Question

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?

Answer 1

The activation energy of the mutant enzyme has increased by approximately 76.56 kJ/mol when compared to the wild-type (unmutated) enzyme.

Step-by-step explanation:

A mutation in the enzyme that results in a significant decrease in the catalytic rate denotes corresponding alterations in the activation energy. Acknowledging that a decrease in catalytic rate signifies that the activation energy has increased, we can resort to the Arrhenius equation which connects the rate constant, activation energy, and temperature. The equation is expressed as k = Ae^-Ea/RT. Here, 'k' represents the rate constant, 'Ea' denotes the activation energy, 'R' is the ideal gas constant (8.314 J/mol/K), 'T' represents the temperature in Kelvin, and 'A' is the frequency factor.

Given that the decrease in the catalytic rate is 4 orders of magnitude, it is essentially saying that the rate has decreased by a factor of 10,000. Logarithmically, this equates to a difference of 9.21 in the natural log (ln) of the rate constant (since ln(10,000) = 9.21). We can utilize this information along with the previously mentioned Arrhenius equation to deduce the change in the activation energy.

Accommodating the transformed form of the Arrhenius equation, lnk = -Ea/RT + lnA, and assuming that factors other than the activation energy (like temperature and the frequency factor) remain unchanged between the wild-type enzyme and the mutated version, the change in the lnk gives us the change in the activation energy. Hence, the increase in activation energy (ΔEa) would be ΔEa = -R*Δ(lnk) = -8.314 J/mol/K * -9.21 = 76.56 kJ/mol.

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