Question

We’ve very briefly covered that an equilibrium constant Keq = [products]/[reactants] can also be expressed as kfwd/krev. To derive this, remember that at equilibrium, kfwd[reactant] = krev[product]. Divide both sides by [reactant] and by krev, and you will find that kfwd/krev = [product]/[reactant].

a. With enzymes that follow the Michaelis‐Menten kinetic mechanism (they don’t all, but in this class we’ll limit our discussion to them), if the kcat is much less than k‐1, the KM is approximately equal to the KD. Explain why this is, and what the KM would tell us about the enzyme in such a circumstance.

b. In the situation described in (a), does the E+S⇌ES reaction (described by the KD) reach equilibrium?

c. With enzymes that follow the Michaelis‐Menten kinetic mechanism, if the kcat is comparable to k‐1, the KM tends to be much larger than the KD. Under these conditions, would the E+S⇌ES reaction be at equilibrium?

d. Under the condition of part (c), would the concentration of [ES] be higher or lower than if kcat << k‐1?

e. When kcat ≈ k‐1, how should we interpret the KM? How is this different from the kcat << k‐1 case?

Answer 1

a)

k_cat is the rate limiting step of any enzyme catalyzed reactions at saturation

The enzyme that follows the Michaelis‐Menten kinetic mechanism

                                                                                k₁                      k₂

                                                      E      +       S           ES          E + P

                                                                                 k_-1

Here k₂ is the rate limiting step

so k_cat = k₂

Km = k₂ + k_-1 / k₁

When k2 << k-1, then

Km = k_-1 / k₁

k_-1 / k₁ is K_D dissociation constant of ES complex.

When the above condition holds then Km represents a measure of the affinity of enzyme for its substrate in the ES complex.

b)

At equilibrium k_fwd[reactant] = k_rev[product]

                          k1

E      +       S           ES

                          k-1

Here

k₁[E][S] = k_-1[ES]

k_-1 / k₁ is K_D

K_D = [ES] / [E][S]

At equilibrium rate of formation of product is zero and k_cat is much less so the product formation is negligible

so the for the above situation equilibrium is reached

c)

When k_cat = k_-1

K_m approximately equal to K_D

at equilibrium

v => dP/dT = 0

when kcat is comparable to k-1 then som of the ES complex will be converted to product

and dP/dT > 0

So the equilibrium is not reached

d)

When k_cat << k_-1 ES complex will dissociate back to E and S

When k_cat ~ k_-1 In addition to dissociating back to E and S , ES complex is converted to product.

the concentration of ES is further reduced.

So the concentration [ES] will be lower than if k_cat << k_-1