Question

kcal/K 210 Which mutation below is the least detrimental to the enzyme function? PH 8.0 variant keat (min-) Km (M) wild-type 3200 HS7A 0.054 H57L 0.075 HS7D 0.78 HSTE 0.69 HSOK 0.83 HSOR 0.017 DIO2N 1.3 H57A, D102N 0.17 H57D, D102N 0.18 H57K, D102N 0.41 H57L, D102N 0.13 S195A 0.079 S19ST 0.011 H57A, D102N, S195A 0.038 kbuffer (min) no enzyme 6.0 x 10-7 21 4.2 87 62 0.0032 0.0037 0.06 0.033 0.020 0.00025 0.30 0.0019 0.0029 0.023 0.0031 0.0019 0.00052 0.00043 H57K H57R S195A H57A,D102N,S195A D102N

Answer 1

You have asked two questions here. As per Chegg Q&A Guidelines, I am answering the first question, both the subparts.

a) The reaction rate is very-well captured by Michelis-Menten kinetics stated by the equation:

$v = \frac{v_{max}[S]}{[S]+K_M}$

where v is the rate of the reaction, v_max is the maximum rate of the reaction, [S] is the concentration of the substrate and K_M is the Michelis' Menten constant.

K_M and v_max are substrate independent and are the characteristics of the enzyme.

K_M determines the concentration of the substrate at which the rate of the reaction is half-maximum, while v_max refers to the maximum reaction rate which the enzyme can undergo.

Now, v_max can be stated as:

where [E]_T is the total enzyme concentration and k_cat is the turnover number of the enzyme which determines the number of product molecules formed from the substrate molecules in one second.

Now, there is one more quantity called the specificity constant specified by

$c = \frac{k_{cat}}{K_M}$

This quantity determines the enzyme efficiency. The higher is the specificity constant, the higher the ability does the enzyme have to react with the substrate.

Now, as can be seen in the table, the mutations are mentioned accompanied by the kinetic constants K_M, k_cat, and c.

A mutation that leads to a lower decrease in specificity constant has the least effect on the enzyme activity and hence is the least detrimental for enzyme function.

Here we can see that the mutation with the highest specificity constant is D102N with a specificity constant c = 0.3.

Hence the correct answer for the first subpart is D102N.

b) The enzyme works by lowering the activation energy for the reaction. It has an active site that binds to the substrate reorients it so that the activation energy is lowered and hence the reaction is facilitated. The active pocket also has a catalytic site that speeds up the conversion of a substrate molecule to the product molecule.

In this enzyme, the mutations H57A, D102N, and S195A result in the loss of the catalytic site. However, these sites do not prevent the binding of the substrate to the active pocket of the enzyme and the reorientation of the enzyme that will still favor the reaction to a large extent.

Hence the answer for the second subpart is Orienting reaction groups favorably.