Download BS11 Final Exam Answer Key Spring `98

1
BS11 Final Exam Answer Key
Spring '98
Question 1. The diagram below depicts the structure of a polypeptide where carbon
atoms are shown as black circles; nitrogen, oxygen and sulfur atoms are shown as
white circles; and hydrogen atoms are not shown.
(See hard copy for figure.)
(3 pt) A) Name the structure shown in the figure.
Ans: antiparallel β sheet
.
(12 pt) B) Write out a possible amino acid sequence of this peptide (from N to C terminus).
Ans: N - Thr - Ile - His - Phe - Glu/Gln - Asp/Asn - Gly - Lys - Val - Val - Gly - Ser/Cys - C
(T - I - H - F - E/Q - D/N - G - K - V - V - G - S/C)
Question 2.
(5 pt) A) The carbonyl and amide groups of the protein backbone are hydrophilic and form
hydrogen bonds with water; they can also hydrogen bond to each other. The free
energy of formation of a hydrogen bond between the atoms of the peptide group in the
interior of a protein is:
==> a. more favorable than it is in water
b. less favorable than it is in water
c. unaffected by its location and the same as in water
Ans: a. the H-bonds are more favorable when isolated in a nonpolar environment due to an
entropic effect.
2
(5 pt) B) The pKa of the imidazolium group of a histidine residue at the NH2-terminus of an
α-helix, compared to that of one at the COOH-terminus, is likely to be:
a. increased
==> b. decreased
c. similar to each other, but not to the free amino acid
d. similar to the free amino acid
Ans: b. since the imidazolium group has a + charge, the + dipole at the N-terminus of a helix will
lower its pKa while the - dipole at the C-terminus will raise its pKa.
(5 pt) C) 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.
(5 pt) D) The following interaction is present in the T (deoxy) state of Hb:
40
Lys
+
_
C-terminus
+
_
Asp 94
146 His
98 Val
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 imidazolium 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
Question 3. Kinesin is a motor protein that "walks" on microtubules (MT) and is
believed to be responsible for movement of some membranous organelles in cells. It
utilizes the energy of hydrolysis of ATP to produce movement. The reaction can be
described as:
k1
kcat
Kinesin + ATP <----------> K.ATP --------> K.ADP.Pi
k -1
The ATPase rate of a specific kinesin (DKH392), in the presence of microtubules,
has been measured as kcat = 93 s-1 and Km = 43 µM.
(4 pt) A) Calculate Vmax when the total kinesin concentration is 50 µM.
Answer: Vmax = kcat [Etotal] = (93 s-1 )(50µM) = 4600µM s -1 or 4.6 x 10 -3 M s -1
(4 pt) B) Draw a V vs. S graph showing the relationship between the velocity of ATP
hydrolysis (v) and the concentration of ATP. Label the graph to show Vmax and Km.
Vmax
v
1/2 Vmax
Km
[ATP]
(3 pt) C) Define Km in terms of the kinetic constants of the reaction.
Answer: Km = (k-1 + kcat )/ k1
(4 pt) D) If kcat>>> k-1 , calculate the value of k1 . Explain in one sentence if diffusion is ratelimiting for this reaction.
Answer: k1 = kcat /Km = (93 s-1 )/(43µM) = 2 µM -1 s-1 or 2 x 10 6 M -1 s-1
No, the rate of a diffusion-limited reaction is 108 or 10 9 M -1 s-1
4
(3 pt) E) The ATPase rate of this kinesin has been measured in the absence of MTs with the
following results :
no MT
MT
1/v
-0.08
-0.02
1/[ATP], µM-1
Considering the data in the presence of MTs as baseline, does the value of each
kinetic term increase, decrease or remain the same when the reaction is in the absence
of MTs.
i) Km
ii) Vmax
iii) kcat/Km
Ans:
Km decreases (43µM to 12.5 µM)
Vmax decreases (higher y-intercept, lower Vmax)
kcat/Km stays the same (the lines are parallel, so the slopes are the same. Remember
that the slope is Vmax/Km and that Vmax = kcat. E tot)
Question 4.
(5 pt) A). The rate of hydrolysis of ATP by one myosin in contact with an actin filament
(measured by phosphate release) is limited by:
1).
the rate of binding of ATP
2).
the rate of the hydrolytic step
==> 3).
the rate of conformational change between the R and N states
4).
the rate of release of Pi
5).
the rate of release of ADP
6).
the rate of interaction between actin and myosin
Ans: 3. The rate-limiting step in the presence of actin is the conformational change between the R
and N states.
