Name: _____________________________________________________________________ 1 CHEM 376 FALL 2015 EXAM 1-­‐ VERSION 2 The Academic Honor Code binds you. This means: 1) you will not give or receive information during this exam, nor will you consult unauthorized sources of information; 2) you will not discuss the contents of this exam with other students until all students have finished the exam; and 3) you will not tolerate violations of academic integrity on the part of other students – if you see someone else cheating, please bring it to the attention of me or a proctor immediately. Please sign: ___________________________________ Time allowed for this exam: 4:10 – 5:25 p.m. Note: 1) For all calculations, it is only necessary to use ONE significant digit beyond a decimal point; to simplify calculations for this exam; YOU ARE ALLOWED TO USE ONLY NON-­‐
PROGRAMABLE BASIC CALCULATORS. 2) A number of the questions have more than one part – please be sure to answer each part! 3) You have a detachable blank sheet at the back of exam to be used for calculations or for scratching out answers because WE WILL ONLY GRADE NEATLY WRITTEN ANSWERS WITHIN THE SPACES PROVIDED; 4) Time is limited, so I recommend that you look over the entire exam (and point value of questions) first, and plan your time accordingly; 5) Show ALL work so that you can receive partial credit for correct calculations even if a final answer is wrong; 6) Use only the space allowed under each question. For any rough calculations, scratch paper is attached at the back; 7) The total points for this exam are 100. There are 25 multiple-­‐choice questions, 3 points each and 2 short answer problems worth 25 points; 8) THE MULTIPLE CHOICE QUESTIONS HAVE TO BE ANSWERED ON THE SCANTRON PROVIDED. PLEASE BUBBLE YOUR LAST NAME AND THEN FIRST NAME WITH A SPACE IN BETWEEN ON THE BACK OF SCANTRON FIRST. 9) PLEASE WRITE THE VERSION/COLOR OF THE EXAM ON THE SCANTRON. 2 1. The standard reduction potentials (Eo’ ) for the following half reactions are given: Fumarate + 2H+ + 2e-­‐ à succinate Eo’= +0.031 V FAD + 2H+ + 2e-­‐ à FADH2 Eo’= -­‐0.219 V If you mixed succinate, fumarate, FAD and FADH2 together, all at 1M concentrations and in the presence of succinate dehydrogenase, which of the following would happen initially? a.
b.
c.
d.
e.
Succinate would become oxidized, FAD would be unchanged
Fumarate would become reduced, FADH2 would become oxidized.
Succinate would become oxidized, FADH2 would be unchanged because it is a cofactor not a
substrate
Both fumarate and substrate would become oxidized, both FAD and FADH2 would become
reduced
No reaction would occur, because all reactants and products are at their standard state.
2. The higher the ____ content of a DNA, the ____ the melting temperature because of higher _______, and the ____ the ionic strength, the ____ the melting temperature. a.
b.
c.
d.
e.
G:C; higher; base stacking energies; lower; lower
G:C; higher; hydrogen bonding; higher; lower
G:C; lower; hydrogen bonding; higher; lower
A:T; higher; base stacking energies; higher; lower
A:T; lower; hydrogen bonding; lower; higher
3. The chemical reasons for the large negative ΔG°’for the hydrolysis of ____ include destabilization of the reactant due to bond strain caused by electrostatic repulsion, stabilization of the products by ionization and resonance, and entropy factors due to hydrolysis and subsequent ionization. a. AMP b. cAMP c. Phosphocreatine d. ATP e. All of the above 4. In eukaryotic cells, a class of ____-­‐ and ____-­‐rich proteins called ____ interact ionically with the anionic phosphate groups in the DNA backbone to form ____. a.
b.
c.
d.
e.
lysine; leucine; prions; ribosomes arginine; alanine; histones; nucleosomes arginine; lysine; histones; nucleosomes arginine; lysine; prions; ribosomes None of the above the normal and the mutant genes. The loss of a restrictio
(1) allows variation in base composition and (2) decreases the
would lead to the replacement of two fragments on the S
likelihood that a substitution for a base will change the encoded
blot by a single longer fragment. Such a finding would n
amino acid. If the degeneracy were equally distributed, each of the
that GTG replaced GAG; other sequence
20 amino acids
would
have
three
codons.
