Quick questions on Enzyme action and rate of reaction: VCE Biology Unit 3

A competitive inhibitor has a shape similar to the substrate and binds the active site. While it occupies the site, the real substrate cannot bind, so rate falls. Adding more substrate displaces the inhibitor and restores Vmax. The apparent Km (substrate concentration for half Vmax) rises.

As temperature rises, kinetic energy and successful collisions increase, so rate rises. Above the optimum (around 37 degrees Celsius for human enzymes), the weak bonds that hold tertiary structure break, the enzyme denatures, and rate falls sharply. Denaturation is usually irreversible.

Each enzyme has an optimum pH at which its R groups carry the charges needed for substrate binding and catalysis. Pepsin works near pH 2 (stomach); trypsin near pH 8 (small intestine); most cytosolic enzymes near pH 7. Changes in pH alter ionic and hydrogen bonding within the enzyme, distort the active site, and reduce rate.

At low substrate concentrations, rate rises linearly with substrate because most active sites are empty. As more substrate is added, more active sites are occupied, and the rate approaches a maximum (Vmax) when all active sites are saturated. Beyond this point, adding more substrate has no further effect.

Provided substrate is in excess, rate rises linearly with enzyme concentration because more active sites are available.

Define an enzyme and explain why each enzyme catalyses only one (or a few) specific reactions. [3 marks]

An enzyme has a measured rate of 25 units at pH 5, 60 units at pH 7, and 10 units at pH 9. (a) State the optimum pH. (b) Explain in terms of protein structure why activity drops at pH 5 and pH 9.

Refer to enzyme inhibition. (a) Distinguish competitive from non-competitive inhibition. (b) Predict the effect of a competitive inhibitor when substrate concentration is very high.