MCAT Bros
AAMC Bio
Question Pack 1 #1-120
Presented by Charlie Hoying
NOW vs. LATER recap
• Some questions are “Now” questions, because they rely exclusively or
primarily on outside-passage information.
• We do now questions before reading, only looking up a line or two max!
• Some questions are “Later” questions, because they require extensive
passage analysis
• Remember, the now vs. later technique is designed to minimize the
number of questions you worry about for AFTER reading a CP/BB passage.
• Conserve brainpower, conserve willpower!
• If a question takes more than 15-30s to determine its now/later
classification, it’s a LATER!
• Let’s analyze passage 1 for now/later questions.
Passage 1 Intro:
-
Na/K ATPase
Erythrocytes
Isotonic
Looks like we have a passage on Na/K ATPase!
Q1: Active sodium pump
Q1:
• Most active sodium pump?
A. Veins
B. Loop of Henle
C. Lungs
D. Bone marrow
Q1:
• Most active sodium pump?
A. Veins
B. Loop of Henle. Correct. The loop of Henle is responsible for the
osmolarity gradient of the nephron. As such, it will need active
transport.
C. Lungs
D. Bone marrow
•
A.
B.
C.
D.
In which region of the loop of Henle would the sodium pump be
most active?
Proximal tubule
Descending limb
Thin ascending limb
Thick ascending limb
• The answer is D.
A. Incorrect. Proximal tubule is not part of the loop of Henle.
B. Incorrect. The descending limb is where H2O is passively
reabsorbed.
C. Incorrect. The thin ascending limb is permeable to water, and salt
is passively reabsorbed.
D. Correct. The thick ascending limb is impermeable to water, and
is the region most responsible for the kidney ion gradient due to
its active transport.
• Descending limb
• Passive H2O reabsorption (?)
• Osmosis
• Water flows down its
concentration gradient
• i.e. up the ion gradient
• Thin ascending limb
• Passive transport of NaCl down
its concentration gradient (?)
• Diffusion
• Thick ascending limb
• Active transport of NaCl,
creating the nephron
concentration gradient
• Thick because impermeable to
H2O
• Active transport
Q2: Osmosis
Q2:
• Reason erythrocyte volume increases upon placement into
distilled water?
A. Ion gradient causes water to enter cells
B. Contractile filaments of the cytosol open pores in PM
C. Na pump transports sodium out of cells more rapidly
D. Erythrocytes produce degradative enzymes
Q2:
• Reason erythrocyte volume increases upon placement into
distilled water?
A. Ion gradient causes water to enter cells. Correct. Osmotic
pressure causes water to go down its concentration gradient,
i.e. H2O travels from lower [ion] to higher [ion]. This increases
the volume of the cells.
B. Contractile filaments of the cytosol open pores in PM. Incorrect.
This mechanism does not exist and would be maladaptive for
cells.
C. Na pump transports sodium out of cells more rapidly. Incorrect.
The Na pump is not regulated by external osmolarity. Na pump
regulation is beyond the scope of the MCAT.
D. Erythrocytes produce degradative enzymes. Incorrect. No reason
why these enzymes would increase Vol of cell.
• Define hypertonic, hypotonic, and isotonic.
• Hypertonic = higher [ion] outside cell
• Hypotonic = lower [ion] outside cell
• Isotonic = same [ion] outside cell
• Aquaporins, channels that facilitate diffusion of H2O, contain an
Arg-rich selectivity filter. This filter most effectively prevents
transport of which species, while allowing H2O movement?
A. Glucose
C. ClB. H+
D. O2
A. Incorrect. Arg could interact with glucose at Phys
pH.
B. Correct. Being positively charged at Phys pH, Arg
will keep H+ out of the channel.
C. Incorrect. Arg(+) and Cl- would be attracted.
D. Incorrect. O2 could diffuse through an Arg-rich
filter as well as a plasma membrane.
Q3: Sodium pump & action potentials
Q3:
• Could the Na pump cause action potentials in neurons?
A. No; APs result in increased PM permeability to Na.
B. No; myelin sheaths prevent movement of ions across neuron PMs.
C. Yes; Na is transported out of neurons during action potentials.
D. Yes; APs are accompanied by ATP hydrolysis.
Q3:
• Could the Na pump cause action potentials in neurons?
A. No; APs result in increased PM permeability to Na. Correct. Na
can diffuse down its concentration gradient into the cell during
an AP because ligand- and voltage-gated channels open. The
sodium pump moves Na out of the cell.
B. No; myelin sheaths prevent movement of ions across neuron PMs.
Incorrect. Myelin sheaths have nothing to do with the Na pump.
C. Yes; Na is transported out of neurons during action potentials.
Incorrect. Na enters neurons during APs.
D. Yes; APs are accompanied by ATP hydrolysis. Incorrect. ATP
hydrolysis is used to set up conc gradients for an AP.
• Which of the following gives the producer of myelin sheaths for
the CNS and PNS, respectively?
A. Schwann cells, oligodendrocytes
B. Oligodendrocytes, Schwann cells
C. Schwann cells, astrocytes
D. Astrocytes, oligodendrocytes
• The answer is B. Myelin sheaths are produced by oligodendrocytes
in the CNS and Schwann cells in the PNS.
• What is the function of a myelin sheath?
• Myelin sheaths are impermeable to ions, and thus increase
conductivity along the axon of a neuron by preventing Na+
from exiting. The AP gets relayed at nodes of Ranvier, which
contain voltage-gated Na+ channels.
Q4: Glycolysis vs Na/K ATPase
Q4:
• Number of Na+ ions removed per molecule of glucose?
A. 3
B. 6
C. 9
D. 12
Q4:
• Number of Na+ ions removed per molecule of glucose?
A. 3
B. 6. Correct. Since glycolysis produces 2 net ATP per glucose, via
reaction A, 6 Na+ could be removed from the cell per glucose.
C. 9
D. 12
• If a 12-C diunsaturated fatty acid were metabolized completely,
how many Na+ could be removed from the cell?
A. 20 Na+
B. 54 Na+
C. 225 Na+
D. 234 Na+
• The answer is C. For a 12-C saturated fatty acid,
• 5 rounds of Beta-ox
• 5 NADH (x2.5 ATP) = 12.5 ATP
• 5 FADH2 (x1.5 ATP) = 7.5 ATP
• 6 acetyl-CoA
• 18 NADH = 45 ATP
• 6 FADH2= 9 ATP
• 6 GTP = 6 ATP
• Cost: -2 ATP for activation
• Cost: 2 FADH2 for diunsaturation = -3 ATP
• Total = 75 ATP (x3 Na+) = 225 Na+
Data interpretation practice: What observations can we
make from this table?
•
•
•
•
High rate of ATP hydrolysis when erythrocyte contains NaCl
Low rate when ecf contains NaCl
Low rate when Mg2+ absent (?)
• Cofactor
Low rate when ATP is outside cell only (?)
• ATP hydrolysis occurs on intracellular side
Passage 1 Takeaways
• Body sodium – region where lots of Na+ is transported
• Nephron – major transport in each area
• Osmosis, diffusion, facilitated diffusion, active transport
• Hypertonicity, hypotonicity, isotonicity
• Amino acids
• Action potentials – phases, permeabilities
• Normal [ion] inside and outside cell
• Myelin sheaths and cells that produce them
• COPS = CNS oligodendrocytes; PNS Schwann Cells
• Beta-oxidation
• Location
• Electron carriers (NADH, FADH2) per round
• Number of acetyl-CoA
• Strategy in analyzing data tables
Passage 2 Intro:
-
Ulcers
H. pylori
vacA
Looks like we have a passage on microbio/ulcers!
Q5: Link to cancer
Q5:
• H pylori infection may  ^cell proliferation in stomach
• How H. pylori infection would lead to cancer
A. Genetic mutations in proliferating germ cells
B. Genetic mutations in proliferating somatic cells
C. Immune system fails to recognize bacterial antigens
D. Crowded mucosal cells remain in interphase
Q5:
• H pylori infection may  ^cell proliferation in stomach
• How H. pylori infection would lead to cancer
A. Genetic mutations in proliferating germ cells. Incorrect. Germ cells are not
present in the stomach.
B. Genetic mutations in proliferating somatic cells. Correct. Cancer arises
mainly from mutations in genes regulating cell cycle.
C. Immune system fails to recognize bacterial antigens. Incorrect. This would
promote cancer due to ^B, but would not lead to cancer on its own.
D. Crowded mucosal cells remain in interphase. Incorrect. Cancer is
uncontrolled cell growth/division; cells in interphase cannot cause cancer.
• CagA is an H. pylori cytotoxin thought to be involved in increased cell
proliferation and cancer. Researchers transfected gastric cells with CagA in
vitro and measured cell growth. What would be the most appropriate
negative control experiment?
A. Measuring apoptosis of the gastric cells
B. Transfecting the cells with vacA, an H. pylori toxin that creates holes in
intracellular membranes
C. Incubating the cells with lenaldekar, a drug that inhibits cell proliferation
D. Transfecting the cells with cDNA for PDGF, which promotes cell division
• The answer is C.
A. Control experiments are IVs, not DVs.
B. Transfection with a different toxin does not give control information
regarding cagA.
C. Preventing the effect of cagA, which is
hypothesized to be cell proliferation,
constitutes a negative control expt.
D. This would be more of a positive control,
since it elicits the expected effect, i.e.
proliferation.
Q6: Reason host antibodies are ineffective
against H. pylori
Q6:
• Host antibodies are ineffective against H. pylori
• Why?
A. Antibodies denatured in stomach
B. Antibodies not effective against bacteria
C. H. pylori suppresses immune system
D. Antibodies not secreted into extracellular spaces
Q6:
• Host antibodies are ineffective against H. pylori
• Why?
A. Antibodies denatured in stomach. Correct. Extreme pH denatures
proteins, and antibodies are proteins.
B. Antibodies not effective against bacteria. Incorrect. Antibodies are
generally effective against bacteria and viruses.
C. H. pylori suppresses immune system. Incorrect. No reason to
assume immune suppression.
D. Antibodies not secreted into extracellular spaces. Incorrect.
Antibodies secreted into extracellular spaces are known as
immunoglobulins.
• Which region of H. pylori antibodies is most likely to be different
from HIV antibodies?
A
The answer is A.
• The variable region of
antibodies is the arms (top-left
C
and top-right).
• The A regions bind to antigens
• The B region binds to host
molecules depending on
antibody function
B
• The C regions are unimportant
on the MCAT
Q7: Diff between bacterial strains
Q7:
• Difference between bacterial strains
A. Attack different hosts
B. Express different genes
C. Diff antibiotic resistance
D. Exist in diff countries
Q7:
• Difference between bacterial strains
A. Attack different hosts. Incorrect. This does not necessarily define a
bacterial strain.
B. Express different genes. Correct. By definition, different strains of
a pathogen express different genes. Otherwise, they would be
the same strain. This is the broadest answer.
C. Diff antibiotic resistance. Incorrect. This may be true, but it does
not define diff strains.
D. Exist in diff countries. Incorrect. Same as A and C.
• H. pylori strains can be categorized as either CagA(+) or CagA(-).
CagA is a highly antigenic toxin that is injected into the cell,
mimicking effects of hepatocyte growth factor and triggering a
morphological cellular change to a more motile phenotype. Given
this, which is LEAST likely seen in CagA(+) strains of H. pylori?
A. Increased metastasis
B. Increased resistance to antibiotics
C. Increased vascular permeability
D. Increased cytoplasmic volume
• The answer is B.
A. Incorrect. Greater motility would mean greater metastasis.
B. Correct. No reason to suspect greater antibiotic resistance.
C. Incorrect. Vascular permeability would increase due to CagA’s high
antigenicity.
D. Incorrect. Question states CagA mimicks hepatocyte growth factor.
Q8: Effects of CagA?
Q8:
• Effects of cagA?
•
A.
B.
C.
D.
CagA  inflammation
Disruption of host cell enzymatic activity.
Disruption of host cell protein synthesis.
Movement of leukocytes into mucosal tissue.
Vasoconstriction of arterioles in mucosal layer.
Q8:
• Effects of cagA?
• CagA  inflammation
A. Disruption of host cell enzymatic activity. Incorrect. Inflammation
can lead to loss of function, but C is better.
B. Disruption of host cell protein synthesis. Incorrect. Inflammation is
not associated with decreased protein synthesis.
C. Movement of leukocytes into mucosal tissue. Correct.
Inflammation is associated with extravasation.
D. Vasoconstriction of arterioles in mucosal layer. Incorrect.
Inflammation leads to local vasodilation.
• Leukocyte extravasation, aka diapedesis, requires WBCs to cross
which type of cell-cell connection?
A. Gap junctions
B. Tight junctions
C. Desmosomes
D. Intercalated disks
• The answer is B.
A. Gap junctions allow ion flow between cells.
B. Tight junctions normally prevent movement esp out of BVs.
C. Desmosomes are permeable to molecular flow.
D. Intercalated disks are present between cardiac myocytes.
Q9: Reason H. pylori doesn’t lead to cancer in
most people
Q9:
• Let’s look for any backup info from the passage:
•
A.
B.
C.
D.
Increased risk of cancer, but why do most not develop cancer?
Do not incorporate bacterial genes into chromosomes
Robust immune systems defeat early cancers
Eradicate infection before tumors develop
Tolerate infection w/o developing tumors
Q9:
• Let’s look for any backup info from the passage:
• Increased risk of cancer, but why do most not develop cancer?
A. Do not incorporate bacterial genes into chromosomes. Incorrect. No
mention of DNA integration (associated with retroviruses).
B. Robust immune systems defeat early cancers. Incorrect. “Early
cancers” would contradict question stem.
C. Eradicate infection before tumors develop. Incorrect. Question only
mentions they don’t get cancer; not eradicate infection.
D. Tolerate infection w/o developing tumors. Correct. This is
somewhat circular, but it addresses the question best.
• The structure of metronidazole, a common antibiotic prescribed for
H. pylori infection, is shown at right. Its methyl group is oxidized to
alcohol during activation. Which of the following would LEAST
inhibit metronidazole activation?
A. LiAlH4
C. Beta-mercaptoethanol
B. NAD+
D. glucose
A. LAH is a reducing agent that could compete for oxidation.
B. Correct. NAD+ is already oxidized; it gets reduced to NADH.
C. Beta-mercaptoethanol gets oxidized when it reduces disulfides.
D. Glucose is oxidized during metabolism, so it can compete with
metronidazole. Glucose is also a reducing sugar (has hemiacetal).
Passage 2 Takeaways
• Somatic vs. germ-line cells
• Cell cycle phases
• Expt’l design
•
•
•
•
Positive control
Negative control
DV
IV
• Antibodies – secreted immune proteins
•
•
•
•
Produced by B cells (plasma cells)
Neutralize and promote phagocytosis
“Y” structure: Fixed and variable region; heavy and light chain
Immunoglobulin – free antibody in blood
• Inflammation
• Redness, heat, swelling, pain, loss of function
• Extravasation – migration of leukocytes (esp. neutrophils) from bloodstream into tissue
• Types of cellular junctions
• Reducing agents – NaBH4, LiAlH4 (LAH), beta-mercaptoethanol, DTT
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes
Passage 3 Intro:
-
Freeze tolerance
Only extracellular water freezes
Hyperglycemia
Looks like we have a passage on freezing point reduction/osmosis!
Q10: Why hyperglycemia  cellular
dehydration
Q10:
• Why does hyperglycemia lead to cellular dehydration?
A. Glucose energy accelerates osmotic work performed by PMs
B. Glucose energy accelerates PM ion-xc pumps
C. Glucose raise osmotic pressure of ecf
D. Glucose molecules exchanged for water molecules across PM
Q10:
• Why does hyperglycemia lead to cellular dehydration?
A. Glucose energy accelerates osmotic work performed by PMs.
Incorrect. Osmosis does not require work.
B. Glucose energy accelerates PM ion-xc pumps. Incorrect. Active
transport of ions does not lead to increased water export.
C. Glucose raise osmotic pressure of ecf. Correct. ^[solute]ecf leads
to increased osmotic water export.
D. Glucose molecules exchanged for water molecules across PM.
Incorrect. This occurs via secondary active transport with Na+.
• Which solution would cause the greatest cellular dehydration?
A. 3 mM glucose
B. 1 mM NaCl
C. 2 mM NaHCO3
D. 2 mM glycerol
• The answer is C. π = iMRT. As R and T are the same, only iM need
to be evaluated. Greatest cellular dehydration would follow from
highest Posm.
A. iM = 1*0.003 = 0.003
B. iM = 2*0.001 = 0.002
C. iM = 2*0.002 = 0.004. Posm will be the highest.
D. iM = 1*0.002 = 0.002
Q11: Change in glucose regulation leading to
hyperglycemia
Q11:
• Glucose regulation mechanisms leading to hyperglycemia
A. Suppression of insulin secretion
B. Suppression of glucagon secretion
C. Slowing of glycogen catabolism in liver
D. Increased sensitivity of all pancreatic responses
Q11:
• Glucose regulation mechanisms leading to hyperglycemia
A. Suppression of insulin secretion. Correct. Insulin lowers blood
glucose. Suppressing it would help increase blood glucose.
B. Suppression of glucagon secretion. Incorrect. Glucagon increases
blood glucose. Suppressing it would not help.
C. Slowing of glycogen catabolism in liver. Incorrect. Glycogen
breakdown increases blood glucose. Similar to B.
D. Increased sensitivity of all pancreatic responses. Incorrect. Some
pancreatic responses raise blood glucose; some lower it.
• A frog has not eaten a meal in 5 hours. Which of the following
patterns of [glucagon]plasma and lipolysis would be expected?
A. ^[glucagon] and ^ lipolysis
B. ^[glucagon] and v lipolysis
C. v[glucagon] and ^ lipolysis
D. v[glucagon] and v lipolysis
• The answer is A.
A. Increased glucagon would mobilize nutrients; increased lipolysis
would mobilize free fatty acids (ffa’s).
B. More nutrient mobilization; decreased lipolysis would inhibit ffa
mobilization.
C. Decreased nutrient mobilization; increased lipolysis would not
occur with decreased glucagon.
D. Decreased nutrient mobilization and decreased lipolysis would not
help nourish cells during fasting.
Q12: Reason for persistence of HR
Q13: Locations of ice
Q13:
• Location of ice?
I.
II.
III.
A.
B.
C.
D.
Cytoplasm
Plasma
Lymph
II
III
I, II
II, III
Q13:
• Location of ice?
I.
II.
III.
A.
B.
C.
D.
•
I.
II.
III.
IV.
V.
A.
B.
C.
D.
•
Cytoplasm
Plasma
Lymph
II
III
I, II
II, III. Correct. Plasma and lymph fluid are extracellular.
Cytoplasm is intracellular.
Which of the following proteins will NOT be present in a frozen
water medium?
Albumin
Tubulin
Prothrombin
Hemoglobin
Cytochrome C
II, IV, V
I, III, IV
I, II, IV
II, IV, V
The answer is D.
I.
II.
III.
IV.
V.
Albumin is extracellular and is the major
osmoregulatory protein in the blood.
Tubulin composes microtubules, which are
intracellular/cytoskeletal.
Prothrombin is a precursor to clotting protein
thrombin present in the blood.
Hemoglobin is located in the cytoplasm of
RBCs.
Cytochrome C is a mitochondrial protein
essential to the ETC.
Q14: Anaerobic respiration period
Q14:
• Time period of exclusive
anaerobic metabolism
A. Before 0 hours
B. B/w 0 and 6 hours
C. B/w 0 and 24 hours
D. B/w 12 and 26 hours
Q14:
• Time period of exclusive anaerobic
metabolism
A. Before 0 hours
B. B/w 0 and 6 hours
C. B/w 0 and 24 hours
D. B/w 12 and 26 hours. Correct.
Heart rate slows to 0 around 12
hours. Aerobic respiration
requires O2. O2 distribution
requires hemoglobin movement
from lungs to tissues, which
requires pumping from the heart.
• Which of the following species is
most highly produced at hour 20?
A. Glucagon
B. Oxyhemoglobin
C. Lactate
A. Incorrect. Since glucagon serves to dehydrate cells via glucose osmosis, and cells are
D. Acetyl-CoA
already at minimum H2O levels near hour 20, glucagon levels are likely not high.
• The answer is A.
B. Incorrect. Since heart activity stops around hour 12, oxygenated hemoglobin is less
likely to be found at hour 20.
C. Correct. Since heart activity and thus aerobic respiration stops around hour 12,
lactate is likely present in blood.
D. Incorrect. Since aerobic respiration stops around hour 12, Acetyl-CoA would be low.
Q15: Results of Fig 2
Now, to read passage 3:
- Freeze tolerance
- Ice in cells = bad
- Slow cooling permits water redistribution out of cells such that only ecf water freezes
- Accelerated glucose release from hepatic glycogen  hyperglycemia
- Hyperglycemia  cellular dehydration
Now, to read passage 3:
- Body temperature – lowers after heart rate lowers
- Heart rate – lowers around hour 12 (?)
- Enable distribution of glucose in blood
- Ice content – reaches 50% around 6 hours
- Tissue glucose – reaches maximum around 3 hours (?)
- Osmotically pull water out of cells
- Tissue water – reaches minimum between 6 and 18 hours (?)
- Makes freezing difficult – freezing point depression
Now, to read passage 3:
- Research question: Does plasma glucose
prevent destruction of RBCs
- Background: hyperglycemia  cellular
dehydration  cryoprotection against
cellular death
- IV: Saline injection vs. 640 mM glucose vs.
1500 mM glucose
- DV1: Survival %
- DV2: Plasma hemoglobin (?)
- Indicates RBC lysis(?)
- From freezing
- Results:
- Lowest [Hb] with higher glucose
injection
- Higher [Hb] with no glucose injection
- Conclusion:
- Glucose prevents RBC death and lysis
- H2O flows out, leading to shrunken
cells but not lysis
Q12:
• Importance of pulse during ice
formation
A. Circulating blood distributes glucose
B. Circulating blood equilibrates
temperature
C. Beating heart warms body tissues
and slows ice formation
D. Beating heart requires glucose
energy source
Q12:
• Importance of pulse during ice
formation
A. Circulating blood distributes
glucose. Correct. Hyperglycemia was
described as the main
cryoprotective mechanism due to
the danger of intracellular ice.
B. Circulating blood equilibrates
temperature. Incorrect.
Hyperglycemia was described as the
main cryoprotective role of glucose.
This answer choice does not
mention glucose.
C. Beating heart warms body tissues
and slows ice formation. Incorrect.
Same as C.
D. Beating heart requires glucose
energy source. Incorrect. This does
not describe why glucose is
cryoprotective.
Q15:
• Do Fig 2 results support glucose cryoprotective hypothesis?
A. Yes; survival and protection against hemolysis are promoted by
exogenous glucose
B. Yes; death by freezing is directly proportional to hemolysis
C. No; glucose lowered [Hb]  survival levels unrelated to treatment
D. No; saline promoted hemolysis, suggesting death related to
circulatory collapse
Q15:
• Do Fig 2 results support glucose cryoprotective hypothesis?
A. Yes; survival and protection against hemolysis are promoted by
exogenous glucose
B. Yes; death by freezing is directly proportional to hemolysis
C. No; glucose lowered [Hb]  survival levels unrelated to treatment
D. No; saline promoted hemolysis, suggesting death related to
circulatory collapse
• The best answer is A
A. Higher glucose leads to higher
survival and lower [Hb]plasma,
i.e. lower hemolysis.
• Plasma Hb = hemolysis
B. Higher death is opposite [Hb],
suggesting lower death is
related to lower hemolysis
C. Injected glucose is related to
survival rates. Higher glucose =
higher survival.
D. Hemolysis = rupture of RBCs.
RBC destruction =/= circulatory
collapse. Circulatory collapse
occurs during v BP, e.g. during
stroke.
Passage 3 Takeaways
• Hyper/hypoglycemia
• Osmotic pressure
• Equation
• Insulin, glucagon; glycogenesis, glycogenolysis; lipogenesis, lipolysis
• Some important MCAT proteins
I.
II.
III.
IV.
V.
Albumin
Tubulin
Prothrombin
Hemoglobin
Cytochrome C
• Aerobic vs. anerobic respiration, relationship to blood flow
• Hypoxia, anoxia, ischemia
• Important metabolites – lactate, acetyl-CoA
• Osmosis and cell lysis
Q16:
•
A.
B.
C.
D.
Most effective method for increasing water lost through skin?
Inhibiting kidney function
Decreasing salt consumption
Increasing water consumption
Raising the environmental temperature
Q16:
• Most effective method for increasing water lost through skin?
A. Inhibiting kidney function. Incorrect. Water excretion through skin
does not depend on excretion through urinary system.
B. Decreasing salt consumption. Incorrect. Water excretion through
skin does depend on plasma salt.
C. Increasing water consumption. Incorrect. Water excretion through
skin does not depend on blood volume.
D. Raising the environmental temperature. Correct. Sweating
occurs to decrease body temperature.
• The reason sweating decreases body temp is evaporation has
A. (-)ΔS°
C. (+)ΔH°
B. (+)ΔS°
D. (-)ΔH°
• The answer is C.
A. (-)ΔS°. Incorrect. Liquid to gas increases entropy.
B. (+)ΔS°. Incorrect. This is true, but increasing entropy does not
decrease Temp.
C. (+)ΔH°. Correct. Breaking the H-bonds in a liquid requires thermal
energy, which decreases surrounding heat (endothermic).
