Answered: estimate of 8.78 for the microscopic…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Transcribed Image Text:An enzyme uses a lysine and a histidine as general acid-base catalytic residues. In one direction of the reaction, the lysine and histidine residues are typically protonated (~NH3*, ~Im), but in the other direction of the reaction, the two residues are "reverse protonated" (~NH2, ~Hlm*) (see below scheme). A pH profile of the reaction at various pH values yielded estimates of 7.57 and 9.30 for the macroscopic ionization constants pk,' and pK2, respectively. A separate set of experiments with site-speciffic variants of the enzyme yielded an estimate of 8.78 for the microscopic ionization constant pKp. Given this information, what is the value of the microscopic ionization constant pKA? Report your answer to the nearest hundredth. Lys-NH3* His-HIm" KB H KA H* Lys-NH2 Lys-NH3* His-HIm" His-Im Kc H* Kp Lys-NH2 His-Im

Expert Solution

Step 1

Given,

pK₁ = 7.57

pK₂ = 9.30

pK_D = 8.78

$p K_{1} = - \log (K_{1}) K_{1} = 10^{- P K_{1}} K_{1} = 10^{- 7.57} = 2.69 \times 10^{- 8}$

$p K_{2} = - \log (K_{2}) K_{2} = 10^{- 9.30} = 5.0 \times 10^{- 10}$

$K_{D} = 10^{- 8.78} K_{D} = 1.65 \times 10^{- 9}$

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