.
1 Long thin extremities loose joints; and long fingers ( arachnodactyly).
2. Ocular abnormalities such as dislocation of lens (ectopia lentis ).
3. Cardiovascular abnormalities including ascending aortic aneurysm, aortic
dissection , and mitral valve prolapse The patient is this question likely
suffered from an aortic dissection
21
n
23
2i
2&
26
27
23
29
(Choice A) Osteopetrosis is caused by impaired osteoclastic bone resorption and is
characterized by increased bone thickness and density . The osteoblasts in patients
with this condition function normally.
(Choice B) Ankylosing spondylitis is a seronegative inflammatory arthritis associated
with HLA-B 27 positivity This disease results in fusion of the spine and sacroiliac
joints as well as arthritis of the hips shoulders and costochondral joints This
disease is not associated with a defect of connective tissue genes .
>
(Choice C) Osteogenesis imperfecta (Ol is a disease resulting from defects in the
genes encoding type I collagen . Type I collagen is a major component of bones.
Thus defects in type I collagen, as seen in Ol , result in reduced bone mass and
increased propensity for fragility fractures. Involvement of other tissues containing
type I collagen results in other manifestations of the disease including blue sclera
dental abnormalities ( dentinogenesis imperfecta } , and hearing loss .
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Newt
Notes
(
tilcufdtor
characterized by increased bone thickness and density The osteoblasts in patients
with this condition function normally.
A
(Choice B) Ankylosing spondylitis is a seronegative inflammatory arthritis associated
with HLA-B 27 positivity. This disease results in fusion of the spine and sacroiliac
joints as well as arthritis of the hips shoulders and costochondral joints This
disease is not associated with a defect of connective tissue genes
(Choice C) Osteogenesis imperfecta (01) is a disease resulting from defects in the
genes encoding type I collagen . Type I collagen is a major component of bones .
Thus , defects in type I collagen, as seen in Ol, result in reduced bone mass and
increased propensity for fragility fractures Involvement of other tissues containing
type I collagen results in other manifestations of the disease including blue sclera ,
dental abnormalities ( dentinogenesis imperfecta ), and hearing loss
(Choice E) EDS results from heritable disorders in the formation of collagen
molecules.
(Choice F) Achondroplasia is one of the most common causes of dwarfism and
results from a genetic defect in the fibroblast growth factor receptor 3 gene The
primary defect is abnormal chondrocyte proliferation at the growth plates of long
bones . This causes short-limbed dwarfism .
Educational Objective:
Marfan' s syndrome is due to a defect in fibnllm, an extracellular glycoprotein that is
abundant in the zonular fibers of the lens the periosteum and the aortic media The
different locations of fibrillin production explains the varied clinical manifestations of
Marfan' s syndrome .
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A 5-month-old boy is brought to the office due to poor feeding His mother says that
he has difficulty holding his head up while breastfeeding and his suckling seems
weaker than it used to be His current weight is between the 5th-10th percentile , and
length and head circumference are tracking along the 25th percentile . Physical
examination shows hepatomegaly and hypotonia in all 4 limbs Cardiac auscultation
reveals a gallop rhythm, and chest x - ray shows severe cardiomegaly Muscle biopsy
shows enlarged lysosomes containing periodic acid- Schiff (PAS ) -positive material
Which of the following enzymes is most likely deficient in this patient?
&
O A. Acid a-glucosidase
C B . Debrancher enzyme
O C Galactokinase
O D Glucose-6-phosphatase
E . Glycogen phosphorylase
O F . Pyruvate kinase
28
29
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4
&
6
7
a
9
10
11
12
13
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19
17
13
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A 5-month-old boy is brought to the office due to poor feeding His mother says that
he has difficulty holding his head up while breastfeeding and his suckling seems
weaker than it used to be His current weight is between the 5th-1 Oth percentile , and
length and head circumference are tracking along the 25th percentile . Physical
examination shows hepatomegaly and hypotonia in all 4 limbs . Cardiac auscultation
reveals a gallop rhythm and chest x - ray shows severe cardiomegaly Muscle biopsy
shows enlarged lysosomes containing periodic acid- Schiff (PAS ) -positive material
Which of the following enzymes is most likely deficient in this patient?
