Most proteins exist usually only in two forms, the native form (N) and the unfolded often referred t — Physical Chemistry — Kinetics Chemistry Question
Protein Folding
Most proteins exist usually only in two forms, the native form (N) and the unfolded often referred to as the denaturation temperature. At temperatures higher than T½, fU increases above ½, but at temperatures lower than T½ , fU decreases below ½. assuming that both the forward and reverse processes are elementary steps that follow first-order kinetics.
What is the equilibrium constant for the process when the native and denatured states are present in equal proportions at equilibrium?
Model Answer
1
What is the standard free energy change of the process (∆G°(T)) when the native and denatured states are present in equal proportions at equilibrium? Express your answer in SI units.
Model Answer
0 kJ mol-1
If (CN)eq and (CU)eq denote the equilibrium concentrations of N and U in solution, respectively, and C is the total concentration of the protein, the fraction of the total
Model Answer
fU = (CU)eq / ((CN)eq + (CU)eq) = ((CU)eq/(CN)eq) / (1 + (CU)eq/(CN)eq) = K / (1+K)
What is the sign of ∆G°(T) at temperatures below and above T½? Select your answer from the following choices.
a) Negative both below and above T½.
b) Positive both below and above T½.
c) Positive below T½, but negative above T½.
d) Negative below T½, but positive above T½..
Model Answer
Correct answer is (c). Positive below T1/2, but negatíve above T1/2
How does the standard Gibbs free energy change for the process vary when the
Model Answer
Correct answer is (d). Decreases above T1/2, but increases below T1/2.
For the simple chemical equation and elementary kinetic steps used to describe the protein folding-unfolding process outlined above, what is the relationship between equilibrium constant K and the rate constants kf and kb?
Model Answer
K = kf / kb
Derive a rate law for the overall process, that is dCU/dt in terms of only rate constants, CU and (CU)eq.
Model Answer
dCU/dt = kfCN - kbCU
= kf(C - CU) - kbCU = kfC - kfCU - kbCU = kfC - (kf + kb) CU (1)
K = kf / kb = (CU)eq / (CN)eq
1 / K = kb / kf = (CN)eq / (CU)eq
⇒ kb / kf + 1 = (CN)eq / (CU)eq + 1
⇒ (kb + kf) / kf = [(CN)eq + (CU)eq] / (CU)eq
⇒ (kb + kf) / kf = C/(CU)eq
C = [(kb + kf) (CU)eq ] / kf (2)
Now substitute C obtained from eq (2) to eq (1).
We get kf { [(kb + kf) (CU)eq] / kf} - (kf + kb) CU
⇒ [(kb + kf) (CU)eq] - (kf + kb) CU
⇒ - (kf + kb) [CU - (CU)eq]
So we get
dCU/dt = - (kf + kb) [CU - (CU)eq]