D. (-)ΔH°. Incorrect. Evaporation is endothermic.
• The spontaneity of vaporization for a substance…
A. Is favorable at all temperatures
B. Is unfavorable at all temperatures
C.
D.
•
•
•
Depends on the substance
Depends on the temperature
The answer is D.
Vaporization, liquid becoming gas, always involves
a (+)ΔS° and (+)ΔH°
Via ΔG° = ΔH° - TΔS°, “like” signs always mean
spontaneity if temperature-dependent.
Q17:
•
A.
B.
C.
Lipases illustrate the fact that
Some enzymes are molecules other than proteins
Most enzymes interact w/ only one specific substrate molecule
Some enzymes interact w/ several different substrate molecules
with similar chemical linkages
D. Some enzymes interact with many diff bioactive substrates with
dissimilar structures/linkages
Q17:
• Lipases illustrate the fact that
A. Some enzymes are molecules other than proteins. Incorrect. Fats
and carboxylic esters are the substrates.
B. Most enzymes interact w/ only one specific substrate molecule.
Incorrect. 2 different substrates are mentioned.
C. Some enzymes interact w/ several different substrate molecules
with similar chemical linkages. Correct. These represent diff
substrates with similar linkages.
D. Some enzymes interact with many diff bioactive substrates with
dissimilar structures/linkages. Incorrect. See C.
• Which of the following does NOT contain an ester?
A. Triacylglycerols
B. Amino acid linkages
C. Phosphoacylglycerols
D. Nucleic acid linkages
• The answer is B.
• Triglycerides, phospholipids, and phosphodiester bonds contain
esters. Peptide bonds do not.
• Which bond is NOT broken/formed during ribozymal RNase activity?
A. P-(sp3)O
B. P-(sp2)O
C. O-H
D. C-O
• The answer is D.
A. The free 2’-OH (sp3) of the ribozyme attacks the
phosphodiester P.
B. The phosphodiester linkage between
ribonucleotides, P-(sp2)O is broken by the RNase.
C. An O-H bond is formed during phosphodiester
hydrolysis.
D. No carbons are involved.
Q18:
• Process that involves F factor plasmid
A. Transformation
B. Transduction
C. Conjugation
D. Translocation
Q18:
• Process that involves F factor plasmid
A. Transformation. Incorrect. Transformation is bacterial uptake of
DNA from the external environment.
B. Transduction. Incorrect. Transduction is when a virus inserts
genes from a prior host into a new host’s DNA.
C. Conjugation. Correct. Conjugation is horizontal gene transfer
between bacteria when one has the F factor (fertility factor).
D. Translocation. Incorrect. Translocation is gene exchange between
chromosomes in the same organism.
• The mechanism of acquired antibiotic resistance in B. cortus from
strain Y is studied using two chemicals, chromothymine (Chr),
which degrades genomic DNA, and mercaptocysteine (MeC),
which degrades circular DNA under 1000bp in size. Media are
prepared using a combination of both chemicals.
• Results: Chr(+) MeC(+): No growth in either strain. Chr(+) MeC(-):
No growth in either strain. Chr(-) MeC(+): Both strains are
observed to grow. Chr(-) MeC(-): Both strains are observed to
grow. What type of fertile bacterium is strain Y?
A. F(+) containing the resistance gene on a plasmid.
B. F(+) containing the resistance gene in the genomic DNA.
C. Hfr containing the resistance gene on a plasmid.
D. Hfr containing the resistance gene in the genomic DNA.
•
A.
B.
C.
D.
The answer is D.
Both strains grow in the presence of MeC.
F(+) is on a plasmid, not genomic.
Hfr is genomic, not on a plasmid.
Correct. Since neither strains grow in the presence
of Chr, which degrades genomic DNA, strain Y is
Hfr.
• F+ = bacteria with the F factor located on a plasmid
• Hfr = bacteria with the fertility genes located on its
chromosome
• F- = non-fertile bacteria
• Conjugation:
• Hfr or F+ cell forms sex pilus with F- cell
• Hfr (shown left): replication of fertility genes
• Sends to F- cell
• May contain Hfr bacterial genes during
longer conjugations (pro+ here, at right)
• Recombination – donor DNA integrated
into host chromosome
• F+ (not shown): replication of F factor plasmid
• Sends to F- cell
Q19:
•
•
A.
B.
C.
D.
Some offspring show recessive traits, some don’t
Original genotypes?
VVEE x vvee
VvEe x VvEe
vvEE x VVee
vvee x vvee
Q19:
•
•
A.
B.
C.
D.
•
A.
B.
C.
D.
•
A.
Some offspring show recessive traits, some don’t
Original genotypes?
VVEE x vvee. Incorrect. All F1 phenotypes dominant.
VvEe x VvEe. Correct. 9:3:3:1 ratio in offspring.
vvEE x VVee. Incorrect. All F1 phenotypes dominant.
vvee x vvee. Incorrect. All F1 phenotypes recessive.
If in a cross of fruit flies, 50 offspring had red eyes, 2 had white
eyes; and 48 offspring had hairy legs, 4 had smooth legs, what
conclusions can most reasonably be made?
• Assume the genotypes were RRhh * rrHH
White eyes are dominant to red, hairy legs are dominant to
smooth, and both genes are on separate chromosomes.
Red eyes are dominant to white, hairy legs are dominant to
smooth, and both genes are 30 map units away.
Red eyes are dominant to white, hairy legs are dominant to
smooth, and both genes are located on the same chromosome.
Red eyes are dominant to white, hairy legs are dominant to
smooth, and both genes are located on separate chromosomes.
The best answer is C.
White eyes are likely not dominant to red.
B. If the genes were 30 map units away, the
recombination frequency would be 30%, not 4-8% as
is observed.
C. Correct. Both genes are likely located close together
on the same chromosome, as seen by the low
recombination frequencies. Red is dominant to
white eyes, and hairy legs is dominant to smooth.
D. The genes are not located on different chromosomes
because they recombine.
Q20:
• Inflation of the lungs in mammals is accomplished by
A. Diffusion of gases
B. Active transport of gases
C. Positive pressure
D. Negative pressure
Q20:
• Inflation of the lungs in mammals is accomplished by
A. Diffusion of gases. Incorrect. Diffusion is passive movement.
Mammals breathe.
B. Active transport of gases. Incorrect. Active transport requires
proximal ATP hydrolysis.
C. Positive pressure. Incorrect. Positive pressure is pushing.
Mammals inhale to breathe.
D. Negative pressure. Correct. Negative pressure is pulling.
Mammals inhale to breathe.
• Positive and negative pressure are analogous to either oncotic
pressure or blood pressure. Mix and match:
• Positive pressure is analogous to blood pressure. Blood
pressure is a pushing force, i.e. a hydrostatic pressure,
against the blood vessel endothelium.
• Negative pressure is analogous to oncotic pressure (blood
osmotic pressure). Osmosis is a pulling force that causes
water to flow from low [solute] to high [solute].
• Between H2 and O2, which would diffuse best through cell
membranes, and because of which property?
A. H2, induced dipole-induced dipole forces
B. O2, polarizability
C. H2, dipole-dipole forces
D. O2, London dispersion forces
• The answer is A.
• H2 and O2 are both small, nonpolar gases. H2 is
smaller
• Polarizability is the tendency of instantaneous
dipole formation due to random electron cloud
dispersion, and increases w/size and # of e-’s
• ^Polarizability = ^LDFs
• LDFs can be thought of as, and are aka
induced dipole-induced dipole forces
• Dipole-dipole forces occur in polar molecules only
• London dispersion forces
• By chance, random electron motion causes an instantaneous
dipole moment to form in Atom 1
• This temporary dipole causes electron movement in Atom 2,
opposed to that of Atom 1
• Cumulatively across many atoms, these temporary forces lead to a
synchronous balancing of (+) and (-) charges of electrons and
nuclei
• This leads to a weak but cumulative attractive force between
nonpolar molecules
• Takeaway:
• LDFs increase via
• ^molecular size
• ^# of e-’s
• What are the standard
phases of matter for Cl2, Br2,
and I2?
• Gas, liquid, solid
• Rank the LDF strength for
these diatomic molecules
• I2 > Br2 > Cl2 because of
atomic size.
FSQ Takeaways
•
•
•
•
•
•
Thermoregulation methods in humans
Thermodynamics of physical changes (state changes)
Common biological substrates
Enzyme mechanisms
RNA structure
Methods of gene transfer
• Transformation, conjugation, transfection, transduction
• Genetics – Mendelian crosses
• Important ratios: 3:1, 9:3:3:1
• Testcross
• Transport – diffusion, facilitated diffusion, secondary active transport, primary active
transport
• Intermolecular forces – LDFs, dipole-dipole, H-bonding
Passage 4 Intro:
-
2 nuclei
Diploid micronucleous
45-ploid macronucleous
Conjugation
Looks like we have a passage on microbiology/genetics!
Q21: Function of extrachromosomal rRNA
genes
Q21:
• Extrachromosomal rRNA genes are most likely?
A. Nonlinear
B. Nonfunctional
C. Self-replicating
D. Rearranged
Q21:
• Extrachromosomal rRNA genes are most likely?
A. Nonlinear. Incorrect. The AAMC explanation literally says “all
nucleic acids are linear rearrangements of their component
nucleotides.”
• I took this out because they mean “linear vs. branched” as
opposed to “linear vs. circular.” mtDNA and plasmids are
circular, for instance.
B. Nonfunctional. Incorrect. rRNA is necessary for translation.
C. Self-replicating. Correct. This is the best remaining answer. The
genes likely replicate independently of chromosomal DNA.
D. Rearranged. Incorrect. Chromosome rearrangement has nothing
to do with chromosomal or extrachromosomal location.
• RNA pol’s I, II, III correspond to which types of RNA in eukaryotes?
• RNA pol I: rRNA
• RNA pol II: mRNA
• RNA pol III: tRNA
• Possible mnemonic: Rate my teachers??
• DNA pol’s I, II, III correspond to which functions in bacteria?
• DNA pol I: replaces RNA primers
• DNA pol II: repair
• DNA pol III: major replicative enzyme
Q22: Reason for extra S phase
Q23: Similarity between macronuclei and ova
Q24: Heterozygous macronucleus undergoing
repeated binary fission
Q25: Purpose of DNA sequences eliminated
Q26: Genetic probabilities of new system
Now, to read passage 4:
• Micronucleus
• Diploid, germ-line
• Macronucleus
• 45-ploid
• Gene expression during
vegetative state
• Sexual reproduction via conjugation
• Step 1: Cell pair formation
• Step 2: Meiosis of micronucleus
• Destruction of 3 other meiotic
nuclei
• Step 3: Mitosis of remaining nucleus
• Genetic exchange – trade 1 of the
mitotic products
• Step 4: fertilization of meiotic nuclei
• Half filled circles = fertilization
• Step 5: 2 mitotic divisions of fertilized
nucleus
How to read: Each diagram is the result of its step
description!!!!
Now, to read passage 4:
• Step 6: Differentiation of 2 new
micronuclei into macronuclei
• Elimination of some micronuclear
DNA sequences
• Fragmentation of original
micronuclei (X circle in step 7)
• Idk what 2 means
• Step 7: Destruction of old
macronucleus (X square) and 1 new
micronucleus (X circle)
• Separation of mating cells
• Each cell now has 2 macronuclei
and 1 micronucleus
• Step 8: 4 progeny cells, each with:
• 1 macronucleus (no
replication = amitotic)
• A copy of the new
micronucleus (mitotic)
How to read: Each diagram is the result of its step
description!!!!
Now, to read passage 4:
• 4 sexually produced progeny divide
by binary fission
• Each daughter cell receives:
• Exact copy of micronucleus
• Uneven (approx. equal) amount
of DNA from amitotic
macronucleus
• To minimize fluctuations in DNA
content
• Small macronuclei = additional S
phase
• Large macronuclei = eliminate S
phase
How to read: Each diagram is the result of its step
description!!!!
Q22:
• Reason for extra S phase during amitotic division?
A.
B.
C.
D.
Low concentrations of DNA in macronucleus
Centromeres in macronucleus
High concentrations of DNA in micronucleus
Mitotic enzymes in micronucleus
Q22:
• Reason for extra S phase during amitotic division?
A. Low concentrations of DNA in macronucleus. Correct. An
additional S phase would increase DNA concentration.
B. Centromeres in macronucleus. Incorrect. No reason why
centromeres would require extra S phase.
C. High concentrations of DNA in micronucleus. Incorrect. Passage
states large macronuclei eliminate an S phase.
D. Mitotic enzymes in micronucleus. Incorrect. No mention in
passage of mitotic enzymes.
• If the [genomic DNA] in a primary spermatocyte in interphase is
288 pg/mL, what is the concentration in a spermatid?
A. 72 pg/mL
B. 144 pg/mL
C. 216 pg/mL
D. 288 pg/mL
• The answer is B. During interphase, the DNA in a primary
spermatocyte is 2n. After meiosis, each spermatid has 1n DNA.
288/2 = 144 pg/mL.
Q23:
• Similarity between macronuclei and ova? Their cytoplasm
A. Undergo uneven division
B. Contain uneven amounts of nuclear material
C. Regulate their contents by adding/skipping an S phase
D. Are apportioned at mitosis
Q23:
• Similarity between macronuclei and cytoplasm of ova?
A. Undergo uneven division. Correct. Ova form polar bodies, in
which the ova contains the cytoplasm of all 4 meiotic products
but the genome of only one.
B. Contain uneven amounts of nuclear material. Incorrect. Ova
contain 1n of all chromosomes.
C. Regulate their contents by adding/skipping an S phase. Incorrect.
Not a thing in ova.
D. Are apportioned at mitosis. Incorrect. Ova cytoplasm are
apportioned during meiosis with formation of the first and second
polar bodies.
• What is the difference between a Barr body and a polar body?
• Polar body – small cells that bud off from oocyte during
meiosis
• Barr body – inactivated X-chromosome in many mammalian
females
Q24:
• Result of heterozygous macronucleus undergoing repeated binary
fission
A. Loss of macronuclear chromosomes
B. Increased rate of crossing over
C. Production of macronucleus with distinct genetic origin
D. Variable allele distribution in macronucleus
Q24:
• Result of heterozygous macronucleus undergoing repeated binary
fission
A. Loss of macronuclear chromosomes. Incorrect. The macronucleus
is 45-ploid, so this will not result in loss of chromosomes for many
divisions.
B. Increased rate of crossing over. Incorrect. Crossing over is during
meiosis, not binary fission.
C. Production of macronucleus with distinct genetic origin. Incorrect.
Distinct genetic origin would mean that the macronucleus came
from a different organism.
D. Variable allele distribution in macronucleus. Correct.
Heterozygosity in a 45-ploid nucleus that undergoes repeated
fission would result in a variable allele distribution.
• E.g. 23 dominant, 22 recessive
• Then 12 dominant, 11 recessive or something
• Etc etc.
Q25:
• Function of eliminated DNA sequences during macronuclear
differentiation
A.
B.
C.
D.
Transcription
Translation
Meiosis
Ribosome production
Q25:
• Function of eliminated DNA sequences during macronuclear
differentiation
A.
B.
C.
D.
•
Transcription. Incorrect. Necessary for gene expression.
Translation. Incorrect. Necessary for gene expression.
Meiosis. Correct. Micronucleus undergoes meiosis.
Ribosome production. Incorrect. Necessary for gene expression.
What are the small, large, and overall ribosomes in the ciliate
described in the passage?
• Small – 40S
• Large – 60S
• Overall – 80S
• Ciliates are protists, not bacteria.
Q26:
• Mating of 2 Tetrahymena strains
• Homozygous macronuclei
• Heterozygous micronuclei
• Recessive gene
• Percentage of F1 generation that will express recessive
phenotype?
A. 0%
B. 25%
C. 50%
D. 100%
Q26:
• Mating of 2 Tetrahymena strains
• Homozygous macronuclei
• Heterozygous micronuclei
• Recessive gene
• Percentage of F1 generation that will express recessive
phenotype?
A. 0%
B. 25%. Correct.
• Only micronucleus is involved in mating (germ-line)
• Macronucleus is destroyed in step 7
• The cross ends up being a monohybrid Aa*Aa cross, where
75% express the dominant phenotype
• 25% express the recessive phenotype
C. 50%
D. 100%
Passage 4 Takeaways
• Genetics – plasmids, linear vs circular DNA
• Punnett squares
• Stages of mitosis and meiosis, what happens during
• X-chromosome inactivation
• Polar bodies – ovum meiosis
• Polyploidy
• Central dogma of molecular biology
• Ribosome structure in prokaryotes vs eukaryotes
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes
Passage 5 Intro:
-
Nucleic acid macromolecules
Antisense nucleic acids
Prevent production
Looks like we have a passage on molecular biology!
Q27: Purpose of antisense drugs
Q27:
•
A.
B.
C.
D.
Purpose of antisense drugs is to prevent
DNA replication
RNA transcription
RNA translation
Cell replication
Q27:
•
A.
B.
C.
D.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Purpose of antisense drugs is to prevent
DNA replication
RNA transcription
RNA translation. Correct. mRNA is produced, but due to
antisense binding, cannot be converted to protein.
Cell replication
Cellular injection of which of the following RNA strands would
prevent translation of the following template strand sequence
5’-GAGCATCC-3’?
5’-GAGCAUCC-3’
3’-CTCGTAGG-5’
5’-CUCGUAGG-3’
3’-CUCGUAGG-3’
The answer is A.
Correct. The template strand is antisense. An antisense strand is
needed to suppress gene expression.
• Compare this to the coding strand vs. mRNA, which also
have the same sequence with T vs. U
RNA does not contain T.
This represents the mRNA produced by that sequence with its 5’
and 3’ ends swapped.
This represents the mRNA produced by that sequence.
Q28: Effectiveness of antisense gene therapy
Q28:
• To be effective, an antisense gene must be
A. On the same chromosome as the target gene but not necessarily
physically adjacent
B. On the same chromosome as the target gene and physically
distant
C. Regulated in a similar manner as the target gene
D. Coded on the same strand of DNA as the target gene
Q28:
• To be effective, an antisense gene must be
A. On the same chromosome as the target gene but not necessarily
physically adjacent. Incorrect. An antisense gene does not need to
be on the same chromosome.
B. On the same chromosome as the target gene and physically
distant. Incorrect. Same as A.
C. Regulated in a similar manner as the target gene. Correct. If the
antisense gene gets expressed whenever the target gene does, it
will always be present to inhibit the target gene’s translation.
D. Coded on the same strand of DNA as the target gene. Incorrect.
• Are the coding and template strands of DNA the same for every
gene on a chromosome?
• No, a coding strand may be a template strand for another
gene.
• Via C), the antisense gene must have the same
A. Promoter
B. Enhancer
C. Activator
C. Activators act on enhancers. A is a better answer.
D. Second messenger
D. Second messengers act broadly on gene expression
• The answer is A.
and are not sufficient to ensure simultaneous gene
A. The best way to ensure similar regulation is using the same
and antisense gene expression.
promoter, which ensures RNA pol is recruited at the same time.
B. Enhancers increase txn, but promoter is more essential for RNA
pol recruitment.
Q29: Antisense drug for phenylketonuria
Q29:
• Phenylketonuria is caused by mutation in phenylalanine
hydroxylase that eliminates enzymatic activity
• Could an antisense drug help?
A. Yes, if it binds to enzyme’s mRNA, preventing translation
B. Yes, if incorporated in chromosome, preventing expression
C. No, b/c mRNA doesn’t persist in cytoplasm
D. No, b/c blockage of enzyme does not remedy disorder
Q29:
• Phenylketonuria (PKA) is caused by mutation in phenylalanine
hydroxylase that eliminates enzymatic activity.
• Could an antisense drug help?
A. Yes, if it binds to enzyme’s mRNA, preventing translation
B. Yes, if incorporated in chromosome, preventing expression
C. No, b/c mRNA doesn’t persist in cytoplasm
D. No, b/c blockage of enzyme does not remedy disorder
• PKA is a disorder in which a normal dietary amount of phenylalanine that
cannot be metabolized leads to competition at blood-brain barrier
transporters with other large, neutral amino acids (LNAAs).
• Transport of which a.a. into the brain will NOT be inhibited in
phenylketonuria?
A. Tryptophan
B. Histidine
C. Methionine
D. Arginine
• The answer is D. W, H, and M are LNAAs. Arg is charged.
• Production of which hormone is inhibited in PKA?
A. Aldosterone
B. Bicarbonate
C. Triiodothyronine
D. Oxytocin
• The answer is C. Thyroid hormones and catecholamines are derived from
tyrosine, which is structurally related to and derived from Phe.
Q30: Hybridization
Q30:
• Which of the following can hybridize with
• 5’-CGAUAC-3’
A. 5’-GCTATG-3’
B. 5’-GCUAUG-3’
C. 3’-GCUAUG-5’
D. 3’-GCAUAG-5’
Q30:
• Which of the following can hybridize with
• 5’-CGAUAC-3’
A. 5’-GCTATG-3’
B. 5’-GCUAUG-3’
C. 3’-GCUAUG-5’. Correct. This would hybridize to the given sequence.
D. 3’-GCAUAG-5’
• The Duchenne muscular dystrophy (DMD) drug, eteplirsen, binds to
exon 51 of the mutant DMD gene, a 30-bp exon. What is the likely
mechanism of treatment for eteplirsen?
A. Restores the improper reading frame in mutant DMD mRNA.
B. Causes deletion of exon 51 from the DNA, leading to a functional
protein.
C. Triggers endogenous DNA repair, which restores the DNA for exon 51 to
its original state.
D. Prevents translation of exon 51, resulting in a truncated but functional
protein.
• The answer is D.
A. 30-bp is a multiple of 3, so reading frame will not change.
B. Eteplirsen cannot bind to dsDNA since hybridization is already complete
with the other DNA strand.
C. Similar to B.
D. Eteplirsen binds to exon 51 in the mRNA, preventing its translation.
Exon 51 is mutated, leading to a nonfunctional protein. By preventing
its translation, the protein is shorter but still functional.
Q31: Antisense gene delivery
Q31:
• Effective, efficient method for gene delivery of antisense gene
A. Orally as an emulsified product
B. Microinjection into individual body cells
C. Intravenously as a nonantigenic, blood-stable product
D. Infection of embryo by virus modified to carry the gene
Q31:
• Effective, efficient method for gene delivery of antisense gene
A. Orally as an emulsified product. Incorrect. The gene would likely be
destroyed in the stomach.
B. Microinjection into individual body cells. Incorrect. This is inefficient
(many injections) and not long-lasting.
C. Intravenously as a nonantigenic, blood-stable product. Incorrect. No
guarantee of uptake into body’s cells.
D. Infection of embryo by virus modified to carry the gene. Correct. At
the embryonic stage, there are fewer cells to transfect, and they would
express the gene for life.
• What type of virus would be the best choice for such a gene therapy?
A. (-)-sense RNA virus
B. (+)-sense RNA virus
C. ssDNA virus
D. dsDNA virus
• The answer is B. Retroviruses integrate DNA into the host cell and are
(+)-sense RNA viruses.
Q32: Purpose of quick RNA degradation
Passage 5 Takeaways
• Central dogma of molecular biology
• Replication, txn, trln, rev trln
• Transcriptional elements – gene transduction, second messenger, enhancer,
activator, promoter
• RNA/DNA or DNA/DNA hybridization
• RNA processing – 5’capping, 3’poly A tail, splicing
• Degradation
• Genetic diseases
• Amino acids – class, structure
• Drug delivery methods – examine physiology of various
organs/compartments
• Viral life cycles
Passage 6 Intro:
- Elevated pulse and ventilation rate
- Diving, skiing
- Looks like we have a “case study” type passage!
Q33: Cause of increased HR and ventilation
rates
Q33:
•
A.
B.
C.
Cause of initial increase in HR and ventilation rates?
Activation of sympathetic NS by new exp
Activation of parasympathetic NS
Hypoxia from inability of blood hemoglobin to supply O2 for
swimming at sea level
D. Elevated core temperature from swimming in warm waters
Q33:
• Cause of initial increase in HR and ventilation rates?
A. Activation of sympathetic NS by new exp. Correct. Since Sarah
was already in excellent physical condition, the excitement of the
new experience likely led to her symptoms.
B. Activation of parasympathetic NS. Incorrect. Opposite.
C. Hypoxia from inability of blood hemoglobin to supply O2 for
swimming at sea level. Incorrect. Since Sarah was in excellent
physical condition, this is likely not the cause.
D. Elevated core temperature from swimming in warm waters.
Incorrect. Neither symptom is related to thermoregulation.
• Which of the following would NOT be related to sympathetic
nervous activation?
A. Dilation of pupils
B. Vasodilation of GI blood supply
C. Vasodilation of skeletal muscle blood supply
D. Bronchodilation
• The answer is B.
A. Pupillary dilation helps vision during sympathetic activation.
B. Vasoconstriction of GI blood supply occurs during sympathetic
activation because digestion is less necessary in fight-or-flight.
C. This helps feed skeletal muscle with more O .
D. Bronchodilation helps get more O2 to the alveoli
during fight-or-flight .
Q34: Cause of increased HR and ventilation
rates during skiing trip
Q34:
• Cause of increase in HR and ventilation rates during skiing trip?
A. Activation of sympathetic NS by new exp
B. Activation of parasympathetic NS
C. Hypoxia from inability of blood hemoglobin to supply O2 for PO2
found at high altitudes
D. Depressed core body temperature
Q34:
• Cause of increase in HR and ventilation rates during skiing trip?
