Why are enzyme-catalysed reactions slower at lower substrate concentration?

Question

Suppose I'm using 200 nmoles of enzyme and 2 mmoles of substrate. The enzyme should be saturated but if I use 50 mmoles of substrate, the reaction will be faster. Why? I just can't get it! Even at lower concentration of substrate the enzyme is saturated (all enzyme molecules are engaged in the reaction) but still the reaction rate is lower.

Answer 1

Alan Boyd's answer covers the mechanistic aspects quite well, but there is another aspect he didn't quite touch on - what's going on at the molecular level. To understand this, you need to think about the actual conditions of the reaction: unless it's taking place inside a cell, which straight biochemical reactions of the kind you're talking about almost never are, then the enzyme, substrate, and product are all floating around at a certain concentration. The enzymatic reaction is proceeding nicely, and as the substrate binding pocket is freed as product dissociates from the enzyme and diffuses away, allowing another substrate molecule to take its place and the next reaction to occur. This is the situation at your 2 mmol [substrate] level. At 50 mmol, the concentration of substrate is much higher, and so even though the actual catalytic portion of the enzymatic reaction is taking place at essentially the same speed, the rate of substrate diffusing into the active pocket is somewhat higher because the local concentration of substrate is higher, allowing for an apparently slightly faster reaction rate.

Answer 2

In terms of Michaelis-Menten kinetics, the rate never reaches the maximum rate:

$v = V_{max} \times \frac{S}{K_m + S}$

where $S$ is substrate concentration. Notice that however large $S$ is, the term on the bottom line ($K_m + S$) will be larger than $S$, so $v$ will be less than $V_{max}$.

You can think about this in terms of the binding equilibrium between the enzyme and the substrate:

$E + S \leftrightarrow ES$

Since it is an equilibrium, it can never lie completely over to the right hand side.

This is why $V_{max}$ is always determined by extrapolation. In practice of course, at very high substrate:enzyme concentrations your measurements would have to be very accurate to see the difference: if $S$ is $100 \times K_m$ you are looking at a 1% difference. In your hypothetical example you don't mention what the value of $K_m$ is.