(5 pt) B). The rate of movement of an actin filament propelled by 100 myosin molecules in
the presence of ATP is limited by:
1).
the rate of binding of ATP
2).
the rate of the hydrolytic step
3).
the rate of conformational change between the R and N states
4).
the rate of release of Pi
==> 5).
the rate of release of ADP
6).
the rate of interaction between actin and myosin
Ans: 5. The rate-limiting step when many myosins are working together is the rate of ADP
release from the A. M. D complex. For the duration of time each myosin is tightly bound to
actin, the surrounding myosins are kept from being able to pull. As soon as the ADP
dissociates from the A. M. D complex, a new ATP can bind and the A. M complex will fall apart.
5
Question 5. The elongation rate constants for actin filaments in acrosomal bundles
determined by Bonder et al. are:
kon(ATP)
koff
barbed end (+)
pointed end (-)
(M -1 sec-1 )
12 x 106
1.2 x 106
(sec-1 )
2
0.6
An actin filament 0.1 µm in length is placed in a solution of 75 mM KCl, 5 mM Mg++
and 2 mM ATP; actin monomer with bound ATP is added at a concentration of 0.197
µM. The size of the filament is measured as a function of time.
(6 pt) A) Determine the value of the critical concentration for the barbed end and for the
pointed end.
Answer: Cc = koff /kon
Barbed end (+) :
Pointed end (-) :
2/12x106 = 0.167 µM
0.6/1.2x106 = 0.5 µM
(6 pt) B) Calculate what happens at the two ends of the filament under these conditions.
What happens to the size of the filament?
Answer: The size of the filament does not change. It grows at the (+) end at the
same rate that it shrinks at the (-) end. This is called treadmilling.
rate = k on [Actin] - koff
(+) end.
(12 µM -1sec-1)(0.197 µM) - 2sec -1 = +0.36 sec-1
(-) end.
(1.2µM -1sec-1 )(0.197 µM) - 0.6 sec -1 = -0.36 sec-1 Net = 0
(6 pt) C) Calculate what happens at the two ends of the filament if the concentration of actin
is increased to 0.3 µM. What happens to the size of the filament?
Answer: The filament grows in length because it grows at the (+) end faster than it
shrinks at the (-) end.
(+) end.
(12 µM -1sec-1)(0.3 µM) - 2sec -1 = 1.6 sec-1
(-) end.
(1.2µM -1sec-1 )(0.3 µM) - 0.6 sec -1 = -0.24 sec-1 Net = + 1.36 sec-1
6
(12 pt) Question 6. When a carbon-carbon bond is broken, one electron might go with each
carbon to form two radicals as follows:
C
C
However, this rarely happens in nature. In metabolism, we have seen an alternate
chemical mechanism for C-C bond cleavage which relies on the stabilization provided
by certain nearby groups.
First, draw fructose 1,6-bisphosphate. Now, draw a diagram showing how the
carbon atoms at the cleavage point can be stabilized during cleavage. (Do NOT include
the active site of an enzyme.)
Answer:
Fructose 1,6-bisphosphate
H 2 -C-OP
C=O
HO-C-H
H-C-OH
H 2 -C-OP
C=O
HO-C-H
H 2 -C-OP
C=O
HO-C-H
..
-
H 2 -C-OP
C-O HO-C-H
H-C-OH
H-C-O-H
+
H-C-O-
H-C=O
H 2 -C-OP
H-C-OH
H-C-OH
H-C-OH
H 2 -C-OP
H 2 -C-OP
H 2 -C-OP
7
Question 7. The substrate hydroxypyruvate, shown below, is metabolized to pyruvate
in a five-step process requiring four intermediates whose structures are labeled A
through D. The order of the intermediates does not necessarily reflect the true order
in which they are formed.
Hydroxypyruvate:
O
C-OC=O
H-C-OH
H
A.
B.
O
C.
O
O
D.
O
C-O-
C-OC-OP
C-OH-C-OH
C-OH-C-OP
H-C-OH
CH2
H-C-OH
H-C-OH
H-C-OP
H
H
H
Conversion of hydroxypyruvate to pyruvate requires NADH and catalytic amounts of
ATP and ADP.
(5 pt) A) What order are the intermediates formed during the metabolism of
hydroxypyruvate to pyruvate?
Hydroxypyruvate --> __B ___ --> __ D__ --> __ C__ --> __ A __ --> pyruvate
(6 pt) B) Write the overall balanced equation for the conversion of hydroxypyruvate to
pyruvate.
Ans: Hydroxypyruvate + NADH + H+ ---> Pyruvate + NAD+ + H 2O
(4 pt) C) Indicate which of the four intermediates are normal intermediates in the glycolytic
pathway.
Ans: A, C and D
(4 pt) D) Why are only catalytic rather than stochiometric amounts of ATP and ADP required?