A. Activation of sympathetic NS by new exp. Incorrect. C is a better
answer.
B. Activation of parasympathetic NS. Incorrect. See C.
C. Hypoxia from inability of blood hemoglobin to supply O2 for PO2
found at high altitudes. Correct. Since air pressure decreases as
altitude increases, initial physiological adjustments include
increased HR and ventilation rate.
D. Depressed core body temperature. Incorrect. C is a better answer.
• Which of the following would be seen at higher altitude?
A. Higher Hb-O2 binding affinity in tissues due to 2,3-BPG
B. Rightward Hb shift due to decreased Temp
C. Slower breathing due to fresher mountain air
D. Lower Hb-O2 binding affinity in tissues due to 2,3-BPG
• The answer is D.
A. Hb-O2 binding affinity decreases in tissues due to 2,3-BPG,
enabling more oxygen delivery.
B. Decreased Temp causes a leftward Hb shift.
C. Breathing increases because of decreased pressure of O2.
D. 2,3-BPG causes decreased Hb-O2 binding affinity in tissues.
• Decreased unloading/increased loading in lung
conditions
• Increased unloading in tissue conditions
• Tissues are metabolically active, produce
acidic waste, higher Temp
Q35: Increased urination
Q35:
• Cause of increased urine during initial dive session
• ^HR and ventilation rate
A. Increased BP from excitement/anxiety
B. Reduced BP from excitement/anxiety
C. Absorption of water from ocean
D. Inability to cool skin through evaporative water loss
Q35:
• Cause of increased urine during initial dive session
• ^HR and ventilation rate
A. Increased BP from excitement/anxiety. Correct. ^BP = ^GFR =
^urine output.
B. Reduced BP from excitement/anxiety. Incorrect. Opposite of A.
C. Absorption of water from ocean. Incorrect. Humans cannot
directly absorb water through their skin.
D. Inability to cool skin through evaporative water loss. Incorrect.
This would lead to increased body temperature and more
sweating, but not more urine excretion.
• As Sarah dives deeper and the water temperature decreases, O2
solubility in ocean water
A. Decreases
B. Increases
• The answer is B. Higher temperatures give O2 greater thermal
energy, which results in greater escape from H2O and lower
solubility.
• If Sarah dives in a mountain hot spring vs. the ocean at sea level,
assuming same Temp, in which will O2 solubility be higher?
A. Mountain spring
B. Ocean at sea level
• The answer is B. At sea level, the atmospheric PO2 is higher, so
more O2 will be dissolved in water.
Q36: Increased urination
Q36:
• Cause of myoglobin presence in urine
I. Broken bone
II. Damaged muscle
III. Damaged kidney
A.
B.
C.
D.
I
III
I and III
II and III.
Q36:
• Cause of myoglobin presence in urine
I. Broken bone
II. Damaged muscle
III. Damaged kidney
A.
B.
C.
D.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
I
III
I and III
II and III. Correct. Myoglobin is usually intact in the muscle, so
myoglobinuria would indicate damaged muscle. Proteins are
usually kept out of the kidney by glomerular filtration, so
myoglobinuria would also indicated damaged kidney.
What is the purpose of myoglobin?
Absorb O2 during periods of high muscle usage
Provide equilibrium for hemoglobin in muscle
Enable aerobic metabolism to occur for longer in active muscle
Protect muscle tissue against damage during long periods of use
The answer is C.
Myoglobin enables O2 delivery during high muscle usage.
This doesn’t really mean anything.
Myoglobin is an O2 storage protein in muscle tissues for when
actively contracting muscles prohibit blood flow.
Myoglobin is not a protective protein.
Q37: Nervous system structure vs. function
Q37:
• Control of HR, muscle coordination, and appetite is maintained by
the
A. Hypothalamus, cerebrum, and brain stem
B. Brain stem, hypothalamus, and cerebrum
C. Cerebellum, hypothalamus, and brain stem
D. Brain stem, cerebellum, and hypothalamus
Q37:
• Control of HR, muscle coordination, and appetite is maintained by
the
A. Hypothalamus, cerebrum, and brain stem
B. Brain stem, hypothalamus, and cerebrum
C. Cerebellum, hypothalamus, and brain stem
D. Brain stem, cerebellum, and hypothalamus. Correct. This is the
only answer choice that correctly matches the nervous system
structure with its function.
• Which of the following pathways does NOT give the correct
hormonal products by organ, from hypothalamus to anterior
pituitary to target organ?
A. Thyrotropin-releasing hormone, thyroid hormone, growth
hormone
B. Corticotropin-releasing hormone, adrenocorticotropic hormone,
cortisol
C. Gonadotropin releasing hormone, FSH/LH, estradiol/testosterone
D. Growth hormone-releasing hormone, GH, insulin-like growth
factor-I
• The answer is A
A. Thyroid hormone results in release of T3, T4
B. CRH  ACTH  cortisol
C. GnRH  FSH/LH  estradiol/testosterone
D. GHRH  GH  IGF-I (this last part is unnecessary to memorize)
Q38: Reason for periodic vasodilation
Q38:
• Reason for periodic vasodilation
A. Maintain normal skin tone
B. Maintain sufficient oxygenation of cells
C. Reduce excessive blood pressure
D. Maintain normal muscle tone
Q38:
• Reason for periodic vasodilation
A. Maintain normal skin tone. Incorrect. Skin tone is usually
understood as skin color, i.e. melanin levels.
B. Maintain sufficient oxygenation of cells. Correct. Vasoconstriction
leads to less oxygenation of cells; vasodilation ensures they are
oxygenated.
C. Reduce excessive blood pressure. Incorrect. Blood pressure is not
regulated directly in this manner.
D. Maintain normal muscle tone. Incorrect. Blood vessel tone is not
understood as muscle tone. Blood vessel muscle tone is regulated
in response to O2 and CO2.
• Similarly to vasodilation, which would occur periodically at high
elevation to keep tissue oxygenation constant?
A. Tachypnea (higher ventilation rate)
B. Polyuria (higher urine output)
C. Hypernatremia (higher serum sodium)
D. Ischemia (lower blood perfusion)
• The answer is A.
A. Periodic tachypnea would help keep blood oxygenated similarly to
how periodic vasodilation would help keep O2 distributed.
B. Incorrect. No reason to suspect higher blood pressure or lower salt
reabsorption at higher elevation.
C. Incorrect. No reason to suspect higher [Na+]serum at high elevation
D. Incorrect. Ischemia would not help keep tissue oxygenation constant.
Passage 6 Takeaways
• Nervous system organization
• Autonomic – sympathetic vs. parasympathetic
• Hemoglobin role and effects
• Gas solubility – concentration and pressure effects
• Myoglobin
• Neurophysio – role of brain parts
• Hormone axes – hypothalamic-pituitary axis
• Cardiovascular-respiratory system – effect of changing one variablef
Q39:
• Placement of additional neurons to represent pathway involved in
feeling pain
A. At II and III
B. At II and IV
C. At III and IV
D. At I and IV
Q39:
• Placement of additional neurons to represent pathway involved in
feeling pain
A. At II and III. Correct. Additional neurons cannot be placed at the
sensory or motor neuron (I and IV, respectively). They can be
placed at the interneuronal junctions (II and III).
B. At II and IV
C. At III and IV
D. At I and IV
•
A.
B.
C.
D.
•
A.
B.
C.
A tyrosine deficiency would most affect signaling by the
Preganglionic neuron of the parasympathetic nervous system
Postganglionic neuron of the parasympathetic nervous system
Motor neuron of the somatic nervous system
Postganglionic neuron of the sympathetic nervous system
The answer is D.
The preganglionic and postganglionic neurons of the PNS use ACh.
Same as A.
ACh is the neurotransmitter at the NMJ of the somatic nervous
system.
D. Correct. Norepinephrine (and epinephrine) are derived from
tyrosine. ACh is the NT released by the preganglionic neuron of
the SNS; NE is the NT released by the postganglionic neuron of
the SNS.
Q40:
• Characteristic that most marks fungi as eukaryotes:
A. Cell walls
B. Ribosomes
C. Mitochondria
D. Sexual reproduction
Q40:
• Characteristic that most marks fungi as eukaryotes:
A. Cell walls. Incorrect. Plants and fungi have cell walls.
B. Ribosomes. Incorrect. All living organisms have ribosomes.
C. Mitochondria. Correct. Only eukaryotes have mitochondria (and any
membrane-bound organelles).
D. Sexual reproduction. Incorrect. Bacterial sexual reproduction is called
conjugation.
• Fungi are remarkable in their ability to grow in a variety of environments,
including irradiated environments that many other organisms cannot.
Fungi growing in environments containing ionizing radiation e.g. gamma
rays are found to contain high levels of melanin. Gamma rays can cause
the photoelectric effect in cellular molecules. The most likely role of
melanin is
A. To absorb charged particles from the photoelectric effect and add them
to melanin atom nuclei, preventing damage.
B. To provide oxidizing agent molecules that can absorb particles from the
photoelectric effect, preventing damage.
C. To scatter photoelectric particles into nearby cells, distributing any
damage across multiple tissues.
D. To promote photoelectric particles into higher-energy states, so they can
be subsequently converted to more harmless photons.
• The answer is B.
A. Electrons cannot be added to nuclei.
B. Oxidizing agents can be reduced by
accepting electrons (e.g. photoelectrons).
Melanin contains stable ROS’s that can be
reduced by photoelectrons, protecting
against gamma radiation.
C. The cumulative damage would not help a
fungus grow against high environmental
gamma radiation.
D. Photoelectrons are ambient electrons and
cannot be promoted, as they do not belong
to an atom.
Q41:
•
A.
B.
C.
D.
Which occurs during mitosis but not meiosis I?
Synapsis
Splitting of centromeres
Pairing of homologous chromosomes
Breaking down of nuclear membrane
Q41:
• Which occurs during mitosis but not meiosis I?
A. Synapsis. Incorrect. Synapsis, where crossing-over takes place,
occurs during meiosis I but not mitosis. Opposite.
B. Splitting of centromeres. Correct. Centromeres connect sister
chromatids. Sister chromatids are split during telophase of
mitosis and telophase of meiosis II. Homologous chromosomes
are split during telophase of meiosis I.
C. Pairing of homologous chromosomes. Incorrect. Homologous
chromosomes are only paired during prophase I of meiosis.
D. Breaking down of nuclear membrane. Incorrect. Breaking down of
nuclear membrane occurs during both.
• During which cell cycle phase is the pentose phosphate pathway
most active?
A. Telophase of mitosis
B. Telophase of meiosis II
C. S before mitosis and meiosis
D. G2 before mitosis and meiosis
• The answer is C. The PPP is responsible for
• NADPH production, which is used in
• Cellular detoxification (esp free radicals)
• Fatty acid/sterol synthesis
• Ribose-5-phosphate synthesis, which can be used in
production of nucleotides.
Q42:
• Which is NOT a function of mammalian skin?
A. Sensation
B. Respiration
C. Protection from disease
D. Protection against internal injury
Q42:
• Which is NOT a function of mammalian skin?
A. Sensation. Incorrect. The skin has sensory receptors.
B. Respiration. Correct. Mammals do not perform gas exchange
upon their skin.
C. Protection from disease. Correct. The skin is a barrier to pathogen
entry as part of innate immunity.
D. Protection against internal injury. Correct. Dead skin serves as a
barrier that protects the more sensitive inner layers and
bloodstream from injury.
• Skin respiration is common in amphibious vertebrates, who often
additionally have pulmonary respiration. Which of the following is
NOT an advantage of skin respiration in these animals?
A. The capability to inhale CO2 underwater during the cold months.
B. The capability to breathe in low-oxygen conditions, e.g. being
coated in hypoxic mud during the hot months.
C. Reduced energy expenditure associated with gas exchange
compared to pulmonary.
D. Greater overall surface area for gas exchange.
• The answer is A.
A. Animals do not require CO2; they exhale CO2 as a metabolic
byproduct.
B. Frogs coat themselves in mud during hot months and use skin
respiration to enhance gas exchange.
C. Skin respiration occurs via diffusion (does not require breathing).
D. Skin respiration adds more surface area
compared to lungs alone.
Q43:
•
A.
B.
C.
D.
Why does inbreeding reduce population fitness? ^ in
Genetic diversity
Levels of aggression
Rate of spontaneous mutations
Incidence of expression of deleterious recessive traits
Q43:
• Why does inbreeding reduce population fitness? ^ in
A. Genetic diversity. Incorrect. Inbreeding decreases genetic diversity.
B. Levels of aggression. Incorrect. This would only be true if aggression were
tied to a specific recessive allele disproportionately present in the
inbreeding population.
C. Rate of spontaneous mutations. Incorrect. The rate of spontaneous
mutations depends mainly on DNA polymerase and DNA repair
mechanisms, not inbreeding.
D. Incidence of expression of deleterious recessive traits. Correct. Any
deleterious recessive trait that is overrepresented in an inbreeding
population will be overexpressed.
• While inbreeding is associated with a reduction in fitness, outbreeding
can sometimes also result in depression of offspring fitness. Which of the
following would represent outbreeding depression?
A. Three generations after a bottleneck event resulting in 1 male and 1
female, a male turkey with a recessive colorblindness allele mates with a
female with the same genotype, resulting in a colorblind turkey.
B. A hummingbird mates with a member of another hummingbird species,
resulting in a non-viable resulting embryo.
C. A horse mates with a donkey, producing a sterile hybrid mule offspring.
D. A sickle-cell human carrier with no symptoms reproduces with an
unaffected partner, resulting in a child who is a sickle-cell carrier.
• The answer is C.
A. This is inbreeding; both parents stem from
the same pair.
B. Outbreeding requires a viable offspring.
C. Correct. Since the organism produced from
2 unrelated parents is sterile, this
represents an extreme case of outbreeding
depression.
D. Since sickle-cell trait usually results in no
symptoms, this case does not represent
outbreeding depression.
FSQ Takeaways
• Functions of skin
• EvoBio – inbreeding vs outbreeding
From last week:
• Nervous structure – interneurons
• Tyrosine-derived hormones: catecholamines, thyroid hormones, dopamine
• Tryptophan-derived hormones: serotonin, melatonin
• Differences between prokaryotes and eukaryotes
• Mitosis and meiosis
•
•
•
•
Steps
Ploidy
Homologous chromosomes vs sister chromatids in each
Synapsis
• Pentose phosphate pathway – products (NADPH and ribose-5-Pi, important intermediates and
enzymes
NOW vs. LATER recap
• Some questions are “Now” questions, because they rely exclusively or
primarily on outside-passage information.
• We do now questions before reading, only looking up a line or two max!
• Some questions are “Later” questions, because they require extensive
passage analysis
• Remember, the now vs. later technique is designed to minimize the
number of questions you worry about for AFTER reading a CP/BB passage.
• Conserve brainpower, conserve willpower!
• If a question takes more than 15-30s to determine its now/later
classification, it’s a LATER!
• Let’s analyze passage 1 for now/later questions.
Passage 7 Intro:
-
Plasma clearance
Tubular transport
Glomerular filtrate
Looks like we have a kidney physiology passage!
Q44: Definition of Tm
Q44:
• Definition of Tm
•
A.
B.
C.
D.
Rate of plasma filtration that just exceeds
Rate of concentration of substance in glomerular filtrate
Rate of concentration of substance in urine
Capacity of kidney tubules to reabsorb substance
Capacity of bladder to store and excrete substance
Q44:
• Definition of Tm
• Rate of plasma filtration that just exceeds
A. Rate of concentration of substance in glomerular filtrate. Incorrect. This is
filtration rate, which may be less than or exceed Tm, leading to the
presence of the substance in the urine.
B. Rate of concentration of substance in urine. Incorrect. See A.
C. Capacity of kidney tubules to reabsorb substance. Correct. Tm represents
the Vmax of transport of a reabsorbed substance at saturation.
D. Capacity of bladder to store and excrete substance. Incorrect. The bladder
stores fluid, not solutes, and excretes everything when it fills and the
person urinates.
• Which cardiovascular variable most determines glomerular filtration rate?
A. Heart rate
B. Systolic blood pressure
C. Capillary resistance
D. Stroke volume
• The answer is B.
A. CO = HR*SV. This determines blood flow, not GFR.
B. BP = CO*TVR (total vascular resistance). Correct. Blood pressure
“squeezes” water and some solutes through the glomerular membrane.
C. BP = CO*TVR. BP determines GFR moreso.
D. CO = HR*SV. B is a better answer.
Q45: Amount of glucose in urine
Q45:
• Amount of glucose in urine when Tm is 125 mg/min
A.
B.
C.
D.
0 mg/min
125 mg/min
195 mg/min
515 mg/min
Q45:
• Amount of glucose in urine when Tm is 125 mg/min
A.
B.
C.
D.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
0 mg/min. Correct. If
125 mg/min
195 mg/min
515 mg/min
If substance B is found in urine at a rate of 12 mg/min but the rate
of glomerular filtration is 14 mg/min, what is its Tm?
2 mg/min
6 mg/min
13 mg/min
26 mg/min
The answer is A. Rateurine can be thought of as Rategf - Tm (where gf
= rate of glomerular filtration).
14 mg/min – 12 mg/min = 2 mg/min. Correct. Tubular maximum
is the maximum for reabsorption, so at a 14 mg/min GFR and a 2
mg/min Tm, 12 mg/min will end up in the urine.
Incorrect. This answer comes from nowhere.
Incorrect. This answer comes from taking the average of GFR and
rate of appearance in urine. See A.
Incorrect. This answer comes from adding the two rates. See A.
Q46: vBP effect on plasma clearance of
Substance A
Q46:
• Effect of vBP on clearance of Substance A
A. Increase, because ^[A]urine
B. Increase, because vADH
C. Decrease, because decreased rate of urine output will allow more
reabsorption
D. Decrease, because ^ADH
Q46:
• Effect of vBP on clearance of Substance A
A. Increase, because ^[A]urine. Incorrect. Water and substance A are
simultaneously filtered, so concentration of A should not change with
changes in blood pressure.
B. Increase, because vADH. Incorrect. ADH affects water reabsorption, not
solute excretion.
C. Decrease, because decreased rate of urine output will allow more
reabsorption. Correct. Via kinetics, lower filtration rate will mean less
saturation and more time for tubular membrane transporters to
reabsorb A.
D. Decrease, because ^ADH. Incorrect. See B.
Q47: Which clears faster – A or glucose?
Q47:
• Given equal [A] and [glucose] in plasma, which clears more rapidly?
A. Substance A, higher slope
B. Substance A, lower Tm
C. Glucose, higher Tm
D. Glucose, lower slope
Q47:
• Given equal [A] and [glucose] in plasma, which clears more rapidly?
A. Substance A, higher slope
B. Substance A, lower Tm. Correct. Glucose will be completely reabsorbed since
it reaches its Tm at 10 mg/mL. Substance A will be partly secreted, as 8
mg/mL surpasses its Tm.
C. Glucose, higher Tm
D. Glucose, lower slope
• At 20 mg/mL of each in plasma, (2 options)
A. Substance A will have a higher clearance fraction
B. Glucose will have a higher clearance fraction
• The answer is B. Since glucose has a higher slope, at a 20 mg/mL plasma
concentration, more glucose will be seen in urine. Glucose will have the higher
clearance fraction.
Q48: [glu]pl when Tm reached
Q48:
• Tm of glucose
A. 6.5 mg/mL
B. 10.0 mg/mL
C. 11.5 mg/mL
D. 12.5 mg/mL
Q48:
• Tm of glucose
A. 6.5 mg/mL
B. 10.0 mg/mL. Correct. Tm is defined as the concentration at which
the substance begins appearing in the urine.
C. 11.5 mg/mL
D. 12.5 mg/mL
• Assuming 100% dietary absorption, how much glucose would an
individual with 4.8 L of blood need to consume to reach the Tm of
glucose?
A. 48 mg
B. 480 mg
C. 4.8 g
D. 48 g
The answer is D.
• 10 mg/mL can be simplified as 10 g/L
• If the individual has 4.8 L of blood, it would take 10
g/L*4.8 L = 48 g to reach the Tm
Q49:
Q49:
• Effect on urinary output of A from high ADH
A. ^GFR
B. ^Tm of solutes
C. ^H2O reabsorption
D. ^concentrating ability of loop of Henle
Q49:
• Effect on urinary output of A from high ADH
A. ^GFR. Correct. Blood pressure increases GFR.
B. ^Tm of solutes. Incorrect. Tm is a property of transport
proteins.
C. ^H2O reabsorption. Incorrect. Higher blood pressure leads to
lower water reabsorption.
D. ^concentrating ability of loop of Henle. Incorrect.
Concentrating ability of such a countercurrent system
increases with length of the system (loop of Henle).
• The major glucose transporter in the kidney is SGLT2, which
cotransports Na+ and glucose in the proximal tubule.
Gliflozins are drugs that inhibit SGLT2. Which best reflects
their physiological outcomes?
A. Increased blood pressure and hyperglycemia.
B. Decreased blood pressure and hyperglycemia.
C. Increased blood pressure and hypoglycemia.
D. Decreased blood pressure and hypoglycemia.
• The answer is D. Due to inhibition of Na+ reabsorption, less
H2O will be reabsorbed, leading to decreased blood
pressure. Due to inhibition of glucose reabsorption,
hypoglycemia is more likely.
Passage 7 Takeaways
• Kidney functions: filtration, secretion, reabsorption
• Structures: glomerulus, proximal/distal tubule, loop of Henle, collecting duct
• Effect *of* blood pressure (filtration) and effect *on* blood pressure (H2O
reabsorption or secretion)
• Interpreting new concept: Tm
• Density = mass/vol
Passage 8 Intro:
-
Vasodilation
Arterial smooth muscle
Acetylcholine
Sounds like we have a passage on cardiovascular/smooth muscle contraction!
Q50: Observation implying endothelium
necessary for smooth muscle relaxation
Q50:
• Observation that implied intact endothelium is necessary for
smooth muscle relaxation in presence of ACh
A. Tension initially changes when norepinephrine is added
B. Tension decrease occurs more quickly in ring w/o endothelium
C. Tension decreases w/10-7 ACh only in ring w/o endothelium
D. Tension decreases during washout in ring w/o endothelium
Q50:
• Observation that implied intact endothelium is
necessary for smooth muscle relaxation in presence of
ACh
A. Tension initially changes when norepinephrine is
added. Incorrect. NE causes contraction.
B. Tension decrease occurs more quickly in ring w/o
endothelium. Incorrect. Opposite.
C. Tension decreases w/10-7 ACh only in ring w/o
endothelium. Correct. Relaxation upon addition of
ACh only occurs in the ring with endothelium.
Relaxation in the ring w/o endothelium requires
washout of NE.
D. Tension decreases during washout in ring w/o
endothelium. Incorrect. This observation supports C,
because washout is removal of NE, not application of
ACh.
Q51: Sensitivity of smooth muscle to ACh is
Q51:
• Sensitivity of aortic smooth muscle to ACH is
A. Decreased by NE
B. Increased by NE
C. Increased by endothelium
D. Greatest at 10-8 M w/ or w/o endothelium
Q51:
• Sensitivity of aortic smooth muscle to ACH is
A. Decreased by NE. Incorrect. NE is applied in both cases.
B. Increased by NE. Incorrect. Same as A.
C. Increased by endothelium. Correct. The ring with endothelium
relaxed at 10-7 M ACh, whereas the ring w/o didn’t relax until
washout after 10-6 M ACh, indicating at least 10x less sensitivity
to ACh without endothelium.
D. Greatest at 10-8 M w/ or w/o endothelium. Incorrect. Both
samples continued to contract beyond 10-8 M ACh.
Q52: Most sensitive to ACh
Q52:
• Concentration range most sensitive to ACh
A. < 10-8 M
B. 10-7 M
C. > 10-6 M
D. Wider in ring without endothelium vs ring with endothelium
Q52:
• Concentration range most sensitive to ACh
A. < 10-8 M
B. 10-7 M. Correct. The ring with endothelium is most sensitive to
ACh around 10-7 M.
C. > 10-6 M
D. Wider in ring without endothelium vs ring with endothelium.
Incorrect. The ring without endothelium is never shown to be
sensitive to ACh at all, only contracting when NE is washed out.
Q53: Factors that determine BP
Q53:
• Factors that determine blood pressure
A.
B.
C.
D.
[L-NMMA] and norepinephrine
Cardiac output and resistance to blood flow
Blood volume and dietary L-arginine
HR and SV
Q53:
• Factors that determine blood pressure
A. [L-NMMA] and norepinephrine. Incorrect. L-NMMA is synthetic
(described as “developed”).
B. Cardiac output and resistance to blood flow. Correct. BP = CO *
TVR.
C. Blood volume and dietary L-arginine. Incorrect. L-arginine supply
does not regulate NO synthase activity.
D. HR and SV. Incorrect. HR * SV = CO.
• The structure of NO is shown at right. NO synthase (NOS) has an
inducible isoform, iNOS. Which of the following is the most likely
function of iNOS?
A. DNA repair
B. Skeletal muscle contraction
C. Immune defense
D. Water balance
• The answer is C.
A. Since NO has a free radical, it is unlikely to aid in DNA repair.
C. Correct. NO can use its free radical to generate ROSs
B. No reason to suspect a role in skeletal muscle contraction. NO has a
e.g. superoxide, which can damage pathogens. iNOS
role in smooth muscle relaxation.
is induced by proinflammatory molecules.