* ® A. Acid a-glucosidase [56%]
O B. Debrancher enzyme [15%]
21
n
21
24
25
25
27
23
29
&
.
C Galactokinase [ 5% ]
D. Glucose - 6-phosphatase [9%]
E. Glycogen phosphorylase [12%]
O F. Pyruvate kinase [2%]
Explanation:
Impairments in glycogenotysis
*
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This patient most likely has glycogen storage disease type II i Pompe
disease ) This condition is caused by deficiency of acid a-glucosidase (acid
maltase ) , an enzyme responsible for breaking down glycogen within the acidic
A
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*
Notes
I tilculalor
© UWorid
A
This patient most likely has glycogen storage disease type II ( Pompe
disease ) This condition is caused by deficiency of acid a-glucosidase ( acid
maltase ) an enzyme responsible for breaking down glycogen within the acidic
environment of lysosomes Although most glycogen is degraded in the cytoplasm, a
small amount is inadvertently engulfed by lysosomes , especially in cells containing
high amounts of glycogen such as hepatocytes and myocytes . As such deficiency
of acid maltase results in pathologic accumulation of glycogen within liver and
muscle lysosomes Cardiac and skeletal muscle are particularly susceptible , as the
ballooning lysosomes interfere with contractile function .
u
The classic form of the disease presents in early infancy with marked
cardiomegaly , severe generalized hypotonia macroglossia . and
hepatomegaly . Blood glucose levels are normal, unlike with glycogen storage
diseases that primarily affect the liver ( eg , von Gierke ) . A key distinguishing feature
is that muscle biopsy will show accumulation of glycogen in lysosomes
(Choices B, D . and E) Other glycogen storage diseases are caused by
deficiencies of glucose-6- phosphatase, glycogen phosphorylase, and debrancher
enzyme . However , glycogen accumulation within lysosomal vacuoles is specific for
acid a-giucosidase deficiency .
(Choice C) Galactokinase catalyzes the phosphorylation of galactose to galactose 1-phosphate in the first committed step of galactose catabolism. Galactokinase
deficiency causes neonatal cataract formation due to accumulation of galactitol in the
lens
.
(Choice F) Pyruvate kinase deficiency causes chronic hemolytic anemia ,
splenomegaly , and iron overload as a result of impaired erythrocyte survival
v
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The classic form of the disease presents in early infancy with marked
cardiomegaly , severe generalized hypotonia macroglossia and
hepatomegaly . Blood glucose levels are normal unlike with glycogen storage
diseases that primarily affect the liver ( eg . von Gierke } . A key distinguishing feature
is that muscle biopsy will show accumulation of glycogen in lysosomes .
20
21
lens
21
24
29
26
21
23
29
(Choice F) Pyruvate kinase deficiency causes chronic hemolytic anemia ,
splenomegaly, and iron overload as a result of impaired erythrocyte survival .
1]
19
n
*
Notes
Iale uldtor
A
(Choices B, D , and E) Other glycogen storage diseases are caused by
deficiencies of glucose-6-phosphatase , glycogen phosphorylase , and debrancher
enzyme . However , glycogen accumulation within lysosomal vacuoles is specific for
acid a-glucosidase deficiency.
(Choice C) Galactokinase catalyzes the phosphorylation of galactose to galactose 1-phosphate in the first committed step of galactose catabolism , Galactokinase
deficiency causes neonatal cataract formation due to accumulation of galactrtol in the
17
Ldb Value
Newt
Educational objective:
Acid mattase ( a-glucosidase ) deficiency presents in early infancy with cardiomegaly ,
macroglossia and profound muscular hypotonia Abnormal glycogen accumulation
within lysosomal vesicles is seen on muscle biopsy.
References:
1 . Lysosomal dysfunction in muscle with special reference to
glycogen storage disease type II.
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Molecular biologists studying signal transduction apply an agent to human cells that
activates G-protein- dependent phospholipase C . Which of the following intracellular
substances is most likely to increase immediately after exposure to this agent?
