Name________________________ Student number______________________ ____________
UNIVERSITY OF GUELPH CHEM 4540 ENZYMOLOGY
Winter 2005 Quiz #1: February 17, 2005, 11:30 – 13:00
Instructor: Prof R. Merrill
ANSWERS
Instructions: Time allowed = 80 minutes. Total marks = 32. This quiz represents 15% of
the final grade. Please write all your answers in ink (not red ink). No examination
materials may be removed from the examination room.
Answers to parts A and B are to be written directly on this question booklet. Answers to part C
are to be written in the answer booklet provided.
Part A. “Multiple-choice” questions; circle the letter corresponding to the best answer.
Ten (10) questions x 1 mark per question = 10 marks total.
No marks will be deducted for incorrect answers.
1. One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under
conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as
the:
A) holoenzyme.
B) prosthetic group.
•C) apoenzyme.
D) coenzyme.
E) substrate.
2. Enzymes differ from other catalysts in that enzymes:
A) lower the activation energy of the reaction catalyzed.
B) fail to influence the equilibrium point of the reaction.
C) form an activated complex with the reactants.
•D) usually display specificity toward a single reactant.
E) are not consumed in the reaction.
3 Briggs and Haldane assumed that the overall scheme for an enzyme-catalyzed reaction
could be written as
E+S
k1
⇔
k−1
k2
ES → P
Using this reaction, the rate of formation of the enzyme-substrate complex can be
described by the expression:
•A) k1([Et] − [ES])[S].
B) k−1[ES] + k2[ES].
C) k2[ES].
D) k−1[ES].
E) k1([Et] − [ES]).
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Name________________________ Student number______________________ ____________
4 To determine Km from the Hanes-Woolf transformation of the Michaelis-Menten
equation you would:
A) take the x-axis intercept where V0 = 1/2 Vmax.
B) take the reciprocal of the x-axis intercept.
•C) multiply the x-axis intercept by −1.
D) take the reciprocal of the y-axis intercept.
E) multiply the y-axis intercept by −1.
5 In the coupled enzyme system shown below, in order to measure the kinetics of E1 the
following conditions should be established:
E1
A→ B
Reaction 1
E2
B→C
Reaction 2
A) B and C must be in excess
B) A must be in excess
•C) E2 must be in excess
D) A and B must be in excess
E) none of the above
6 The term, molecular activity, refers to:
A) the number of moles of substrate transformed into product per unit time per mole of
enzyme active site
B) the maximum velocity of a reaction over the number of moles of substrate
•C) the number of moles of substrate transformed into product per unit time per mole of
enzyme
D) the initial velocity of the catalyzed reaction at low [S].
E) none of the above.
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A continuous indirect assay involves:
A) stopping the reaction followed by the addition of a detection reagent
B) monitoring the reaction progress in a spectrophotometer by following a change in the
absorbance of the substrate upon conversion to product
C) monitoring the reaction using a continuous flow system
D) using an immobilized enzyme attached to a matrix to monitor the reaction progress
•E) none of the above
8 In the simple pH dependence model for Michaelis-Menten enzymes as described in the
lecture notes, vo = Vmax'[S]/(KM' + [S]) then Vmax' = Vmax/f2 where f2 =
A) [H+]/KE1 + 1 + KE2/[H+]
•B) [H+]/KES1 + 1 + KES2/[H+]
C) [H+]/KES1 + KES2/[H+]
D) [H+]/KE1 + KE2/[H+]
E) none of the above.
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Name________________________ Student number______________________ ____________
9.
The specificity constant, kcat/KM is a useful parameter for comparing the catalytic
efficiencies of enzymes with substrates because:
•A) it is a measure of how rapidly an enzyme can work at low [S]
B) it is a first order rate constant that relates [S] to reaction velocity
C) it is a measure of how rapidly an enzyme can work at high [S]
D) it indicates how rapidly an enzyme collides with its substrate
E) none of the above
10.
A noncompetitive inhibitor is:
•A) a substance that binds either to the free enzyme or to the E-S complex
B) a substance that binds to the E-S complex only
C) a substance that binds to the free enzyme only
D) a substance that causes a change in Vmax and KM for the reaction
E) none of the above
Part B. “Short answer/calculation” questions. Answer the following (five) questions with a
short answer consisting of a few sentences or less (your answer can be in point
form). Each question is worth 2 marks, except Question 11, which is worth 4 marks.
