Enzymes: rates, plateaus and conclusions

Initial rate rises from 0.12 µmol min⁻¹ at 0.5 mmol dm⁻³ to 0.50 at 4. Doubling substrate from 4 to 8 only increases rate to 0.52, a difference of 0.02 µmol min⁻¹. The measured relationship is approaching a plateau.

• State that initial rate rises across the measured concentrations.
• Compare 0.12 at 0.5 with 0.50 µmol min⁻¹ at 4 mmol dm⁻³.
• Identify the small 0.02 µmol min⁻¹ increase between 4 and 8 mmol dm⁻³.

Common confusion: Doubling substrate does not double the rate throughout the data.

120 seconds = 2.00 minutes. Product formed = rate × time = 0.50 × 2.00 = 1.00 µmol. This uses the supplied assumption that the rate remains constant over that interval.

• Convert 120 seconds to 2.00 minutes.
• Multiply 0.50 µmol min⁻¹ by 2.00 minutes.
• Give 1.00 µmol and recognise the stated constant-rate assumption.

Common confusion: Multiplying a per-minute rate by a time in seconds without conversion gives a sixtyfold error.

At high substrate concentration, many active sites may already be occupied. With a fixed enzyme amount, catalytic turnover limits how quickly product can be formed. Adding substrate then has little effect because available enzyme capacity, rather than substrate supply, is limiting under this interpretation.

• Connect the enzyme amount to a finite number of active sites.
• Explain increasing occupancy and a limit on simultaneous catalytic processing.
• Relate near-saturation to substrate no longer being the main limiting factor.

Common confusion: A plateau in these data does not by itself demonstrate denaturation.

Under the supplied proportional model, the predicted rate is 0.52/2 = 0.26 µmol min⁻¹. Fewer active enzyme molecules provide fewer active sites and lower total catalytic capacity. The prediction should be checked experimentally because other limiting conditions can affect proportionality.

• Use the stated proportional relationship rather than an unrestricted rule.
• Halve 0.52 to predict about 0.26 µmol min⁻¹.
• Explain the reduction in available catalytic capacity at high substrate.

Common confusion: This is a conditional prediction; halving any variable in an enzyme experiment does not always halve the rate.

The rate at 8 remains high and slightly exceeds that at 4; the plateau alone does not establish permanent damage. Saturation is an alternative. Compare treated and untreated enzyme under a common assay condition, controlling enzyme amount and handling, to test whether activity remains impaired. Repeat assays and their spread are also needed to judge the small difference.

• Recognise that the measured rate is still high and not a direct damage measurement.
• Offer substrate saturation as an alternative explanation for the plateau.
• Propose testing recovered enzyme under a shared lower-substrate condition with a suitable untreated comparison.

Common confusion: A mechanism must be supported by evidence that discriminates it from plausible alternatives.