Question

Problem 8.4 The folding and unfolding rate constants for a myoglobin mutant have been determined. The...

Problem 8.4 The folding and unfolding rate constants for a myoglobin mutant have been determined. The unfolding rate constant kF→U = 3.62×10−5s−1 and the folding rate constant kU→F=255s−1, where F is the folded protein and U is the unfolded (denatured) protein.

For wild-type myoglobin, ΔG∘′F→U=+37.4kJ/mol. Part A Which myoglobin is more thermodynamically stable, the mutant or the wild-type?

Homework Answers

Answer #1


Mutant Protein Equilibria: F = U ; K(mutant)
forward unfolding rate constant, kf = 3.62×10−5s−1
reverse folding rate constant, kr =255s−1

Equillibrium constant, K(mutant)F→U = kf/kr = 3.62×10−5/255 = 1.42×10-7
Temperature T = 298 K; Gas constant R = 8.314 J/mol K
ΔG∘(mutant)F→U = -2.303*R*T*Log(K) = -2.303*8.314*298*Log(1.42×10−7) = +39702 J/mol K = +39.702kJ/mol

For wild-type myoglobin, ΔG∘(wild)F→U=+37.4kJ/mol

ΔG∘(wild)F→U < ΔG∘(mutant)F→U

So, mutant-type myoglobin is more thermodynamically stable in the Folding state as it requires higher ΔG to change to Unfolding(denatured) state.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions