Hexokinase can use glucose or fructose as a substrate. The Km forglucose is 0.15 mM and for fructose Km = 1.5 mM.

which of the following statement is true?

a. For which substrate does hexokinase have the highest affinity? Why?

b. Calculate V as a percentage of Vmax for each substate when [S]equals 0.15mM.

a) glucose because Km is less for glucose.

b) For glucose, V as a percentage of Vmax = 50 %

For fructose, V as a percentage of Vmax = 9 %

Explanation:

a) Hexokinase has highest affinity for glucose because it's Km for glucose is low.

When enzyme [E] binds a substrate [S] an [ES] complex is formed. It is represent as under:

[E] + [S] ⇌ [ES] ⇌ [E] + [P]

As soon as [ES] is formed it may either form [E] & [P] in a forward reaction or dissociate back into [E] & [S]. Km is known as dissociation constant which represents the extent to which [E] & [S] are being formed. A higher value of Km represents formation of more [E] & [S] back from [ES]. It also means that binding affinity of enzyme with substrate is less or we can say that Km is inversely proportional to binding affinity.

Km ∝ 1 / binding affinity.

So, it simply means that lower the Km more will be the affinity of the enzyme for the substrate. In the given question, Km for glucose is 0.15 mM which is less than fructose (Km for fructose = 1.5 mM) that is why glucose is the answer.

b) The formula for Michaelis-Menten equation is as under:

V = Vmax * [S] / ([S] + Km)

V / Vmax = [S] / ([S] + Km)

Here, for glucose Km = 0.15 mM and [S] = 0.15mM.

So V as a percentage of Vmax for glucose will be as under:

V / Vmax = 0.15/0.15 + 0.15

→ V / Vmax = 0.15/0.30

→ V / Vmax = 0.5 = 1/2 = 50 %

Here, for fructose Km = 1.5 mM and [S] = 0.15mM.

So V as a percentage of Vmax for fructose will be as under:

V / Vmax = 0.15/0.15 + 1.5

→ V / Vmax = 0.15/1.65

→ V / Vmax = 0.09 = 9/100 = 9 %