11. Which class (name) of enzyme of the six main classes catalyzes the following reactions
(0.5 marks each)? Give the specific name of the enzyme (0.5 marks each). Total 1 mark
for each of a) through d).
a) isomerase (Alanine racemase)
b) lyase (pyruvate decarboxylase)
c) oxidoreductase or dehydrogenase (lactate dehydrogenase)
d) ligase (glutamine synthetase)
12.
For the reaction E + S ⇔ ES → P the Michaelis-Menten constant, Km, is actually a
summary of three terms. What are they? How is Km determined graphically?
Ans:
KM = (k2 + k–1)/ k1, where k–1and k1 are the rate constants for the breakdown and
association, respectively, of the ES complex and k2 is the rate constant for the breakdown
of ES to form E + P. KM can be determined graphically on a plot of vo vs. [S] by finding
the [S] at which vo = 1/2 Vmax. More conveniently, on a double-reciprocal plot, the xaxis intercept = –1/ KM.
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Name________________________ Student number______________________ ____________
13. For a single-substrate enzyme-catalyzed reaction, double reciprocal plots were
determined for three different enzyme concentrations. Which of the following three
families of curves (curves 1, 2 or 3) would you expect to be obtained for an enzyme that
obeys Michaelis-Menten kinetics? Explain.
1)
2)
3)
Plot #2 would be expected for an enzyme that obeys Michaelis-Menten kinetics for three
different enzyme concentrations. If [E]T is increased, then Vmax will also increase because
Vmax = k2[E]T. But, KM = (k-1 + k2)/k1, i.e., KM is independent of [E]T. Plot 1 shows that both
KM and Vmax change and plot 3 shows that Vmax doesn't change while KM does. Plot 2 is
correct because it shows that KM doesn't change while Vmax does.
14. A simple Michaelis-Menten enzyme, in the absence of any inhibitor, displayed the
following kinetic behaviour. The expected value of Vmax is shown on the y-axis.
(a) Sketch a double-reciprocal plot that corresponds to the velocity-versus-substrate curve.
(b) Provide an explanation for the kinetic results.
This behaviour is substrate inhibition. At high [S], the substrate forms unproductive
complexes at the active site. The example given in class was succinate dehydrogenase
where two molecules of succinate may simultaneously bind to the active site in an
improper orientation. Recall the sheep crowding through the gate analogy.
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Name________________________ Student number______________________ ____________
15. Answer each of the following points (a) and (b), worth 1 mark each.
(a) Draw the structure of the cofactor lipoic acid
(b) Define the transition state stabilization theory used to describe enzyme-substrate
interactions.
An enzyme increases the rate of the reaction by stabilizing the transition state. This occurs by
intimate interactions between the residues within the active site of the enzyme and the atoms of
the substrate. The greater the degree of transition-state stabilization, the faster the reaction will
proceed relative to the uncatalyzed reaction.
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Name________________________ Student number______________________ ____________
Part C. “Problems-based” questions. Answer the following three (3) questions in the
examination booklet provided. Please write your answers in pen (not red ink).
Each question is worth 3, 3, and 4 marks, respectively. Three (3) questions (3, 3,
and 4 marks) = 10 marks total.
16.
The hydrolysis of pyrophosphate to orthophosphate is important in driving forward
biosynthetic reactions such as the synthesis of DNA. This hydrolytic reaction is catalyzed in
E. coli by a pyrophosphatase that has a mass of 120 kDa and consists of six identical
subunits. For this enzyme, a unit of activity is defined as the amount of enzyme that
hydrolyzes 10 µmol of pyrophosphate in 15 minutes at 37°C under standard assay
conditions. The purified enzyme has a Vmax of 2800 units per milligram of enzyme.
(a) How many moles of substrate are hydrolyzed per second per milligram of enzyme when
the substrate concentration is much greater than KM?
Given definition: 1 U = 10 µmol PPi/15 min (for this enzyme)
Given: Vmax = 2800 U/mg enzyme
10umolPPi
2800U
1 min
x 15 min X
= 31.1 µmol PPi s-1 mg-1
mg
1U
60s
(b) How many moles of active site are there in 1 mg of enzyme?
1 mg x
1.mmol
6.mmol .active.sites
x
= 5.0 x 10-5 mmol = 5.0 x 10-8 mol
120,000.mg
1.mmol .enz
(c) What is the turnover number of the enzyme?
120,000.mg 1.mmol
) = 3732 s-1 per mol
x 3
1.mmol
10 .umol
enzyme but there are 6 active sites per enzyme
1.mol.enz
∴ 3732 s-1 mol enz-1 x
= 622 s-1 (per active site).
