Question

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?

Answer 1

Step-by-step explanation:

We know that relation between
`pK_(a)` and
`pK_(b)` is as follows.

`pK_(a) + pK_(b)` = 14

As it is given that
`pK_(a)` is 8.18. Therefore, calculate the value of
`pK_(b)` as follows.

`pK_(a) + pK_(b)` = 14

`8.18 + pK_(b)` = 14

`pK_(b)` = 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.

`B + H^(+) \rightarrow BH^(+)` (protonated active enzyme)

Hence, pOH =
`pK_(b) + log([BH^(+)])/([B])`

6.82 = 5.82 +
`log_(10) ([BH^(+)])/([B])`

`([BH^(+)])/([B])` = 10

Therefore, percentage of active enzyme = %
`[BH^(+)]` =
`(10)/(10 + 1) \rimes 100`

%
`[BH^(+)]` = 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.