Ans: One ATP is consumed to create compound D (3-phosphoglycerate), but one
ATP is formed upon the conversion of compound A (phosphoenolpyruvate) to
pyruvate.
8
(12 pt) Question 8. For a certain cell, the cytoplasmic Cl- concentration is 50 mM and there is
500 mM Cl- in the extracellular solution. The membrane potential is -60 mV (negative
inside). The membrane of this cell has a chloride channel that is normally closed but
that, when open, allows only Cl- ions to move.
Assume that the Cl- channel flickers open and that during this time neither the
concentration gradient of Cl- nor the membrane potential change. Indicate in which
direction Cl- will move and calculate the free energy change for movement of Cl- in
this situation.
Ans: Cl- is at electrochemical equilibrium and there is no net movement of the ion.
∆G = RT ln ([Cl- in]/[Cl- out]) + zF (ψi - ψo)
= (600 cal/mol) ln (54/540) + (-1) (23 kcal V-1 mol -1 ) (-0.06 - 0)V
= -1.38 kcal/mol + 1.38 kcal/mol
= 0 kcal/mol
∆G = 0, so system is at Cl- equilibrium.
(12 pt) Question 9. The rates of uptake of low density lipoprotein (LDL) and horse radish
peroxidase (HRP) by human fibroblasts are shown in the Figure.
(A) HRP Uptake
(B) LDL Uptake
(See hard copy for figures.)
Explain the difference in the shapes of the uptake curves.
Ans: The uptake of HRP does not saturate, and a 100-fold higher concentration of
HRP compared to LDL is required to get equivalent rates of uptake. This indicates
that HRP does not bind to a specific cellular receptor and is being taken up only by
fluid-phase endocytosis. Since endocytosis is a continuous (constitutive) process,
HRP gets taken up steadily and does not saturate. By contrast, LDL uptake occurs by
a saturable process at much lower concentrations than is seen for HRP. This
indicates that LDL binds to a specific LDL receptor and is internalized by receptormediated endocytosis. The limit to the rate at which LDL gets taken up is set by the
number of receptors on the cells. When the receptors are saturated with LDL, no
further increase in the rate of uptake occurs (except at enormously high
concentrations where fluid-phase endocytosis would become significant.)
9
Question 10. Patients with Hunter's syndrome or Hurler's syndrome rarely live
beyond their teens because they accumulate undigested glycosaminoglycans in their
lysosomes. Patients with Hunter's syndrome have a defective gene for, and are hence
unable to make, a certain lysosomal enzyme, enzyme A, which is necessary for the
degradation of glycosaminoglycans. Patients with Hurler's syndrome have a defective
gene for, and are hence unable to make, a different lysosomal enzyme, enzyme B,
which is also necessary for the degradation of glycosaminoglycans. When cells from a
patient with Hurler's syndrome are fused with cells from a patient with Hunter's
syndrome, glycosaminoglycans are degraded properly in the fused cells.
Surprisingly, even if cells from the two patients are just cultured together, they still
correct each other's defects. Most surprising of all, when the overlaying medium from
a culture of Hurler's cells is placed over a culture of Hunter's cells, it will correct the
defects in Hunter's cells (and vice versa). The corrective factors in the media are
inactivated by treatment with proteases, by treatment with periodate (which destroys
carbohydrate), or by treatment with alkaline phosphatase (which removes phosphates).
(5 pt) A) What do you think the corrective factors are? Ans: Lysosomal enzymes A and B themselves.
(5 pt) B) Why are these corrective factors in the cell culture medium?
Ans: The process of protein sorting is not perfect. Instead of always being sent to the lysosome,
sometimes the enzymes are exported from the cell by constitutive secretion.
(6 pt) C) Do you suppose the corrective factor in the medium from Hurler's syndrome cells
is taken up by the Hunter's syndrome cells? If yes, how? If no, how does the
corrective factor work? Include in your answer why the above treatments inactivate
the corrective factors.
Ans: Yes. The periodate and phosphatase treatments inactivate the corrective factors, so the
secreted lysosomal proteins only correct the other cell's defects when they possess their original
sugar and phosphate configurations. This must be the M-6-P tag that normally directs a protein to
the lysosme. Some of the M-6-P receptors must also be missorted to the plasma membrane where
they can bind to the secreted enzymes A or B and cause their endocytosis. Then the endocytic
pathway carries them ultimately to the lysosome.
10
Question 11. The transition from metaphase to anaphase can be studied using a cell-free
extract from unfertilized frog eggs. When nuclei are added to the extract, they
spontaneously form spindles with condensed chromosomes aligned on the metaphase
plate. Anaphase and the separation of sister chromatids can be triggered by the addition
of Ca2+, which, in this case, activates APC.
