Question

Hydrogen bonding of the –OH group of Ser 110 of a hydrolytic enzyme was shown by several experimental techniques to contribute 2.2 kcal/mol to the stabilization of the transition state and 0.48 kcal/mol to the stabilization of the substrate complex. The kinetic characterization of the wild-type enzyme determined the kcat = 105 s-1 and the Km = 10-6 M. note: T = 27°C, R = 2 kcal/°K.mol; Show work for all calculations; Include units. Calculate the change in the rate of reaction (fold change) for the Ala 110 mutant with respect to the wild-type at: i) [S] >> Km; Rate of reaction for the mutant is ______ times the rate of reaction for the wildtype.

Answer 1

Hydrogen bonding of the –OH group of Ser 110 of a hydrolytic enzyme = 2.2 kcal/mol stabilization of the transition state

- 0.48 kcal/mol to the stabilization of the substrate complex.

For wild-type enzyme : kcat = 10⁵ s-1 and Km = 10^-6 M.

T = 27°C = 300 K ,

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I)

When [S]₀ >> K_m, the rate reaches its maximum value and is independent of [S]₀:
v (rate) = v_max = kcat [E]₀

So, rate of reaction is decided by enzyme conc. and kcat.

we assume cataysis follows transition state theory :

Being other thing similar except transition state energy and stabilization of substrate complex :

total stabilization : 2.68 kcal/mol

rate(mut.) / rate(wild) = e^{−(ΔG )/RT} /  e^{−(ΔG-2.68 kcal/mol)/RT}

=  exp (.( -2.68 kcal)/RT)

= exp (( -2680 cal/mol) /2 cal/K-mol* 300K)

rate(mut.) / rate(wild) = 0.01145

Rate of reaction for the mutant is _0.01145_ times the rate of reaction for the wildtype.