D. No reason to suspect a role in water balance.
Q54: Effect of vasoconstriction
Q54:
• Effect of vasoconstriction
A. Decrease in BP associated with fainting
B. ^blood flow to muscle during exercise
C. ^blood flow to skin during blushing
D. Maintaining blood pressure during hemorrhage
Q54:
• Effect of vasoconstriction
A. Decrease in BP associated with fainting. Incorrect.
Vasoconstriction increases blood pressure and counteracts
fainting (syncope).
B. ^blood flow to muscle during exercise. Incorrect. Vasoconstriction
would decrease blood flow to muscle.
C. ^blood flow to skin during blushing. Incorrect. Same as B.
D. Maintaining blood pressure during hemorrhage. Correct.
Vasoconstriction would ^BP, maintaining oxygenation during
hemorrhage.
• Which of the following would MOST lead to D)
A. Increase in liver products such as albumin
B. Increase in inflammatory molecules such as TNF-alpha
C. Consumption of diuretic molecules such as caffeine
D. Administration of beta-blockers, which are antagonists for betaadrenergic receptors
• The answer is A.
A. Albumin is an osmoregulatory protein that maintains blood Vol.
B. Inflammation leads to increased vascular permeability, i.e. greater
water loss from the bloodstream.
C. Diuretics cause water loss and decrease in BP.
D. Beta-adrenergic receptors cause vasoconstriction, so beta-blockers
would decrease BP. They are commonly used to treat
hypertension.
Q55: NO, L-NMMA, ring tension
Q55:
• Saturating L-NMMA on relaxing aortic ring w/endothelium
A.
B.
C.
D.
^ACh sensitivity
Ring dilation
Ring contraction
Prevent NE from increasing ring tension
Q55:
• Saturating L-NMMA on relaxing aortic ring w/endothelium
A. ^ACh sensitivity. Incorrect. NO mediates relaxation downstream
from ACh. L-NMMA would decrease ACh sensitivity by inhibiting
NO synthase.
B. Ring dilation. See C.
C. Ring contraction. Correct. If NO is relaxing and L-NMMA is an NO
synthase inhibitor, L-NMMA would cause contraction.
D. Prevent NE from increasing ring tension. Incorrect. By inhibiting
NO, NE would have no opposition to increasing tension.
• Supplementation of which of the following amino acids would
LEAST decrease local blood pressure?
A. Asp
B. Glu
C. Lys
D. Ala
• The answer is C. Lys would most compete with Arg on the enzyme
that produces NO. Less NO would lead to higher blood pressure.
Passage 8 Takeaways
• Graphical interpretation: NE/ACh in ring w/ and w/o endothelium
• Cardiovascular equations
• CO = HR * SV
• BP = CO * TVR
• ROSs
• Vascular dynamics: vasoconstriction vs. vasodilation
• Albumin – osmoregulatory protein
• Inflammation – increase vascular permeability
• Effects of sympathetic vs. parasympathetic nervous system
• Inferring relationships from a passage: L-NMMA vs. NO synthase vs. NO vs.
blood vessel diameter
• Amino acids
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes
Passage 9 Intro:
-
Uncouples oxidative metabolism
Accumulates in adipose tissue
Na+ diffuse thru axonal membranes
Looks like we have a kidney physiology passage!
Q56: Result of testicular uncoupling
Q56:
• Uncoupling of oxidative metabolism from ATP production  v
fertility because reduces
A. Glucose concentration of semen
B. Testosterone concentration of semen
C. Blood circulation in testes
D. Sperm motility
Q56:
• Uncoupling of oxidative metabolism from ATP production  v
fertility because reduces
A. Glucose concentration of semen. Incorrect. Though aerobic ATP
production will be reduced by uncoupling, aerobic metabolism will
not switch to anaerobic.
B. Testosterone concentration of semen. Incorrect. ATP production is
less of a limiting factor in testosterone production than D.
C. Blood circulation in testes. Incorrect. Blood circulation does not
depend on tissue-specific ATP production.
D. Sperm motility. Correct. Sperm need lots of ATP for flagella.
• Semen carries energetic substrate in the form of fructose. When in
the cell, fructose is converted by fructokinase to fructose-1phosphate, which is then broken down into 2 molecules. What are
they?
A. Glyceraldehyde-3-phosphate and dihydroxyacetone
phosphate
B. 3-phosphoglycerate and dihydroxyacetone phosphate
C. Phosphoenolpyruvate and glycerol
D. Dihydroxyacetone phosphate and glyceraldehyde
• The answer is D.
A. Too many phosphates
B. Same as A
C. Less analogous to fructose-1,6-bisphosphate than D.
D. Analogous to aldolase reaction with 1 less phosphate.
Q57: Reason for skin sensation
Q57:
• Cause of burning, itching, pain from DDT absorption
A.
B.
C.
D.
Motor neurons are depolarized
Motor neurons are hyperpolarized
Sensory neurons are depolarized
Sensory neurons are hyperpolarized
Q57:
• Cause of burning, itching, pain from DDT absorption
A. Motor neurons are depolarized. Incorrect. Motor neurons don’t
cause burning/itching/pain.
B. Motor neurons are hyperpolarized. Incorrect. See B.
C. Sensory neurons are depolarized. Correct. Depolarization of
sensory neurons could cause inappropriate sensations.
D. Sensory neurons are hyperpolarized. Incorrect. See C.
• Rheobase is the minimum voltage stimulus needed to reach a
depolarization threshold. Which of the following best describes
the effect of pro-inflammatory mediator bradykinin and morphine
withdrawal, respectively, on nociceptor (pain receptor) rheobase?
A. Increase, increase
B. Increase, decrease
C. Decrease, increase
D. Decrease, decrease
• The answer is D.
• Inflammation is associated with pain. Bradykinin decreases
rheobase, i.e. lowers voltage requirement to reach threshold,
making nociceptor AP easier to reach (i.e. more pain)
• A person chronically on morphine (an analgesic) may also have a
lower pain threshold in withdrawal, i.e. lower rheobase, due to
tolerance.
Q58: Liver function
Q58:
• If DDT accumulates in liver, all functions impaired EXCEPT
A. Absorption of fats in small int
B. Bile production
C. Detoxification
D. BP regulation
Q58:
• If DDT accumulates in liver, all functions impaired EXCEPT
A. Absorption of fats in small int. Incorrect. Bile production would be
impaired, so fat absorption would also be impaired.
B. Bile production. Incorrect. See A.
C. Detoxification. Incorrect. Detoxification is a major liver fxn.
D. BP regulation. Correct. BP regulation is primarily via aldosterone,
ADH, and atrial natriuretic peptide.
• Albumin is an osmoregulatory protein produced by the liver,
but this is still the best (worst) answer choice.
• The structure of DDT and one of its metabolites, DDA, are shown
at right. What would be the best solvent system for extraction?
A. Cyclohexanol, diethyl ether
B. 5% aqueous HCl, water
C. Ethyl acetate, 5% bicarbonate
D. Ammonia, water
• The answer is C.
• Since DDA is acidic, the best system is polar/basic. DDT will end up
in the polar solvent, and DDA will end up in the bicarbonate layer.
• Ammonia/H2O is not a better system than C because DDT will have
poor solubility in both and DDA will have good solubility in both
due to H-bonding.
DDT
DDA
Q59: Cancer
Q59:
• Damage of which most leads to cancer/cause mutation
A. Nuclear envelope
B. Chromosome
C. Ribosome
D. Histone
Q59:
• Damage of which most leads to cancer/cause mutation
A. Nuclear envelope. Incorrect. This could lead to DNA damage if
reactive cytoplasmic molecules enter nucleus, but see B.
B. Chromosome. Correct. DNA is carried on the chromosomes.
C. Ribosome. Incorrect. Ribosome damage would impair translation.
D. Histone. Incorrect. Though this could lead to chromosomal
instability, see B.
• Global distillation is the process by which a pollutant shifts from
region to region, often affecting animals far from where the
chemical was originally released. DDT is an example of global
distillation. In which animal population would DDT most likely be
found 40 years after being outlawed? (Structure is seen at right)
A. Polar bears
B. Arctic seals
C. Toucans (seen close to equator)
D. Galapagos turtles (seen in warm water)
• The answer is B.
• Land vs. aquatic: DDT is found more in aquatic species due to its
poor solubility in their habitat, water
• I.e. it is uptook into the animal because insoluble in the
medium
• Arctic vs. equatorial: Distillation is evaporation of a substance in
higher temperature and condensation in lower
• DDT is found in Arctic despite never being used there!
Passage 8 Takeaways
• Sperm function – motility largest energy consumption
• Glycolysis intermediates and enzymes
• Sensory vs. motor neurons
• Tolerance vs. sensitization
• Liver functions
• Extractions
• Cancer – DNA mutations
• Distillations
• IMFs
Q60:
•
A.
B.
C.
D.
Why is pepsin inhibited in excess acid
Pepsin is feedback-inhibited
Pepsin synthesis is reduced
Peptide bonds in pepsin are more stable
3-D structure of pepsin changed
Q60:
• Why is pepsin inhibited in excess acid
A. Pepsin is feedback-inhibited. Incorrect. Pepsin does not produce H+, so no
reason to suspect feedback inhibition at low pH.
B. Pepsin synthesis is reduced. Incorrect. “Pepsin catalysis occurs very
slowly” does not necessarily mean that pepsin levels are lower.
C. Peptide bonds in pepsin are more stable. Incorrect. Peptide bonds are
likely less stable and potentially prone to hydrolysis.
D. 3-D structure of pepsin changed. Correct. Extreme pH denatures protein.
•
A.
B.
C.
D.
•
•
A.
B.
C.
D.
Which amino acid is most likely in the active site of pepsin?
Tryptophan
Aspartic acid
Phenylalanine
Glutamine
The answer is B. Pepsin is a protease, and an acidic amino acid is most
likely involved in hydrolysis to donate a proton to the peptide bond.
Pepsin is implicated in damage caused by acid reflux. Why?
The esophageal pH is normally acidic with a pH of 1.5-2.
Pepsin diffuses through mucosal cell membranes and degrades cellular
proteins.
Pepsin is reactivated by mucus secretions and degrades phospholipids.
Mucus secretions contain protein, so pepsin leads to damage by H+.
• The answer is D.
A. The esophagus has a pH ~7.0 close to
physiological pH. The stomach has a pH of
1.5-2.
B. Pepsin is a protein and cannot diffuse
through cell membranes.
C. Pepsin is a protease.
D. Pepsin degrades mucin, a glycoprotein,
which exposes the esophageal epithelium.
Q61:
• ^Albumin 
A. ^Immune response
B. ^Tissue albumin
C. Outflow of blood fluid to tissues
D. Influx of tissue fluid to bloodstream
Q61:
• ^Albumin 
A. ^Immune response. Incorrect. Albumin should not trigger an immune
response.
B. ^Tissue albumin. Incorrect. Albumin is a protein that remains in the
bloodstream under normal conditions.
C. Outflow of blood fluid to tissues. Incorrect. An increased osmotic
pressure would not lead to more fluid entering tissues.
D. Influx of tissue fluid to bloodstream. Correct. Increased osmotic
pressure would lead to an influx of tissue fluid into the bloodstream.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Increased serum albumin is known as hyperalbuminemia. Which of the
following would cause hyperalbuminemia?
A hormone-secreting adrenal tumor
Prolonged hepatitis
Dehydration
3rd-degree burns leading to plasma loss
The answer is C.
Increased aldosterone would lead to increased blood volume, so
hypoalbuminemia.
Hepatitis would impair albumin secretion, similar effect as A.
Dehydration would lead to decreased blood volume, so
hyperalbuminemia.
Plasma loss from burns would result in loss of fluids and albumin. No
change in albumin concentration.
Q62:
• Would ^NaCl ingestion  ^aldosterone
A. No; aldosterone causes Na+ reabsorption.
B. No; aldosterone causes Na+ secretion.
C. Yes; aldosterone causes Na+ reabsorption.
D. Yes; aldosterone causes Na+ secretion.
Q62:
• Would ^NaCl ingestion  ^aldosterone
A. No; aldosterone causes Na+ reabsorption. Correct. Aldosterone causes
increased reabsorption of Na+ (and therefore H2O) to raise blood
pressure. Decreased [NaCl]serum would lead to ^aldosterone.
B. No; aldosterone causes Na+ secretion.
C. Yes; aldosterone causes Na+ reabsorption.
D. Yes; aldosterone causes Na+ secretion.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Addison’s disease is characterized by insufficient secretion of adrenal
cortex hormones. Which of the following would NOT be a symptom of
Addison’s disease?
Increased adrenocorticotropic hormone levels
Dizziness upon standing
Hyponatremia (low serum sodium)
Decreased immune system activity
The answer is D.
ACTH levels are expected to be higher due to low adrenal cortex
hormones.
Decreased aldosterone leads to decreased blood pressure (orthostatic
hypotension), which lead to decreased blood flow to the brain when
standing due to blood pooling in the lower body.
Decreased aldosterone would lead to less Na+ reabsorption in kidney.
Decreased cortisol would not lead to decreased immune system
activity.
Q63:
• Liver regeneration is from
A. Fission
B. Meiosis
C. Mitosis
D. Cell growth
Q63:
• Liver regeneration is from
A. Fission. Incorrect. Fission is prokaryotic cell division.
B. Meiosis. Incorrect. Only germ-line cells do meiosis.
C. Mitosis. Correct. Many organs can have cell growth (hypertrophy), but
the liver is somewhat unique in its ability to have increased cell
division (hyperplasia) to regenerate.
D. Cell growth. Incorrect. Many organs can have cell growth.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Which of the following organs or tissues canonically exhibits hyperplasia
(increased cell division)?
Nervous
Adipose
Muscle
Epithelium
The answer is D.
Nervous tissue is canonically terminally differentiated and amitotic.
Same as A.
Muscle tissue is terminally differentiated and amitotic.
Epithelium. Epithelial cells regularly slough off and are replaced by
division of stem cells.
Q63:
• Mechanism for cyclin oscillation
A. Replication of cyclin gene during S phase
B. Segregation of chromosomes carrying cyclin
genes
C. Translation of cyclin mRNA in interphase,
degradation during mitosis
D. Translation of cyclin mRNA in mitosis,
degradation during interphase
Q63:
• Mechanism for cyclin oscillation
A. Replication of cyclin gene during S phase.
Incorrect. Replication would not lead to
^cyclin protein.
B. Segregation of chromosomes carrying cyclin
genes. Incorrect. Same as A.
C. Translation of cyclin mRNA in interphase,
degradation during mitosis. Correct.
Translation would lead to higher cyclin in
interphase, proteolysis would lead to lower
cyclin during mitosis.
D. Translation of cyclin mRNA in mitosis, degradation during interphase.
Incorrect. Opposite of C.
• Match the following with G1 cyclins, G1/S cyclins, S cyclins, M cyclins:
A. Promote centrosome duplication
B. Induce DNA replication
C. Coordinate cell growth
D. Induce spindle assembly
A. G1/S cyclins
B. S cyclins
C. G1 cyclins
D. M cyclins
FSQ Takeaways
• Protein expression and structure
• Effects of pH, osmolarity, heat, reducing agents, chaotropic agents (disrupt Hbonding)
• Reason a protein conc increases/decreases
• Blood osmotic pressure (oncotic pressure)
• Role of albumin
• Water/salt balance
• Role of aldosterone
• Types of tissues that don’t exp cell division
Passage 10 Intro:
-
Stomach ulcers
HCl
H. pylori
Looks like we have a gastric/microbio passage!
Q65: Prevention of stomach ulcers
Q66: Evidence for H. pylori in ulcers
Q66:
• Proof H. pylori causes ulcers
A. People with stomach ulcers have antibodies to H. pylori
B. Healthy individuals have antibodies to H. pylori
C. Ulcers induced by infection with H. pylori
D. Organism can be passed from mother to fetus
Q66:
• Proof H. pylori causes ulcers
A. People with stomach ulcers have antibodies to H. pylori. Incorrect. Not
as direct evidence as C.
B. Healthy individuals have antibodies to H. pylori. Incorrect. This would
not support the H. pylori theory of ulcers.
C. Ulcers induced by infection with H. pylori. Correct. The change in an
independent variable (H. pylori infection) leading to a change in
dependent variable (ulcer formation) is best causal evidence.
D. Organism can be passed from mother to fetus. Incorrect. No connection
to ulcers.
• Which of the following would provide the best additional evidence for
the H. pylori hypothesis of stomach ulcers?
A. Taking antibiotics leads to healing of the stomach lining.
B. H. pylori contains urease, which converts urea to NH3.
C. 5% of people with stomach ulcers do not have H. pylori in their
stomach.
D. H. pylori has a flagella used to bury into stomach epithelial cells.
• The answer is A.
A. Eradication of bacteria leading to ulcers healing is the best evidence.
B. Survival mechanisms for living in acidic environments is weak evidence.
C. This does not support the role of H. pylori in ulcer formation.
D. Same as B.
Q67: Explanation for results of figure 1
Q68: Comparison b/w bacteria and human cells
Q68:
• Similarity between bacterial and human cells
A. Ability to produce ATP via ATP synthase
B. Chemical composition of their ribosomes
C. Enclosure within cell walls
D. Shape of DNA
Q68:
• Similarity between bacterial and human cells
A. Ability to produce ATP via ATP synthase. Correct. Both use ATP
synthase to produce ATP.
B. Chemical composition of their ribosomes
C. Enclosure within cell walls
D. Shape of DNA
• Follow-up:
A. Where do bacteria vs. human cells produce ATP in their cells?
• Bacteria: cell membrane
• Humans: mitochondrial inner membrane
B. What is the composition of the ribosomes in each?
• Bacteria: 30S/50S = 70S
• Humans: 40S/60S = 80S
C. What types of eukaryotes have cell walls?
• Plants, fungi
D. Describe the DNA structure in both
• Bacteria: 1 circular chromosome
• Humans: 46 chromosomes
Q69: Confirm urease required for stomach
colonization
Q69:
• Which expt best tests hypothesis of urease as required for H. pylori
infection?
A. Exposing ulcer patients to antibodies to urease
B. Exposing uninfected animals to H. pylori lacking urease
C. Exposing ulcer patients to radioactive urea
D. Measuring urease activity in ulcer biopsies
Q69:
• Which expt best tests hypothesis of urease as required for H. pylori
infection?
A. Exposing ulcer patients to antibodies to urease. Incorrect. Antibodies
likely degraded in stomach environment.
B. Exposing uninfected animals to H. pylori lacking urease. Correct. If
urease is necessary for stomach infection, the animals should not be
susceptible.
C. Exposing ulcer patients to radioactive urea. Incorrect. Radioactive urea
metabolites (NH3, CO2) would show urease activity not necessity.
D. Measuring urease activity in ulcer biopsies. Incorrect. Presence of
urease does not show necessity of urease to infection.
• In a urea breath test, either radioactive C-14 or nonradioactive C-13
isotopically labeled urea is administered, and the presence of isotopic C
is detected by scintillation counting or isotopic ratio mass spectrometry,
respectively. C-14 can be detected as it emits subatomic particles as it
becomes N-14. What type of radioactivity does C-14 exhibit?
A. Alpha-decay
B. Beta-decay
C. Gamma-decay
D. Positron emission
• The answer is B. If C-14 decays to N-14, resulting in an increase in
charge (protons), a corresponding decrease in charge must occur (loss
of e-).
Now, to read the passage:
• Ulcers prev thought to be from ^HCl
• Drugs that neutralize/reduce acid prod  ulcer recurrence
• H. pylori found in >95% of ulcer patients
• Combination of antibiotics and anti-acid drugs  ulcer cure
Now, to read the passage:
• H. pylori produces urease
• Urea  NH3 and CO2
• Test for H. pylori: radioactive urea  radioactive CO2 exhalation
• H. pylori has
• Inhibitor of acid secretion
• Enzyme destroying mucin lining
• Protein cytotoxic to gastric cells
Now, to read the passage:
• H. pylori strains A & B
• Culture initiation  2
hours
• Subcultures – one of
each exposed to
streptomycin
• Research question: Does
streptomycin inhibit
growth of strains A/B?
• IV1: Time
• IV2: Strain
• IV3: Presence of
streptomycin
• DV: Number of cells
• Results: Strain A and B in
absence of streptomycin
grow.
• Strain A grows faster
• Strain A inhibited by
streptomycin
• Strain B grows
normally w/antibiotic
Q65:
• Which method prevents stomach ulcer recurrence
A. Drugs preventing acid production
B. Drugs neutralizing stomach acid
C. Drugs inhibiting bacterial protein synthesis
D. C + A
Q65:
• Which method prevents stomach ulcer recurrence
A. Drugs preventing acid production. Incorrect. Passage states ulcers recur.
B. Drugs neutralizing stomach acid. Incorrect. Same as A.
C. Drugs inhibiting bacterial protein synthesis. Incorrect. Passage doesn’t
specify about antibiotics alone.
D. C + A. Correct. Passage states that antibiotics + anti-acid drugs  cure
ulcers.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Prilosec (generic name omeprazole) is a proton pump inhibitor used in
conjunction with chlarithromycin to treat stomach ulcers. It binds to the
stomach’s H+/K+ ATPase pump in a site away from the active site.
Prilosec’s effects on the H+/K+ ATPase most likely are
^Km, =Vmax
v Km, v Vmax
=Km, v Vmax
v Km, =Vmax
The answer is C.
This describes a competitive inhibitor.
This describes an uncompetitive inhibitor.
This describes a noncompetitive inhibitor. Since Prilosec binds to a site
away from the active site, it is most likely a noncomp inhibitor.
Noncompetitive inhibitors do not alter enzyme-substrate affinity but
do decrease enzyme activity.
Not a thing, this would not be an inhibitor.
Q67:
• Which accounts for figure 1 results?
A. Strain A growth > strain B.
B. Strain A growth > strain B, strain A streptomycin-resistant.
C. Strain B growth > strain A, strain B streptomycin-resistant.
D. Strain A growth > strain B, strain B streptomycin-resistant.
Q67:
• Which accounts for figure 1 results?
A. Strain A growth > strain B.
B. Strain A growth > strain B, strain A streptomycin-resistant.
C. Strain B growth > strain A, strain B streptomycin-resistant.
D. Strain A growth > strain B, strain B streptomycin-resistant. Correct.
Strain A results in higher cell count earlier than strain B, and strain B
growth is not inhibited by streptomycin.
Passage 10 Takeaways
• Experimental design:
• Best way to establish a causal relationship is to change only one variable
directly tied to the hypothetical cause & observe the effects on the
organism/system
• For any given complex experiment, define research question, IVs, DVs,
methods, results, conclusion
• Prokaryotic/eukaryotic cells is high yield – know all major differences
• Enzyme kinetics and types and effects of inhibitors
Passage 11 Intro:
-
Bone tissue
Resorption/formation
PTH, VitD, Calcitonin, VitC
Looks like we have a bone passage!
Q70: Which conditions produce rickets?
Q70:
• Which could cause rickets? (Look up rickets in passage)
I. v PTH
II. Impairment of vitD conversion  active form
III. Inability of active vitD  act on target tissue
A.
B.
C.
D.
I
I, II
I, III
II, III
Q70:
• Which could cause rickets? (Look up rickets in passage)
A.
B.
C.
D.
•
A.
B.
C.
D.
•
I. v PTH
II. Impairment of vitD conversion  active form
III. Inability of active vitD  act on target tissue
I.
I, II
I, III
II, III. Correct. Both would result in insufficient vitamin D activity.
In advanced rickets, PTH and calcitonin levels would be expected to
be…
Low, high
High, normal
High, high
Normal, low
The answer is B. Since vitD activity is insufficient, plasma [Ca] is low due
to inadequate absorption, leading to rickets symptoms such as bowed
legs, stunted growth, and bone pain. PTH is expected to be high,
although its effect is weaker due to lack of vitD. Calcitonin is normal,
since [Ca]serum does not need to be lowered.
Q71: Why Ca supplements contain VitD
Q71:
• Why do calcium supplements often include vitD?
A. VitD needed to prevent rickets
B. Activated vitD stimulates Ca absorption
C. Activated vitD enhances calcitonin action
D. Activated vitD enhances uptake of Ca by bone tissue
Q71:
• Why do calcium supplements often include vitD?
A. VitD needed to prevent rickets. Incorrect. This may be true, but doesn’t
explain its inclusion in Ca supplements as well as B.
B. Activated vitD stimulates Ca absorption. Correct. VitD stimulates Ca
uptake in the small int.
C. Activated vitD enhances calcitonin action. Incorrect. Should be PTH.
D. Activated vitD enhances uptake of Ca by bone tissue. Incorrect. Uptake
is into the blood. Osteoblasts uptake Ca into bone tissue.
•
A.
B.
C.
D.
•
Calcium supplements include calcium (Ca2+) and a counterion. Which of
the following calcium compounds would have the highest solubility?
Calcium carbonate (CaCO3, Ksp = 3.3x10-9)
Calcium hydroxide (Ca(OH)2, Ksp = 5.5x10-6)
Calcium sulfate (CaSO4, Ksp = 9.1x10-6)
Calcium fluoride (CaF2, Ksp = 5.3x10-9)
The answer is B. The problem includes two AB salts (Ksp = x2) and two
AB2 salts (Ksp = 4x3). The best strategy is to narrow down to the highest
solubility AB and AB2 salts, B and C. The solubility, x, of calcium sulfate
is √(9.1x10-6), which is ~3x10-3 M. The solubility of calcium hydroxide is
3√(5.5x10-6/4), which is ~1x10-2 M.
