Enzyme Kinetics

How come enzymes with a lower $$K_\mathrm{m}$$ are more easily saturated with substrate than an enzyme with higher $$K_\mathrm{m}?$$

Is it because enzymes with higher $$K_\mathrm{m}$$ are able to bind substrate and hold onto it tighter and therefore have a higher chance of becoming a product?

I'm confused because shouldn't a lower $$K_\mathrm{m}$$ indicate that an enzyme is able to catalyze a reaction quicker?

The Michaelis–Menten kinetics model formula is: $$V= \frac{(d¦P¦)}{dt} = \frac{(Vmax)¦S¦}{Km + ¦S¦}$$ Reaction rate V is the rate of formation of product, ¦P¦ is the concentration of product and the ¦S¦ the concentration of a substrate S.

In a saturated regime V = Vmax and this occours only if: $$\frac{¦S¦}{Km + ¦S¦} \approx 1$$

This is nearly equal to 1 only when $$¦S¦ >> Km$$

In conclusion: If Km is small then a lower concentration of substrate ¦S¦ is required to saturate the enzyme.

• ahh, you're right, thank you! the two comments on my question really helped. – Ahmer Imam Aug 2 at 19:48

Application of the Michaelis-Menten equation generally assumes steady state concentration of the enzymatic complex intermediate. Then $$K_M$$ allows two interpretations$$^1$$:

• an apparent dissociation constant describing the equilibrium $$\ce{ES<=>E +S}\tag{1}$$ where $$E$$ and $$S$$ are the concentrations of enzyme and substrate, respectively
• the concentration of substrate at which the rate of formation of products $$\nu=\frac{V_{max}}{2}$$ where $$V_{max}$$ is the maximum (limiting) rate at saturation with substrate.

Note that in the case of Michaelis-Menten kinetics, $$V_{max} = k_{cat}[E]_0\tag{2}$$ where $$k_{cat}$$ is the rate constant for conversion of substrate into product: $$\ce{ES->[k_{cat}]E + P}\tag{3}$$ Equation (2) follows from the general Michaelis-Menten expression at saturation since all of the enzyme is then converted into intermediate ($$[ES]=[E]_0$$).

In light of these definitions, an enzyme with a lower $$K_M$$ saturates at a lower concentration of substrate, ie the intermediate complex is more favored.

Is it because enzymes with higher Km are able to bind substrate and hold onto it tighter and therefore have a higher chance of becoming a product?

Beware how you think about this. Larger $$K_M$$ means enzyme-substrate complex is less favored.

When the intermediate complex $$ES$$ is more thermodynamically stable than the free enzyme, the equilibrium is shifted toward the intermediate, leading to faster turnover. There is more of the complex to "leak" into product. If this is what you mean by "tighter", then yes, "tighter" leads to more product.

I'm confused because shouldn't a lower Km indicate that an enzyme is able to catalyze a reaction quicker?

All else being equal, and at steady-state, yes.

References

1. Fersht, Alan. Structure and Mechanism in Protein Science. New York: Freeman, 1999.
• this was a good answer! thank you! it helped me look at Km in a different light – Ahmer Imam Aug 2 at 19:48