To determine the relationship between MPF inactivation and sister chromatid
separation, you make use of two mutant forms of cyclin B. Cyclin B∆90 is missing the
destruction box, a sequence of amino acids required for inactivation by APC, but cyclin
B∆90 retains its ability to bind to Cdc2 kinase and make functional MPF. Cyclin B13-110
retains the destruction box but cannot bind to Cdc2 kinase. When either cyclin B∆90 or
cyclin B13-110 is added in excess to the extract, MPF kinase activity remains high after
the activation of APC. The two proteins differ, however, in their effects on chromatid
separation. In the presence of cyclin B∆90, sister chromatids separate normally; in the
presence of cyclin B13-110, sister chromatids remain linked.
Wild-type cyclin B
destruction
box
Cdc2 binding region
Cyclin B ∆ 90
Cyclin B 13-110
(6 pt) A) Why does MPF remain active, after the activation of APC, when cyclin B∆90 is added
to the extract?
Ans: MPF activity remains high because cyclin B∆90 can activate Cdc2 kinase and cyclin B∆90 won't
be destroyed by the APC pathway since it lacks the destruction box.
(6 pt) B) Why does MPF remain active, after the activation of APC, when cyclin B13-110 is
added to the extract?
Ans: MPF activity remains high because an excess of cyclin B13-110 must be overwhelming the APC
pathway so that the wild-type cyclin B does not get destroyed.
(6 pt) C) How is the separation of sister chromatids related to MPF inactivation? Explain.
Ans: MPF inactivation is not necessary for for sister chromatid separation since cyclin B∆90 prevents
MPF inactivation but not chromatid separation. Since cyclin B13-110 prevents both MPF inactivation
and chomatid separation, the APC proteolysis pathway must be responsible for both. The proteins in
the kinetochore which hold the chromatids together are destroyed in an APC-dependent manner.
11
Question 12. Mevalonate is synthesized from β-OH-methylglutaryl-CoA (HMGCoA) in a reaction catalyzed by HMG-CoA reductase:
HO CH3
O
NADP+
+ NADPH + H+
S CoA
Intermediate + NADPH + H+
NADP+
HO CH3 H H
OH
CoASH
COO-
COO-
(4 pt) A). Write the structure of the "Intermediate" in the above reaction.
Ans:
HO CH3
O
H
COO(4 pt) B). Incubation of the intermediate with the enzyme in the presence of 4-[3H]NADPH
results in the formation of the following compound.
Is the 3H in the proR or proS position?
proR .
HO CH3 H 3H
OH
COO(7 pt) C). Determine the standard free energy change for the overall reaction:
HMG-SCoA + 2 NADPH +2H+ ---------> Mevalonate + CoASH + 2NADP+
The standard redox potentials for the half-reactions are:
Eo' (volt)
HMG-SCoA + 2H+ + 2e - -------> [Intermediate] + CoASH -0.41
[Intermediate] + 2H+ + 2e - -------> Mevalonate
-0.16
NADP+ + H + + 2e - --------> NADPH
-0.32
Ans: ∆G°’ = -nF∆Eo’ = -2(23kcal/V.mol)[-.41+.32]V + (-2)(23kcal/V.mol)[-.16+.32]V
= 4.14 kcal/mol - 7.36 kcal/mol = -3.2 kcal/mol
OR
= -2(23kcal/V .mol)[-0.41 - 0.16 + 2(0.32)]V
= -2(23kcal/V.mol)(-0.07V) = -3.2 kcal/mol
12
Question 13. A novel protein, Protein X, that binds to the Gαi subunit of heterotrimeric
G-proteins has recently been isolated. In an attempt to find the function of Protein X,
the following experiment is performed:
1) Purified Gαi is incubated with excess γ-32P-GTP under conditions which promote
nucleotide exchange.
2) 32P release from Gαi is then measured over time in the presence and absence of
Protein X.
The following results were seen:
Gαi + Protein X
% Pi
released
Gαi only
time
(5 pt) A) What is the most likely function of Protein X?
Ans: It activates the GTPase activity of Gα i. (It is a GAP protein.)
(5 pt) B) What effect would you predict Protein X would have on cell signaling through
dopamine 2 receptors (a Gαi -coupled receptor)?
Ans: Protein X would decrease the signal since Gα i would stay in its active GTP-bound state for a
shorter period of time.
(5 pt) C) What is the normal effect of Gαi on adenylate cyclase? How will the presence of
Protein X affect [cAMP] when an agonist binds to the dopamine 2 receptor?
Ans: Gα i normally inhibits adenylate cyclase, so Protein X should cause cAMP levels to fall less
sharply.