Problem Set 6 Answers
1. a) Enzyme: Vmax= 10 Km = 3.3
Inhibitor 1: Vmax= 10 Km = 7.7
Inhibitor 2: Vmax = 3.33 Km = 3.33 competitive mixed
2.
3.
4.
5.
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 y = 0.9989x + 0.3001
y = 0.7541x + 0.1
y = 0.3017x + 0.0987
0.5
1 1.5
enzyme inhibitor 1 inhibitor 2
Linear (inhibitor 2)
Linear (inhibitor 1)
Linear (enzyme) b) [ES]/[E]
T
can be estimated by v o
/v max
which is equal to [S]
Km+[S]
62.5%
Vmax = 43
µ
M for the enzyme. vo = 43
µ
M This occurs because the Km value is so low it’s negligible and the enzyme is fully saturated at both substrate concentrations. a) L- lactate has been oxidized to pyruvate; therefore, enzyme is an oxidoreductase and NAD+, NADP+, FAD, or FMN could be cofactors. b) A phosphate group has been transferred to glucose; therefore, the enzyme is a transferase (specifically a kinase) and ATP must be the cofactor which supplies the phosphate group. c) These are isomers of each other; therefore, the enzyme is an isomerase.
K class; Km
Specificity of the enzyme for substrate resides in the amino acids comprising the substrate binding site. The amino acids determine the folding of the site which determines the size, shape, and non covalent forces available in the binding site to bind the substrate. These could be secondary amino acids in the active site.
↓
and Km=breakdown of ES/formation of ES
The formation of ES must have increased, so this is an activator.

7.
6.
The specificity of the enzyme resides in the amino acids present in the active site. These amino acid participate in the bond-breaking and bond- making reactions that convert the substrate to product.
The active form of the enzyme contains the thiolate anion (S
-
) of Cys. The increased pK would increase the nucleophilicity of the thiolate, thereby increasing the rate of the reaction catalyzed by the active form of the enzyme.
However, the amount of enzyme in the active form depends on the pH of the solution; therefore, the pH of the solution would now have to be above 10 which would be difficult in the body, but easy in the lab.
Isotopically labeled A will appear only if the reaction follows a Ping Pong mechanism, since only a double displacement reaction can exchange an isotope P back to A in the absence of B.