Question

A researcher synthesizes a new molecule called inhibitor X that inhibits the activity of phosphofructokinase (PFK), the enzyme that phosphorylates fructose 6−phosphate during glycolysis. Inhibitor X competes with ATP for binding to the PFK active site. The researcher determines that the Michaelis–Menten constant ( K m ) with respect to ATP is 40 µM. When 1.5 µM inhibitor X is added to the purified PFK reaction, the apparent K m for ATP increases to 52 µM. Calculate the dissociation constant of the PFK‑inhibitor complex ( K I ) .

Answer 1

Answer : The dissociation constant of the PFK‑inhibitor complex is, 5 µM

Explanation :

The expression for reversible competitive inhibition when apparent Km affected by addition of the inhibitor is:

$K_m_a=K_m[1+\frac{I}{K_i}]$

where,

$K_m_a$ = apparent value = 52 µM

$K_m$ = Michaelis–Menten constant = 40 µM

I = inhibitor concentration = 1.5 µM

$K_i$ = dissociation constant of the PFK‑inhibitor complex

Now put all the given values in the above formula, we get:

$52\mu M=40\mu M[1+\frac{1.5\mu M}{K_i}]$

$K_i=5\mu M$

Therefore, the dissociation constant of the PFK‑inhibitor complex is, 5 µM