Potato tissue contains the enzyme catalase, which breaks hydrogen peroxide down into water and oxygen. In this investigation equal-sized discs of potato wer...

Assessment: Biology 0610 | Paper 3 Mock 01 | Theory (Core) Subject: Biology - 0610

Question 1 Report

Potato tissue contains the enzyme catalase, which breaks hydrogen peroxide down into water and oxygen. In this investigation equal-sized discs of potato were placed in hydrogen peroxide buffered to hold it at a chosen pH. The oxygen produced formed a foam, and the height of this foam after two minutes was measured with a ruler. The experiment was repeated at a range of pH values, and the results are given in the table. Use the diagram and the table to answer the questions.

diagram
pH357911
Height of foam / mm52045258

(a) State the pH at which catalase is most active. [1]
(b) Describe the effect of pH on the activity of catalase shown by these results. [3]
(c) Measure the difference in foam height between pH 5 and pH 7. [1]
(d) Explain why catalase is important in living cells. [2]
(e) Suggest why discs of equal size were used. [2]
(f) Before the experiment some discs were left in a concentrated salt solution and became soft and floppy. Explain this change in terms of osmosis. [3]
(g) State the class of biological molecule that catalase is made from. [1]

Answer Details

This question tests how pH affects enzyme activity, using catalase breaking down hydrogen peroxide. Taller foam means faster oxygen release, so foam height is a measure of enzyme activity.

(a) pH of greatest activity [1] - pH 7, where the foam is tallest (45 mm).

(b) Effect of pH on catalase activity [3]

  1. As the pH increases from 3 to 7, the activity (foam height) increases (1).
  2. Activity is highest at the optimum, pH 7 (1).
  3. Above pH 7 the activity decreases again as the pH rises to 11 (1).

(c) Difference in foam height between pH 5 and pH 7 [1]

\( 45 - 20 = 25 \ \text{mm} \) (1).

(d) Why catalase is important in living cells [2]

Hydrogen peroxide is a toxic waste product of metabolism (1); catalase breaks it down into harmless water and oxygen, protecting the cells from damage (1).

(e) Why discs of equal size were used [2]

Discs of equal size have the same surface area, and so the same amount of enzyme exposed (1); this keeps the test fair so that only pH affects the result (1).

(f) Why discs in concentrated salt solution became soft and floppy [3]

  1. The salt solution has a lower water potential than the cell contents (1).
  2. Water moves out of the cells by osmosis through the partially permeable membrane (1).
  3. The cells lose turgor and become flaccid, so the tissue becomes soft (1).

(g) Class of molecule catalase is made from [1] - protein.

Exam tip: enzyme activity peaks at an optimum pH and falls either side because extreme pH changes the shape of the active site (denaturation). Describe the trend on both sides of the optimum for full marks.

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