Enzymes: Question 3

Worked solution

Part (a): The substrate concentration graph and why it plateaus

Plotting initial rate against starch concentration would give a curve that rises steeply between 0.5% and about 2.0%, then flattens into a plateau from 2.0% to 3.0% (the rate barely changes, from 7.5 to 7.6 arbitrary units).

At low substrate concentrations, many of the enzyme’s active sites are free at any given moment. Increasing the starch concentration increases the chance of a starch molecule colliding with a free active site, so more enzyme-substrate complexes form per unit time and the rate rises.

As substrate concentration keeps increasing, however, a point is reached where essentially all of the active sites are occupied at any instant. The enzyme is said to be saturated. Beyond this point, adding more starch cannot increase the rate any further, because it is now the number of available enzyme molecules (and how quickly each one can process a substrate molecule and release product) that limits the reaction, not the substrate concentration. This maximum rate, reached once the enzyme is saturated, is called Vmax.

Part (b): Doubling the enzyme concentration

Doubling the concentration of amylase doubles the number of enzyme molecules present, and so doubles the number of active sites available in the reaction mixture. Even at high starch concentrations, there are now twice as many active sites able to bind substrate and form enzyme-substrate complexes before all of them become occupied.

This means the plateau (the maximum rate, Vmax) would be reached at a higher rate than in the original experiment. (The substrate concentration needed to reach that new, higher plateau might also increase, since there are now more active sites to fill, but the key result asked for here is that the maximum rate itself increases.)

Part (c): Repeating the experiment at a low pH

Every enzyme has an optimum pH at which its rate of reaction is greatest. This is because the active site’s precise shape depends partly on ionic bonds between charged R-groups (side chains) of amino acids, and the correct ionisation of R-groups directly in the active site is often needed for the substrate to bind properly.

At a pH well below the enzyme’s optimum, there is a much higher concentration of hydrogen ions than usual. This changes the ionisation state of acidic and basic R-groups both in and around the active site, disrupting some of the ionic bonds (and hydrogen bonds) that hold the enzyme’s tertiary structure in shape. As a result, the active site becomes less complementary to the starch substrate, it either binds substrate less readily or catalyses the reaction less effectively.

Because this effect on active site shape is independent of how much starch is present, the initial rate would be reduced at every substrate concentration tested, not just at one point on the curve. If the pH were extreme enough, the ionic and hydrogen bonds could be disrupted so severely that the enzyme becomes permanently denatured.

Final answers

• (a) The graph rises steeply then plateaus at Vmax; the plateau occurs because at high substrate concentration nearly all active sites are occupied (saturated) at any moment, so the enzyme, not the substrate, limits the rate.
• (b) The maximum rate (Vmax) would increase, because doubling the enzyme concentration doubles the number of active sites available.
• (c) The initial rate would be lower at every starch concentration, because the low pH disrupts ionic/hydrogen bonds maintaining the active site’s shape, making it less complementary to the substrate.