Original application practice · 13 indicative marks
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An enzyme assay uses the same enzyme amount, pH and temperature (30 °C) at each substrate concentration. Initial rates are measured over a short interval. Each value is a mean of three independent assays; their spread is not supplied.
| Substrate concentration / mmol dm⁻³ | Initial rate / µmol min⁻¹ |
|---|---|
| 0.5 | 0.12 |
| 1 | 0.22 |
| 2 | 0.38 |
| 4 | 0.50 |
| 8 | 0.52 |
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Initial rate rises from 0.12 to 0.50 µmol min⁻¹ as substrate increases from 0.5 to 4 mmol dm⁻³. From 4 to 8, it rises by only 0.02 µmol min⁻¹.
Alternatives: Other accurate data pairs are acceptable. The description should distinguish the levelling trend from a rate that is exactly constant.
A small difference between means is not automatically a statistically significant difference.
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.
120 s = 2 min. Product = 0.50 µmol min⁻¹ × 2 min = 1.00 µmol.
Alternatives: Working in seconds with 0.50/60 µmol s⁻¹ is equivalent. Accept 1 µmol, 1.0 µmol or 1.00 µmol.
This calculation depends on the stated constant-rate assumption; an initial rate need not persist throughout a real assay.
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.
Most active sites may already be occupied. With a fixed enzyme amount, catalytic capacity limits the rate, so extra substrate causes little further increase.
Alternatives: A clear saturation explanation using maximum turnover or limited available active sites is acceptable. This is a plausible interpretation, not proof that no other factor limits the assay.
Do not infer enzyme denaturation solely from a plateau.
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.
The predicted rate is 0.26 µmol min⁻¹. Halving active enzyme amount halves the available catalytic capacity under the stated model.
Alternatives: Equivalent descriptions of half the active enzyme capacity are acceptable. The proportionality is supplied as an assumption.
Do not present the supplied proportional model as an experimentally established result.
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.
The enzyme remains active at 8 mmol dm⁻³, and saturation can explain the plateau. Assay exposed and untreated enzyme under the same subsequent conditions, controlling enzyme amount and handling. Replicates would show whether any persistent activity difference exceeds repeat variation.
Alternatives: Diluting or removing the original substrate before a common assay can help separate persistent damage from reversible assay effects. Other controls are acceptable if they address that distinction.
Lack of supplied spread prevents a significance claim about the small difference between the means.
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.
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