Q72: Low Ca  increase…
Q72:
• Low [Ca]serum would trigger increase of
I. Osteoclast activity
II. PTH
III. VitC
A.
B.
C.
D.
I
I, II
I, III
II, III
Q72:
• Low [Ca]serum would trigger increase of
I. Osteoclast activity
II. PTH
III. VitC
A.
B.
C.
D.
I
I, II. Correct. PTH acts to ^Caserum by stimulating osteoclast activity.
I, III. Incorrect. No clear role of vitamin C.
II, III
•
To resorb bone matrix, osteoclasts secrete H+ ions, collagenase, and
hydrolytic enzymes. The process involves 2 steps: 1) creation of an
acidic microenvironment increasing solubility of bone mineral, which is
then taken up by osteoclasts and delivered to the capillaries; and 2)
degradation of organic matrix materials such as collagen, which are
absorbed in cellular transport vesicles, degraded, and released back to
the bloodstream. Based on this information, which cell type are
osteoclasts most similar to?
Astrocytes
Macrophages
Renal tubular cells
Chemoreceptor cells
A.
B.
C.
D.
• The answer is B. The secretive function (acid
hydrolases, proteinases) and phagocytic
function (uptake of organic matrix molecules)
resemble macrophages.
• In fact, osteoclasts are considered a
member of the mononuclear phagocyte
family.
Q73: What would cause ^calcitonin
Q73:
• What would cause ^calcitonin?
A. Dietary deficiency of Ca
B. Dietary deficiency of vitD
C. High [Ca]serum
D. Low PTH
Q73:
• What would cause ^calcitonin?
A. Dietary deficiency of Ca. Incorrect. This would cause increased PTH.
B. Dietary deficiency of vitD. Incorrect. This would cause [Ca]serum to
decrease, which would lead to increased PTH activity.
C. High [Ca]serum . Correct. Calcitonin acts to decrease [Ca]serum.
D. Low PTH. Incorrect. This would lead to decreased [Ca]serum, so calcitonin
would not be able to compensate.
• Calcitonin is a 32-amino acid protein with sequence as follows:
• Cys-Gly-Asn-Leu-Ser-Thr-Cys-Met-Leu-Gly-Thr-Tyr-Thr-Gln-AspPhe-Asn-Lys-Phe-His-Thr-Phe-Pro-Gln-Thr-Ala-Ile-Gly-Val-Gly-AlaPro
• What is the charge of calcitonin at physiological pH?
A. -1
B. 0
C. +1
D. +2
• The answer is B.
• N-Cys-Gly-Asn-Leu-Ser-Thr-Cys-Met-Leu-Gly-Thr-Tyr-Thr-Gln-AspPhe-Asn-Lys-Phe-His-Thr-Phe-Pro-Gln-Thr-Ala-Ile-Gly-Val-Gly-AlaPro-C
• Two acidic groups = -2
• Two basic residues = +2. Net charge = 0.
• His has a neutral charge at phys pH (pKa = 6) – the MCAT expects
you to know this!
Q74: Effect of removal of PTH glands
Q74:
• Effect of PTH glands removal
A. Severe neural and muscular problems due to deficiency of plasma Ca
B. Increase in calcitonin production to compensate for calcium deficiency
C. Drastic change in ratio of mineral to matrix tissue in bones
D. Calcification of some organs due to Ca accumulation in plasma
Q74:
• Effect of PTH glands removal
A. Severe neural and muscular problems due to deficiency of plasma Ca.
Correct. Ca is involved in neurons (neurotransmitter release) and
muscle contraction, and lack of PTH would lead to inability to raise
Caserum.
B. Increase in calcitonin production to compensate for calcium deficiency.
Incorrect. Calcitonin acts to decrease plasma Ca.
C. Drastic change in ratio of mineral to matrix tissue in bones. Incorrect.
Answer doesn’t specify the direction of the change; A is more specific.
D. Calcification of some organs due to Ca accumulation in plasma.
Incorrect. If anything, plasma Ca would be lower.
• Parathyroidectomy is sometimes necessary in secondary
hyperparathyroidism. What might be the primary cause?
A. Calciphylaxis (calcium deposition in small blood vessels)
B. Osteoporosis
C. Chronic kidney disease
D. Nephrolithiasis (kidney stones)
• The answer is C.
A. This can be a result of hyperparathyroidism, not a cause.
B. Similar to A.
C. Insufficient Ca reabsorption in kidney disease can lead to secondary
hyperparathyroidism to raise serum Ca.
D. Similar to A.
Passage 11 Takeaways
• Bone
• Composition – hydroxyapatite (Ca, OH-, PO43-)
• Effect of osteoclasts and osteoblasts
• Ca regulation (small intestine, kidney)
• VitD, calcitonin, PTH
• Class of hormone, organ secreted by, effect on blood Ca and bone,
mechanism
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes
Passage 12 Intro:
-
Visual communication
UV photoreceptors
5 lizard species
Looks like we have a genetics passage!
Q75: Disadvantage from mutation
Q76: Type of inheritance
Q77: Speciation
Q78:
A.
B.
C.
D.
Neighboring lizard populations are diff species if
One lived in forest, other lived in field
One population had dewlap, other did not
They did not communicate
They did not interbreed and produce fertile offspring
Q78:
A.
B.
C.
D.
Neighboring lizard populations are diff species if
One lived in forest, other lived in field
One population had dewlap, other did not
They did not communicate
They did not interbreed and produce fertile offspring. Correct. The other
answer choices may be true regarding different species, but this is the
definition of separate species.
•
Anolis is a very large genus of American lizards, and some come to occupy
separate ecological niches, especially regarding the types of vegetation
they forage. Morphological changes accompany changes in nutritional
niches, e.g. twig ecomorphs have short limbs while trunk ecomorphs have
long limbs. This is an example of
Speciation
Directional selection
Adaptive radiation
Outbreeding
The answer is C.
Speciation = populations that cannot breed to produce viable offspring
Directional selection = evolution towards an extreme trait e.g. height
Morphological changes with divergent niche describes adaptive radiation
Outbreeding = mating with non-relatives within a species
A.
B.
C.
D.
•
A.
B.
C.
D.
Q78: Conclusion about dewlap reflectance
Q79: Natural selection
Q79:
A.
B.
C.
D.
UV-reflecting dewlaps would evolve by natural selection if
Individuals with dewlaps produced more offspring
Individuals with dewlaps were better able to communicate
Individuals with dewlaps were less susceptible to predation
Individuals with dewlaps mated more frequently
Q79:
- UV-reflecting dewlaps would evolve by natural selection if
A. Individuals with dewlaps produced more offspring. Correct. Natural
selection favors the genotype of individuals that have produce the
most number of offspring who likewise can successfully produce *fit*
offspring.
B. Individuals with dewlaps were better able to communicate
C. Individuals with dewlaps were less susceptible to predation
D. Individuals with dewlaps mated more frequently
•
The UV-reflective dewlap most likely evolved when (2 choices)
A. Anolis lizards with a dewlap mutated to express a UV-reflective
pigment in it
B. Anolis lizards with a UV-reflectance gene mutated to express a
dewlap
• Dewlaps are only found in males and are brightly colored, making the
males more visible to predators. The evolutionary explanation for this is
• The sexual advantage of the dewlap must outweigh the
disadvantage of increased predation.
• Dewlaps in different species are colored differently. The purpose of this is
most likely
• To discourage inter-species mating, which would be futile
Now, to read passage 12:
•
•
•
•
5 species of Anolis lizard
UV photoreceptors discovered in lizards – hypothesis = involved in communication
3 species (A, B, C) live in open unshaded fields, 2 (D, E) live in forest understory
Male lizards have dewlap – fanlike skin flap
• Functions in communication – territory, warning signals, courtship
Now, to read passage 12:
• Cameras to measure UV
reflectance
• 2 species reflected UV
light highly (?)
• A and B – live in open
fields
• 2 species reflected UV
light lowly (?)
• D and E – live in forest
• 1 species = intermediate
• C – live in open fields
• Conclusion: relationship
between UV reflectance
and habitat
Q75:
A.
B.
C.
D.
Which species least needs UV light reflectance for communication?
Species A
Species B
Species D
Species E
Q75:
A.
B.
C.
D.
•
A.
B.
C.
D.
Which species least needs UV light reflectance for communication?
Species A
Species B
Species D
Species E. Correct. This species has the lowest reflectance for UV
light (<400 nm).
Which species’s dewlap has the highest reflectance for green light?
Species A
Species B
Species D
Species E
The answer is D. Green light is around
550 nm.
• Which species’s dewlap has the
lowest reflectance for red light?
• Species D. Red light is around
700 nm.
• Note: The MCAT expects you to
know the visible light spectrum is
between
• 400 (violet)
• 700 (red)
• And they do expect you to know that
green is in the middle (550 nm).
Q76:
•
•
A.
B.
C.
D.
Only male lizards have dewlaps
Locus for gene for UV reflectance pigment is on
X-chromosome
Y-chromosome
Autosome
Sex chromosome or autosome
Q76:
•
•
A.
B.
C.
D.
Only male lizards have dewlaps
Locus for gene for UV reflectance pigment is on
X-chromosome
Y-chromosome
Autosome
Sex chromosome or autosome. Correct. Only males express genes
located on the Y-chromosome, but certain genes on other
chromosomes may also be only expressed by males.
•
If female Anolis lizards also contain the dewlap UV reflectance gene,
it most likely exists in their
Euchromatin
Heterochromatin
mtDNA
Telomeric DNA
The answer is B.
Euchromatin is open (histones are not methylated) and likely to be
actively expressed
Heterochromatin is not actively expressed (histones are
methylated)
Nothing to suggest a mitochondrial locus
Telomeres do not contain active genes
A.
B.
C.
D.
•
A.
B.
C.
D.
Q78:
- Conclusion about dewlap reflectance
A. Lizard habitat is determined by dewlap reflectance
B. High UV dewlap reflectance is most important in bright habitats
C. High dewlap reflectance is most important in dim habitats
D. Dewlap reflectance is highest at the blue end of the visible spectrum
Q78:
- Conclusion about dewlap reflectance
A. Lizard habitat is determined by dewlap reflectance. Incorrect. It is
likely the lizards evolved different reflectances based on their
habitats.
B. High UV dewlap reflectance is most important in bright habitats.
Correct. Lizards A and B had the highest reflectance, followed by C;
all live in open fields. Lizards D and E had the lowest reflectances
and lived in forest understory.
C. High dewlap reflectance is most important in dim habitats. Incorrect.
See B.
D. Dewlap reflectance is highest at the blue end of the visible spectrum.
Incorrect. Only some dewlaps are UV-reflective, and being UVreflective does not necessarily mean reflecting blue light.
• The dewlaps of Anolis lizards with high dewlap absorbance of visible
red light would appear
A. Violet in color
B. Blue in color
C. Green in color
D. Red in color
• The answer is C.
• The color(s) most strongly reflected by a pigment = its perceived
color
• The color(s) most strongly absorbed by a pigment =
complementary to its perceived color
• Tip: Memorize the
color wheel with
“ROY G. BV”
• E.g. if a pigment
absorbs orange
light, it will appear
blue
Passage 12 Takeaways
• Evolutionary biology – less represented on the new MCAT but still
important
•
•
•
•
Speciation
Fitness
Natural selection
Several other terms – see Khan Academy and Jack Westin articles
• Light – visible light spectrum, reflectance vs. absorbance
• Genetics – sex-specific gene expression does not = gene locus on sex
chromosome
Q80:
A.
B.
C.
D.
Aldosterone deficiency would lead to
^V ^P
^V vP
vV ^P
vV vP
Q80:
A.
B.
C.
D.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Aldosterone deficiency would lead to
^V ^P. Incorrect. Would more indicate elevated aldosterone
^V vP. Incorrect. Internally inconsistent.
vV ^P. Incorrect. Internally inconsistent.
vV vP. Correct. Aldosterone deficiency would lead to decreased Na+
absorption and decreased H2O retention, which would lead to
decreased blood volume and pressure.
Hypoaldosteronism is often associated with diabetic patients. What is the
most likely cause of this?
Kidney damage from chronic hyperglycemia leads to impaired renin
secretion.
Excess urination leads to excretion of aldosterone.
Hyperglycemia leads to osmotic retention of water and sodium,
eliminating the need for aldosterone secretion.
Hyperglycemia exceeds kidneys’ reabsorption capacity leads to osmotic
excretion of water, causing the patient to drink water excessively.
The answer is A.
Chronic hyperglycemia damages the kidney (diabetic kidney disease,
DKD), leading to impaired renin secretion and hypoaldosteronism.
Aldosterone in urine would indicate hyperaldosteronism.
Hyperglycemia would not lead to osmotic sodium retention.
If the patient drank water to replace urinary water loss, this would lead
to normal aldosterone levels.
Q81:
A.
B.
C.
D.
Which organs coordinate in the menstrual cycle?
Hypothalamus-thyroid-ovary
Hypothalamus-pituitary-ovary
Pituitary-thyroid-ovary
Pituitary-adrenal glands-ovary
Q81:
- Which organs coordinate in the menstrual cycle?
A. Hypothalamus-thyroid-ovary
B. Hypothalamus-pituitary-ovary. Correct. The hypothalamus secretes
GnRH, the pituitary secretes FSH and LH, and the ovary responds.
C. Pituitary-thyroid-ovary
D. Pituitary-adrenal glands-ovary
•
A.
B.
C.
D.
•
A.
B.
C.
D.
GnRH is released by the hypothalamus in a pulsatile fashion. Too frequent
GnRH secretion is seen in polycystic ovarian syndrome (PCOS), leading to
a LH/FSH ratio that is too high and causes many PCOS symptoms. What is
the most likely reason for pulsatile secretion of GnRH?
Fluctuating need for FSH and LH throughout one menstrual cycle.
Sensitization of target organs to GnRH.
Negative feedback from FSH and LH on the hypothalamus.
Delay between GnRH synthesis and secretion.
The answer is B.
The fluctuating need for FSH and LH is periodic, not pulsatile.
GnRH is secreted in a pulsatile fashion to avoid receptor
downregulation, i.e. to sensitize the pituitary to receive it.
GnRH is regulated by negative feedback by the sex hormones, not by FSH
and LH. This would not explain “pulsatile” secretion.
GnRH synthesis and secretion could be coordinated synchronously, so this
not explain “pulsatile” secretion.
Q82:
- If a dsDNA with a 3:1 AT:GC ratio is
replicated, what is the AT:GC ratio of
the daughter dsDNA?
A. 1:3
B. 1:1
C. 2:1
D. 3:1
Q82:
- If a dsDNA with a 3:1 AT:GC ratio is
replicated, what is the AT:GC ratio of
the daughter dsDNA?
A. 1:3
B. 1:1
C. 2:1
D. 3:1. Correct. The daughter dsDNA
should be identical to the parent.
• If a ssDNA molecule with a 3:1 AT:GC ratio is treated with DNA polymerase
and dNTPs, what would its AT:GC ratio be?
• 3:1.
• A’s will be converted to T’s and vice versa; G’s will be converted to C’s
and vice versa. The AT:GC ratio will remain the same.
• If a dsDNA containing 30% adenine is replicated, what will be the
percentage cytosine of the daughter dsDNA?
• 20%.
• The % adenine should = % thymine, i.e. 30% A and 30% T.
• The remaining 40% should be equally C and G.
• C and G will be 20% each.
Q83:
A.
B.
C.
D.
In which organelle is uracil incorporated into nucleic acid?
Nucleus
Golgi body
Ribosomes
ER
Q83:
- In which organelle is uracil incorporated into nucleic acid?
A. Nucleus. Correct. Uracil is incorporated into nucleic acid during
transcription, which occurs in the nucleus.
B. Golgi body
C. Ribosomes
D. ER
•
A.
B.
C.
D.
•
In which of the following cellular compartments will tRNAGly NOT be
found?
I. Nucleus
II. Cytoplasm
III. Rough ER
IV. Mitochondria
I and IV
I, II, and III
I and III
III and IV
The answer is C. tRNAGly will not be found in the nucleus or in the rough
ER (translation occurs in the cytoplasm, rough ER membrane, and
mitochondria)
• tRNAGly will be found in the cytoplasm and mitochondria for
translation.
Q83:
A.
B.
C.
D.
Which is NOT derived from mesoderm?
Circulatory
Bone
Dermal
Nerve
Q83:
A.
B.
C.
D.
Which is NOT derived from the mesoderm?
Circulatory
Bone
Dermal
Nerve. Correct. Nerve tissue is derived from the ectoderm.
•
A.
B.
C.
D.
•
Which of the following is derived from the ectoderm?
Triceps muscle
Renal tubule
Alveoli
Pituitary epithelium
The answer is D. All epithelium are derived from the endoderm or
ectoderm. Muscle and kidney are derived from the mesoderm. Alveoli
(and respiratory lining) are derived from the endoderm.
•
A.
B.
C.
D.
•
Which of the following is NOT derived from the endoderm?
Testes
Hepatocytes
Pancreatic Beta-cells
Parathyroid
The answer is A. Other than the germ cells, the reproductive system is
derived from the mesoderm. The liver, pancreas, and parathyroid (as well
as thyroid) are derived from the endoderm.
FSQ Takeaways
• Hormones
• Aldosterone – renin-angiotensin-aldosterone system
• Menstrual cycle – role of GnRH, FSH, LH, estrogen; cellular and physical
changes
• DNA – Watson-Crick-Franklin model
• A=T, C=G
• Be able to handle basic math problems using the bases
• Cell structure – organelle function
• Relationship to central dogma of molecular biology
• Germ layers – high-yield
Passage 13 Intro:
-
Mitochondria – endosymbiotic hypothesis/theory
DNA, tRNAs, ribosomes
Inner membranes – ETC
Looks like we have a passage about mitochondria!
Q85: Function of mitochondria
Q85:
- What function were mitochondria able to carry out that primitive
eukaryotic cells could not?
A. Glycolysis
B. Krebs cycle & ETC
C. Cell division
D. Txn and trln
Q85:
- What function were mitochondria able to carry out that primitive
eukaryotic cells could not?
A. Glycolysis
B. Krebs cycle & ETC. Correct. These functions are carried out in the
mitochondria.
C. Cell division
D. Txn and trln
•
A.
B.
C.
D.
•
Mitochondria have their own DNA and carry out transcription and
translation from it. Which of the following are LEAST likely coded for by
mtDNA?
I. rRNA
II. Pyruvate carboxylase
III. Lactate dehydrogenase
IV. Cytochrome c oxidase
I. rRNA – mitochondria code rRNA because they have their own ribosomes.
V. Fatty acid synthase
II. Pyruvate carboxylase – this gluconeogenic enzyme is not coded by mtDNA
I, II, V
(gluconeogenesis occurs in the cytoplasm)
I, III, V
III. Lactate dehydrogenase – this fermentation enzyme is not coded by mtDNA
II, III, IV
(fermentation occurs in the cytoplasm)
II, III, V
IV. Cytochrome c oxidase – this ETC enzyme is coded by the mtDNA.
The answer is D.
V. Fatty acid synthase – this fatty acid synthesis enzyme is not coded by mtDNA
(fatty acid synthesis occurs in the cytoplasm)
Q86: Endosymbiotic hypothesis
Q86:
- Can the fact that most mitochondrial proteins are made using nuclear DNA and
cytoplasmic ribosomes be reconciled with the endosymbiotic hypothesis?
A. Yes – transfer of genes to eukaryotic nucleus could have occurred
B. Yes – similarities outweigh the contradiction
C. No – this fact is convincing that mitochondria do not have bacterial origin
D. No – bacteria can make all their proteins, so this disproves the hypothesis
Q86:
- Can the fact that most mitochondrial proteins are made using nuclear DNA and
cytoplasmic ribosomes be reconciled with the endosymbiotic hypothesis?
A. Yes – transfer of genes to eukaryotic nucleus could have occurred during
evolution. Correct. This answer choice doesn’t attempt to dismiss the
discrepancy but provides an answer, making it better than B.
B. Yes – similarities outweigh the contradiction. Incorrect. While this might be
true, A goes a step further by attempting to explain the contradiction.
C. No – this fact is convincing that mitochondria do not have bacterial origin.
Incorrect. This answer choice means the examinee is rejecting the
endosymbiotic theory, which is widely accepted.
D. No – bacteria can make all their proteins, so this disproves the hypothesis.
Incorrect. Similar to C.
• Which of the following best suggests mitochondria entered eukaryotic cells
evolutionarily long ago?
A. Mitochondria have a double membrane structure
B. Mitochondria, like bacteria, possess a circular DNA
C. Mitochondrial ribosomes have 28S and 39S subunits, totaling 55S.
D. Mitochondria divide by binary fission
• The answer is C.
A. The outer mitochondrial membrane is hypothesized to be of host origin.
B. This is a similarity between mitochondria and bacteria.
C. Since most modern bacteria have 30S/50S subunits and 70S ribosomes, it is
plausible that bacteria continued to evolve after endosymbiosis occurred.
D. This is a similarity between mitochondria and bacteria.
Q87: Gramicidin effects on ATP synthesis
Q87:
- What effect would a chemical that provides an alternative H+ pathway have on
ATP synthesis?
A. Increase – increased proton movement back into matrix
B. Decrease – decreased rate of H-atom donation by NADH
C. Decrease – proton gradient will rapidly reach equilibrium
D. Not change – sufficient proton gradient will remain to generate ATP
Q87:
- What effect would a chemical that provides an alternative H+ pathway have on
ATP synthesis?
A. Increase – increased proton movement back into matrix. Incorrect. Proton
movement only generates ATP if it is through ATP synthase.
B. Decrease – decreased rate of H-atom donation by NADH. Incorrect. Hydride (H:) is donated by NADH, not H+.
C. Decrease – proton gradient will rapidly reach equilibrium. Correct. A pathway
that doesn’t generate ATP will be more thermodynamically favorable.
D. Not change – sufficient proton gradient will remain to generate ATP. Incorrect.
The gradient will be dissipated since an easier H+ pathway is available.
• Gramicidin is a peptide molecule. Which amino acids likely line its pore?
• Acidic amino acids, D/E. The negative charge will favor H+ transport.
• Gramicidin was the first commercial antibiotic and is a nonribosomal peptide
(NRP), meaning that no gene encodes for it. What can be inferred about the
enzyme that synthesizes gramicidin?
A. The enzyme only synthesizes one peptide molecule, gramicidin.
B. The gene for the enzyme is located extrachromosomally on a plasmid.
C. Daughter cells lack the enzyme to synthesize gramicidin.
D. Bacteria do not have enzymes to synthesize gramicidin, only eukaryotes.
• The answer is A.
A. NRP synthetases contain their own template and only synthesize 1 peptide.
B. The gene for the enzyme cannot be inferred to be extrachromosomal.
C. Since chromosomal DNA codes for the enzyme, it should be heritable.
D. Bacillus brevis originally produced gramicidin as defense against other bacteria.
Q88: Mitochondria similarity to bacteria
Q87:
- Which of the following should be true about mitochondria?
A. 80S ribosomes
B. Incapable of binary fission
C. Circular DNA
D. Capable of anaerobic respiration
Q87:
A.
B.
C.
D.
Which of the following should be true about mitochondria?
80S ribosomes. Incorrect. This describes eukaryotic ribosomes.
Incapable of binary fission. Incorrect. Bacteria do binary fission.
Circular DNA. Correct. Bacteria have circular DNA.
Capable of anaerobic respiration. Incorrect. The eukaryotic cells that
mitochondria entered were already capable of anaerobic respiration.
•
If all of a eukaryotic cell’s mitochondria were removed, which of the following
would NOT be true?
The cell would produce much fewer ATP per molecule of glucose.
The cell would synthesize additional mitochondria to offset the metabolic
disadvantage.
Its daughter cell would lack mitochondria.
The cell would lose the ability to metabolize fats.
The answer is B.
Since the cell would only be capable of anaerobic respiration, it would
produce 2 net ATP per glucose instead of 32 ATP.
Mitochondria divide by binary fission. Eukaryotic cells have no means to
synthesize mitochondria, particularly because they contain unique DNA.
Since no mitochondria would be present in the parent cell, the daughter cell
would have no mitochondria.
Beta oxidation occurs in the mitochondria.
A.
B.
C.
D.
•
A.
B.
C.
D.
Q89: Effects of valinomycin
Q89:
- What effect on ATP synthesis would movement of K+ into the mitochondria
have?
A. Decrease – K+ compete with protons in ATP synthase
B. Decrease – K+ will disrupt proton movement
C. Increase – increase in positive charge will cause increased H+ movement.
D. Increase – additional positive charge will activate ATP synthase.
Q89:
- What effect on ATP synthesis would movement of K+ into the mitochondria
have?
A. Decrease – K+ compete with protons in ATP synthase. Incorrect. K+ is much
larger than H+ and would not be expected to occupy the ATP synthase active
site.
B. Decrease – K+ will disrupt proton movement. Correct. The mitochondrial
membrane gradient is electrochemical, i.e. movement of H+ resulting in ATP
synthesis relies on differences in [H+] as well as positive charge. Increased
positive charge will diminish the electrical gradient  less ATP synthesis.
C. Increase – increase in positive charge will cause increased H+ movement.
Incorrect. Electrostatic repulsion between H+ and K+ will not outweigh the
disruption the electrical part of the gradient described in B.
D. Increase – additional positive charge will activate ATP synthase. Incorrect. ATP
synthase runs on H+ movement, not presence of positive charge.