Both
benefits
(1
and
2)
are
3 changes at th
tion site could yield the same result.
maximized by the assignment of more codons to prevalent amino
7. Although the two enzymes cleave the same recognitio
acids than to less frequently used ones.
5. Estimate the pH of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate each break different bonds within the 6-bp sequence. C
40. Phe-Cys-His-Val-Ala-Ala
1.25 process
mole of hydrochloric The pKa values phosphoric acid on
are 7.2 and 2HPOis
4) aand by KpnIfor yields
an overhang
the2.1, 39 strand,
whereas
41. Exon(Na
shuffling
molecular
that
can lead to theacid. genera12.4. by Acc65I produces an overhang on the 59 strand. Th
tion of new
proteins by the rearrangement of exons within genes.
ends do not overlap.
Because
many exons encode functional protein domains, exon shuffling is aa.rapid
and
efficient
means
of
generating
new
genes.
pH < 2.1 42. It shows
that
GGTACC
GGTACC
b. pH = 2the
.1 genetic code and the biochemical means of
interpreting
the
code
are
common
to
even
very
distantly
related
life
CCATGG
CCATGG
c. pH = 7.2 forms. It
also
testifies
to
the
unity
of
life;
that
all
life
arose
from
a
d. 7.2 < pH < 12.4 Cleave with
Cle
common
e. ancestor.
2.1 < pH < 7.2 KpnI
Acc
43. (a) A codon for lysine cannot be changed to one for aspartate
by the mutation of a single nucleotide. (b) Arg, Asn, Gln, Glu, Ile,
C process and G
GTA
6. ssDNA will renature ____ into a dsDNA . This process, called ____ is aGGTAC
________ order Met, or Thr.
C
CATGG
CCATG
occurs code
best isif degenerate.
the temperature is _amino
___. acids, 18 are
44. The genetic
Of the 20
specified
by more than one codon. Hence, many nucleotide changes
a. inCooperatively; irst; increased slowly (especially
the third base ofmaelting; codon)fdo
not
alter the nature
of
b. Cooperatively; annealing; first; decreased slowly the encoded
amino acid. Mutations
leading
to an
altered amino
GGTAC
GTACC
Cooperatively; melting; second; ncreased lowly acid arec.usually
more deleterious
than those
that ido
not and shence
C
G
d. toCooperatively; elting; second;decreased slowly are subject
more stringent m
selection.
Incompatible sticky ends
45. GCe.baseCooperatively; pairs have threeahydrogen
compared withstwo
nnealing; bonds
second;decreased lowly for AT base pairs. Thus, the higher content of GC means more
hydrogen
bonds
greater
helix stability.
8. A simple
for
generating
7. A
s the and
phase changes from water to ice, the decreased density of the strategy
water is due to: many mutants is to s
essentially
corresponds
to
the
complexity
of
the
DNA
46. C0a.t value
a
degenerate
set
of
cassettes
by using a mixture of activat
the loss of hydrogen bonds sequence—in
other
words,
how
long
it
will
take
for
a
sequence
of
sides in particular rounds of oligonucleotide synthesis. Su
b. the gain of ionic bonds DNA c.
to find
its
complementary
strand
to
form
a
double
helix.
the 30-bp coding region begins with GTT, which encode
the gain of more van der Waals contacts The more complex the DNA, the slower it reassociates to make the
mixture a
ofs all
fourform nucleotides
is used in the first and
d. the altered spatial arrangement of the asymmetric water If
maolecules they the maximum double-stranded form.
rounds of synthesis, the resulting oligonucleotides will be
number of hydrogen bonds the sequence XYT (where X and Y denote A, C, G, or T
e.
