A titratable group is in the active site of an enzyme and must be protonated for

Question

A titratable group is in the active site of an enzyme and must be protonated for the enzyme to function. Assume that the group has a pKa of 6.5 and that the activity is proportional to the fraction of molecules in the protonated state. Which of the following are true? (more than one is correct)

At pH 9.0 close to 100% of the enzyme molecules are active.

At pH 9.0 close to 100% of the enzyme molecules are inactive.

At pH 6.5 the enyzme is 50% active.

At pH 6.8 the enzyme is 66.6% inactive.

At pH 7.2 there are about 5 inactive enzymes for every active enzyme.

At pH 9.0 close to 100% of the enzyme molecules are active.

At pH 9.0 close to 100% of the enzyme molecules are inactive.

At pH 6.5 the enyzme is 50% active.

At pH 6.8 the enzyme is 66.6% inactive.

At pH 7.2 there are about 5 inactive enzymes for every active enzyme.

Explanation / Answer

If fraction of molecules are protonated then it act as acidic buffer. We know for which

pH=Pka+log(protonated/in protonated)

Given Pka=6.5 so at

a)pH=6.5, protonated/unprotonated=1 i.e half molecules are active

b) pH=7.2,  protonated/unprotonated=5 i.e 80% molecules are active

c) pH=9, protonated/unprotonated=316.3 i.e close to 100% molecules are active

D)pH=6.8, protonated/unprotonated=2 i.e 66.66% molecules are active

So options A and C are correct

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