Question 1
Paper 2Hard16 marksThe reaction S → P occurs in a cell and is catalysed by enzyme E. To investigate the reaction, equal volumes containing the same initial amount of S and no P were held at the same temperature and pH, either with E or without E. Table 1 shows the measured concentration of P at different times. Values are reported to the nearest 0.1 mmol dm⁻³.
| Time / min | [P] with E / mmol dm⁻³ | [P] without E / mmol dm⁻³ |
|---|---|---|
| 0 | 0.0 | 0.0 |
| 1 | 2.8 | 0.1 |
| 2 | 5.1 | 0.2 |
| 4 | 6.9 | 0.4 |
| 8 | 7.8 | 0.8 |
| 16 | 8.0 | 1.7 |
| 32 | 7.9 | 3.2 |
| 64 | 8.0 | 5.5 |
| 128 | 8.0 | 7.8 |
Figure 1 shows schematic energy profiles for the conversion of S to P.
Calculate the mean rate of formation of P during the first 2.0 min, with E and without E. Include units.
Describe the change in [P] in each mixture during the first 8 min.
Using Figure 1, calculate how much E lowers the activation energy.
Explain how E causes the faster formation of P and state one benefit of this increase in reaction rate for cells.
A cell process Q requires a supply of P at an average rate of at least 1.2 mmol dm⁻³ min⁻¹ during the first 2.0 min. Using your results from (a), determine whether the reaction with E and the reaction without E meet this requirement.
A student concludes that “E increases the total amount of P that can be produced from a fixed starting amount of S, not just the rate at which P appears”. Evaluate this conclusion using the data at 2 min and 128 min, and any other relevant data in Table 1.