Question

At the isoelectric of an enzyme, which is 5, it was observed that there are 10 positively charged groups and 10 negatively charged groups. when the enzyme was titrated with alkali starting from pH 5 to give pH 7,3 positively charged groups were deprotonated. The net charge on the protein at pH 7would be

Answer 1

Answer:

-3

Explanation:

At the isoelectric point, net charge is zero. (10 pos and 10neg).

Then, 3 groups were deprotonated, i.e., now the net charge is (-3).