Question

8. You discover an enzyme that catalyzes the reaction: AMP →

→→ → IMP + NH2. Based
on your experiments, you determine that the KM for substrate AMP is

3.0 µ

µµ µM and the kcat = 0.5 sec-1. At 8.0 mM AMP, you determine that the velocity of

the reaction is 0.6 µ

µµ µM/min. What was the concentration of total enzyme, [E]total,

used in this experiment?

Answer 1

AMP → IMP + NH₂

E + S ↔ ES → E + P

Rate of formation of ES = k₁[E][S]

Rate of breakdown of ES = (k_-1 + k₂) [ES]

and set equal to each other (Note that the brackets represent concentrations). Therefore:

k₁[E][S] = (k_-1 + k₂) [ES]

Rearranging terms,

[E][S]/[ES] = (k_-1 + k₂)/k₁

The fraction [E][S]/[ES] has been coined K_m, or the Michaelis constant.

According to Michaelis-Menten's kinetics equations, at low concentrations of subtrate, [S], the concentration is almost negligible in the denominator as K_M >> [S], so the equation is essentially

V₀ = V_max [S]/K_M

which resembles a first order reaction.

At High substrate concentrations, [S] >> K_M, and thus the term [S]/([S] + K_M) becomes essentially one and the initial velocity approached V_max, which resembles zero order reaction.

v= v₀[s]/([s]+k_m)

v= 0.6 µ

[s]= 0.008 µ

v₀ = 0.6*(3+0.008)/0.008

v₀ = 225.6 µ

v₀ = k_cat*[E_t]

[E_t] = 225.6/0.5 = 451.2 µ