Chapter 5the hydrophobic effect, which is driven by electrostatic interactions 16 different versions of the cassette will encode proteins c
1. (a) 59-GGCATAC-39
either Phe, Leu, Ile, Val, Ser, Pro, Thr, Ala, Tyr, His, As
(b) The
dideoxydideoxy
method of
sequencing
would
the gel gel pattern.
8. Sanger
The Sanger
method
shows
the give
following
WhatorisGly
theatsequence
of the Likewise,
templatedegenera
Cys, Arg,
the first position.
pattern shown
here.
strand?
can be made in which two or more codons are simultaneo
9. Because PCR can amplify as little as one molecule of
Direction of
electrophoresis
statements claiming the isolation of ancient DNA need
ed with some skepticism. The DNA would need to be s
Is it similar to human, bacterial, or fungal DNA? If so,
Dideoxy ATP
tion is the likely source of the amplified DNA. Is it sim
of birds or crocodiles? This sequence similarity would s
Dideoxy CTP
the case that it is dinosaur DNA because these species a
tionarily close to dinosaurs.
Dideoxy TTP
10. PCR amplification is greatly hindered by the prese
Dideoxy GTP
G–C-rich regions within the template. Owing to their h
temperatures, these templates do not denature easily, pr
the initiation of an amplification cycle. In addition, rigi
3'
A
C
G
T
T
A
C
C
G
5'
a.
b.
c.
d.
e.
5’-­‐ACGTTACCG-­‐3’ 5’-­‐GCCATTGCA-­‐3’ 5’-­‐TGCAATGGC-­‐3’ 5’-­‐CGGTAACGT-­‐3’ none of the above 4 9. Hydrolysis of 1M glucose-­‐6-­‐phosphate at 25 °C is catalyzed by glucose-­‐ 6-­‐phosphatase and is 99% complete at equilibrium. Which of the statements is most nearly correct? ANSWERS GIVEN WITHOUT SHOWING CALCULATIONS WILL BE GIVEN ZERO CREDIT. a. ΔG°’ is incalculably large. b. ΔG°’= 0 c. ΔG°’= + 11 KJ/mol d. ΔG°’= +16 KJ/mol e. ΔG°’=-­‐11 KJ/mol ΔG°’= -­‐RT Ln Keq = -­‐5.6 * log (0.99*0.99)/(0.01) = -­‐5.6* 1.99 =-­‐11.15 KJ/mol 10. Which atoms, or groups attached to them, face into the major groove of B-­‐form DNA? H
O
N
8
N
5
9
deoxyribose
H
N
7
6
4
1 N
3
H
N
H
5
2
1
N 3
2
N
4
6
N
O
deoxyribose
H
purine 6 and 7; pyrimidine 4 and 5 purine 1 and 2; pyrimidine 2 and 3 purine 2 and 3; pyrimidine 2 purine 8 and 9; pyrimidine 1 and 6 purine 6 and 7; pyrimidine 2 a.
b.
c.
d.
e.