-
O A . Ca
to B . cAMP
&
O C. cGMP
O D Cl
O E. mRNA
O F NO
.
21
2
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t ole uftftlor
4
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9
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19
20
A
Molecular biologists studying signal transduction apply an agent to human cells that
activates G-protein- dependent phospholipase C . Which of the following intracellular
substances is most likely to increase immediately after exposure to this agent?
*
k
•A. Ca
[74%]
O B. cAMP [11%J
C C . cGMP [11% j
O D. Cl' [0%]
O E. mRNA [1%]
O F . NO [2%]
21
n
21
24
2&
25
Explanation:
Phosphatidylinositol 2nd messenger system
21
28
20
Hormone
Extracellular fluid
Phospholipase C
Receptor
1
j
V
,«
GOP
P v
G protein
v
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*
Notes
£ dlculdtor
Explanation:
&
6
A
Phosphatidylinositot 2nd messenger system
i
8
9
10
11
Phospholipase C
Hormone
Extracellular ffu d
'
12
[,
13
14
15
IS
n
19
G protem
20
21
22
21
24
25
26
Protein kinase C
activation
27
28
29
Phosphorylaied proteins
Physiologic effects
( eg,
Endoplasm : reticulum
Cytosol
*
smooth muscle
contraction )
© UWwld
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Notes
Ialculdtor
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e
7
8
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u
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2b
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A
A variety of hormone receptors are known to exert their intracellular effects via the
phosphoinositol system Examples include a -adrenergic . M. and M cholinergic .
V ( vasopressin ), H . (histamine ), oxytocin angiotensin II TRH . and GnRH
receptors . This signal transduction pathway proceeds through the following steps :
1. Binding of a ligand to its cell surface receptor causes the exchange of GDP
for GTP on the o-subumt of a G.-protein associated with the receptor . The
activated o -subunit undergoes a conformational change and exposes a
phospholipase C ( PLC ) activating site .
2. After activation PLC hydrolyzes phosphatidyl inositol bisphosphate (PIPJ
into diacylglycerol ( DAG) and inositol triphosphate ( IPJ.
3 . DAG is able to directly stimulate protein kinase C (PKC ) , but the major
activator of PKC is increased intracellular Ca 1 * that occurs due to IP
mediated- release of intracellular Ca * stores from the endoplasmic
reticulum . PKC is the major effector molecule in this pathway: it directly
modulates the activity of other proteins via phosphorylation .
&
-
(Choices B and C ) Intracellular cAMP and cGMP concentrations increase during
activation of adenylate or guanylate cyclase second messenger systems,
respectively . Levels can also increase following cyclic nucleotide
phosphodiesterase inhibition , as seen on exposure to sildenafil, which selectively
inhibits cGMP phosphodiesterease and results in smooth muscle relaxation in blood
vessels
(Choice D ) Intracellular Cl concentration increases slightly after inhibitory
neurotransmitters leg GABA , glycine ) act on the neuron to increase Cl membrane
conductance (hyperpolarization ).
-
fChoice El The intracellular concentration of mRNA increases durina cellular states
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phosphorylation
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Notes
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£ alculdtor
A
(Choices B and C) Intracellular cAMP and cGMP concentrations increase during
activation of adenylate or guanylate cyclase second messenger systems
respectively . Levels can also increase following cyclic nucleotide
phosphodiesterase inhibition , as seen on exposure to sildenafil , which selectively
inhibits cGMP phosphodiesterease and resutts in smooth muscle relaxation in blood
vessels .
(Choice D) Intracellular Cl concentration increases slightly after inhibitory
neurotransmitters ( eg GABA glycine ) act on the neuron to increase Cl membrane
conductance (hyperpolarization).
(Choice E) The intracellular concentration of mRNA increases during cellular states
of elevated protein synthesis (eg. during cell division).
