Enzymes: Question 2
Syllabus 5.1
A food scientist is developing a protein-tenderising powder based on papain, a protease enzyme extracted from papaya fruit. The scientist places identical raw chicken breast cubes into five water baths at different temperatures, adds the same mass of papain powder to each cube, and leaves each cube in its water bath for 20 minutes. The scientist then blots each cube dry and measures how much mass it has lost, since papain breaks down some of the protein in the chicken into smaller fragments that dissolve into the surrounding water.
| Temperature of water bath / °C | 10 | 25 | 40 | 55 | 70 |
|---|---|---|---|---|---|
| Mass lost by chicken cube / g | 0.2 | 0.9 | 2.6 | 1.1 | 0.1 |
(a) State the independent variable and the dependent variable in this investigation. [2]
(b) Describe the pattern shown by the results in the table. [2]
(c) The scientist repeats the investigation, but this time first heats one chicken cube, with papain powder already added, to 90°C for 5 minutes, then cools it back down and leaves it at 40°C for the remaining 20 minutes. Predict how the mass lost by this cube will compare with the mass lost by the cube kept at 40°C throughout, and explain your answer. [2]
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Worked solution
Part (a): Identifying the variables
The scientist deliberately changes the temperature of the water bath between the five trials, so this is the independent variable. The scientist then measures how much mass each cube loses, so the mass lost by the chicken cube is the dependent variable.
Part (b): Describing the pattern in the results
Reading along the table from left to right:
- at 10°C, only 0.2 g of mass is lost
- the mass lost rises through 25°C (0.9 g) up to a peak of 2.6 g at 40°C
- above 40°C, the mass lost falls away again, down to 1.1 g at 55°C and just 0.1 g at 70°C
So the mass lost increases with temperature up to 40°C, then decreases as the temperature rises further. This bell-shaped pattern, rising to a peak and then falling, is typical of enzyme activity across a range of temperatures, with the peak close to the enzyme’s optimum temperature.
Part (c): Predicting the effect of pre-heating the enzyme
Heating the cube (with papain already added) to 90°C is far above papain’s optimum temperature of around 40°C. This very high temperature will have denatured the papain.
Denaturation cannot be undone just by cooling the enzyme back down. Even though the cube spends the remaining 20 minutes at 40°C, the temperature that worked best in part (b), the papain has already stopped working normally, so very little further breakdown of the protein can occur during that time.
The pre-heated cube will therefore lose less mass than the cube kept at 40°C for the full 20 minutes.
Final answers
- (a) Independent variable: temperature of the water bath. Dependent variable: mass lost by the chicken cube.
- (b) Mass lost increases with temperature up to a peak of 2.6 g at 40°C, then decreases as temperature rises further, a bell-shaped pattern with an optimum close to 40°C.
- (c) The pre-heated cube loses less mass, because heating it to 90°C denatures the papain, and the enzyme does not work normally again even once cooled back to 40°C.