11. Which of the following statements about H-­‐bond donors or acceptors of Watson-­‐Crick pairing is not true: a. the carbonyl group at position 2 of thymine and cytosine, and position 6 of guanine are H bond acceptors. b. the unprotonated imino nitrogens of adenine and cytosine are acceptors. c. the hydrogens at position 8 of purines and position 5 of pyrimidines are donors. d. the amino group of the cytosine, adenine and guanine are all donors. e. the ring nitrogens at position 3 of uracil and 1 of guanine are donors 5 12. The Eo’ of the NAD+/NADH half reaction is –0.32 V. The Eo’ of the oxaloacetate/malate half reaction is –0.17 V. When the concentrations of NAD+, NADH, oxaloacetate and malate are all 0.1 μM, which one of the following is the “spontaneous” reaction? Calculate the standard free energy change (ΔG°’) for the spontaneous reaction involving 2e-­‐ transfer. F is 96.48 KJ/VŸmol. ANSWERS GIVEN WITHOUT SHOWING CALCULATIONS WILL BE GIVEN ZERO CREDIT. ΔG°’= -­‐nFΔE = -­‐2 * 96.5 *(0.15) = -­‐28.95 KJ/mol + a. NAD +Oxaloacetate àmalate + NADH, ΔG°’=+53 KJ/mol b. Oxaloacetate + NADHàmalate + NAD+, ΔG°’=-­‐29 KJ/mol c. NAD+ +NADH àmalate + oxaloacetate, ΔG°’=-­‐20.1 KJ/mol d. Malate + NADHàoxaloacetate + NAD+, ΔG°’=-­‐3.5 KJ/mol e. Malate +NAD+ à oxaloacetate + NADH, ΔG°’=+52.5 KJ/mol 13. RNA is _________ stable to alkaline hydrolysis than DNA because RNA's vicinal ____ group makes the 3'-­‐
phosphodiester bond susceptible to ____________ cleavage. This group does not allow RNA to adopt a ______form helix because of _______ clashes with ________. a. less; 3'-­‐OH; nucleophilic; A; steric; solvent molecules b. more; 2'-­‐OH; electrophilic; B;backbone; nucleotides c. more; 2'-­‐OH; nucleophilic; Z; backbone; solvent molecules d. more; 3'-­‐OH; electrophilic; ;B; steric; phosphate backbone e. less; 2'-­‐OH; nucleophilic; B; steric; phosphate backbone 14. If a weak acid is 25% deprotonated at pH 4, what would the pKa be? SHOW ALL WORK TO GET FULL CREDIT. a. 4.48 b. 3.40 pKa= 4-­‐log (0.25/0.75) = 4+log 3 = 4.48 c. 3.52 d. 4.60 e. cannot be determined from the given information 15. Which of the following statements are false? a. Some cyclic nucleotides are second messengers. b. Propellor twist increases the area of contact between bases and water. c. RNA is transcribed as a single stranded biopolymer. d. Adenosine functions as a neuromodulator and is an autacoid. e. None of the above. 6 16. Which of the following statements are true? (i) For an endothermic reaction to become spontaneous, the entropy must increase substantially. (ii) The equilibrium constant of a reaction changes when concentration of reactants and products change. (iii) A thermodynamically unfavorable reaction can become favorable if the mass action ratio is less than the equilibrium mass action ratio. (iv) ΔH can be determined by experimentally measuring temperature-­‐dependence of the ΔS using a van’t hoff plot. a. (ii), (iii) and (iv) b. (ii) and (iii) only c. (i) and (iii) only d. (i), (iii) and (iv) e. (i), (ii) and (iv) 17. Which of the following statements are False: (i) Base-­‐catalyzed cleavage of RNA molecules mechanism involves a trigonal bipyramidal transition state. (ii) In PCR reactions, the absence of products or presence of non-­‐specific products indicate incorrect cycling conditions (often wrong annealing temperature) (iii) “A” type phosphodiesterases cleave the 5’ phosphate bond producing 5’ phosphate products. (iv) The stability of the DNA helix is mainly due to stacking and hydrophobic interactions. a. (ii) and (iii) only b. (i) , (ii) and (iii) only c. (iv) only d. (iii) only e. None of the above 18. Which of the following statements are False? (i) Adaptibility is the capacity of individual representatives of an organism to respond to environmental changes. (ii) Amino acids and simple organic compounds like carbon dioxide and hydrogen cyanide cannot be produced without the use of enzymes. (iii) Many biochemical reactions that form biopolymers from monomeric units involve the removal of water. (iv) Lysozymes are used for storage and replication. a. (i) and (iii) only b. (i) , (ii) and (iii) only c. (iv) only d. (i) only e. (ii) and (iv) only 7 19. If a circular DNA containing no supercoils is incubated with ethidium bromide (Hint: EtBr intercalates between DNA base pairs): a. the number of twists will increase, positive (LH) supercoils will develop, and the linking number will remain the same. b. the number of twists will increase, negative (RH) supercoils will develop, and the linking number will remain the same. c. the number of twists will decrease, negative (RH) supercoils will develop, and the linking number will remain the same. d. The linking number will be altered, and both the writhe and twist will adjust accordingly. e. the number of twists will decrease, positive (LH) supercoils will develop, and the linking number will remain the same. 20. Calculate the pH of a weak acid that is 0.2% ionized in a 0.2 M solution: a.
