Question 1 Report
A food-production laboratory checks whether a lactase preparation will work in chilled milk. Lactase breaks down the milk sugar lactose into smaller sugars. A technician places equal volumes of milk and lactase in water baths at different temperatures. The pH is kept constant. Fig. 1 shows one set of test tubes being kept in a water bath. After 8 minutes, the technician uses glucose test strips to find the concentration of glucose produced.
Table 1 shows the results.
| Temperature / degrees C | Glucose concentration after 8 minutes / mmol dm-3 |
|---|---|
| 5 | 4 |
| 20 | 18 |
| 35 | 41 |
| 45 | 29 |
| 60 | 3 |
(a) Name the substrate in this investigation. [1]
(b) Use Table 1 to state the temperature at which the lactase has its highest rate of activity. [1]
(c) Describe the change in glucose concentration between 5 degrees C and 35 degrees C. [2]
(d) Explain why very little glucose is produced at 60 degrees C. [3]
(e) Give two variables, other than temperature, that the technician should control to make this a valid comparison. [2]
(f) Calculate the mean rate of glucose production at 35 degrees C. Give your answer in mmol dm-3 min-1. [1]
(a) The substrate is lactose. [1 mark]
(b) Lactase has its highest activity at 35 °C, because the greatest glucose concentration after the same 8 minutes is recorded there. [1 mark]
(c) Glucose concentration increases. [1 mark] It rises from \(4\ \text{mmol dm}^{-3}\) at 5 °C to \(41\ \text{mmol dm}^{-3}\) at 35 °C, an increase of \(37\ \text{mmol dm}^{-3}\). [1 mark]
(d) At 60 °C, high temperature denatures lactase or changes its overall shape. [1 mark] The active site changes shape. [1 mark] Lactose no longer fits the active site, so few enzyme-substrate complexes form and little glucose is produced. [1 mark]
(e) Any two suitable control variables are acceptable, for example:
[1 mark each, maximum 2 marks]
(f)
\[\text{mean rate}=\frac{41\ \text{mmol dm}^{-3}}{8\ \text{min}}=5.125\ \text{mmol dm}^{-3}\text{ min}^{-1}\]
To an appropriate precision, the mean rate is \(5.1\ \text{mmol dm}^{-3}\text{ min}^{-1}\). [1 mark]
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