In biological systems, it is very common for proteins such as enzymes or receptors to bind to multip — Physical Chemistry — Kinetics Chemistry Question
Enzyme-substrate interaction
In biological systems, it is very common for proteins such as enzymes or receptors to bind to multiple ligands or substrates at the same time. Binding of the first ligand usually affects the binding of the second ligand either positively or negatively. In this question, imagine that there is a protein P which can bind to two different ligands L and MH+ as shown in the figure given below.
[VISUAL]
For simplicity, assume that the binding of these two ligands are independent of each other; i.e. binding of the first ligand does not change the binding constant (complex formation constant) for the second ligand.
Equal volumes of 100 µM solutions of ligand L and protein P are mixed in a buffer solution of pH 9.50. The formation constant is Kf(P-L) = 2.22 ⋅ 104. Calculate molar concentration for all the species present in this solution. What percentage of the protein P is complexed with the ligand L?
Model Answer
Reaction: P + L ⇌ PL
Since equal volumes of 100 µM solutions are mixed, the initial concentrations before reaction (diluted by half) are:
[P]initial = 5.0 ⋅ 10^-5 mol dm^-3
[L]initial = 5.0 ⋅ 10^-5 mol dm^-3
Let x be the concentration of PL at equilibrium:
[PL] = x
[P] = 5.0 ⋅ 10^-5 - x
[L] = 5.0 ⋅ 10^-5 - x
Kf(P-L) = [PL] / ([P][L]) = x / (5.0 ⋅ 10^-5 - x)^2 = 2.22 ⋅ 10^4
Solving the quadratic equation gives:
x = 2.0 ⋅ 10^-5 mol dm^-3
Equilibrium concentrations:
[PL] = 2.0 ⋅ 10^-5 mol dm^-3
[P] = 3.0 ⋅ 10^-5 mol dm^-3
[L] = 3.0 ⋅ 10^-5 mol dm^-3
[H+] = 3.2 ⋅ 10^-10 mol dm^-3 (from pH 9.50)
[OH-] = 3.2 ⋅ 10^-5 mol dm^-3
Percentage of protein P complexed with ligand L:
[(2.0 ⋅ 10^-5) / (5.0 ⋅ 10^-5)] × 100 = 40% of protein is bound to the ligand L.
Ligand M has a free amine group and only its protonated form, i.e. MH+, can bind to protein P. What percentage of the ligand M is protonated at pH 9.50? pKa(MH+) = 10.00.
Model Answer
The acid dissociation equilibrium for protonated ligand MH+ is:
MH+ + H2O ⇌ M + H3O+
Ka = [M][H3O+] / [MH+] = 1.0 ⋅ 10^-10
The fraction of protonated ligand (α_MH+) is:
α_MH+ = [MH+] / ([MH+] + [M]) = 1 / (1 + [M]/[MH+]) = 1 / (1 + Ka/[H3O+])
At pH 9.50, [H3O+] = 10^-9.50 = 3.16 ⋅ 10^-10 mol dm^-3
α_MH+ = 1 / (1 + 1.0 ⋅ 10^-10 / 3.16 ⋅ 10^-10) = 1 / (1 + 0.316) = 0.76
Therefore, 76% of the ligand M is protonated at pH 9.50.
Equal volumes of 100 µM solutions of ligand M and protein P are mixed in a buffer solution of pH 9.50. Calculate molar concentration for all the species present in this solution. What percentage of the protein P is complexed with the ligand MH+? Kf(P-MH+) = 5.26 ⋅ 105.
Model Answer
Reaction: P + MH+ ⇌ P-MH+
Initial concentrations after mixing (diluted by half):
[P]initial = 5.0 ⋅ 10^-5 mol dm^-3
[M]total,initial = 5.0 ⋅ 10^-5 mol dm^-3
Since only 76% of M is protonated as MH+:
[MH+]initial = 0.76 × 5.0 ⋅ 10^-5 = 3.8 ⋅ 10^-5 mol dm^-3
Let x be the concentration of P-MH+ at equilibrium:
[P-MH+] = x
[P] = 5.0 ⋅ 10^-5 - x
[MH+] = 3.8 ⋅ 10^-5 - x
Kf(P-MH+) = [P-MH+] / ([P][MH+]) = x / [ (5.0 ⋅ 10^-5 - x) * (0.76 * 5.0 ⋅ 10^-5 - x) ] = 5.26 ⋅ 10^5
Solving this quadratic equation gives:
x = 4.0 ⋅ 10^-5 mol dm^-3
Equilibrium concentrations:
[P-MH+] = 4.0 ⋅ 10^-5 mol dm^-3
[P] = 1.0 ⋅ 10^-5 mol dm^-3
[MH+] = 7.6 ⋅ 10^-6 mol dm^-3
[M] = 2.4 ⋅ 10^-6 mol dm^-3
[H3O+] = 3.2 ⋅ 10^-10 mol dm^-3
[OH-] = 3.2 ⋅ 10^-5 mol dm^-3
Percentage of protein complexed with ligand MH+:
[(4.0 ⋅ 10^-5) / (5.0 ⋅ 10^-5)] × 100 = 80% of protein is bound to the ligand MH+.
100 µL of 100 µM protein P, 50 µL of 200 µM of ligand L and 50 µL of 200 µM of ligand M are mixed in a buffer solution of pH 9.50. What percentage of the protein P is bound to (i) only L, (ii) only MH+ and (iii) both L and MH+? Calculate the concentration (mol dm-3) for all the species present in this solution.
Model Answer
Total final volume = 100 µL + 50 µL + 50 µL = 200 µL
Initial concentrations after mixing:
[P]initial = 100 µM × (100 µL / 200 µL) = 50 µM = 5.0 ⋅ 10^-5 mol dm^-3
[L]initial = 200 µM × (50 µL / 200 µL) = 50 µM = 5.0 ⋅ 10^-5 mol dm^-3
[M]initial = 200 µM × (50 µL / 200 µL) = 50 µM = 5.0 ⋅ 10^-5 mol dm^-3
These initial concentrations are identical to those in 24.1 and 24.3.
Since the binding of L and MH+ to protein P are completely independent of each other:
- Probability of P binding L = 40% (from 24.1)
- Probability of P binding MH+ = 80% (from 24.3)
Therefore, statistically:
(iii) Percentage of P bound to both L and MH+:
0.40 × 0.80 × 100 = 32%
(i) Percentage of P bound to only L:
(0.40 - 0.32) × 100 = 8%
(ii) Percentage of P bound to only MH+:
(0.80 - 0.32) × 100 = 48%
Percentage of free protein P:
100% - 32% - 8% - 48% = 12%
Equilibrium concentrations of all species:
[P-L-MH+] = 5.0 ⋅ 10^-5 × 0.32 = 1.6 ⋅ 10^-5 mol dm^-3
[P-L] = 5.0 ⋅ 10^-5 × 0.08 = 4.0 ⋅ 10^-6 mol dm^-3
[P-MH+] = 5.0 ⋅ 10^-5 × 0.48 = 2.4 ⋅ 10^-5 mol dm^-3
[P] = 5.0 ⋅ 10^-5 × 0.12 = 6.0 ⋅ 10^-6 mol dm^-3
[MH+] = 7.6 ⋅ 10^-6 mol dm^-3 (same as in 24.3)
[M] = 2.4 ⋅ 10^-6 mol dm^-3 (same as in 24.3)
[L] = 3.0 ⋅ 10^-5 mol dm^-3 (same as in 24.1)
[H3O+] = 3.2 ⋅ 10^-10 mol dm^-3
[OH-] = 3.2 ⋅ 10^-5 mol dm^-3