Download BS 11 First Mid-Term Answer Key Spring 1998

1
BS 11
First Mid-Term Answer Key
Spring 1998
Question 1. (48 pts) Note: These questions may be identical to ones that you have seen
before, but they may have been changed. READ the questions carefully!
(2 pt) A) An H-bond between two H2O molecules in isolation is about 10 times stronger
energetically (∆H) than a van der Waals contact between two xenon atoms.
H
O
Xe
H
Xe
O
H
H
The number of H2O dimers and xenon dimers in the gas phase are:
=> a. about the same
b. 10x more H2O dimers
c. 10x more Xe dimers
Ans: a The numbers are about the same due to the entropic constraint that an H2O
dimer can only form in a restricted set of orientations whereas a Xe dimer can form
with any possible orientation.
(2 pt) B) Ala-(Ala)8-Ala, (a decapeptide of L-Ala), is soluble in H2O and in octanol.
a. The structure of the peptide is the same in both solvents.
=> b. The peptide is an α-helix in octanol and a random coil in H2O.
c. The peptide is a random coil in octanol and forms β-structure in H2O.
Ans: b. The peptide bond carbonyls and amides will H-bond more favorably with
one another in a non-polar environment, and hence will form an α -helix in octanol.
In water, these groups can form H-bonds with water molecules and hence the
decapeptide will be a random coil.
(2 pt) C) The following peptide was isolated from a wasp.
KELVISISHERETS
At pH 7, it has the following net charge:
a. positive
=>
b. negative
c. zero
Ans: b. There will be a positive charge on the terminal amino group and a negative
charge on the terminal carboxyl. There are three negative E (Glu) and one positve K
(Lys), and R (Arg). Since the pK of H (His) is around 6.5 to 7, it will have 0.25 to 0.5
positive charge. Thus, the peptide will have a net negative charge.
2
(2 pt) D) The hemoglobin tetramer dissociates into (αβ) dimers more readily in the oxygenated
(R) state than in the deoxygenated (T) state. Therefore, dilution of a hemoglobin solution
has the following effect on O2 binding:
=> a. increases O2 affinity
b. no effect
c. decreases O2 affinity
Ans: a. By decreasing the concentration, you increase the amount dissociated and
hence increase the affinity for O2.
(2 pt) E) Binding protons stabilizes the T (deoxy) state of Hb. Therefore, increasing the pH of the
solution has the following effect on O2 binding:
=> a. increases O2 affinity
b. no effect
c. decreases O2 affinity
Ans: a. By increasing the pH, you decrease the proton concentration and hence
destabilize the T state, and so this increases the affinity for O2.
(2 pt) F) The following interaction is present in the T (deoxy) state of Hb:
40
Lys
+
_
C-terminus
+
_
146 His
98 Val
Asp 94
Tyr 145
-OH---O=C
beta 2
beta 2
beta 2
His 146 does not interact with Asp 94 in the R (oxy) state. The pK of the imidizolium of
His 146 is:
a. the same in the T and R states
=> b. pKT > pKR
c. pKR > pKT
Ans: b. The negative charge on Asp 94 serves to stabilize the protonated state of His
146; this raises the pK of the imidizolium of His 146 above what would be seen in the
absence of this interaction.
3
(6 pt) G) For enzyme A, kcat/KM = 105 M -1s-1 and [S] = 10-2 M. What are the smallest and largest
possible values of v/[Etot]?
Ans: v/[Etot] = kcat [S]/(Km + [S]) When Km >> [S], v/[Etot] = (105 M -1s-1)( 10-2 M)
= 103 s-1 which is the upper limit. There is no definable lower limit except zero; if
Km = 10-5 M, then v/[Etot] = 1 s -1, but if Km = 10-7 M, then v/[Etot] = 10 -2 s-1, etc.
(2 pt) H) Enzyme B catalyzes the reaction E + S <===> ES ----> E + P. For one substrate S1, the
KS is 10 -5 M, while for S2, KS is 10 -7 M. Which of these statements is true?
a) S1 binds more tightly than S2
=> b) S2 binds more tightly than S1
c) S1 and S2 bind the same
d) cannot be determined
Ans: b. Ks is the dissociation constant for the formation of the ES complex. The
substrate with the smaller Ks binds more tightly.