Passage 13 Takeaways
• Mitochondria – Function – Citric acid cycle, ETC and ATP synthesis
• Endosymbiotic theory – similarities to bacterial cells
• Ends up being the classic bacteria vs. eukaryote type question
• H+ electrochemical gradient, effects that would increase/decrease ATP
synthesis
• Passage-based reasoning
• If an answer choice leads you to refute a known theory, triple-check it against
the other answer choices
• Better answer choices will attempt to explain discrepancies rather than ignore
them
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes
Passage 14 Intro:
-
X-linked genes
Sex ratios
Chromosomal inversions
Looks like we have a (Drosophila) genetics passage!
Q90: Hardy-Weinberg?
Q90:
- Sounds like H-W equilibrium – look for more on these alleles
- Assuming all genotypes equally fit
- Fate of population with 50% Xi and 50% Xs?
•
These alleles seem to be the main focus of the passage…
A.
B.
C.
D.
Extinction
Stable population size, predominance of females
Stable population size, all individuals produce 50:50 sex ratio
Stable population size, some individuals produce >females, some
produce >males
Q91: Proportion of XiY in next gen
Q92: Explanation for Xi genetic behavior
Q93: Another question about Xi and Xs
Q94: Another question about Xi and Xs
Now, to read passage 14:
• e and f genes are X-linked
• Affect sex ratios of individuals’ offspring if brought together by inversion
• XsY male = standard – equal male/daughter offspring
• XiY male = “sex ratio trait” – offspring only daughters
• Both sire equal number of offspring
Now, to read passage 14:
• If no Xi genotypes are selected against, Xi  100%
• Unless other genes act to suppress expression of e and f
• XiY males occasionally sire viable but sterile sons (normal appearance)
• These sons are XO (inherit X from mom, no Y)
• Similar to what condition in humans?
• Turner syndrome (not MCAT prerequisite knowledge)
Q90:
- Passage: If no Xi genotypes are selected against, Xi  100%
- Passage: Both XiY and XsY males sire equal number of offspring
- Assuming all genotypes equally fit
- Fate of population with 50% Xi and 50% Xs?
A.
B.
C.
D.
Extinction
Stable population size, predominance of females
Stable population size, all individuals produce 50:50 sex ratio
Stable population size, some individuals produce >females, some
produce >males
Q90:
- Passage: If no Xi genotypes are selected against, Xi  100%
- Passage: Both XiY and XsY males sire equal number of offspring
- Assuming all genotypes equally fit
- Fate of population with 50% Xi and 50% Xs?
A. Extinction. Correct. If Xi increases to 100%, then all males will only
produce daughters, prohibiting mating.
B. Stable population size, predominance of females. Incorrect. This
accounts for the “equal number of offspring” but not the Xi  100%
C. Stable population size, all individuals produce 50:50 sex ratio.
Incorrect. Same as B.
D. Stable population size, some individuals produce >females, some
produce >males. Incorrect. Same as B.
Q91:
- Males: XsY (equal numbers of sons and daughters)
- Females:
- 15% XiXi
- 50% XiXs
- 35% XsXs
- What proportion of male flies in next generation will be XiY?
A. 12%
B. 30%
C. 40%
D. 65%
Q91:
- Males: XsY (equal numbers of sons and daughters)
- Females:
- 15% XiXi
- 50% XiXs
- 35% XsXs
- What proportion of male flies in next generation will be XiY?
A. 12%
B. 30%
C. 40%. Correct. Males get their X chromosome from mom. All male
progeny from the XsY * XiXi will inherit the Xi chromosome, so XiY =
15%.
• Half the male progeny from the XsY * XiXs will inherit the Xs
chromosome from mom and be XsY = 25%; and half will be XiY
= 25%
• No XiY males will be produced from the XsY * XsXs mating = 0%.
• 15% + 25% = 40%
D. 65%
Q91, cntd.:
- Follow-up: Eye color in Drosophila is determined by a single gene, of
which R denotes red eyes and r denotes brown eyes. If 9% of a
Drosophila population has red eyes, what percentage has brown
eyes?
- The answer is 49%. This is a Hardy-Weinberg equilibrium
question. Recall:
- p + q = 1, where p is the frequency of the dominant allele
and q that of the recessive allele
- p2 + 2pq + q2 = 1
- p2 denotes dominant homozygotes
- 2pq denotes homozygotes
- q2 denotes recessive heterozygotes
- p2 = 0.09, so p = √0.09 = 0.3. Therefore, q must be 0.7. The
frequency of p is 30%, and the frequency of q is 70%.
- Brown eyes = recessive heterozygotes = q2 = 0.72 = 0.49 = 49%.
- What is the population percentage that are heterozygotes?
- 2pq = 2(0.3)(0.7) = 2(0.21) = 0.42 = 42%.
- Note: The math can be checked by adding up 9% + 49% + 42% = 58%
+ 42% = 100%.
Q92:
- Which explains why Xi has the potential to increase to 100%
frequency?
A. XiXs flies have the highest fitness
B. XiXi flies migrate and introduce the Xi chromosomes into new
populations
C. XiXi flies pass X chromosomes to all offspring, but XsXs flies pass to
only half of offspring
D. XiY flies pass X chromosome to all offspring, but XsY flies pass X
chromosome to only half their offspring
Q92:
- Which explains why Xi has the potential to increase to 100%
frequency?
A. XiXs flies have the highest fitness. Incorrect. Heterozygote advantage
does not explain Xi  100%.
B. XiXi flies migrate and introduce the Xi chromosomes into new
populations. Incorrect. Question specifies within gene pools.
C. XiXi flies pass X chromosomes to all offspring, but XsXs flies pass to
only half of offspring. Incorrect. Passage never states any funky Xinheritance ratios for females.
D. XiY flies pass X chromosome to all offspring, but XsY flies pass X
chromosome to only half their offspring. Correct. Since females get
1 chromosome from mom and 1 from dad, and XiY males only sire
females, then all female progeny will have at least 1 Xi. Since XsY
males produce equal sons and daughters, and since sons only
inherit a Y chromosome from dad, only half the progeny will get an
Xi chromosome from dad (i.e., all the daughters).
Q92:
- Follow-up: ABO blood typing in humans is based on A and B antigens
on RBCs’ cell surface. Blood type A, for instance, indicates the
presence of an A antigen that is tolerated by people with type A and
AB, but not type B or type O. The latter individuals will experience
an immune reaction if exposed to blood containing the A antigen.
Type O individuals have neither antigen, so their blood can be
tolerated by all blood types. The A and B alleles are inherited
autosomally.
- What kind of inheritance pattern describes the ABO system?
A. Imprinting
B. Incomplete dominance
C. Codominance
D. Epistasis
• The answer is B.
A. Imprinting. Incorrect. The question never mentions parent-specific
inheritance.
B. Incomplete dominance. Incorrect. This describes an intermediate
property (e.g. red + white = pink).
C. Codominance. Correct. Since both A and B can be simultaneously
expressed with each producing a different effect (tolerance to A
and B blood, respectively), ABO blood typing is an example of
codominance. Note: O describes a recessive homozygote.
D. Epistasis. Incorrect. This describes gene expression that is
dependent on expression of another gene.
Q92:
- Follow-up: ABO blood typing in humans is based on A and B antigens
on RBCs’ cell surface. Blood type A, for instance, indicates the
presence of an A antigen that is tolerated by people with type A and
AB, but not type B or type O. The latter individuals will experience
an immune reaction if exposed to blood containing the A antigen.
Type O individuals have neither antigen, so their blood can be
tolerated by all blood types. The A and B alleles are inherited
autosomally.
- If a female has blood type B, which of the following could NOT be
the genotypes of her parents?
- (IA = A antigen, IB = B antigen, and i = neither antigen. Blood
type O results from the ii genotype).
A. IAIA * IBi
B. IAIB * ii
C. IAi * IBIB
D. IAIB * IAIB
• The answer is A.
A. IAIA * IBi. Correct. This cross can only produce type AB or A.
B. IAIB * ii. Incorrect. This cross can produce type A or type B.
C. IAi * IBIB. Incorrect. This cross can produce type A or type B.
D. IAIB * IAIB. Incorrect. This cross can produce type A, type B, or type
AB.
Q93:
- Female Drosophila produces 34 daughters and 38 sons.
- 18 of the sons produce only daughters.
- Remainder produce equal daughters/sons.
- Genotype of original female and male?
A. XiXs and XsY
B. XiXs and XiY
C. XiXi and XsY
D. XsXs and XiY
Q93:
- Female Drosophila produces 34 daughters and 38 sons.
- 18 of the sons produce only daughters.
- Remainder produce equal daughters/sons.
- Genotype of original female and male?
A. XiXs and XsY. Correct. Since half the sons sire only daughters, they
must be XiY. The sons who sire equal sons/daughters must be XsY.
Since males get their X-chromosome from mom, she must
therefore be a heterozygote.
B. XiXs and XiY. Incorrect. There would be no sons from the original
cross.
C. XiXi and XsY. Incorrect. All of the sons from the original cross would
produce only daughters.
D. XsXs and XiY. Incorrect. There would be no sons from the original
cross.
Q94:
- What prevents the Xi chromosome from reaching 100% frequency?
A. XsXs flies have the lowest fitness
B. XsXs flies have the highest fitness
C. XiY flies and XsY flies have equal fitness
D. XiXi flies and XsXs flies have equal fitness
Q94:
- What prevents the Xi chromosome from reaching 100% frequency?
A. XsXs flies have the lowest fitness
B. XsXs flies have the highest fitness. Correct. If Xi is selected against,
this would explain why it can persist in the population without
extinction occurring, as in the earlier question.
C. XiY flies and XsY flies have equal fitness
D. XiXi flies and XsXs flies have equal fitness
•
A.
B.
C.
D.
•
A.
B.
Which of the following best explains why the Xi genotype is selected
against?
When the inversion occurs, the SRY gene gets interrupted
When the inversion occurs, a gene important for flight gets
interrupted
When the inversion occurs, the reading frame for several important
genes becomes unreadable to DNA polymerase
When the inversion occurs, the reading frame for several important
genes becomes unreadable to RNA polymerase
The answer is B.
The SRY gene, which determines male sex, is located on the Y
chromosome. This would not explain decreased fitness.
Correct. This would decrease fitness.
C. DNA polymerase would still be able to read and
replicate the DNA.
D. RNA polymerase would still be able to
transcribe RNA from the DNA.
Passage 14 Takeaways
Passage 15 Intro:
-
BMI
Leptin
Two hypotheses
Looks like we have a metabolism/2 competing hypotheses passage!
Q95: Type of chemical messenger
Q95:
- Class of chemical messenger traveling in the blood, links brain with
digestive tract/fat cells
A. NTs
B. Digestive enzymes
C. Protein receptors
D. Hormones
Q95:
- Class of chemical messenger traveling in the blood, links brain with
digestive tract/fat cells
A. NTs. Incorrect. NTs do not travel in the blood.
B. Digestive enzymes. Incorrect. Do not travel in blood.
C. Protein receptors. Incorrect. Do not travel in blood.
D. Hormones. Correct. Hormones link organs and travel in blood.
•
A.
B.
C.
D.
•
•
A.
B.
C.
D.
•
Which of the following correctly describes the release route of a
peptide hormone?
Rough ER  exosome  extracellular fluid (ecf)
Rough ER  Golgi apparatus  vesicle  ecf
Smooth ER  Golgi apparatus  exosome  ecf
Cytoplasm  rough ER  Golgi  vesicle  ecf
The answer is B.
The presence of digestive enzymes in the blood is associated with
acute pancreatitis. Which of the following enzymes would NOT be
found in the blood during pancreatitis?
Pepsin
Trypsin
Amylase
Lipase
The answer is A. Pepsin is a gastric enzyme, not a pancreatic
enzyme.
Q96: Hypotheses from passage
Q97: Hypotheses from passage
Q98: Organ that breaks down glycogen
Q98:
- Which organ breaks down glycogen?
A. Stomach
B. Liver
C. Pancreas
D. Small intestine
Q98:
- Which organ breaks down glycogen?
A. Stomach
B. Liver
C. Pancreas
D. Small intestine
•
A.
B.
C.
D.
•
•
•
A.
B.
C.
D.
•
Loss of function mutations to which of the following enzymes would
most directly impact glycogenolysis?
Hexokinase
Glucokinase
Glycogen branching enzyme
Glycogen phosphorylase
The answer is D. Glycogen phosphorylase hydrolyzes glycogen to
free individual glucose-1-phosphate molecules.
Between hexokinase and glucokinase, which likely has a higher Km?
• Glucokinase. Glycolysis needs to operate even in conditions of
low glucose, whereas glycogenesis operates in high glucose.
Hexokinase and glucokinase both catalyze the conversion of glucose
to glucose-6-phosphate. They are most likely...
Structural isomers
Isozymes
Isoforms
Zymogens
The answer is B.
A. Structural isomers are molecules with the same net
molecular formula but diff connectivity.
B. Isozymes are different enzymes that catalyze the
same reaction.
C. Isoforms are alternatively spliced proteins from the
same gene.
D. Zymogens are enzyme precursors.
Now, to read passage 15:
• BMI = w/h2
• Genes account for 40% of factors determining BMI
• Leptin – hormone required for maintaining normal weight
• Ob = codes for leptin
• Db = leptin receptor
• Stable weight regulated by metabolic feedback loops
• Brain, fat cells, digestive tract, muscles involved
Now, to read passage 15:
• Set point hypothesis
• Dictated by inheritance
• Metabolism and behavior adjusted by brain to maintain predetermined weight
• Diet and exercise cannot change set point
• Settling point hypothesis
• Metabolism and genes interact with environment to determine body weight
• Feedback loops may allow weight to be stabilized at different level
Q96:
- What do the hypotheses attempt to explain?
- Set point: Dictated by inheritance, metabolism/behavior adjusted by
brain, exercise cannot change set point
- Settling point: Metabolism and genes interact with environment to
determine body weight
A.
B.
C.
D.
How multiple, interacting factors determine body weight
How individual factors alone influence body weight
How metabolism and environment influence body weight
How environment and behavior influence body weight
Q96:
- What do the hypotheses attempt to explain?
- Set point: Dictated by inheritance, metabolism/behavior adjusted by
brain, exercise cannot change set point
- Settling point: Metabolism and genes interact with environment to
determine body weight
A. How multiple, interacting factors determine body weight. Correct.
Both hypotheses mention multiple factors that interact.
B. How individual factors alone influence body weight. Incorrect.
Neither hypothesis looks at independent factors.
C. How metabolism and environment influence body weight.
Incorrect. Only the settling point hypothesis argues that
metabolism/environment can influence body weight.
D. How environment and behavior influence body weight. Incorrect.
Both hypotheses mention genes and metabolism.
•
•
According to the settling point hypothesis, what might happen to a
person’s body weight when they live in a cold climate?
• It might decrease, as metabolism increases in response to cold
temperatures in order to keep body temperature up.
According to the set point hypothesis, what might happen to an
individual who begins a strenuous exercise routine?
• They may eat more to maintain the same set body weight.
Q96:
- Which hypothesis implies a person can deliberately alter their body
weight?
- Set point: Dictated by inheritance, metabolism/behavior adjusted by
brain, exercise cannot change set point
- Settling point: Metabolism and genes interact with environment to
determine body weight
A. Set point – a thermostat can be reset
B. Set point – the set point can change with age
C. Settling point – with correct genotype, one’s metabolism may allow
weight to stabilize at a different level
D. Settling point – diet and exercise cannot reset the set point
Q96:
- Which hypothesis implies a person can deliberately alter their body
weight?
- Set point: Dictated by inheritance, metabolism/behavior adjusted by
brain, exercise cannot change set point
- Settling point: Metabolism and genes interact with environment to
determine body weight
A. Set point – a thermostat can be reset. Incorrect. The set point
hypothesis says body weight is predetermined.
B. Set point – the set point can change with age. Incorrect. A person
cannot deliberately alter their age.
C. Settling point – with correct genotype, one’s metabolism may
allow weight to stabilize at a different level. Correct. The settling
point hypothesis says weight is a result of interaction between
metabolism and genes with environment.
D. Settling point – diet and exercise cannot reset the set point.
Incorrect. This describes the set point hypothesis.
Passage 15 Takeaways
Q99:
- Which of the following would NOT interfere with repeated impulse
transmission at the NMJ?
A.
B.
C.
D.
Cholinesterase blocker
Toxin that blocks ACh release
^ACh receptor sites on motor end plate
Substance that binds ACh receptor sites
Q99:
- Which of the following would NOT interfere with repeated impulse
transmission at the NMJ?
A. Cholinesterase blocker. Incorrect. High ACh would interfere with
repeated transmission by inhibiting ACh-receptor dissociation.
B. Toxin that blocks ACh release. Incorrect. This would impair impulse
transmission.
C. ^ACh receptor sites on motor end plate. Correct. This might
increase the signal, but it would not interfere with transmission.
D. Substance that binds ACh receptor sites. Incorrect. This could
impede transmission (e.g. if it were an antagonist).
• Black widow spiders inject latrotoxin in their bites, which causes
ACh release, leading to muscle pain and spasms. The effects can be
mediated by the muscle relaxant tubocurarine. What is the most
likely mechanism of tubocurarine?
A. Facilitates Ca2+ entry into the postsynaptic terminal.
B. Degrades latrotoxin.
C. Increases nicotinic receptors at the NMJ
D. Blocks ACh receptors at the motor end plate.
• The answer is D.
A. This would cause more ACh release, which would not relax muscles.
B. This would not explain its role as a muscle relaxant.
C. Activation of nicotinic ACh receptors lead to
muscle contraction, so this would not cause muscle
relaxation.
D. Correct. Prevention of ACh binding its NMJ
receptors would cause muscle relaxation.
Q100:
- Which has striated muscle fibers?
A. Heart
B. Uterus
C. Arteries and veins
D. Small intestine
Q100:
- Which has striated muscle fibers?
A. Heart. Correct. Cardiac muscle is striated (has organized, repeating
sarcomere units).
B. Uterus. Incorrect. The uterus is smooth muscle.
C. Arteries and veins. Incorrect. Smooth muscle.
D. Small intestine. Incorrect. Smooth muscle.
•
Considering smooth, cardiac, and skeletal muscle:
• Which is/are striated?
• Cardiac and skeletal
• Which is/are multinucleated?
• Skeletal
• Which contain(s) gap junctions?
• Cardiac and smooth. They are described as
“functional syncytia” – i.e., functionally they are
multiple cells acting together
• Which contain(s) troponin?
• Cardiac and skeletal. Smooth contain calmodulin
and myosin light chain kinase.
• Which rely on calcium for contraction?
• All
• Which rely on autonomic innervation?
• Cardiac and smooth
• Which contain(s) sarcoplasmic reticula/T-tubules?
• Cardiac and skeletal
• Which contain(s) intercalated disks?
• Cardiac
Q101:
- Respiratory dead space vs. ventilation:
- What is dead space?
- Volume of air that reaches trachea, bronchioles, but not
alveoli.
- Breathing rate = 10 breaths/min
- Tidal volume = 800 mL/breath
- Dead space = 150 mL
A.
B.
C.
D.
65 mL
95 mL
6,500 mL
7,850 mL
Q101:
- Respiratory dead space vs. ventilation:
- What is dead space?
- Volume of air that reaches trachea, bronchioles, but not
alveoli.
- Breathing rate = 10 breaths/min
- Tidal volume = 800 mL/breath
- Dead space = 150 mL
A. 65 mL
B. 95 mL
C. 6,500 mL.
• Air reaching alveoli = intake – dead space = 800 mL – 150 mL
= 650 mL
• Breathing rate = 10 breaths/min
• 650 mL/breath * 10 breaths/min = 6,500 mL/min
D. 7,850 mL
Q102:
A.
B.
C.
D.
Which describes actin and myosin?
Both shorten, causing contraction
Both catalyze reactions that result in muscle contraction
Actin is disassembled by myosin  shortening of sarcomeres
Bridges form, break, and re-form  shortening of sarcomeres
Q102:
A.
B.
C.
D.
•
A.
B.
C.
D.
•
A.
B.
C.
D.
Which describes actin and myosin?
Both shorten, causing contraction
Both catalyze reactions that result in muscle contraction
Actin is disassembled by myosin  shortening of sarcomeres
Bridges form, break, and re-form  shortening of sarcomeres
Myosin storage myopathy is a rare genetic disease caused by a
mutation in the myosin heavy chain beta gene (MYH7). The mutant
myosin can form clumps that crowd out sarcomeres, leading to
delayed walking during development, waddling gait, and
generalized weakness. Which would be the best treatment?
Administration of siRNA targeting the MYH7 gene.
Administration of restriction enzymes that target a specific amino
acid sequence in the mutant myosin protein for degradation.
Administration of antibodies specific to the mutant myosin-myosin
binding interface.
Gene therapy that inserts a wild-type MYH7 sequence into the DNA.
The answer is C.
This would result in no production of the myosin protein.
Restriction enzymes target DNA sequences, not proteins.
Correct. This would dissolve the myosin clumps, which the
question implies are the root cause of the pathology.
This doesn’t take care of the myosin clumps; additionally, no
promoter was mentioned.
Q103:
A.
B.
C.
D.
When the optic cup fails to develop, the lens also fails. This implies
Neurulation follows gastrulation
The eye develops early in morphogenesis
Cells may induce neighboring cells to differentiate
Cell differentiation is an “all or none” phenomenon
Q103:
- When the optic cup fails to develop, the lens also fails. This implies
A. Neurulation follows gastrulation. Incorrect. Formation of nervous
tissue was not mentioned.
B. The eye develops early in morphogenesis. Incorrect. No information
on timeline was mentioned.
C. Cells may induce neighboring cells to differentiate. Correct. The
question stem implies that optic cup differentiation is necessary
for that of the lens.
D. Cell differentiation is an “all or none” phenomenon. Incorrect. “All
or none” does not imply to differentiation of adjacent tissue.
• Which of the following could be corrected by a diverging lens?
A. Myopia
B. Hyperopia
C. Astigmatism
D. Presbyopia, in which the eye loses the ability to focus on nearby
objects with age.
• The answer is A.
A. In myopia, the lens focuses the image prior to the retina. This can
be corrected by a diverging lens.
B. In hyperopia, the lens focuses the image behind the retina. This is
corrected with a converging lens.
C. In astigmatism, the optical power of the eye differs
for light coming from different directions. This is
corrected using a lens with the opposite irregularity of
the eye.
D. Presbyopia is corrected with a converging lens,
similar to hyperopia.
Passage 16 Intro:
-
Ectopic pregnancy
Human chorionic gonadotropin
Hemorrhaging
Looks like we have a passage on ectopic pregnancy!
Q104: Tubal prengancy
Q104:
• This a NOW question because we studied normal
implantation in content review.
• Drug that increases risk of tubal pregnancy most likely
inhibits
A. Contraction of the uterus.
B. Secretion of FSH.
C. Onset of menstruation.
D. Transport of the ovum from ovary to uterus.
Q104:
• This a NOW question because we studied normal implantation in
content review.
• Drug that increases risk of tubal pregnancy most likely inhibits
A. Contraction of the uterus. Incorrect.
• Uterine contractions occur during vaginal birth.
B. Secretion of FSH. Incorrect.
• FSH causes follicular development before an egg gets
released, so inhibiting FSH secretion shouldn’t increase the
risk of tubal pregnancy.
C. Onset of menstruation. Incorrect.
• Menstruation is monthly shedding of unfertilized egg and
endometrial lining.
• Inhibition of the onset of menstruation would not increase
the chances of a tubal pregnancy.
D. Transport of the ovum from ovary to uterus. Correct.
• If a drug inhibits transport of the ovum from ovary to uterus,
which occurs via the fallopian tube, a fertilized egg could get
stuck in the fallopian tube
• This would cause an ectopic pregnancy in the fallopian tube.
Q104 Challenge Question:
• Benzodiazepine use before conception is associated with a
47% increase in the risk of ectopic pregnancy. One
hypothesis suggests that the mechanism involves binding to
the GABA-A receptor in fallopian smooth muscle.
• Which best describes the effect of benzodiazepines binding
GABA receptors in the fallopian tube?
A. Depolarization and contraction.
B. Repolarization and relaxation.
C. Hyperpolarization and relaxation.
D. Graded depolarization and relaxation.
Q104 Challenge Question:
• Benzodiazepine use before conception is associated with a 47% increase in the
risk of ectopic pregnancy. One hypothesis suggests that the mechanism involves
binding to the GABA-A receptor in fallopian smooth muscle.
• Which best describes the effect of benzodiazepines binding GABA receptors in
the fallopian tube?
A. Depolarization and contraction. Incorrect.
• If benzodiazepines caused fallopian smooth muscle contraction, they
would not increase the risk of ectopic pregnancy in the fallopian tube.
B. Repolarization and relaxation. Incorrect.
• Repolarization occurs after potassium voltage-gated channels open,
causing K+ ions to flow out of the cell.
• Repolarization is not due to ligand-gated channels.
C. Hyperpolarization and relaxation. Correct.
• For the MCAT, GABA receptors cause an increase in permeability to Cl-,
allowing Cl- to flow down its concentration gradient into the cell.
• Causing hyperpolarization, which makes action potential initiation
more difficult.
• In a muscle, this would promote relaxation.
• If benzodiazepines are GABA agonists in fallopian smooth muscle, it
would follow that they would cause relaxation, which would increase the
risk of an egg getting stuck.
D. Graded depolarization and relaxation. Incorrect.
• Graded depolarization promotes contraction, and GABA-gated ion channels
(i.e. GABA receptors) do not cause depolarization.
