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Reg. No…………………
KIGALI INSTITUTE OF SCIENCE AND TECHNOLOGY
INSTITUT DES SCIENCES ET TECHNOLOGIE
Avenue de l'Armée, B.P. 3900 Kigali, Rwanda
INSTITUTE EXAMINATIONS – ACADEMIC YEAR 2012-1213
END OF SEMESTER MAIN EXAMINATION
FACULTY OF SCIENCE
SCIENCE-4-CHEMISTRY
THIRD YEAR, SEMESTER II
CHE 3324 PROTEINS & ENZYMES I/ ENZYMOLOGY
DATE: 03/05/2013
TIME:
2 HOURS
MAXIMUM MARKS = 60
INSTRUCTIONS
1. This paper contains TWO sections
2. Answer Question ONE in Section A, and any TWO out of
questions in Section B
3. No written materials allowed into the examination room.
4. Write all answers in the booklet provided.
5. Do not forget to write your Registration Number.
6. All Questions carry a maximum of 20 Marks each
THREE
Section A
1.
i. Give at least one example of the coenzyme that (5 marks)
a. participates as oxidation–reduction reagents.
b. act as acyl carriers.
c. transfer methyl groups.
d. transfer groups to and from amino acids.
e. are involved in carboxylation or decarboxylation reactions.
ii. Outline the role of Metal ion cofactors in the enzyme catalyzed reactions. (5 marks)
iii. List the four general ways in which enzymes are controlled. (4 marks)
iv. Assume that you have a solution of 0.1 M glucose-6-phosphate. To this solution, you add
the enzyme phosphoglucomutase, which catalyzes the reaction:
Glucose-6-Phosphate
Glucose-1-Phospate
The G°′ for the reaction is 7.5 kJmol1
a. Demonstrate whether the reaction proceeds as written.
(3 marks)
b. Calculate the final concentrations of glucose-6-phosphate and glucose1-phosphate?
(3 marks)
The value of Gas constant = 8.314JMol.−1K−1
Section B
2.
i. The Km for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8  102 M,
and the Km for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 
104 M.
a. Which substrate has the higher apparent affinity for the enzyme? Explain. (3 marks)
b. Which substrate is likely to give a higher value for Vmax? Explain.
(3 marks)
ii. For an enzyme-catalyzed reaction, the presence of 5 nM of a reversible inhibitor yields a
Vmax value that is 80% of the value in the absence of the inhibitor. The Km value is
unchanged.
a. What type of inhibition is likely occurring? (2 marks )
b. What proportion of the enzyme molecules have bound inhibitor? (2 marks)
iii. What is feedback inhibition? Why is it a useful property?
1
(4 marks)
iv. For an enzymatic reaction, draw curves that show the appropriate relationships between
the variables in each plot below. (6 marks)
3.
i. Differentiate between homotropic and heterotropic effectors (modulator). (4 marks)
ii. Which of the curves in Figure below exhibits the greatest cooperativity? Explain.
(3 marks)
iii. For an enzyme that displays Michaelis–Menten kinetics, what is the reaction velocity,
ν (in terms of Vmax), observed at the following values? (5 marks)
(a) [S] = 0.5Km
(d) [S] = 2Km
2
iv. Treatment of carbonic anhydrase with high concentrations of the metal chelator
EDTA (ethylenediaminetetraacetic acid) results in the loss of enzyme activity.
Provide an explanation. (2 marks)
v. An enzyme that follows Michaelis–Menten kinetics has a Km of 1 μM. The initial
velocity is 0.1 μM min-1 at a substrate concentration of 100 μM. Determine the
initial velocity when [S] is equal to
(6 marks)
a. 1 mM
b. 2 μM?
4. The kinetics of an enzyme are measured as a function of substrate concentration in the
presence and absence of 100 M inhibitor.
a. Construct a Lineweaver–Burk plot.
(6 marks)
b. Determine the values of Vmax and Km in the absence of this inhibitor? (4 marks)
c. What type of inhibition is it? (2 marks )
d. Calculate the dissociation constant of this inhibitor? (3 marks)
e. If [S] = 30 M, determine the fraction of the enzyme molecules that have a bound
substrate in the presence and in the absence of 100 M inhibitor? (5 marks)
[S] (M)
3
5
10
30
90
Velocity (mol min1)
No inhibitor
Inhibitor
10.4
2.1
14.5
2.9
22.5
4.5
33.8
6.8
40.5
8.1
3