The rate, V, of an enzyme reaction is often described by the Michaelis–Menten equation, V =

VmaxS K + S

, where Vmax and S are constants. This is an equation we’ve used a lot in this book. However, the rate of an enzyme reaction is also commonly described by the Hill equation, V =

VmaxS n

Kn + S n

, for some constant n > 1. Typically n is 2 or 3; for haemoglobin we use n = 4, as previously discussed on page 265 and Fig.

12.9.

a. Just to make things a bit simpler, set n = 2 in the Hill equation. Then show that the Hill equation has an inflection point but the Michaelis–Menten equation doesn’t.

b. Show that, for a general n, the inflection point occurs at S =

n q

n−1 n+1 K. Where is the inflection point in the limit as n → ∞?

Don’t calculate the second derivatives or the limit by hand! Use a computer.

c. Show that the inflection point occurs at the steepest part of the Hill function.

To do part c you don’t need to do any additional complicated calculations.

d. Now let n be any number. Draw Hill functions for n = 2, 3 and 4. How does the slope of the curve at the inflection point depend upon n?

Don’t calculate this analytically, just describe approximately.

e. When n is a large number (n = 100, say), what does the Hill equation look like? What is the scientific interpretation of the Hill function when n is large?