Q105: Cause of death from ectopic pregnancy
Q105:
• This could be a NOW question provided we can find some
passage info about complications of ectopic pregnancy.
•
A.
B.
C.
D.
Cause of death from ectopic pregnancy is
Severe hormonal imbalance.
Loss of blood when the fallopian tube ruptures.
Infection in the region of the pregnancy.
Inadequate nutrition due to fetal use of maternal nutrients.
Q105:
• This could be a NOW question provided we can find some
passage info about complications of ectopic pregnancy.
• Cause of death from ectopic pregnancy is
A. Severe hormonal imbalance. Incorrect.
• While this can be an issue during pregnancy, it’s not
directly life-threatening.
B. Loss of blood when the fallopian tube ruptures. Correct.
• Passage states that internal hemorrhage can follow
tubal pregnancy.
C. Infection in the region of the pregnancy. Incorrect. Not
specific enough/no passage support.
D. Inadequate nutrition due to fetal use of maternal nutrients.
Incorrect.
• Fallopian tube rupture will occur before the embryo
grows large enough to require substantial nutrition.
Q105 Challenge Question:
• Following tubal rupture, the embryo can be found in which
body cavity?
A. Peritoneal cavity
B. Uterine cavity
C. Thoracic cavity
D. Vertebral cavity
Q105 Challenge Question:
• Following tubal rupture, the embryo can be found in which
body cavity?
A. Peritoneal cavity. Correct.
• The peritoneal cavity, also known as the
abdominopelvic cavity, contains 2 regions:
• Abdominal cavity, which contains the GI and GI
accessory organs, as well as spleen and kidneys.
• Pelvic cavity, which contains the bladder, lower
colon, and internal reproductive organs.
B. Uterine cavity. Incorrect.
• The uterine cavity is inside of the uterus, so this would
not result from a tubal rupture.
C. Thoracic cavity. Incorrect.
• The thoracic cavity contains the heart and lungs, and is
above the peritoneal cavity.
D. Vertebral cavity. Incorrect.
• Also known as the spinal cavity or spinal canal, this
cavity contains the spinal cord.
Q106: Hormone involved in delayed ovulation
Q106:
• This should be a NOW question because we studied these
hormones in content review, and we should be able to look
up delayed ovulation in the passage..
•
A.
B.
C.
D.
Delayed ovulation associated with delayed secretion of
which hormone?
Progesterone
Estrogen
HCG
LH
Q106:
• This should be a NOW question because we studied these
hormones in content review, and we should be able to look
up delayed ovulation in the passage..
•
A.
B.
C.
D.
Delayed ovulation associated with delayed secretion of
which hormone?
Progesterone. Incorrect.
• Progesterone prepares the uterus for implantation of a
fertilized egg in the second half of the cycle, after
ovulation.
• Progesterone does not cause ovulation.
Estrogen. Incorrect.
• Estrogen causes the LH surge that triggers ovulation,
so this is a better answer choice than A.
• Estrogen also promotes thickening of the
endometrium in the luteal phase.
• D is better.
HCG. Incorrect.
• hCG would indicate fertilization and uterine
implantation have already occurred.
LH. Correct.
• LH directly causes ovulation, so delayed secretion of
LH is the best answer causing delayed ovulation.
Q106 Challenge Question:
• In November 2012, a man posted on Reddit that he had
jokingly took a pregnancy test for hCG left behind by an exgirlfriend, which came back positive. hCG is normally
secreted by the placenta during pregnancy. Upon urging by
other users, he underwent a screening for and was
diagnosed with testicular cancer.
• The most likely reason testicular tumors can secrete hCG is
A. Tumors can randomly overexpress genes unrelated to their
tissue type.
B. The male testes are homologous structures to the female
ovaries.
C. Tumors can express genes from another tissue type in order
to disguise themselves from immune system recognition.
D. Dysregulation of the cell cycle can lead to differentiation
into another tissue type.
Q106 Challenge Question:
• In November 2012, a man posted on Reddit that he had jokingly took a
pregnancy test for hCG left behind by an ex-girlfriend, which came back
positive. hCG is normally secreted by the placenta during pregnancy. Upon
urging by other users, he underwent a screening for and was diagnosed with
testicular cancer.
• The most likely reason testicular tumors can secrete hCG is
A. Tumors can randomly overexpress genes unrelated to their tissue type.
Correct.
• Tumors often overproduce normal tissue secretions and express genes
unrelated to the tissue type whose transcription would normally be
downregulated.
B. The male testes are homologous structures to the female ovaries. Incorrect.
• While this is true, it does not explain hCG secretion.
C. Tumors can express genes from another tissue type in order to disguise
themselves from immune system recognition. Incorrect.
• While it is true that tumors express “self” cell surface antigens that make
immune recognition difficult,
• hCG is a secreted protein, so it would not help disguise the tumor.
D. Dysregulation of the cell cycle can lead to differentiation into another tissue
type. Incorrect.
• While tumors by definition do have a dysregulated cell cycle, they cannot
de-differentiate and re-differentiate into a different type of tissue type.
Q107: Common aspect to ectopic pregnancy
Q107:
• This should be a NOW question because we studied normal
implantation in content review and should be able to look
up any additional necessary info.
•
A.
B.
C.
D.
One aspect common to all causes of ectopic pregnancy is
that the zygote fails to
Implant in the uterus.
Leave the ovary.
Reach the fallopian tube.
Begin its development.
Q107:
• This should be a NOW question because we studied normal
implantation in content review and should be able to look
up any additional necessary info.
•
A.
B.
C.
D.
One aspect common to all causes of ectopic pregnancy is
that the zygote fails to
Implant in the uterus. Correct.
• By definition, all ectopic pregnancies involve
implantation in a location other than the uterus.
Leave the ovary. Incorrect.
• The zygote is formed in the fallopian tube, and can
only form after the egg has left the ovary.
Reach the fallopian tube. Incorrect.
• The egg enters the fallopian tube after release by the
ovary.
• The problem with most ectopic pregnancies is that
implantation occurs in the fallopian tube, not the
uterus.
Begin its development. Incorrect.
• Development in the ectopic location does begin.
Q107 Challenge Question:
• hCG pregnancy tests are used to rule out other conditions
during diagnosis of ectopic pregnancy. A urine test draws
urine through a series of regions, giving 1 or 2 “lines,”
depending on whether the user is pregnant.
• A typical hCG urine test is composed of the following:
• A sample pad for application of urine,
• A reaction zone containing mobile enzyme-linked
antibodies that bind hCG,
• A test zone containing fixed antibodies that bind hCG,
• A control zone antibodies that bind other antibodies
• Dye molecules in both test and control zones that
react with the enzyme on the mobile antibodies
• Which of the following does NOT correctly describe a urine
hCG test?
A. 1 line only in the control zone allows ectopic pregnancy to
be ruled out.
B. 2 faint lines indicate pregnancy and may indicate
dehydration.
C. The antibodies in the control zone bind to excess mobile
antibodies that did not bind hCG.
D. The mobile antibodies and the fixed antibodies in the test
zone must bind to different epitopes of hCG.
Q107 Challenge Question:
• hCG pregnancy tests are used to rule out other
conditions during diagnosis of ectopic pregnancy.
A urine test draws urine through a series of
regions, giving 1 or 2 “lines,” depending on
whether the user is pregnant.
• A typical hCG urine test is composed of the
following:
• A sample pad for application of urine,
• A reaction zone containing mobile enzymelinked antibodies that bind hCG,
• A test zone containing fixed antibodies that
bind hCG,
• A control zone antibodies that bind other
antibodies
• Dye molecules in both test and control zones
that react with the enzyme on the mobile
antibodies
Q107 Challenge Question:
• hCG pregnancy tests are used to rule out other conditions during diagnosis of ectopic pregnancy. A urine test draws urine
through a series of regions, giving 1 or 2 “lines,” depending on whether the user is pregnant.
• A typical hCG urine test is composed of the following:
• A sample pad for application of urine,
• A reaction zone containing mobile enzyme-linked antibodies that bind hCG,
• A test zone containing fixed antibodies that bind hCG,
• A control zone antibodies that bind other antibodies
• Dye molecules in both test and control zones that react with the enzyme on the mobile antibodies
• Which of the following does NOT correctly describe a urine hCG test?
A. 1 line only in the control zone allows ectopic pregnancy to be ruled out. Incorrect.
• If a line only appears in the control zone, there is no hCG detected and the individual is not pregnant, so this would
allow ectopic pregnancy to be ruled out.
B. 2 faint lines indicate pregnancy and may indicate dehydration. Correct.
• It is true that 2 lines indicates pregnancy because hCG was detected in the test zone.
• Faint lines on a pregnancy test more likely indicate overhydration, as the concentration of hCG is low.
C. The antibodies in the control zone bind to excess mobile antibodies that did not bind hCG. Incorrect.
• This is true. The purpose of the control zone is to ensure proper use of the test by the user.
• Excess mobile antibodies from the reaction zone that do not contain hCG, and consequently pass over the test zone,
will bind to antibodies in the control zone.
D. The mobile antibodies and the fixed antibodies in the test zone must bind to different epitopes of hCG. Incorrect.
• This is also true. In this antibody “sandwich,” the mobile antibody has bound to one epitope of hCG.
• In order for the test zone fixed antibodies to also bind hCG, they must be specific to a different epitope.
Passage 16 Takeaways
• Fertilization, pregnancy and birth - anatomy, oogenesis, hormones,
placental development, zygote/embryo/fetus, uterine contractions
• Reproductive hormones - FSH, LH, GnRH, inhibin,
testosterone/androgens, estrogen, progesterone, hCG
• Menstrual cycle - hormonal and structural changes
• GABA
• Anatomy - body cavities and organs that reside in them
• Tumors and cancer - overexpression of genes, dysregulation of cell
cycle, immune system
• Biotechnology - diagnostic antibodies, antigens/epitopes
Passage 17 Intro:
-
Vasodilation
Homeostasis
Sweating
Looks like we have a passage on thermoregulation!
Q108: Vasodilation
Q108:
• This is a NOW question because we studied
thermoregulation in content review.
• When environmental temp is 33°C, vasodilation of
cutaneous blood vessels helps to regulate the body
temperature of a human by?
A. Slowing blood flow through the skin.
B. Maintaining an even distribution of heat throughout the
body.
C. Radiating excess body heat into the environment.
D. Preventing needed body heat from being lost to the
environment.
Q108:
• This is a NOW question because we studied thermoregulation
in content review.
• When environmental temp is 33°C, vasodilation of cutaneous
blood vessels helps to regulate the body temperature of a
human by?
A. Slowing blood flow through the skin. Incorrect.
• This answer choice doesn’t attempt to explain how body
temperature is regulated as a result.
B. Maintaining an even distribution of heat throughout the body.
Incorrect.
• Since 33°C approaches physiological temperature, warmblooded animals that radiate heat to the environment
must find strategies to not overheat as the temperature
gradient between body and environment approaches 0.
C. Radiating excess body heat into the environment. Correct.
• Vasodilation in hot temperatures assists
thermoregulation by allowing blood to more closely
contact the cooler outside environment.
D. Preventing needed body heat from being lost to the
environment. Incorrect.
• This is true in cold temperatures, not hot.
Q108 Challenge Question:
• Which of the following terms does NOT correctly describe a
phenomenon of electromagnetic radiation (EMR)?
A. Diffraction – light waves spread out after passing through a
small gap, leading to constructive or destructive
interference.
B. Dispersion – different frequencies of EMR refract to
different degrees after passing to another medium.
C. Convection – heat is transferred from one place to another
due to movement of fluid.
D. Wave-particle duality – the photoelectric effect
demonstrates EMR as a particle, while refraction
demonstrates EMR as a wave.
Q108 Challenge Question:
• Which of the following terms does NOT correctly describe a
phenomenon of electromagnetic radiation (EMR)?
A. Diffraction – light waves spread out after passing through a
small gap, leading to constructive or destructive
interference. Incorrect.
• This is the textbook definition of diffraction.
B. Dispersion – different frequencies of EMR refract to
different degrees after passing to another medium.
Incorrect.
• This is the textbook definition of dispersion.
C. Convection – heat is transferred from one place to another
due to movement of fluid. Correct.
• While this correctly describes convection, convection
does not apply to light/EMR.
D. Wave-particle duality – the photoelectric effect
demonstrates EMR as a particle, while refraction
demonstrates EMR as a wave. Incorrect.
• This is true. Light does have wave-particle duality.
• The photoelectric effect is an example of light acting as
a particle (photon).
• Refraction is a property of waves, and light refraction
demonstrates light acting as a wave.
Q109: Highest body temperature
Q109:
• This is a NOW question because we may have background
knowledge of these animals and/or we should be able to
look them up in the passage easily.
• When the environmental temperature is 45°C, which
organism would have highest body temp?
A. Human
B. Kangaroo rat
C. Camel
D. Lizard
Q109:
• This is a NOW question because we may have background
knowledge of these animals and/or we should be able to
look them up in the passage easily.
• When the environmental temperature is 45°C, which
organism would have highest body temp?
A. Human
B. Kangaroo rat
C. Camel
D. Lizard. Correct.
• This answer choice can be arrived at a few different
ways:
1. Knowing that lizards are cold-blooded, so their body
temperature fluctuates according to the outside
environment.
2. “One of these things is not like the others.”
3. Passage states that lizards have an impermeable
integument (skin), meaning they cannot sweat.
• Thus, they have less ways to cool down than
mammals in hot temps.
Q109 Challenge Question:
• Warm-blooded animals, endotherms, regulate their body
temperature by metabolic processes. Cold-blooded
animals, ectotherms, rely on environmental heat sources.
Which of the following correctly describes endotherms or
ectotherms?
A. Endotherms have constant body temperature across their
circadian clocks.
B. Ectotherms require more nourishment than endotherms.
C. Endotherms are more likely to engage in active foraging
behavior, while ectotherms are more likely to be ambush
predators.
D. A local habitat can have a higher carrying capacity for
endotherms than ectotherms.
Q109 Challenge Question:
• Warm-blooded animals, endotherms, regulate their body temperature by
metabolic processes. Cold-blooded animals, ectotherms, rely on
environmental heat sources. Which of the following correctly describes
endotherms or ectotherms?
A. Endotherms have constant body temperature across their circadian clocks.
Incorrect.
• Endotherms such as humans have near-constant body temperatures,
but they do fluctuate.
• “Constant” is an extreme word here
• Humans have lower body temp in the morning than in the evening.
B. Ectotherms require more nourishment than endotherms. Incorrect.
• Endotherms require more nourishment, as the question states they
regulate their body temperature by metabolic processes.
C. Endotherms are more likely to engage in active foraging behavior, while
ectotherms are more likely to be ambush predators. Correct.
• Since ectotherms have lower metabolic activity, they can lie in wait for
prey. Endotherms have greater metabolic requirements and must
actively forage.
D. A local habitat can have a higher carrying capacity for endotherms than
ectotherms. Incorrect.
• Similar to C, since endotherms have greater metabolic requirements, a
region will have a lower carrying capacity for endotherms.
Q110: Kidney failure
Q110:
• Cause of kidney failure during severe dehydration?
A. Inadequate blood volume for effective filtration.
B. Inability to produce sufficient urine.
C. Buildup of salts in the distal tubules.
D. Increased body temperature.
Q110:
• Cause of kidney failure during severe dehydration?
A. Inadequate blood volume for effective filtration. Correct.
• During severe dehydration, a person will go into
hypovolemic shock.
• Glomerular filtration, which relies on blood pressure,
will be ineffective, and the kidneys will fail to perform
most normal functions.
B. Inability to produce sufficient urine. Incorrect.
• This is not a cause of kidney failure but a result.
C. Buildup of salts in the distal tubules. Incorrect.
• Similar to B.
D. Increased body temperature. Incorrect.
• This is a result of decreased blood volume during
dehydration, but not a cause of kidney failure.
Q110 Challenge Question:
• Which of the following does NOT correctly describes a mechanism for regulating
blood pressure?
A. Prorenin is cleaved to renin and released into the bloodstream by the
juxtaglomerular apparatus in response to low pressure of the filtrate.
B. Angiotensinogen is secreted by the liver and cleaved into angiotensin I by renin,
then into angiotensin II by ACE2 receptors in the kidneys, brain, endothelial cells,
and lungs.
C. Atrial natriuretic peptide is released by atrial cardiac muscle cells in response to
mechanical stretching and increases sodium excretion in the nephron.
D. Aldosterone causes release of vasopressin from the posterior pituitary, which
causes increased Na+ reabsorption in the distal convoluted tubule and cortical
collecting duct of the nephron.
Q110 Challenge Question:
• Which of the following does NOT correctly describes a mechanism for regulating
blood pressure?
A. Prorenin is cleaved to renin and released into the bloodstream by the
juxtaglomerular apparatus in response to low pressure of the filtrate. Incorrect.
• This is true.
B. Angiotensinogen is secreted by the liver and cleaved into angiotensin I by renin,
then into angiotensin II by ACE2 receptors in the kidneys, brain, endothelial cells,
and lungs. Incorrect.
• This is true.
C. Atrial natriuretic peptide is released by atrial cardiac muscle cells in response to
mechanical stretching and increases sodium excretion in the nephron. Incorrect.
• This is true.
• “Natri” = sodium, “ur” = urine, “etic” = pertaining to an action of process
D. Aldosterone causes release of vasopressin from the posterior pituitary, which
causes increased Na+ reabsorption in the distal convoluted tubule and cortical
collecting duct of the nephron. Correct.
• Angiotensin II causes vasopressin release from the posterior pituitary.
• Vasopressin causes increased synthesis of aquaporins in the distal tubule
and collecting duct, not increased Na+ reabsorption.
• Aldosterone itself causes increased Na+ reabsorption in the distal
tubule and collecting duct.
Q111: Heat stroke and sweat glands
Q111:
• People born w/o sweat glands are likely to die of heat stroke in the
tropics.
• This indicates that under tropical conditions, the human body may
A. Gain, rather than lose, heat by evaporation.
B. Gain, rather than lose, heat by radiation.
C. Need to use different mechanisms than in temperate zones to
maintain body temperature.
D. Be better able to regulate body temperature than under temperate
conditions.
Q111:
• People born w/o sweat glands are likely to die of heat stroke in the
tropics.
• This indicates that under tropical conditions, the human body may
A. Gain, rather than lose, heat by evaporation. Incorrect.
• Evaporation is endothermic, so it will not cause the body to
gain heat.
B. Gain, rather than lose, heat by radiation. Correct.
• Since people without sweat glands cannot use evaporative
cooling, the fact that they are likely to die of heat stroke
indicates that they are unable to sufficiently radiate heat by
vasodilation.
C. Need to use different mechanisms than in temperate zones to
maintain body temperature. Incorrect.
• The question stem does not indicate that humans need to rely
on different mechanisms in temperate vs. tropical conditions.
• Instead, the mechanisms are the same, but people
without sweat glands are unable to use evaporation.
D. Be better able to regulate body temperature than under temperate
conditions. Incorrect.
• The fact that people without sweat glands are likely to die of
heat stroke in tropical conditions suggests that the human
body is less able to regulate temperature under these
conditions.
Q111 Challenge Question:
• Anhidrosis is a congenital or acquired inability to sweat.
Sjogren’s syndrome, the second-most common rheumatic
disorder, is an autoimmune anhidrotic condition caused by
inflammation of lacrimal (sweat) and salivary glands. It is
associated with a significantly higher risk of non-Hodgkin
lymphoma.
• Which of the following does NOT describe Sjogren’s
syndrome?
A. ACTH and cortisol levels are lower than in healthy
individuals.
B. Treatment with immune suppressing drugs can alleviate
symptoms of dryness and decrease risk of lymphoma.
C. Sjogren’s patients must see a dental hygienist frequently for
teeth cleanings.
D. Viral infection may help trigger the onset of Sjogren’s
syndrome by molecular mimicry toward antigen-presenting
cells.
Q111 Challenge Question:
• Anhidrosis is a congenital or acquired inability to sweat. Sjogren’s syndrome, the second-most common
rheumatic disorder, is an autoimmune anhidrotic condition caused by inflammation of lacrimal (sweat)
and salivary glands. It is associated with a significantly higher risk of non-Hodgkin lymphoma.
• Which of the following does NOT describe Sjogren’s syndrome?
A. ACTH and cortisol levels are lower than in healthy individuals. Incorrect.
• ACTH and cortisol levels are lower in many autoimmune diseases such as Sjogren’s.
• Cortisol downregulates immune system activity.
B. Treatment with immune suppressing drugs can alleviate symptoms of dryness and decrease risk of
lymphoma. Correct.
• While immune suppressing drugs are useful for decreasing immune activity towards the lacrimal
and salivary glands, and this can alleviate symptoms,
• The problem with using them is an increased risk of cancer, since the immune system is less able to
respond to tumors.
C. Sjogren’s patients must see a dental hygienist frequently for teeth cleanings. Incorrect.
• Decreased salivation provides a more ideal environment for bacteria.
• This is true.
D. Viral infection may help trigger the onset of Sjogren’s syndrome by molecular mimicry toward antigenpresenting cells. Incorrect.
• This is the hardest answer choice to eliminate, but answer choice B is more clearly wrong.
• In fact, viral infections are implicated in many autoimmune disorders, as viral antigens can sometimes
look similar to self antigens.
Passage 17 Takeaways
• Thermoregulation - hair, vasodilation/constriction, metabolic changes,
sweating
• Properties of light - reflection, refraction, diffraction, dispersion, waveparticle duality, photoelectric effect
• Warm-blooded/cold-blooded, maybe?
• Circadian clock basics
• Kidneys - filtration, osmolarity gradient, effect of hormones
• Blood pressure regulatory hormones - RAAS system, vasopressin/ADH,
atrial natriuretic peptides
• ACTH and cortisol
• Cancer and the immune system
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes
Q111: Heat stroke and sweat glands
Q111:
• People born w/o sweat glands are likely to die of heat stroke in the
tropics.
• This indicates that under tropical conditions, the human body may
A. Gain, rather than lose, heat by evaporation.
B. Gain, rather than lose, heat by radiation.
C. Need to use different mechanisms than in temperate zones to
maintain body temperature.
D. Be better able to regulate body temperature than under temperate
conditions.
Q111:
• People born w/o sweat glands are likely to die of heat stroke in the
tropics.
• This indicates that under tropical conditions, the human body may
A. Gain, rather than lose, heat by evaporation. Incorrect.
• Evaporation is endothermic, so it will not cause the body to
gain heat.
B. Gain, rather than lose, heat by radiation. Correct.
• Since people without sweat glands cannot use evaporative
cooling, the fact that they are likely to die of heat stroke
indicates that they are unable to sufficiently radiate heat by
vasodilation.
C. Need to use different mechanisms than in temperate zones to
maintain body temperature. Incorrect.
• The question stem does not indicate that humans need to rely
on different mechanisms in temperate vs. tropical conditions.
• Instead, the mechanisms are the same, but people
without sweat glands are unable to use evaporation.
D. Be better able to regulate body temperature than under temperate
conditions. Incorrect.
• The fact that people without sweat glands are likely to die of
heat stroke in tropical conditions suggests that the human
body is less able to regulate temperature under these
conditions.
Q111 Challenge Question:
• Anhidrosis is a congenital or acquired inability to sweat.
Sjogren’s syndrome, the second-most common rheumatic
disorder, is an autoimmune anhidrotic condition caused by
inflammation of lacrimal (sweat) and salivary glands. It is
associated with a significantly higher risk of non-Hodgkin
lymphoma.
• Which of the following does NOT describe Sjogren’s
syndrome?
A. ACTH and cortisol levels are lower than in healthy
individuals.
B. Treatment with immune suppressing drugs can alleviate
symptoms of dryness and decrease risk of lymphoma.
C. Sjogren’s patients must see a dental hygienist frequently for
teeth cleanings.
D. Viral infection may help trigger the onset of Sjogren’s
syndrome by molecular mimicry toward antigen-presenting
cells.
Q111 Challenge Question:
• Anhidrosis is a congenital or acquired inability to sweat. Sjogren’s syndrome, the second-most common
rheumatic disorder, is an autoimmune anhidrotic condition caused by inflammation of lacrimal (sweat)
and salivary glands. It is associated with a significantly higher risk of non-Hodgkin lymphoma.
• Which of the following does NOT describe Sjogren’s syndrome?
A. ACTH and cortisol levels are lower than in healthy individuals. Incorrect.
• ACTH and cortisol levels are lower in many autoimmune diseases such as Sjogren’s.
• Cortisol downregulates immune system activity.
B. Treatment with immune suppressing drugs can alleviate symptoms of dryness and decrease risk of
lymphoma. Correct.
• While immune suppressing drugs are useful for decreasing immune activity towards the lacrimal
and salivary glands, and this can alleviate symptoms,
• The problem with using them is an increased risk of cancer, since the immune system is less able to
respond to tumors.
C. Sjogren’s patients must see a dental hygienist frequently for teeth cleanings. Incorrect.
• Decreased salivation provides a more ideal environment for bacteria.
• This is true.
D. Viral infection may help trigger the onset of Sjogren’s syndrome by molecular mimicry toward antigenpresenting cells. Incorrect.
• This is the hardest answer choice to eliminate, but answer choice B is more clearly wrong.
• In fact, viral infections are implicated in many autoimmune disorders, as viral antigens can sometimes
look similar to self antigens.
Passage 17 Takeaways
• Thermoregulation - hair, vasodilation/constriction, metabolic changes,
sweating
• Properties of light - reflection, refraction, diffraction, dispersion, waveparticle duality, photoelectric effect
• Warm-blooded/cold-blooded, maybe?