6.mol.active.sites
T.N. = Vmax/[E]T = 31.1 µmol PPi s-1 mg-1 x (
17.
Human carbonic anhydrase was tested at a bicarbonate concentration of 3 x 10-7 M, the KM
for the substrate is 2 x 10-5 M. After 4 minutes, 13% of the substrate was converted to
product. (a) What percent of the substrate will be CO2 product after 8 minutes? (b) What is
the Vmax given these conditions?
[S]0 = 3 x 10-7M; KM = 2 x 10-5M; After 4 min 13% of substrate has been converted to
product.
(a) [S]0 = 3 x 10-7M << KM (2 x 10-5) therefore first order and v0 ∝ [S]0.
2.303 log([S]0/[S]t) = kt
Calculate k since after 4 min 87% of the original substrate remains.
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Name________________________ Student number______________________ ____________
2.303 log (100/87) = k(4 min)
k = 3.48 x 10-2 min-1
Calculate [S]t at t = 8 min
2.303 log (100/[S]t) = 3.48 x 10-2 min-1 (8 min)
2.303 log (100 - log[S]t) = 0.279
4.606 - 2.303 log[S]t = 0.279
-2.303 log[S]t = -4.327
log[S]t = 1.879
[S]t = 75.7% of original [S], thus after 8 min 24.3% of substrate has been converted to
product.
(b) Vmax can be estimated since we know KM:
At [S]0 = 3 x 10-7M the velocity for the first min is: v0 = k[S]
v0 = (3.48 x 10-2 min-1)(3 x 10-7M) = 1.04 x 10-8M min-1
v0/Vmax = [S]0/(KM + [S]0)
1.04 x 10-8M min-1/Vmax = (3 x 10-7M)/(2 x 10-5M + 3 x 10-7M)
Vmax = [(1.04 x 10-8M-1) (2 x 10-5M + 3 x 10-7 M)]/(3 x 10-7M) = 7.03 x 10-7 M min-1.
18.
Malonate, a competitive inhibitor of succinate dehydrogenase, was found to cause 95%
inhibition of the enzyme’s activity. (a) If the succinate (substrate) concentration for the
enzyme was 3.5 x 10-5 M and the KM for this succinate is 4.4 x 10-6 M, what was the initial
malonate concentration (Ki = 2.4 x 10-7 M)? (b) What concentration must succinate be
raised to in order to restore the velocity of the enzyme to its uninhibited value?
(a) Work the problem in terms of vo and Vmax
vo = Vmax[S]/KM + [S] = Vmax(3.5 x 10-5M)/(4.4 x 10-6M + 3.5 x 10-5M) = Vmax(3.5 x 105
M)/3.94 x 10-5M = 0.89Vmax
∴ vo = 0.89Vmax and substitute for vo where at 95% inhibition: vi = 0.05 vo ∴ vi =
0.05(0.89 Vmax) = 0.045 Vmax
[S ]
3.5. x.10−5 M
=
vi/Vmax =
[I ]
[I ]
K M (1 + ) + .[S ] [(4.4. x.10−6 M )(1 +
) + 3.5. x.10−5 M ]
−7
Ki
2.4. x.10 M
0.045 =
3.5. x.10 −5 M
3.5. x.10 −5 M
=
4.4. x.10 −6 M * [ I ]
18.33[ I ] + 3.94. x.10 −5 M
−5
[( 4.4. x.10 −6 M ) + (
)
3
.
5
.
x
.
10
M
]
+
2.4. x.10 −7 M
0.825[I] + 1.77 x 10-6M = 3.5 x 10-5M
0.825[I] = 3.32 x 10-5M
∴ [I] = 4.02 x 10-5M
(b) To overcome the inhibition then vi = vo ∴ vi = 0.89 Vmax and vi/Vmax = 0.89
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Name________________________ Student number______________________ ____________
vi/Vmax =
[S ]
= 0.89
4.02. x.10 −5 M
−6
4.4. x.10 M (1 +
) + [S ]
2.4. x.10 −7 M
[S] = 0.89 (4.4 x 10-6M (
1 + 4.02. x.10 −5 M
) + [S])
2.4. x.10 −7 M
[S] = 0.89 (4.4 x 10-6M + 7.37 x 10-4M + [S])
[S] = 4.22 x 10-6M + 6.56 x 10-4M + 0.89[S]
0.11[S] = 6.60 x 10-4M
[S] = 6.0 x 10-3 M
Have a nice Break Week!
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