Question

Serine 14 of glycogen phosphorylase was replaced by glutamate in an experiment. The Vmax of the mutant enzyme was compared to wild type phosphorylase a and b.

Wild type phosphorylase a 100 ± 5
Wild type phosphorylase b 25 ± 0.4
Serine (S) to Glutamate (E) mutant 60 ± 3

Explain the results obtained with the S to E mutant glycogen phosphorylase.

a) Glutamate positively charged R group mimics the phosphoryl group on serine. There is a reduced response because the carboxyl group is smaller and less charged than the phosphate.
b) Glutamate negatively charged R group mimics a phosphorylated serine residue. The VmaxVmax is reduced because the carboxyl group is smaller and less charged than a phosphate group.
c) The S to E mutant would mimic the phosphorylase in the T state. The reduced response is due to blockage of phosphorylase kinase’s catalytic site.
d) The S to E mutant would mimic the phosphorylase in the R state. The reduced response is due to blockage of phosphorylase kinase’s catalytic site.

Answer 1

b) Glutamate's negatively charged R group mimics a phosphorylated serine residue. The Vmax is reduced because the carboxyl group is smaller and less charged than a phosphate group.

Step-by-step explanation:

Vmax is the maximum velocity of a reaction. It is achieved when all the binding sites of an enzyme are occupied. Glycogen phosphorylase enzyme has a serine residue on position 14. Serine is important for its activity because serine's phosphorylation activates it. Hence, phosphorylated Glycogen phosphorylase is active.

Serine has a hydroxyl group which acts as a nucleophile leading to its phosphorylation and addition of negative charge. Glutamate is already a negatively charged amino acid so it repels the phosphoryl group. However because of its negative charge it acts like phosphorylated serine, making the enzyme functionally active. But due to lesser negative charge and smaller size as compared to phosphoryl group, it is not able to fully restore the wild type Vmax.