• Circadian clock basics
• Kidneys - filtration, osmolarity gradient, effect of hormones
• Blood pressure regulatory hormones - RAAS system, vasopressin/ADH,
atrial natriuretic peptides
• ACTH and cortisol
• Cancer and the immune system
Passage 18 Intro:
-
Toxic shock
Staph aureus
Superantigens
Looks like we have a passage on microbio and bacterial infections!
Q112: Main cause of problems in TSS
Q112:
• This should be a NOW question because we studied these
properties of bacteria in content review and/or should be
able to easily look for context clues from the passage.
•
A.
B.
C.
D.
Staphylococcus and Streptococcus bacteria cause problems
in acute infections such as toxic shock syndrome primarily
by
Multiplying to produce large numbers of bacteria.
Stimulating exaggerated immune responses.
Causing autoimmune reactions.
Inhibiting metabolic enzymes with toxins.
Q112:
• This should be a NOW question because we studied these
properties of bacteria in content review and/or should be
able to easily look for context clues from the passage.
•
A.
B.
C.
D.
Staphylococcus and Streptococcus bacteria cause problems
in acute infections such as toxic shock syndrome primarily
by
Multiplying to produce large numbers of bacteria. Incorrect.
• This may occur, but is not the direct cause of
complications.
Stimulating exaggerated immune responses. Correct.
• Superantigens nonspecifically activate polyclonal T
cells, causing a cytokine storm that the passage says
is responsible for many of the problems in TSS.
Causing autoimmune reactions. Incorrect. No mention of
autoimmunity or attacking self antigens.
Inhibiting metabolic enzymes with toxins. Incorrect. No
mention of toxins inhibiting metabolic enzymes.
Q112 Challenge Question:
• Bafilomycin A1 is a macrolide antibiotic produced from
fungi that in eukaryotes inhibits vacuolar-type H+-ATPases,
which are present in the plasma and lysosomal membranes,
and itself also serves as a K+ ionophore in the mitochondrial
inner membrane, conducting K+ into the matrix.
• Which of the following describes the effects of bafilomycin
A1?
A. Acidification of the lysosome and mitochondrial swelling.
B. Basification of the lysosome and mitochondrial shrinkage.
C. Acidification of the cytoplasm and mitochondrial swelling.
D. Acidification of the extracellular fluid and mitochondrial
swelling.
Q112 Challenge Question:
• Bafilomycin A1 is a macrolide antibiotic produced from fungi that in
eukaryotes inhibits vacuolar-type H+-ATPases, which are present in the
plasma and lysosomal membranes, and itself also serves as a K+
ionophore in the mitochondrial inner membrane, conducting K+ into the
matrix.
• Which of the following describes the effects of bafilomycin A1?
A. Acidification of the lysosome and mitochondrial swelling. Incorrect.
• The lysosome has an acidic pH. H+-ATPases should normally pump
protons into the lysosome. If the ATPase is inhibited, the lysosome
will not get acidified.
B. Basification of the lysosome and mitochondrial shrinkage. Incorrect. See
C.
C. Acidification of the cytoplasm and mitochondrial swelling. Correct.
• If plasma membrane H+-ATPases are inhibited, H+ from metabolism
cannot be exported to the ecf, resulting in acidification of the
cytoplasm.
• If more K+ enters the matrix, this would lead to swelling not
shrinkage, as H2O follows ions osmotically.
• The extra K+ also disrupts the mitochondrial membrane
potential, decreasing ATP synthesis, and even leading to
apoptosis.
D. Acidification of the extracellular fluid and mitochondrial swelling.
Incorrect.
• The ecf will not receive protons via the H+-ATPase.
Q113: Organ systems
Q113:
• This is a NOW question because we studied these organ
systems in content review and should be able to connect
the cause of TSS to one of them.
•
A.
B.
C.
D.
In addition to skin and circulatory systems, which organ
system affected by TSS?
Musculoskeletal system
Digestive system
Lymphatic system
Respiratory system
Q113:
• This is a NOW question because we studied these organ
systems in content review and should be able to connect
the cause of TSS to one of them.
•
A.
B.
C.
D.
In addition to skin and circulatory systems, which organ
system affected by TSS?
Musculoskeletal system
Digestive system
Lymphatic system. Correct. High levels T-cell activation
would involve the lymphatic system.
Respiratory system
Q113 Challenge Question:
• Mucormycosis is a rare but serious opportunistic infection
by a family of molds called Mucormycetes. Following a
Streptococcus pyogenes TSS, a previously healthy 53-year
old French woman’s condition deteriorated despite no
detection of residual S. pyogenes bacteria. During ileostomy,
a black margin suggested mucormycosis, and a culture of
peritoneal fluid examined microscopically revealed
characteristic spores of Rhizopus.
• Why did the woman most likely develop mucormycosis?
A. Antibiotics used to treat her TSS only worked on bacteria,
not fungi.
B. The woman had an underlying immune compromising
disorder.
C. Her immune system was weakened by the TSS, allowing
colonization by Rhizopus.
D. The infection was nosocomial (hospital-acquired) due to
improper operating room sterilization before the ileostomy.
Q113 Challenge Question:
• Mucormycosis is a rare but serious opportunistic infection by a family of molds called Mucormycetes.
Following a Streptococcus pyogenes TSS, a previously healthy 53-year old French woman’s condition
deteriorated despite no detection of residual S. pyogenes bacteria. During ileostomy, a black margin
suggested mucormycosis, and a culture of peritoneal fluid examined microscopically revealed
characteristic spores of Rhizopus.
• Why did the woman most likely develop mucormycosis?
A. Antibiotics used to treat her TSS only worked on bacteria, not fungi. Correct.
• Antibiotic treatment likely killed off normal gut bacteria, allowing for the opportunistic fungal
infection.
B. The woman had an underlying immune compromising disorder. Incorrect.
• This contradicts the question stem, as the woman was described as “previously healthy.”
C. Her immune system was weakened by the TSS, allowing colonization by Rhizopus. Incorrect.
• The immune system is not weakened by infections unless the targeted cell/tissue by the pathogen
is immune.
• On the contrary, the immune system may be stronger towards non-Strep infections as well,
due to buildup of innate immune cells.
D. The infection was nosocomial (hospital-acquired) due to improper operating room sterilization before
the ileostomy. Incorrect.
• The question states that the mucormycosis was found during the ileostomy procedure. This answer
choice contradicts the timeline.
Q114: Calculation of activated T cells
Q114:
• This is a NOW question because we should be able to find
numbers in the passage to easily crunch for the answer.
•
A.
B.
C.
D.
Superantigens increase number of activated T cells by a
factor of?
20
5,000
20,000
100,000
Q114:
• This is a NOW question because we should be able to find
numbers in the passage to easily crunch for the answer.
•
Superantigens increase number of activated T cells by a
factor of?
A. 20
B. 5,000
C. 20,000. Correct.
• 20% = 1/5
•
𝟏
𝟓
𝟏
𝟏𝟎𝟎,𝟎𝟎𝟎
D. 100,000
=
𝟏𝟎𝟎,𝟎𝟎𝟎
𝟓
=
𝟏𝟎∗𝟏𝟎,𝟎𝟎𝟎
𝟓
= 𝟐 ∗ 𝟏𝟎, 𝟎𝟎𝟎 = 𝟐𝟎, 𝟎𝟎𝟎
Q114 Challenge Question:
• Theralizumab is an immunomodulatory drug developed at the
University of Würzburg in Germany that showed in vitro activation of Tregulatory cells, which downregulate T-cell activity. It is a potent
monoclonal antibody agonist of CD28, a protein involved in
costimulation during T-cell activation. Researchers thought it could be
used to treat autoimmune diseases like rheumatoid arthritis.
• However, in the first human clinical trial, subjects exhibited systemic
inflammation, cytokine release syndrome, angioedema, and
catastrophic systemic organ failure, and theralizumab development was
halted.
• What was the most likely cause of the complications?
A. Theralizumab caused downregulation of the immune system, leading to
superantigens exerting powerful inflammatory activity on the body.
B. T-reg cells actually stimulated effector T cells instead of downregulating
them.
C. T cells did not receive the second costimulatory signal for activation.
D. Once T-regulatory activity is induced, T cells eventually downregulate
their regulatory capabilities and become effector T cells.
Q114 Challenge Question:
• Theralizumab is an immunomodulatory drug developed at the University of Würzburg in Germany that showed in vitro
activation of T-regulatory cells, which downregulate T-cell activity. It is a potent monoclonal antibody agonist of CD28, a
protein involved in costimulation during T-cell activation. Researchers thought it could be used to treat autoimmune
diseases like rheumatoid arthritis.
• However, in the first human clinical trial, subjects exhibited systemic inflammation, cytokine release syndrome,
angioedema, and catastrophic systemic organ failure, and theralizumab development was halted.
• What was the most likely cause of the complications?
A. Theralizumab caused downregulation of the immune system, leading to superantigens exerting powerful inflammatory
activity on the body. Incorrect.
• The immune system operates in an inflammatory environment. The purpose of inflammation is to assist in the
immune response.
• If the immune system were downregulated, there would not be inflammation. Further, as described in the
passage, superantigens cause harm specifically by overstimulating the immune system.
B. T-reg cells actually stimulated effector T cells instead of downregulating them. Incorrect.
• While this is a tempting answer choice because it could explain the symptoms, it directly contradicts information in
the question stem.
• D is a better answer.
C. T cells did not receive the second costimulatory signal for activation. Incorrect.
• If this were true, the high immune activation described in the symptoms would not have occurred.
D. Once T-regulatory activity is induced, T cells eventually downregulate their regulatory capabilities and become
effector T cells. Correct.
• This is the only answer choice that both explains the symptoms observed and does not contradict the question
stem. Given 2 answer choices like B and D, always go with the one that doesn’t contradict information given.
• This is actually the hypothesized cause of the complications in the human trial.
Q115: Strain potency
Q115:
• This is a NOW question because it’s a classic
pseudodiscrete.
• Strain A dose to cause infection: 1x105 bacteria
• Strain B dose to cause infection: 5x104 bacteria
• Which statement describes the relative potencies?
A. Strain A is five times as potent as Strain B.
B. Strain A is one-fifth as potent as Strain B.
C. Strain A is twice as potent as Strain B.
D. Strain A is half as potent as Strain B.
Q115:
• This is a NOW question because it’s a classic
pseudodiscrete.
• Strain A dose to cause infection: 1x105 bacteria
• Strain B dose to cause infection: 5x104 bacteria
• Which statement describes the relative potencies?
A. Strain A is five times as potent as Strain B.
B. Strain A is one-fifth as potent as Strain B.
C. Strain A is twice as potent as Strain B.
D. Strain A is half as potent as Strain B. Correct.
• Strain A requires twice as many bacteria to cause
infection, so it is half as potent.
• Strain A bacteria conjugate with Strain B bacteria at a rate of
8 per mL per second. Suppose 1 x 105 Strain A bacteria are
inoculated in enough LB liquid agar to produce a
concentration of 1000 bacteria/mL, and the procedure is
repeated to the same concentration for 5 x 104 Strain B
bacteria. How many conjugations will occur in one minute?
Q115 Challenge Question:
• Strain A bacteria conjugate with Strain B bacteria at a rate of 8 per mL per second. Suppose 1 x 105 Strain A bacteria
are inoculated in enough LB liquid agar to produce a concentration of 1000 bacteria/mL, and the procedure is
repeated to the same concentration for 5 x 104 Strain B bacteria. How many conjugations will occur in one minute?
• This is a complicated calculation question, more than you would have to do for an average MCAT quantitative
question.
• The conjugation rate was described as “8 per mL per second” and we were given a time of one minute, 60 s.
• So, to find the number of conjugations, we need the volume of solution.
• Volume for Strain A:
•
1 𝑚𝐿
1000 𝑏𝑎𝑐𝑡𝑒𝑟𝑖𝑎
1𝑥105 𝑏𝑎𝑐𝑡𝑒𝑟𝑖𝑎
1
105
103
1 𝑚𝐿
1000 𝑏𝑎𝑐𝑡𝑒𝑟𝑖𝑎
5𝑥104 𝑏𝑎𝑐𝑡𝑒𝑟𝑖𝑎
1
5∗104
103
∗
=
= 100 𝑚𝐿
• We had to divide the number of bacteria by the (bacteria/mL) to cancel bacteria and get mL as the
final unit.
• At this point, we could repeat the calculation for Strain B, though if we consider that there are half as many
bacteria for Strain B, we can simply divide the necessary volume by 2 to get the same concentration.
•
∗
=
= 5 ∗ 10 = 50 𝑚𝐿
• So, our total volume is 150 mL.
8 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑖𝑜𝑛𝑠 60 𝑠 150 𝑚𝐿
•
∗ 1 ∗ 1 = 8 ∗ 60 ∗ 150 = 4 ∗ 60 ∗ 300 = 60 ∗ 1200 = 72 ∗ 1000 =
1 𝑚𝐿∗𝑠
72,000 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑖𝑜𝑛𝑠
Now, to read the passage:
- Toxic shock syndrome (?)
- Shock is when the circulatory system cannot adequately perfuse tissues, causing ischemia and hypoxia and
tissue death
- TSS involves high fever, hypotension, and a rash resulting in skin desquamation
- Affects at least three organ systems
- Increased risk linked to high-absorbency tampons (?)
- Left in the user too long, allowing bacterial overgrowth
- Two bacteria that cause TSS are Staphylococcus aureus and Streptococcus pyogenes
- These bacteria produce superantigen proteins
Now, to read the passage:
- Superantigens bypass a processing step normally performed by APCs
- Bind to T cells outside standard antigen binding site (?)
- T-cell receptor
- Activates ~20% of the T-lymphocytes (i.e. nonspecifically)
- Negative effects of superantigens occur b/c ^^T-cell activation  ^^cytokines
- Increased cytokine release cause for many acute problems seen in TSS
- As well as some autoimmune diseases
Passage 18 Takeaways
• Immune system - important cells, be familiar with cytokines and
interleukins as messenger molecules, inflammation, lymphatic system
• Lysosome - acidic
• Osmosis
• Gut microbiome
• Concentrations
• Dimensional analysis
Q116:
• If a person’s gallbladder is removed, the person should restrict the
consumption of
A. Proteins.
B. Polysaccharides.
C. Triglycerides.
D. Lactose.
Q116:
• If a person’s gallbladder is removed, the person should restrict the
consumption of
A. Proteins.
B. Polysaccharides.
C. Triglycerides. Correct. The gallbladder stores bile, which is used to
emulsify lipids.
D. Lactose.
•
•
A.
B.
C.
D.
Lactose intolerance occurs as most mammals wean off mother’s
milk due to genetic programming, although lactase persistence is
common in many human populations. The most accurate test for
lactose intolerance is the hydrogen breath test, which examines an
individual’s breath for bacterially produced H2 and CH4 after taking
lactose on an empty stomach.
Which of the following does NOT correctly describe the hydrogen
breath test?
Gut bacteria can metabolize lactose since they do not downregulate
their lactase gene with age.
The test would be less accurate if the individual were not fasting
since bacteria would downregulate their lactase enzyme.
Hydrogen and methane are produced via reduction and oxidation,
respectively.
The electrons extracted from glucose and galactose are given to
hydrogen and carbon since the bacterial ETC is inactive in the gut.
Q116 Challenge Question:
• Lactose intolerance occurs as most mammals wean off mother’s milk due to genetic programming, although
lactase persistence is common in many human populations. The most accurate test for lactose intolerance is the
hydrogen breath test, which examines an individual’s breath for bacterially produced H2 and CH4 after taking
lactose on an empty stomach.
• Which of the following does NOT correctly describe the hydrogen breath test?
A. Gut bacteria can metabolize lactose since they do not downregulate their lactase gene with age. Incorrect.
• This is true. Unlike mammals, which wean off their mother’s milk and downregulate their lactase gene,
bacteria would not lose ability to digest lactose.
B. The test would be less accurate if the individual were not fasting since bacteria would downregulate their lactase
enzyme. Incorrect.
• This is true. For instance, we know that bacteria prefer glucose from our background on the lac operon, and
will downregulate their lactase enzyme if glucose is present.
C. Hydrogen and methane are produced via reduction and oxidation, respectively. Correct.
• Methane, CH4, is the most reduced form of carbon. It could not be produced by oxidation.
D. The electrons extracted from glucose and galactose are given to hydrogen and carbon since the bacterial ETC is
inactive in the gut. Incorrect.
• This is true. The gut is an anaerobic environment, so ETCs are shut down in gut microbes. Thus, gut bacteria
rely mainly on anaerobic fermentation.
• The H2 and CH4 in this question are fermentation byproducts.
• Gut bacteria also provide the host with nutrients such as short-chain fatty acids (SCFAs – propionate and
butyrate) from food molecules indigestible to the host.
Q117:
• The posttranslational modification of some of the
eukaryotic cell’s most abundant proteins is thought to
affect the ability of these proteins to condense DNA into
30-nm fibers.
• Given this, these proteins are most likely
A. Tubulins.
B. Histones.
C. Transcription activators.
D. DNA polymerase subunits.
Q117:
• The posttranslational modification of some of the
eukaryotic cell’s most abundant proteins is thought to
affect the ability of these proteins to condense DNA into
30-nm fibers.
• Given this, these proteins are most likely
A. Tubulins.
B. Histones. Correct. Histones are the proteins that DNA
wraps around to condense DNA into nucleosomes and
then chromosomes.
C. Transcription activators.
D. DNA polymerase subunits.
•
A.
B.
C.
D.
The DNA polymerase(s) in prokaryotes with 5’ to 3’
exonuclease activity is/are:
I. DNA pol I
II. DNA pol II
III. DNA pol III
I only
I and III
III only
I, II, and III
Q117 Challenge Question:
• The DNA polymerase(s) in prokaryotes with 5’ to 3’ exonuclease activity is/are:
I. DNA pol I. Correct.
• DNA pol I is the enzyme that removes RNA primers before DNA ligase joins Okazaki fragments.
• In order to do this, it excises ribonucleotides while laying down new deoxyribonucleotides, both in
the 5’ to 3’ direction.
• This is exonuclease activity, as well.
• DNA pol I also has 3’ to 5’ exonuclease activity for proofreading.
II. DNA pol II. Incorrect.
• DNA pol II is primarily involved in DNA repair, and it has 3’ to 5’ exonuclease activity for proofreading.
III. DNA pol III. Incorrect.
• DNA pol III is the major replicative enzyme in prokaryotes, and it has 3’ to 5’ exonuclease activity for
proofreading.
A. I only. Correct.
B. I and III
C. III only
D. I, II, and III
Q118:
• When viewing an X ray of the bones of a leg, a doctor can
tell if the patient is a growing child, because the X ray
shows:
A. Cartilaginous areas in the long bones.
B. Bone cells that are actively dividing.
C. The presence of haversian cells.
D. Shorter-than-average bones.
Q118:
• When viewing an X ray of the bones of a leg, a doctor can tell if
the patient is a growing child, because the X ray shows:
A. Cartilaginous areas in the long bones. Correct.
• Children have cartilage that hasn’t fully ossified yet.
B. Bone cells that are actively dividing. Incorrect.
• Osteoprogenitor cells can produce new osteoblasts in
childhood and adulthood.
C. The presence of haversian cells. Incorrect.
• Child and adult bones have haversian canals for
nourishment of bone cells. Haversian cells are not a thing.
D. Shorter-than-average bones. Incorrect.
• This is a weird answer choice, but “average” should be
taken for a person’s age. This would just be a short person.
•
A.
B.
C.
D.
Which of the following statements does NOT correctly describe
bones and bone marrow?
Mobility is primarily facilitated by the long bones, and
protection of the internal organs by the flat bones.
B- and T-lymphocyte development occurs in hematopoietic
bone marrow.
Most hematopoiesis in adult humans occurs in flat bones such
as the ribs and skull.
Yellow marrow is mostly fat and red marrow contains
hematopoietic stem cells.
Q118 Challenge Question:
• Which of the following statements does NOT correctly describe
bones and bone marrow?
A. Mobility is primarily facilitated by the long bones, and
protection of the internal organs by the flat bones. Incorrect.
• This is true.
B. B- and T-lymphocyte development occurs in hematopoietic
bone marrow. Correct.
• B-cell development occurs in hematopoietic bone marrow,
which is red marrow.
• T-cell development occurs in the thymus.
C. Most hematopoiesis in adult humans occurs in flat bones such
as the ribs and skull. Incorrect.
• This is true. Hematopoiesis in adults occurs in flat bones
(and the ends of long bones).
D. Yellow marrow is mostly fat and red marrow contains
hematopoietic stem cells. Incorrect.
• This is true.
Q119:
• In eukaryotic cells, the process of incorporating uridine nts
into nucleic acid polymers occurs in which of the following
structures of the cell?
A. Nucleus
B. Lysosome
C. Ribosome
D. Golgi body
Q119:
• In eukaryotic cells, the process of incorporating uridine nts
into nucleic acid polymers occurs in which of the following
structures of the cell?
A. Nucleus. Correct.
• Uridine nts incorporating into polymers describes
transcription. Transcription occurs in the nucleus.
B. Lysosome
C. Ribosome. Incorrect.
• Ribosomes use mRNA as a template for protein
synthesis, but uridine isn’t incorporated into polymers
here.
D. Golgi body
•
A.
B.
C.
D.
Which of the following structures would NOT contain
uridine in the absence of substrate/ligand?
Histones
The immediate precursor of glycogen
Ribosomes
Telomerase
Q119 Challenge Question:
• Which of the following structures would NOT contain
uridine in the absence of substrate/ligand?
A. Histones. Correct.
• Histones are entirely made of protein. Their ligand is
DNA, which also does not contain uridine.
B. The immediate precursor of glycogen. Incorrect.
• The immediate precursor of glycogen is UDP-glucose.
C. Ribosomes. Incorrect.
• The catalytic portion of ribosomes is rRNA.
D. Telomerase. Incorrect.
• This may be hard to eliminate, though A is the more
obvious best answer.
• Telomerase reverse transcribes a repeating sequence
of DNA at the ends of telomeres.
• It carries around its own RNA template, which contains
uridine in organisms that have adenosine in their
telomere sequence.
Q120:
• The outer layers of human skin are composed of dead cells
impregnated with keratin and oil, which make the
epidermis relatively impermeable to water
• Yet humans sweat freely in hot temperatures. This occurs
because
A. The salt in sweat allows it to diffuse through the skin.
B. Sweat glands have special channels through the skin.
C. An osmotic gradient in sweat moves it through the skin.
D. Sweating occurs in only those areas of the body where the
skin is water permeable.
Q120:
• The outer layers of human skin are composed of dead cells
impregnated with keratin and oil, which make the
epidermis relatively impermeable to water
• Yet humans sweat freely in hot temperatures. This occurs
because
A. The salt in sweat allows it to diffuse through the skin.
Incorrect. This is not a thing.
B. Sweat glands have special channels through the skin.
Correct.
• Sweat pores extend to the outer layer of the skin.
C. An osmotic gradient in sweat moves it through the skin.
Incorrect.
• This doesn’t solve the problem of keratin and oil
making epidermis impermeable to water.
D. Sweating occurs in only those areas of the body where the
skin is water permeable. Incorrect.
• This contradicts the question stem and common sense,
since we sweat in a lot of different places.
Q120 Challenge Question:
• Hypohidrotic epidermal dysplasia is inherited through three
separate genetic patterns: autosomal dominant, autosomal
recessive, and X-linked recessive.
• Karen’s dad has the disorder, and concerned about their
potential children, her fiancée gets screened without telling
her. The lab informs him he has two wild-type alleles. They
have three children, with one affected with the condition.
• The pedigree is shown. What is the inheritance pattern?
A. Autosomal dominant
B. Autosomal recessive
C. X-linked recessive
D. Not enough information to determine
Q120 Challenge Question:
• Hypohidrotic epidermal dysplasia is inherited through three
separate genetic patterns: autosomal dominant, autosomal
recessive, and X-linked recessive.
• Karen’s dad has the disorder, and concerned about their
potential children, her fiancée gets screened without telling
her. They have three children, with one affected with the
condition. The pedigree is shown. What is the inheritance
pattern?
A. Autosomal dominant. Incorrect.
• If the inheritance was autosomal dominant, a child
could never have the condition unless a parent had it.
B. Autosomal recessive. Incorrect.
• If Karen’s fiancée was not a carrier, and the inheritance
was autosomal recessive, there would be no way for
their son to have the condition.
C. X-linked recessive. Correct.
• This checks out with Karen and her fiancée’s children.
A female can carry an X-linked allele and pass it on to
her male children only.
• Note that the grandmother also must be a carrier for
the male in the middle to have the condition.
• The affected female (homozygote) on the right would
pass the condition on to all her male children.
D. Not enough information to determine
FSQ Takeaways
•
•
•
•
•
•
•
•
•
•
•
•
Gallbladder and bile
Lac operon
Fermentation
Metabolism - redox
Histones
3 prokaryotic DNA pols, 3 eukaryotic RNA pols
Bone - cartilage in childhood, haversian canals, long vs flat bones, red vs yellow marrow,
hematopoiesis
Transcription and translation - locations, enzymes
Telomeres and telomerase
Glycogenesis and glycogenolysis
Skin - epidermis, keratin, sweat glands
Pedigree analysis
Thank you!
Questions?
Source: Biochemistry Memes Depicting Intracellular Scenes