Unformatted text preview: Biochemistry 100B Winter 2007
Bogomolni Problem set H #3 1 The enzyme catalase hydrolyses hydrogen peroxide, H202 into H20 and O2 . This is
one of the nearly perfectly evolved enzymes with a k CAT/ KM value of 4. 107 M'1 s']. If the turnover number ( k CAT) for catalase is 5.0 x 105 s-', and the molar mass of catalase
is 240 kDa, calculate the speciﬁc activity of catalase in mol min.:mg-t. If 1.0 mg of
catalase is added to l L of a 0.1 M solution of hydrogen peroxide in water, determine
initial velocity of the reaction in mol s'l. 2. An extract of E.coli containing all the soluble enzymes present in the bacteria was
assayed for phosphatase activity (an enzyme that hydrolyses phosphate esters), giving a
value of 0.30 umol hydrolysis per minute per mg of total protein. After puriﬁcation of the
enzyme, the speciﬁc activity was again measured, and was found to be 0.96 umol
hydrolysis per minute per [Lg of the pure enzyme. What does this tell you about the % of
phosphatase among all the proteins in E.coli? 3. If I is a competitive inhibitor, what ratio of [I]/Kt gives KMtappiof 4.0 Km? 1.5 KM?;
Show how this would appear plotted as a Woolf-Hanes plot. 4. If I is a non-competitive inhibitor, what ratios of [I]/Kt gives:
a-anx[app) Of 0.33 b-anx? 0.80 Vmax?
Show how this would appear plotted as a Lineweaver—Burk plot. 5. If I is a competitive inhibitor, what ratio [I]/Kt causes a 50% reduction in actual rate v
when substrate concentration is equal to the KM of the uninhibited enzyme?
Show how this would appear plotted as a Lineweaver-Burk plot. 6. The enzyme succinate dehydrogenase has KM: 2.0 x 10'3 mol/L for its substrate
succinate, and Vnm= 1.5 umol/min/mg protein). In the presence of 1.0 x 10'2 mol/L
malonate, inhibition occurs so apparent Kmis 5.0 x 10'3 moI/L, Vmaxis unchanged. In the
presence of 3.0 x 10'5 mol/L of the drug amytal, KMlS unchanged, but the apparent Vm is
0.5 umol/rnin/mg. Determine the Ki of each inhibitor, and identify the mode of inhibition
in each case. 7. The enzyme lactate dehydrogenase (KM for lactate = 6.8 x 10'3M) is inhibited
competitively be malate (Ki = 2.5 x 10 GM) and inhibited non-competitively by oxalate
(K1: 5.0 x 10 ' M). Compare the effects of each inhibitor in turn, given: [inhibitor] = 5.0 x 10'3 M, a) [lactate] = 1.7 x 10 '3 M b) [lactate] = 1.7 x 10'2 M?
Look up the structures of lactate and malate in your textbook and suggest a reason why
malate may act as a competitive inhibitor. ...
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- Winter '10