Download Chem464 Abrol Spring2017 FlippedReview4

CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
Step1: Solve these problems on your own; Step2: Solve them and discuss them as a group during
or before review; Step3: Present your solutions in class to other groups.
Problems for Review and Discussion (with Solutions)
1.
Which of the following is not a feature of signal transduction?
(a) Integration of multiple pathways toward the same downstream response
(b) Signal amplification
(c) Covalent binding between the ligand and the receptor
(d) Desensitization or adaptation of the receptor
(e) Variable affinity for different signaling components
2.
Auto-phosphorylation of receptor tyrosine kinases depends on which of the following?
(a) Dimerization of the receptor
(b) ATP
(c) Ligand binding
(d) Transmission of conformational changes through the membrane
(e) All of the above
3.
For the reaction A → B, ∆G'° = –60 kJ/mol. The reaction is started with 10 mmol of A; no B is
initially present. After 24 hours, analysis reveals the presence of 2 mmol of B, 8 mmol of A. Which
is the most likely explanation?
(a) A and B have reached equilibrium concentrations.
(b) An enzyme has shifted the equilibrium toward A.
(c) B formation is kinetically slow; equilibrium has not been reached by 24 hours.
(d) Formation of B is thermodynamically unfavorable.
(e) The result described is impossible, given the fact that ∆G'° is –60 kJ/mol.
4.
All of the following contribute to the large, negative, free-energy change upon hydrolysis of “highenergy” compounds except:
(a) electrostatic repulsion in the reactant.
(b) low activation energy of forward reaction.
(c) stabilization of products by extra resonance forms.
(d) stabilization of products by ionization.
(e) stabilization of products by solvation.
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
5. The standard reduction potentials (E'°) for the following half reactions are given.
Fumarate + 2H+ + 2e– → Succinate
+
–
E'° = +0.031 V
E'° = –0.219 V
FAD + 2H + 2e → FADH2
If you mixed succinate, fumarate, FAD, and FADH2 together, all at 1 M concentrations and in the
presence of succinate dehydrogenase, which of the following would happen initially?
(a) Fumarate and succinate would become oxidized; FAD and FADH2 would be reduced.
(b) Fumarate would become reduced; FADH2 would become oxidized.
(c) No reaction would occur because all reactants and products are already at their standard
concentrations.
(d) Succinate would become oxidized; FAD would become reduced.
(e) Succinate would become oxidized; FADH2 would be unchanged because it is a cofactor.
6. Which of the following reactions in glycolysis requires ATP as a substrate?
(a)
(b)
(c)
(d)
(e)
Hexokinase
Glyceraldehyde-3-phosphate dehydrogenase
Pyruvate kinase
Aldolase
Phosphoglycerate kinase
7. Which of the below is not required for the oxidative decarboxylation of pyruvate into acetyl-CoA?
(a) ATP
(b) CoA-SH
(c) FAD
(d) Lipoic acid
(e) NAD+
8. During oxidative phosphorylation, the proton motive force that is generated by electron transport is
used to:
(a) create a pore in the inner mitochondrial membrane.
(b) generate the substrates (ADP and Pi) for the ATP synthase.
(c) induce a conformational change in the ATP synthase.
(d) oxidize NADH to NAD+.
(e) reduce O2 to H2O
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
9. (Chapter 12, Problem 4) Explain how mutations in the R or C subunit of cAMP-dependent protein
kinase (PKA) might lead to:
A. a constantly active PKA or
B. a constantly inactive PKA.
(A)
(B)
A mutation in the R subunit could prevent it from binding to the C subunit, which will leave
the C subunit uninhibited and constantly active.
A mutation in the R subunit could prevent the binding of cAMP to the R subunit but still
allow for normal R-C interaction to take place. This will leave the C subunit constantly
inhibited and inactive.
10. (Chapter 12, Problem 22) For each of the following situations, provide a plausible explanation
for how it could lead to unrestricted cell division.
A. Colon cancer cells often contain mutations in the gene encoding the prostaglandin E2
receptor. PGE2 is a growth factor required for the division of cells in the gastrointestinal
tract.
B. Kaposi sarcoma, a common tumor in people with untreated AIDS, is caused by a virus
carrying a gene for a protein similar to the chemokine receptors CXCR1 and CXCR2.
Chemokines are cell-specific growth factors.
C. Adenovirus, a tumor virus, carries a gene for the protein E1A, which binds to the
retinoblastoma protein, pRb. (Hint: See Fig. 12–49.)
D. An important feature of many oncogenes and tumor suppressor genes is their cell-type
specificity. For example, mutations in the PGE2 receptor are not typically found in lung
tumors. Explain this observation. (Note that PGE2 acts through a GPCR in membrane.)