(Choice F) Nitric oxide (NO ) is a paracrine signaling molecule with a lifetime of a
few seconds It can freely cross cell membranes and functions as a critical
component of endothelium-mediated vasodilation . NO is synthesized from arginine
and O. by the enzyme NO -synthase .
b
Educational objective:
The phosphoinositol second messenger system begins with ligand-receptor binding
and G -protein activation leading to activation of phospholipase C (PLC ) PLC then
hydrolyzes phosphatidyl inositol bisphosphate and forms diacylglycerol and inositol
triphosphate (IP . ) Finally, IP, activates protein kinase C via an increase in
intracellular Ca;'.
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*
Notes
(
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b
6
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1?
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is
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20
Precursor mRNA undergoes substantial post-transcriptional processing before it
becomes the finalized mRNA template Because of this extensive processing , the
final mRNA sequence can be very different from that of its encoding DNA An
mRNA molecule transcribed from a eukaryotic gene is shown schematically below
Which of the following portions is not transcribed from the DNA template ?
8
A
D
C
5‘-
AUGUUCCCA
CGGUAAGUU
21
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23
24
25
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4
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7
a
9
10
11
12
A
Precursor mRNA undergoes substantial post-transcriptional processing before it
becomes the finalized mRNA template Because of this extensive processing , the
final mRNA sequence can be very different from that of its encoding DNA An
mRNA molecuse transcribed from a eukaryotic gene is shown schematically below
Which of the following portions is not transcribed from the DNA template?
13
8
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19
E
D
17
20
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5-
AUGUUCCCA
CGGUAAGUU
*
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22
23
24
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26
27
28
29
O A A [6%]
O B. B [25%]
O C. C [ 2% ]
O D D [9%J
* B . = [58%J
•
Explanation:
Nucleus
DNA
5
EiOft I
i
i
hflRNA
(pro mRNA }
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a
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AAAAA
3’
3'
RNA polymerase II forms precursor mRNA (pre -mRNA ) from the DNA template
Posttranscriptional processing of pre-mRNA forms mature mRNA which is ready for
export from the nucleus and translation Posttranscriptional processing involves the
following:
1. 5' capping: A 7-methyl-guanosine cap is added to the 5’ end of the mRNA
2. Polyadenylation: A poly-A tail is added to most eukaryotic mRNA molecules
by poly-A polymerase. Poly-A tails are not transcribed from the DNA
template Instead , a consensus sequence (usually "AAUAAA ” ) found within
thft V Anri
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(Notes
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alculator
4
5
6
7
8
9
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1?
u
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15
15
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20
21
n
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25
/%
RNA polymerase II forms precursor mRNA tpre -mRNA ) from the DNA template .
Posttranscriptional processing of pre -mRNA forms mature mRNA , which is ready for
export from the nucleus and translation Posttranscriptional processing involves the
following:
1.5 * capping: A 7-methyl-guanosine cap is added to the 5‘end of the mRNA,
2. Polyadenylation: A poly -A tail is added to most eukaryotic mRNA molecules
by poly-A polymerase. Poly-A tails are not transcribed from the DNA
template. Instead a consensus sequence (usually "AAUAAA") found within
the 3’ end of the gene being transcribed directs the addition of the poly-A tail
onto the mRNA . This tail protects the mRNA from degradation within the
cytoplasm after it exits the nucleus Segments of mRNA downstream from the
,
,
consensus sequence ( 'AAUAAA ’ ) are likely part of the poly A tail ( Choice E)
3 . Splicing: The initial mRNA , called pre -mRNA or heterogeneous nuclear RNA
(hnRNA ) contains sequences from coding and non-coding regions of DNA ,
known as exons and introns, respectively , Removal of introns (non-coding
mRNA segments ) occurs dunng splicing
-
.
26
27
28
29
(Choice A) This region contains the AUG start codon , and is therefore part of the
first exon in the transcribed gene ,
(Choice B) This is an intron being spliced out of the mRNA molecule Introns are
not translated during protein synthesis , but they are transcribed by RNA polymerase
II during RNA production.
(Choices C and D) These are other exonic regions of mRNA that were transcribed
from the DNA template. There is a stop codon f UAA* ) present between the C and
D regions . Thus , region C will be translated while region D will not.
