Enzymes: Question 8

Worked solution

Part (a): Describing the relationship

Looking at the data, every 1% increase in enzyme concentration adds exactly 8 arbitrary units to the initial rate: 0, 8, 16, 24, 32, 40 for 0%, 1%, 2%, 3%, 4%, 5%. This is a directly proportional relationship, the rate is always $8 \times$ the enzyme concentration, so a graph of initial rate against enzyme concentration would be a straight line through the origin.

Part (b): Why the rate is directly proportional to enzyme concentration

The number of active sites available in the reaction mixture depends on how many enzyme molecules are present, so it is directly proportional to enzyme concentration. Because substrate is kept in large excess throughout this experiment, there is always plenty of substrate available, far more than there are active sites, so essentially every active site can be occupied by a substrate molecule as soon as it is free, and no active site is ever left waiting for substrate.

This means the rate of the reaction is limited only by the number of active sites present, not by substrate availability. Doubling the enzyme concentration doubles the number of active sites, so it doubles the number of enzyme-substrate complexes that can form per second, and therefore doubles the rate, giving the directly proportional relationship seen in the data.

Part (c): Predicting the rate at 6% enzyme concentration

Since the relationship is directly proportional, the gradient of the graph gives the rate produced per percent of enzyme concentration:

$\text{gradient} = \frac{40 - 0}{5 - 0} = 8 \text{ (arbitrary units per \% enzyme concentration)}$

Assuming the substrate is still in large excess at 6% enzyme concentration (so the direct proportionality still holds), the expected initial rate is:

$8 \times 6 = 48$

So the expected initial rate at 6% enzyme concentration is 48 arbitrary units.

Part (d): Repeating the experiment with limited substrate

If the substrate concentration is fixed and limited rather than in large excess, the straight-line relationship can only hold while there is still enough substrate to keep every active site occupied. This will still be true at low enzyme concentrations, where active sites are relatively scarce compared with substrate.

As enzyme concentration keeps increasing, however, the fixed amount of substrate eventually becomes too little to occupy all of the (now much more numerous) active sites at any one moment. Substrate availability, not the number of active sites, then limits the rate. So instead of continuing upward as a straight line, the graph would curve over and level off into a plateau at high enzyme concentrations, similar in shape to the plateau seen when substrate concentration itself is the variable being increased and the enzyme becomes saturating.

Final answers

• (a) The initial rate is directly proportional to enzyme concentration, a straight line through the origin.
• (b) With substrate in excess, the number of active sites (proportional to enzyme concentration) is the limiting factor, so rate scales directly with enzyme concentration.
• (c) Gradient $=8$; expected rate at 6% $= 8 \times 6 = \boxed{48}$ arbitrary units.
• (d) With limited substrate, the graph would follow the same line only at low enzyme concentrations, then curve and plateau once substrate, not active sites, becomes limiting.