(A) These mutations might lead to permanent activation of the PGE2 receptor. Such
mutants are called constitutively active mutants (CAMs) as they don’t need an agonist
ligand for activation. The mutant cells would behave as though stimulatory levels of
PGE2 were always present, leading to unregulated cell division and tumor formation.
(B) The viral gene could code for a constitutively active mutant of the receptors like
CXCR1 and CXCR2, such that the cells send a constant signal for cell division. This
unrestrained division would lead to tumor formation.
(C) E1A protein might bind to pRb and prevent E2F from binding, so E2F is constantly
active as a transcription factor. So, it constantly activates genes that trigger cell
division, causing cells to divide uncontrollably.
(D) Lung cells do not normally respond to PGE2 because they do not express the PGE2
receptor. So, mutations resulting in a constitutively active PGE2 receptor do not affect
lung cells, unless its genes are also triggered in those cells.
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
11. (Chapter 13, Problem 10) Using Equation 13–4 (∆G = ∆G’⁰ + RT ln Q), plot ∆G against ln Q (massaction ratio) at 25 ⁰C for the concentrations of ATP, ADP, and Pi in the table below. ∆G’⁰ for the
reaction is –30.5 kJ/mol. Use the resulting plot to explain why metabolism is regulated to keep
the ratio [ATP]/[ADP] high.
Concentration (mM)
ATP
5.0
3.0
1.0
0.2
5.0
ADP
0.2
2.2
4.2
5.0
25
Pi
10
12.1
14.1
14.9
10
The hydrolysis reaction is:
ATP  ADP + Pi
Using Equation 13–4, with definition for mass action ratio Q = [ADP][Pi]/[ATP], expressed as molar
concentrations, the free-energy change for this hydrolysis reaction is given by:
∆G=∆G’o + RT ln ([ADP][Pi]/[ATP])
Let us first calculate ln Q for each of the five conditions listed in the data table above:
ln Q1 = ln [(2 x 10-4)(1.0 x 10-2)/(5 x 10-3)] = -7.8
ln Q2 = ln [(2.2 x 10-3)(1.21 x 10-2)/(3 x 10-3)] = -4.7
ln Q3 = ln [(4.2 x 10-3)(1.41 x 10-2)/(1 x 10-3)] = -2.8
ln Q4 = ln [(5.0 x 10-3)(1.49 x 10-2)/(2 x 10-4)] = -1.0
ln Q5 = ln [(2.5 x 10-2)(1.0 x 10-2)/(5 x 10-3)] = -3.0
Substituting each of these values for ln Q, -30.5 kJ/mol for ∆G’o, and 2.48 kJ/mol for RT into Eq. 13–4:
∆G1=-30.5 kJ/mol + (2.48 kJ/mol)(-7.8) = -50 kJ/mol
∆G2=-30.5 kJ/mol + (2.48 kJ/mol)(-4.7) = -42 kJ/mol
∆G3=-30.5 kJ/mol + (2.48 kJ/mol)(-2.8) = -38 kJ/mol
∆G4=-30.5 kJ/mol + (2.48 kJ/mol)(-1.0) = -33 kJ/mol
∆G5=-30.5 kJ/mol + (2.48 kJ/mol)(-3.0) = -38 kJ/mol
Plot of ∆G vs ln Q using these data is:
The ∆G for ATP hydrolysis is smaller when
[ATP]/[ADP] is low (<<1 or same as ln Q
being higher based on mass action ratio
definition) than when [ATP]/[ADP] is high
(same as when ln Q is lower).
So, the cells have the incentive to keep
higher reserves of ATP to be able to get
more ∆G from their hydrolysis.
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
12. (Chapter 14, Problem 4) When grown anaerobically on glucose, yeast (S. cerevisiae) converts
pyruvate to acetaldehyde, then reduces acetaldehyde to ethanol using electrons from NADH. Write
the equation for the second reaction, and calculate its equilibrium constant at 25 ⁰C, given the
standard reduction potentials in Table 13–7 (see last page of this study guide).
The second reaction is:
CH3CHO + NADH + H+ ⇌ CH3CH2OH + NAD+
Acetaldehyde
Ethanol
We can break it into its individual redox reactions:
CH3CHO + 2H+ 2e-  CH3CH2OH ;
NADH  NAD+ + H+ + 2e-;
E1’o=-0.197V
E2’o=+0.32V
(Notice the sign change for potential for the second reaction compared to the provided Table 13-7).
To calculate equilibrium constant at 25oC, K’eq, we could use ∆G’⁰ to connect K’eq with E’o (redox
potentials):
∆G’⁰ = -RT ln K’eq =-nF ∆E’o,
… (1)
where, ∆E’o is the sum of the reduction potentials of the two individual redox reactions written above.