(2 pt) I) When substance Y is added to an enzyme-catalyzed reaction, v increases:
no Y
1
V
[Y]
1/[S]
Which is true?
a) Y is a noncompetitive inhibitor
b) Y binds only to E
=> c) Y binds to E and ES
d) Y binds only to ES
Ans: c. The graph is similar to that for a noncompetitive inhibitor, but Y activates
rather than inhibits.
k1
(4 pt) J) In an uncatalyzed reaction of the type : A <====> B , the ratio of
k-1
k1/k-1 is 10. When the same reaction is catalyzed by an enzyme, the ratio of kcatf/kcatr is
100. Indicate which result is likely to be incorrect, or whether both can be correct. Explain
your answer with equations.
Ans: Both can be correct. The equilibrium constant is the same for both the
catalyzed and the uncatalyzed reactions. But for the catalyzed reaction, Keq is related to
the kcat's and the Km's as follows: Keq = [B]/[A] = k1/k-1 = (kcat/Km)f/(kcat/Km)r
So, for this enzyme, Kmf/Km r = 10.
4
(2 pt) K) Chymotrypsin is inactivated by diisopropyl fluorophosphate (DIFP). A new protease is
isolated that is also inactivated by the reaction of one mole of DIFP per mole of enzyme.
The peptide fragment that contains the phosphorylated residue is:
WGLEQPAMQNF
Circle the residue(s) which is likely to be phosphorylated.
Ans: The E. Of these amino acids, only the Glu is a good nucleophile.
(2 pt) L) Mutation of S195, H57 and D102 of chymotrypsin to alanine residues yields an
enzyme that enhances proteolysis rates by ~5x104 over the uncatalyzed reaction. The
native enzyme has a rate enhancement factor of ~1010. The explanation for the rate
enhancement by the mutated enzyme is:
a) The mutated residues are not the only ones involved in the acid-base catalysis.
b) The remaining activity is due to non-specific effects of the protein on proteolysis;
all proteins have a similar effect.
=> c) Part of the rate enhancement of the enzyme is attributable to its preferential
binding of the transition state of the substrate.
Ans: c. One of the main ways that enzymes achieve rate enhancement is by
preferential binding of the transition state of the substrate.
(2 pt) M) The values of ∆G°' for the hydrolysis of creatine phosphate and ATP are -10.3 and 7.3 kcal/mol, respectively. What are the expected relative rates of hydrolysis of these
compounds?
Ans: No way to know. Values of ∆G°' are not related to the rates of hydrolysis.
(6 pt) N) Creatine kinase catalyzes the reaction:
ATP + creatine <===> ADP + creatine phosphate
Using the ∆G°' values given in part (N), what is the ratio at equilibrium of ATP/ADP
when an equal number of moles of ATP, ADP, creatine and creatine phosphate are
present at the beginning of the reaction. (RT = 600 cal/mol)
Ans: For this reaction, ∆G°’ = (-7.3) - (-10.3) = 3 kcal/mol. Since ∆G°’ = RT lnKd, then
Kd = 1.5 x 102 = [ATP] [Creatine]/[ADP] [Creatine-P]
We started out with equal numbers of moles, so at equilibrium, [ATP]/[ADP] =
[Creatine]/[Creatine-P]. Therefore, the ratio of [ATP]/[ADP] at equilibrium is the square
root of the Kd, or 12.
5
(2 pt) O) Phosphoarginine is found in the muscles of some invertebrates. Write the structure of
the compound, and indicate whether it is likely to have a high or low phosphoryl group
transfer potential.
Ans: High.
H
H
H
O-
+NH - C - (CH ) - N - C - N - P = O
3
2 3
+NH
COO
O2
(6 pt) P) A substance, I, is a noncompetitive inhibitor of an enzyme. How would one determine
the value of Ki? Show what data are required, how they would be used, and how one
would obtain the value of Ki.
Ans: Measure v as a function of [S], plus and minus I. Make a Lineweaver-Burke
plot to determine Vmax and Vmaxapp as shown by plotting 1/v vs. 1/[S]. Use the
equation Vmax app = Vmax/(1 + ([I]/Ki)) and solve for Ki.