b.
c.
d.
e.
4.76 5.82 5.02 4.82 3.39 21. Base analysis of DNA from Triticum aestivum (wheat) shows that it has 18% cytosine. Therefore, the percentage of adenine would be: a. 32% b. 68% c. 18% d. 0% e. insufficient information to answer question 22. Hydrogen bonds are approximately ____% of the bond strength of covalent C-­‐C or C-­‐H bonds? a. 95% b. 50% c. 20% d. 5% e. 1% 23. In Z-­‐form DNA, the sugar pucker and N-­‐glycosyl bond conformation is a.
b.
c.
d.
e.
C2’-­‐endo, syn C2’-­‐endo, anti for pyrimidines and C3’-­‐endo, syn for purines C2’-­‐endo, anti for purines and C3’-­‐endo, syn for pyrimidines C3’-­‐endo, syn C3’-­‐endo, anti 8 24. Which of the following is not true about water? a. The electron-­‐rich oxygen atom of one water molecule can interact with the electron-­‐
poor proton on another water molecule to form a hydrogen bond b. Liquid water is only ~15% less hydrogen bonded than ice c. Water can form caged, highly ordered structures around small non-­‐polar molecules like methane that are called clathrates d. Amphiphilic detergents often form micelles with the polar groups on the outside exposed to the water (solvent) and the non-­‐polar groups sequestered in the interior e. Water is a non-­‐polar molecule with a bent molecular geometry 25. If the pBR322 were treated with BamHI in order to insert a gene into the plasmid, which of the following
would be correct regarding cells that were transformed with the resulting vector? a. cells that contained the desired vector would be resistant to tetracycline but sensitive to ampicillin b. cells that contained the desired vector would be resistant to both ampicillin and tetracycline c. cells that contained the desired vector would be sensitive to both ampicillin and tetracycline d. cells that contained the desired vector would be resistant to ampicillin but sensitive to tetracycline e. none of the above CH3
H
H
O ....... H N
N
N
N
N
N
.......
H
NPOINTS) Short answer type Questions. (25 N+
H ....... O
H
9 H
N
N
H ....... N
N
N
O
O
N
N
1. (a) Draw the base pairs formed between thymidine-­‐5′-­‐monophosphate and deoxyadenosine-­‐5′-­‐
phosphate (A), thymidine-­‐5′-­‐monophosphate (Hint: This is a nucleotide triplet formation (T:A:T) using both watson-­‐crick pT:A
airing and Hoogsteen pairing)(5 points) C +:G
You have to draw the sHoogsteen
ugars and hosphates long with the bases. (3 points total) for the (right).
FIGURE
11.16
baseppairs:
A!T (left) aand
C!!G
nucleotides and 2 points for the pairing. CH
3
T
...
.
O
...
..
O
H
..
...
N
...
O ....... H N
H
H
N
N
...
CH3
H
H ....... O
.