∆E’o = E1’o + E2’o = -0.197 + 0.32 = 0.123 V
Eq. (1) above can be transformed into:
ln K’eq =-nF ∆E’o/RT
So, with n=2, F =96480 J/mol, RT = 8.315 J/(mol.K) x 298 K,
ln K’eq = 2 x 96480 J/(V.mol) x 0.123 V / (8.315 x 298) (J/mol) = 9.58
So,
K’eq = e9.58 = 1.45 x 104
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
13. (Chapter 14, Problem 6) A “pulse-chase“ experiment using 14C-labeled carbon sources is carried
out on a yeast extract maintained under strictly anaerobic conditions to produce ethanol. The
experiment consists of incubating a small amount of 14C-labeled substrate (the pulse) with the
yeast extract just long enough for each intermediate in the fermentation pathway to become
labeled. The label is then “chased” through the pathway by the addition of excess unlabeled
glucose. The chase effectively prevents any further entry of labeled glucose into the pathway.
a. If [1-14C]glucose (glucose labeled at C-1 with 14C) is used as a substrate, what is the
location of 14C in the product ethanol? Explain.
b. Where would 14C have to be located in the starting glucose to ensure that all the 14C
activity is liberated as 14CO2 during fermentation to ethanol? Explain.
(a) Figure 14–7 illustrates the fate of the carbon atoms of glucose.
C-1 (or C-6) of Glucose becomes C-3 of glyceraldehyde 3-phosphate and subsequently
pyruvate. When pyruvate is decarboxylated and reduced to ethanol, C-3 of pyruvate becomes
the C-2 of ethanol (14CH3—CH2—OH).
(b) If all the labeled carbon from glucose is converted to 14CO2 during ethanol fermentation,
the original label must have been on C-3 and/or C-4 of glucose, because these are converted
to the (C-1) carboxyl group of pyruvate and lost as CO2.
14. (Chapter 16, Problem 5) The nicotinamide coenzymes (see Fig. 13–24) can undergo reversible
oxidation-reduction reactions with specific substrates in the presence of the appropriate
dehydrogenase. In these reactions, NADH + H+ serves as the hydrogen source, as described in
Problem 3. Whenever the coenzyme is oxidized, a substrate must be simultaneously reduced:
Substrate + NADH + H+ ⇌ Product + NAD+
Oxidized
Reduced
Reduced
Oxidized
For each of the reactions in (a) through (f), determine whether the substrate has been oxidized
or reduced or is unchanged in oxidation state (see Problem 3). If a redox change has occurred,
balance the reaction with the necessary amount of NAD+, NADH, H+, and H2O. The objective is to
recognize when a redox coenzyme is necessary in a metabolic reaction.
The changes in oxidation state for going from the substrate to the product are shown on the
next page. A refresher of the oxidation states is also included at the end of this problem. If the
oxidation state goes up, it is oxidation. If the oxidation state goes down, it is reduction.
If the substrate is reduced to the product, then NADH is oxidized to NAD+. If the substrate is
oxidized to the product, then NAD+ is reduced to NADH. If no oxidation state change takes
place, then no NADH/NAD+ reaction needs to be coupled to the substrate -> product reaction.
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
(-1  +1)
Oxidation
(+3  +1)
Reduction
(+2+3  +1+4)
Unchanged
(+2+3  +3+4)
Oxidation
(+2  0)
Reduction
(-2+3  -3+4)
Unchanged
(a)
(b)
(c)
(d)
(e)
(f)
CH3CH2OH + NAD+  CH3CHO + NADH + H+
2O3POCH2-CHOH-CO2PO32- +NADH + H+  2-O3POCH2-CHOH-CHO + HPO42- + NAD+
CH3COCO2- + H+  CH3CHO + CO2
CH3COCO2- + NAD+ + H2O  CH3CO2- + CO2 + NADH + H+
O2C-CH2CO-CO2- + NADH + H+  -O2C-CH2CHOH-CO2- + NAD+
CH3CO-CH2CO2- + H+  CH3CO-CH3 + CO2
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
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Wednesday, 10 May 2017
CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
15. (Chapter 16, Problem 15) Cellular respiration can be studied in isolated mitochondria by
measuring oxygen consumption under different conditions. If 0.01 M sodium malonate is added
to actively respiring mitochondria that are using pyruvate as fuel source, respiration soon stops
and a metabolic intermediate accumulates.
A. What is the structure of this intermediate?
B. Explain why it accumulates.
C. Explain why oxygen consumption stops.
D. Aside from removal of the malonate, how can this inhibition of respiration be
overcome? Explain.
