Question

In hemoglobin Rainier, Tyr 145β is replaced by Cys, which forms a disulfide bond with another Cys residue in the same subunit. This prevents the formation of ion pairs that normally stabilize the T state. How does hemoglobin Rainier differ from normal hemoglobin with respect to (a) oxygen affinity, (b) the Bohr effect, and (c) the Hill constant?

Short answer

Hemoglobin Rainier shows an increase in (a) oxygen affinity, (c) the Hill constant, and a decrease in (b) the Bohr effect than the normal.

Step-by-step solution

Destabilization of the T state

The mutation in the hemoglobin Rainier prevents the formation of ion pairs that normally stabilize the T state of hemoglobin. This means that the mutation promotes the destabilization of the T state conformation of hemoglobin.

Hemoglobin exists in an equilibrium between two conformations: T state⇌ R state. The destabilization of the T state would shift the equilibrium towards the right side, that is, towards the R state of hemoglobin.

(a) oxygen affinity

The R state of hemoglobin represents the oxyhemoglobin state. Hence, shifting the equilibrium towards the R state increases the affinity of the hemoglobin to oxygen.

(b) the Bohr effect

The Bohr effect involves the destabilization of the R state of hemoglobin and stabilizing of the T state of the same.

As the mutation in hemoglobin Rainier destabilizes the T state, thus the Bohr effect would be reduced.

(c) the Hill constant

The Hill constant represents the degree of interaction between hemoglobin and Oxygen.

Since the mutation in hemoglobin Rainier increases the affinity of hemoglobin for oxygen, the degree of interaction between hemoglobin and oxygen should be increased, and thus the Hill constant would also increase.