1
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I dltuldlor
i w T T T
A
template Instead , a consensus sequence ( usually "AAUAAA " ) found within
the 3 ’ end of the gene being transcribed directs the addition of the poly-A tail
onto the mRNA This tail protects the mRNA from degradation within the
cytoplasm after it exits the nucleus Segments of mRNA downstream from the
consensus sequence ( "AAUAAA " ) are likely part of the poly- A tail ( Choice E).
3 . Splicing: The initial mRNAt called pre - mRNA or heterogeneous nuclear RNA
(hnRNA ) . contains sequences from coding and non-coding regions of DNA ,
known as exons and introns respectively . Removal of introns (non-coding
mRNA segments ) occurs during splicing .
,
(Choice A) This region contains the AUG start codon, and is therefore part of the
first exon in the transcribed gene .
I-
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(Notes
(Choice B) This is an intron being spliced out of the mRNA molecule Introns are
not translated during protein synthesis , but they are transcribed by RNA polymerase
II during RNA production
,
(Choices C and D) These are other exonic regions of mRNA that were transcribed
from the DNA template . There is a stop codon (MUAAM ) present between the C and
D regions . Thus, region C will be translated while region D will not
Educational objective:
The polyadenylation signal sequence at the 3 ’ end of the mRNA transcript is
responsible for addition of the poly-A tail The poly-A tail is not transcribed from
DNA , but rather added as a posttranscriptional modification downstream of a
consensus sequence (usually " AAUAAA " ) This tail protects the mRNA from
degradation within the cytoplasm after it exits the nucleus
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In a laboratory experiment , the enzyme reverse transcriptase uses a specific mRNA
template to synthesize a strand of complementary DNA ( cDNA ) The cDNA is then
integrated into a plasmid containing a bacterial promoter , which allows the cDNA to
be expressed < n bacterial cells . Large quantities of the target protein are obtained
and subsequently identified using a radiolabeled DNA probe The target protein is
b
most likely which of the following?
A. Insulin-like growth factor-1
O B. Insulin receptor
C C Protein kinase A
O D . Calmodulin
O E . N-myc protein
O F. k -RAS
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In a laboratory experiment the enzyme reverse transcriptase uses a specific mRNA
template to synthesize a strand of complementary DNA ( cDNAi The cDNA is then
integrated into a plasmid containing a bacterial promoter , which allows the cDNA to
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u
be expressed in bacterial cells . Large quantities of the target protein are obtained
and subsequently identified using a radiolabeled DNA probe The target protein is
most likely which of the following?
*
A Insulin-like growth factor -1 [27%]
O B . Insulin receptor [11%]
C C . Protein kinase A [13%]
O D. Calmodulin [6%]
E, N- myc protein [28%]
O F. k RAS [14%]
-
Explanation:
The process descnbed is called expression cloning Expression cloning is a type of
DNA cloning where the signals necessary for transcription and translation are
included in the cloned DNA . This process allows bacteria to be used to produce
large amounts of a protein of interest
First . mRNA is used as a template by reverse transcriptase to produce a cDNA
strand containing promoter sequences ( such as the Pnbnow box . or -35 sequence in
prokaryotes ) and translation stimulatory sequences (Shine -Dalgamo ) The gene is
then incorporated into a plasmid and subsequently transcribed and translated into
nrnfpin mnct ho ahlo
nrntpin In nrrl r tn thpn ho r HinUholArl hv a HNA nrnho
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Explanation:
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The process described is called expression cloning Expression cloning is a type of
DNA cloning where the signals necessary for transcription and translation are
included in the cloned DNA This process allows bacteria to be used to produce
large amounts of a protein of interest .
*
k
First mRNA is used as a template by reverse transcriptase to produce a cDNA
strand containing promoter sequences ( such as the Pribnow box , or -35 sequence in
prokaryotes ) and translation stimulatory sequences (Shine - Dalgamo ). The gene is
then incorporated into a plasmid and subsequently transcribed and translated into
protein In order to then be radiolabeled by a DNA probe the protein must be able
to bind DNA . Examples of proteins that are able to bind DNA include transcription
factors steroids , thyroid proteins , vitamin D receptors retinoic acid receptors, DNA
transcription an