Question

An enzyme catalyzes a reaction with an initial velocity of 50 micromoles/litre-seconds when the substrate concentration is 5 micromolar and 80 micromoles/litre-seconds when the substrate concentration is 10 micromolar. The Vmax and Km of this enzyme are:

A) 50 micromoles/litre-seconds; 5 micromolar

B) 80 micromoles/litre-seconds; 10 micromolar.

C) 200 micromoles/litre-seconds; 15 micromolar

D) 100 micromoles/litre-seconds; 12.5 micromolar

E) 10 micromoles/litre-seconds; 1 micromolar

Answer 1

V0 = 50 microM/L-s

[S] = 5 microM

and

V0 = 80  microM/L-s

[S] = 10 microM

Michaelis–Menten equation is used typically in enzymatic reactions.

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

Where

Vmax = max rate velocity

[S] = substrate concentration

Km = Michaelis–Menten constant

V = reaction rate

choose any point

V1 = Vmax*[S1] / (Km + [S1])

V2 = Vmax*[S2] / (Km + [S2])

substitute

80 = Vmax*10/(Km+10)

50 = Vmax*5/(Km+5)

divide

80/50 = 10/5 * (Km+5) / (Km+10)

1.6 = 2  (Km+5) / (Km+10)

1.6* (Km+10) = 2*(Km+5)

1.6Km + 16 = 2Km + 10

(2-1.6)Km = 6

Km = 6/(0.4) = 15

substitute

80 = Vmax*10/(Km+10)

80 = Vmax*10/(15+10)

Vmax = 80*(15+10)/10 = 200

from the list, choose C