The active site of an enzyme contains a critical cysteine residue. the enzyme is active only when the cysteine side chain is in its protonated form. the pka for the r-group of cysteine is 8.18. what percentage of the enzyme is active when in a buffer at ph 7.18?

Explanation:

We know that relation between and is as follows.

                      = 14

As it is given that is 8.18. Therefore, calculate the value of as follows.

                            = 14

                            = 14

                                = 14 - 8.18

                                           = 5.82

Similarly, as value of pH is given as 7.18. Therefore, value of pOH will be as follows.

                                pH + pOH = 14

                               7.18 + pOH = 14

                                         pOH = 6.82

Let us take that B represents the enzyme. Hence, its reaction with proton will be as follows.

                               (protonated active enzyme)

Hence,     pOH =

                 6.82 = 5.82 +

                  = 10

Therefore, percentage of active enzyme = % =

                    % = 90.9%

Thus, we can conclude that 90.9% is the percentage of the enzyme which is active in a buffer at pH 7.18.

the answer is a and have a great day