Malonate (-O2C-CH2-CO2-) is a structural analog of succinate (-O2C-CH2-CH2-CO2-) and hence a
competitive inhibitor of succinate dehydrogenase.
(a) Structure of accumulating intermediate Succinate is = -O2C-CH2-CH2-CO2(b) When succinate dehydrogenase is inhibited by Malonate, succinate is not consumed and hence
accumulates.
(c) Inhibition of any reaction in a pathway causes the substrate of that reaction to accumulate.
Because this substrate is also the product of the preceding reaction, its accumulation changes the
effective ∆G of that reaction, and so on for all the preceding steps in the pathway. The net rate of the
pathway (or cycle) slows and eventually becomes almost negligible. In the case of the citric acid cycle,
ceasing to produce the primary product, NADH, has the effect of stopping electron transfer and
consumption of oxygen, the final acceptor of electrons derived from NADH.
(d) Because malonate is a competitive inhibitor, the addition of large amounts of succinate will
overcome the inhibition. If enough succinate accumulates naturally, it could potentially outcompete
malonate, thereby restarting the citric acid cycle until succinate’s concentration drops enough that
malonate takes over. This cyclical behavior could potentially happen by itself if malonate is added.
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
16. (Chapter 19, Problem 3) All the dehydrogenases of glycolysis and the citric acid cycle use NAD+
(E’⁰ for NAD+/NADH is -0.32 V) as electron acceptor except succinate dehydrogenase, which uses
covalently bound FAD (E’⁰ for FAD/FADH2 in this enzyme is 0.050 V). Suggest why FAD is a more
appropriate electron acceptor than NAD+ in the dehydrogenation of succinate, based on the E’⁰
values of fumarate/succinate (E’⁰ = 0.031), NAD+/NADH, and the succinate dehydrogenase
catalyzed FAD/FADH2.
Let us first write out all the redox reactions involved:
NAD+ + H+ + 2e-  NADH ;
(Enzyme-bound FAD) E-FAD + 2H+ + 2e-  E-FADH2;
E’o=-0.32V
E’o = +0.050 V
This reduction potential is different from free FAD, whose E’o = -0.219 V
Fumarate2- + 2H+ + 2e-  Succinate2- ;
E’o = +0.031 V
We are interested in the reverse reaction:
Succinate2-  Fumarate2- + 2H+ + 2e- ;
E’o = -0.031 V
These released electrons can be taken up by NAD+ or E-FAD.
For coupling it with NAD+:
∆G’⁰ (NAD+) =-nF ∆E’o = -2 x 96.5 kJ/(V.mol) x (-0.320 V – 0.031 V) = 68 kJ/mol
For E-FAD:
∆G’⁰ (E-FAD) =-nF ∆E’o = -2 x 96.5 kJ/(V.mol) x (+0.050 V – 0.031 V) = -3.7 kJ/mol
Succinate dehydrogenation is thermodynamically favorable for E-FAD but highly unfavorable for
NAD+.
Page 10 of 13
CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
Wednesday, 10 May 2017
17. (Chapter 19, Problem 13) When O2 is added to an anaerobic suspension of cells consuming
glucose at a high rate, the rate of glucose consumption declines greatly as the O2 is used up, and
accumulation of lactate ceases. This effect, first observed by Louis Pasteur in the 1860s, is
characteristic of most cells capable of aerobic and anaerobic glucose catabolism.
a. Why does the accumulation of lactate cease after O2 is added?
b. Why does the presence of O2 decrease the rate of glucose consumption?
c. How does the onset of O2 consumption slow down the rate of glucose consumption?
Explain in terms of specific enzymes.
The addition of oxygen to an anaerobic suspension allows cells to convert from
fermentation to oxidative phosphorylation as a mechanism for reoxidizing NADH and
making ATP. Because ATP synthesis is much more efficient under aerobic conditions,
the amount of glucose needed will decrease (the Pasteur effect). This decreased
utilization of glucose in the presence of oxygen can be demonstrated in any tissue that
is capable of aerobic and anaerobic glycolysis.
(a) Oxygen allows the tissue to convert from lactic acid fermentation to respiratory
electron transfer and oxidative phosphorylation as the mechanism for NADH
oxidation.
(b) Aerobic respiration is more efficient than anaerobic respiration. Cells produce much
more ATP per glucose molecule that is oxidized aerobically, so less glucose is needed.
(c) As [ATP] rises, phosphofructokinase-1 is inhibited, thus slowing the rate of glucose
entry into the glycolytic pathway.
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CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
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Wednesday, 10 May 2017
CHEM 464 Spring 2017
Final Flipped Review
Abrol Section
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Wednesday, 10 May 2017