N
T
O
N
N
H ....... N
A
N
N
N
C
N+
H ....... N
O
T:A:T
G
N
C+:G:C
FIGURE 11.17 Base triplets formed when a purine interacts with one pyrimidine by Hoogsteen base pairing and
(b) Some DNA binding proteins, such as restriction endonucleases, are highly specific in their another by Watson–Crick base pairing.
interactions with certain DNA sequences, whereas as others, such as helicases or histones, are very nonspecific. What is the molecular basis for recognition between the specific binding proteins and the nonspecific proteins? (i.e. what parts of DNA recognize what parts of proteins – be as specific as Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party conte
possible in this space Please limit answer to learning
100 experience.
words or Learning
no more 10 additional
Editorial
review hasapprox. deemed that any
suppressed content
doesyour not materially
affect the overall
Cengage
reserves thethan right to remove
sentences). (4 points) Specific: formation of hydrogen bonds and/or van der Waals interactions between R groups of
06296_ch11_0336-0377.indd 351
an α- helix or β-strand of a protein with complementary acceptors in the major groove of the
DNA helix. (2 points)
Nonspecific: formation of ion pairs between lysine or arginine within the protein (no specific
structure required) and the negatively charged oxygen atoms of the phosphate backbone (or
dipole-charge interactions between a positive dipole of a R group with the phosphate). (2 points)
10 (c) Specific binding of multivalent metal ions is more important for structure and function of RNA than for double-­‐stranded DNA. What are two roles that specifically bound metal ions may play in RNA structure and function? (Please try to limit your answer in 100 words)(4 points) 1) structure: metal ions are required in certain locations to neutralize negative charge of phosphates and to stabilize higher order structure. (2 points) 2) function: specifically placed metal ions can aid catalysis by facilitating the deprotonation of a 2ʹOH group, stabilizing charge separation, and/or stabilizing a leaving group (the latter two as part of the transition state). (2 points) 2. a) The enzyme alcohol dehydrogenase catalyzes the oxidation of ethyl alcohol by NAD+ to give acetaldehyde plus NADH and a proton: CH3CH2OH + NAD+ à CH3CHO + NADH + H+ Calculate the pH at the end of the above reaction if it were carried out in a 110 mM TRIS buffer, pH 7.0 and 5 mM of protons were produced during the reaction? (pKa of Tris = 8.0) SHOW ALL WORK TO GET FULL CREDIT. (7 points) [A-­‐]= Tris Base and [HA] = Tris-­‐ H+ pH= pKa + log [A-­‐]/[HA] so [Tris Base]/[ Tris-­‐ H+]=10^ -­‐1 =0.1 Total concentration = 0.11M, so [Tris Base] =0.1/1.1*0.11M =0.01M(2 points) [ Tris-­‐ H+]= 1/1.1*0.11M =0.1M (2 points) When H+ is produced, Tris base
H+
Tris- H+
Initial
0.01 M
0.005 M
0.1 M
Change
−0.005 M
−0.005 M
+0.005 M
Equilibrium
0.005 M
0
0.105
pH= 8+ log [0.01 -­‐0.005]/[0.1+ 0.005] (1 point) ; = 8+ log 0.005/0.105 =8 + log 0.048 = =8-­‐1.3 =6.67(2 points) 11 b) PEP drives ATP synthesis at pH 7.0, a reaction that is catalyzed by pyruvate kinase upon consumption of a hydrogen ion as shown below: Phosphoenolpyruvate + ADP + H+ -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐>Pyruvate + ATP If the ΔG°’ for this reaction is -­‐31.7 kJ/mol, what is ΔG° (that is, the free energy change for the same reaction with all components, including H+, at a standard state of 1 M and standard temperature)? SHOW ALL WORK TO GET FULL CREDIT. (5 points) In reactions where H+ is consumed, ΔG°= Δ G°ʹ′ +RT ln [H+] (2 points) = -­‐31.7+(0.008314kJ/mol .K* 298K* ln 10-­‐7) =-­‐31.7-­‐ 39.93 (1 point) =-­‐71.93kJ/mol (2 points) 12