Copper sulfate and salivary amylase: inhibition or denaturation?
Marko is studying salivary amylase. He mixes 1.0 mL of diluted saliva with 5.0 mL of starch solution (10 g/L) at pH 7.0 and 37 °C and measures the time taken for starch to disappear using iodine drops. He repeats the assay at pH 4.0, 7.0 and 9.0, and at 20, 40, 60 and 80 °C. Activity is highest at pH 7.0 and 40 °C. At pH 4.0 activity is 14% of the maximum; at pH 9.0 it is 38%. At 60 °C activity is 45% and at 80 °C it is 4%. He then repeats the assay at pH 7.0 and 40 °C with 0.5 mmol/L copper(II) sulfate added. Activity falls to 28% of the uninhibited value. To identify the type of inhibition he measures the initial rate at five starch concentrations with and without copper sulfate. Without copper sulfate Vmax = 12.0 µmol min−1 and Km = 2.4 g/L; with copper sulfate Vmax = 3.4 µmol min−1 and Km = 2.5 g/L. Marko concludes that copper sulfate denatures amylase.
1. Using the Km and Vmax values, identify the type of inhibition. Explain why Marko's conclusion that copper sulfate denatures amylase is not supported by these data.
Read the worked answer
Kinetic diagnosis
Km is essentially unchanged (2.4 → 2.5 g/L), while Vmax falls from 12.0 to 3.4 µmol min−1, a 72% reduction. This is the signature of non-competitive (or mixed) inhibition: the inhibitor binds at a site other than the active site and reduces the number of enzyme molecules able to catalyse, but the affinity of the remaining active enzyme for starch is unchanged. Competitive inhibition would have raised Km and left Vmax unchanged.
Why denaturation is not proved
Partial denaturation would also remove active enzyme molecules and lower Vmax without changing Km, so the kinetic pattern alone cannot distinguish reversible non-competitive inhibition from irreversible denaturation. Marko needs a reversibility test: pre-incubate amylase with copper sulfate, then dilute or dialyse out the copper and remeasure activity. Recovery means reversible inhibition; no recovery means denaturation. He should also show that the effect depends on copper concentration and that adding substrate does not protect the enzyme.
2. Predict what would happen to the reaction rate if Marko doubled the starch concentration in the inhibited assay, and support your answer with a calculation.
Read the worked answer
At [S] = 4.8 g/L (twice the uninhibited Km):
- Uninhibited: v = 12.0 × 4.8 / (2.4 + 4.8) = 8.0 µmol min−1
- Inhibited: v = 3.4 × 4.8 / (2.5 + 4.8) = 2.24 µmol min−1
That is 28% of the uninhibited rate, the same fraction as before. At very high starch concentration the inhibited rate approaches 3.4 µmol min−1, not 12.0. Doubling the substrate therefore gives only a small absolute increase and cannot restore the original rate, because non-competitive inhibition lowers Vmax and is not overcome by more substrate.
3. A classmate says the 4% activity at 80 °C proves that amylase is denatured at that temperature. Evaluate this claim.
Read the worked answer
Low activity at 80 °C is consistent with denaturation but is not proof. The assay measures activity over a fixed time, so the result confounds the catalytic rate at 80 °C with the rate at which the enzyme loses activity during the assay. The enzyme might be rapidly inactivated after a few seconds, or it might retain a low catalytic rate.
Marko should pre-incubate the enzyme at 80 °C for 0, 1, 2, 5 and 10 minutes, cool each sample to 40 °C and then assay under standard conditions. If activity does not return after cooling, irreversible denaturation is demonstrated. He should also check that pH has not drifted at the higher temperature and that the iodine endpoint is still reliable.