Enzyme kinetics question

[salemstein]

Given E + S ⇄ ES -> E + P, will lowering k-1 increase kcat? Will it decrease Km?

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[aldol16]

Km = (k-1 + k2)/k1. Lowering k-1 thus lowers Km. Whether it lowers kcat depends on whether k1/k-1 or k2 is slow - one would have to apply either the pre-equilibrium or steady-state approximation and work out the kinetics.

 

[salemstein]

Km = (k-1 + k2)/k1. Lowering k-1 thus lowers Km. Whether it lowers kcat depends on whether k1/k-1 or k2 is slow - one would have to apply either the pre-equilibrium or steady-state approximation and work out the kinetics.

So you're saying if k2 is faster than k-1 or k1, then you'd have a higher kcat?

 

[aldol16]

So you're saying if k2 is faster than k-1 or k1, then you'd have a higher kcat?

I actually don't know. The calculation is a bit more involved but I think if you apply either the pre-equilibrium approximation or the steady-state assumption, the math works out that k-1 will affect the kcat. You just have to work it out and see. Intuitively, I would say it does because if k2 is rate-limiting, then k-1 definitely affects rate because it's before the RDS. If k2 is not rate-limiting, then k-1 may or may not affect the rate, depending on how big it is. In that case, since the first step would have to be rate-limiting, it's more likely that k-1 will not affect the rate because if the first step is rate-limiting, then k-1 must be necessarily << k1.