1
+I
Vmax app
no I
1
V
-1
KM
1
Vmax
1
[S]
(2 pt) Q) Under what conditions is Km equal to the dissociation constant of the enzymesubstrate complex.
Ans: Km = Ks = k-1 /k1 when k-1 >>k2
6
Question 2. (18 pts)
(12 pt) A) The enzyme triose phosphate isomerase (TIM) catalyzes the interconversion of
dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (GAP) by way of a
cis-enediol intermediate. Make a sketch (one diagram) of the active site of the enzyme
containing the transition state between the cis-enediol intermediate and DHAP. Be sure
to include the important amino acid side chains and structural elements and show how
the transition state would collapse to form DHAP.
Ans:
Glu
(proR H)
C
O
O
H
H
C
C
O
HCH
O
O-
H-N
H3N+
P
O
OH
O-
His
N:
+ end of
α-helix
Lys
+ end of
α-helix
(6 pt) B) The pH dependence of the enzymatic activity of triose phosphate isomerase indicates
that two ionizable groups are involved in the reaction: a base with pKa 7 and an acid
with pKa 9.5. At what pH will the enzyme have optimum activity? 8.25
Briefly
explain why this is the optimum pH.
Ans: The optimum pH is the pH at which the amounts of unprotonated base and
protonated acid are at a maximum. This happens only at pH = (pKA + pK B )/2 and here
the amounts of unprotonated base and protonated acid are equal to each other. This can
also be worked out quantitaively: log (B/A)B = pH - pKB and log (A/B)A = pKA - pH.
At pH = (pK A + pK B )/2 , log (B/A)B = log (A/B)A.
At pH < or > (pKA + pK B )/2 , log (B/A)B < or > log (A/B)A and either (B)B or (A)A
are less then their maximum.
7
Question 3. (18 pts) The Pompeii worm lives on the sides of underwater volcanoes at
temperatures of up to 81° C. The blood of these worms contains a hemoglobin of 3780
kDa. Each molecule of this hemoglobin is comprised of 120 globin chains (each chain of
16.5 kDa and with one heme group) surrounding a core of 72 linker chains (each of 25
kDa).
The P50 of the Pompeii worm Hb (Hbp) is 0.4 torr and the Hill coefficient, n, is 2.8.
The P50 for human Hb (Hbh ) is 26 torr and the P50 of human myoglobin (Mbh ) is 4 torr.
(12 pt) A) Draw on a single figure the relative Hill plots for Hbp, Hbh and Mb h . Be sure to label
each curve and the axes of your graph.
Ans:
Hbp
log (Y/1-Y)
Hbh
n=2.8
Mbh
n=1
0
log(0.4)
log(4)
log(26)
log(pO2)
(6 pt) B) How many molecule(s) of oxygen per molecule of Hbh protein are released when the
O2 concentration is changed from 100 torr to 26 torr? 2
.
How many molecule(s) of oxygen per molecule of Mbh protein are released when the O2
concentration is changed from 100 torr to 4 torr?
0. 5
.
How many molecule(s) of oxygen per molecule of Hbp protein are released when the O2
concentration is changed from 100 torr to 0.4 torr?
60
.
8
Question 4. (16 pts) Order the following states or processes that occur during muscle
contraction according to what you have learned in class. Once you have determined the
correct order, fill in the cycle below in a clockwise direction. For uniformity, we have
positioned item (a) for you. (Note: Regrading will not be an option for this question.)
a. The S1 heads of myosin are largely dissociated from the thin filament and
contain bound ADP and Pi.
b. Pi is released and myosin undergoes a conformational change.
c. Myosin dissociates from actin.
d. The thick filament moves with respect to the thin filament.
e. ADP dissociates from myosin.
f. Myosin hydrolyzes ATP to ADP and Pi and undergoes a conformational change.
g. ATP binds to myosin.
h. Myosin-ADP-Pi binds tightly to actin forming a high affinity complex.
a
h
f
c
b
d
g
e
+
-
power stroke -
ADP
(d) conf. change
fast
+
(b)
ADP (rigor mortis)
D
Pi
ADP
DPiN
(g)
(c)
(h)
(f)
slow
rate-limiting
conf. change to
cock myosin head
fast
ATP
fast
productive
interaction
with actin
(e)
-
(a)
DPiR
non-productive
interaction
with actin
DPiR
ATP
fast