Enzymes: Question 9

Worked solution

Part (a): Why use the tangent at $t=0$, not the average rate

Across the whole 10 minutes, the reaction does not proceed at a constant rate: it is fastest right at the start and gets progressively slower as the experiment continues (as later parts of this question explore). The tangent to the curve at $t=0$ gives the gradient of the curve at that one instant. The true initial rate, before the rate has had any chance to fall.

The average rate over the full 10 minutes, in contrast, is found from the total product formed divided by 10 minutes. This blends the fast initial rate together with the much slower rate later on (once the curve has nearly flattened out), so it gives a value lower than, and not representative of, the actual rate at the start of the reaction. Using the initial rate (from the tangent at $t=0$) is also the standard way of comparing rates between different experiments (for example, at different substrate or enzyme concentrations), since it captures the rate before substrate depletion begins to complicate the comparison.

Part (b): Calculating the initial rate from the tangent

The tangent at $t=0$ is given as passing through the points $(0, 0)$ and $(2, 16)$. Its gradient, which is the initial rate of reaction, is:

$\text{initial rate} = \frac{16 - 0}{2 - 0} = \frac{16}{2} = 8$

So the initial rate of reaction is 8 mmol dm-3 per minute.

Part (c): Why the rate decreases over time

Substrate is continuously converted into product as the reaction proceeds, so the concentration of substrate remaining steadily decreases through the experiment. Because temperature and pH are constant, this fall in rate cannot be due to a change in the enzyme’s activity through denaturation or altered ionisation. The enzyme itself is chemically unchanged and reusable throughout.

Instead, with less substrate present, substrate molecules encounter free active sites less often. The frequency of successful collisions between enzyme and substrate falls, so fewer enzyme-substrate complexes form per minute. A greater proportion of active sites are left unoccupied at any given moment as substrate becomes scarcer, so the rate of product formation progressively slows.

Part (d): Why the curve becomes horizontal

By around 8 to 10 minutes, almost all of the substrate originally present has already been converted into product. With no meaningful amount of substrate left, active sites have nothing left to bind, so no further enzyme-substrate complexes can form and no more product is produced. The concentration of product therefore stays constant, producing the flat, horizontal section of the curve. Because temperature and pH have been kept constant throughout, this plateau reflects the substrate being used up, not the enzyme becoming denatured.

Final answers

• (a) Rate is not constant; the tangent at $t=0$ gives the true initial rate, whereas the average rate over 10 minutes is diluted by the much slower later rate and underestimates the initial rate.
• (b) Initial rate $=\dfrac{16-0}{2-0}=\boxed{8}$ mmol dm-3 per minute.
• (c) Falling substrate concentration means fewer enzyme-substrate collisions and complexes form per minute, slowing the rate, even though the enzyme itself and the conditions are unchanged.
• (d) The plateau occurs because the substrate has essentially all been converted to product, leaving none for further reaction, not because the enzyme has denatured.