Suppose that a mutation of the above enzyme decreases the catalytic rate by 4 orders of magnitude. the decreased rate indicates that the mutation has caused δg° to increase. by how many kj/mol has the activation energy increased in the mutant enzyme compared to the wild-type (unmutated) enzyme?

We know that reate constant (-Ea/RT)

K = Ae^(-Ea/RT)

A = frequency factor

Ea = activation energy

R = constant

T = temperature

Let us take two rate constants

K1 =  Ae^(Ea1/RT)

K2 = Ae^-(Ea2/RT)

K2 = K1/H

H = e^ (Ea2/RT  -  Ea1/RT)

1.386 = 1/RT (Ea2 – HG Kj/mol)

Substituting everything to the equation, we get 3.43 kJ/mol

the answer to your question is sperm

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