CHK 395G Kinetics Homework
1. Graphical Analysis of Vmax and Km Values The following experimental data were collected
during a study of the catalytic activity of an intestinal peptidase capable of hydrolyzing the
dipeptide glycylglycine.
Glycylglycine + H2O → 2 glycine
[S] (mM)
1.5
2.0
3.0
Product formed
(µmol/min)
0.21
0.24
0.28
[S] (mM)
4.0
8.0
16.0
Product formed
(µmol/min)
0.33
0.40
0.45
From these data determine by graphical analysis (see Box 8-1) the values of Km and Vmax for this
enzyme preparation and substrate.
2. Irreversible Inhibition of an Enzyme Many enzymes are inhibited irreversibly by heavy-metal
ions such as Hg2+, Cu2+, or Ag+, which can react with essential sulfhydryl groups to form
mercaptides:
Enz−SH + Ag+ → Enz−S−Ag + H+
The affinity of Ag+ for sulfhydryl groups is so great that Ag+ can be used to titrate –SH groups
quantitatively. To 10 mL of a solution containing 1.0 mg/mL of a pure enzyme was added just
enough AgNo3 to completely inactivate the enzyme. A total of 0.342 µmol of AgNO3 was
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required. Calculate the minimum molecular weight of the enzyme. Why does the value
obtained in this way give only the minimum molecular weight?
3. pH Optimum of Lysozyme The enzymatic activity of lysozyme is optimal at pH 5.2 (see graph
below). The active site of lysozyme contains two amino acid residues essential for catalysis:
Glu35 and Asp52. The pKa values of the carboxyl side chains of these two residues are 5.9 and 4.5,
respectively. What is the ionization state (protonated or deprotonated) of each residue at the
pH optimum of lysozyme? How can the ionization states of these two amino acid residues
explain the pH-activity profile of lysozyme shown below?
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4. Two different enzymes are able to catalyze the same reaction, A → B. They both have the same
Vmax, but differ in their Km for the substrate A. For enzyme 1, the Km is 1.0 mM; for enzyme 2,
the Km is 10 mM. When enzyme 1 was incubated with 0.1 mM A, it was observed that B was
produced at a rate of 0.0020 mmoles/minute.
a. What is the value of the Vmax of the enzymes?
b. What will be the rate of production of B when enzyme 2 is incubated with 0.1 mM A?
c. What will be the rate of production of B when enzyme 1 is incubated with 1M (i.e., 1000
mM) A?
Answer a. 0.022 mmol/min; b. 0.00022 mmol/min; c. 0.022mmol/min
5. An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was
found to be 2.0 mM, and the Km for S2 was found to be 20 mM. A student determined that the
Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and
needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other
with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which
substrate. Determine the value of Vmax from the results he obtained:
Tube Number
1
2
Rate of formation of product
0.5
4.8
Answer Vmax = 101
6. An enzyme catalyzes a reaction at a velocity of 20 µmol/min when the concentration of
substrate (S) is 0.01 M. The Km for this substrate is 1 x 10-5 M. Assuming that Michaelis-Menton
kinetics are followed, what will the reaction velocity be when the concentration of S is (a) 1 x 105
M and (b) 1 x 10-6 M?
Answer The velocity of 20 µmol/min is essentially Vmax because it is measured at [S] >> Km.
a. When [S] = 10-5 M = Km, V = ½ Vmax, 10.01 µmol/min.
b. When [S] = 10-6 M, velocity can be calculated from the Michaelis-Menton equation:
V0 = Vmax [S]/(Km + [S]) = (20 µmol/min)(10-6 M)/(10-5 + 10-6) = 1.8µmol/min.
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7. An enzyme catalyzes the reaction A → B. The initial rate of the reaction was measured as a
function of the concentration of A. The following data were obtained:
[A], micromolar
0.05
0.1
0.5
1
5
10
50
100
500
1,000
5,000
10,000
20,000
V0 nmoles/min
0.08
0.16
0.79
1.6
7.3
13
40
53
73
76
79
80
80
Be sure to include units in your answers.
a. What is the Km of the enzyme for the substrate A?
b. What is the value of V0 when [A] = 43?
The above data was plotted as 1/V0 vs. 1/[A], and a straight line was obtained.
c. What is the value of the y-intercept of the line?
d. What is the value of the x-intercept of the line?
Answer a. 50 micromolar; b. 37 nmoles/min; c. 0.0125 (nmole/min)-1; d. -0.02 micromolar-1.
8. When 10 µg of an enzyme of Mr 50,000 is added to a solution containing its substrate at a
concentration one hundred times the Km, it catalyzes the conversion of 75 µmol of substrate
into product in three minutes. What is the enzyme’s turnover number?
Answer Because the velocity measured occurs far above Km, it represents Vmax. Ten µg of the
enzyme represents 10 x 10-6 g/(5 x 104 g/mol), or 2 x 10-10 mol of enzyme. In three minutes, this
amount of enzyme produced 75 µmol of product, equivalent to 25 x 10-6 mol of product per
minute. The turnover number is, therefore, (25 x 10-6 mol/min)/(2 x 10-10 mol) = 12.5 x 104 min-1.
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9. It has been proposed that for enzymes that obey Michaelis-Menton kinetics, the Km value for an
enzyme may represent the approximate physiological substrate concentration in the cell.
Examine a typical plot of V versus [S] and explain why this appears to be a reasonable
suggestion.
Answer
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