Biology 9700 · AS & A Level · Protein synthesis

Protein synthesis — practice question

The enzyme catalase occurs in many plant and animal tissues. It catalyses the breakdown of hydrogen peroxide, a poisonous metabolic waste product. The reaction is: $\text{2H}_2\text{O}_2 \xrightarrow{\text{catalase}} \text{O}_2 + 2\text{H}_2\text{O}$. A research team examined the activity of two catalase forms, P and Q, taken from \emph{Anopheles gambiae}, an important malaria vector. The team studied how increasing hydrogen peroxide concentrations affected the activity of these two catalase forms. The results are shown in Fig. 3.1.
(a)[5]

With reference to Fig. 3.1, describe and explain the effect of increasing the concentration of hydrogen peroxide on the activity of catalase P.

(b)[2]

Each catalase molecule is made of four identical polypeptides. The two catalase forms in $A.\ gambiae$ differ by only one amino acid at position 2 in the amino acid sequence. Catalase P has serine and catalase Q has tryptophan. Suggest how the difference in one amino acid is responsible for the lower activity of catalase Q compared with catalase P.

(c)[2]

Female mosquitoes feed on blood so that they can produce their eggs. After feeding, the metabolic rate rises for egg production. The researchers allowed female mosquitoes to feed on blood. They found that female mosquitoes with only catalase P produced more eggs than those with only catalase Q. Suggest why there is a difference in egg production between the two types of $A.\ gambiae$.

(d)[2]

Metal ions can also act as a non-competitive inhibitor of catalase. Explain how copper ions can act as a non-competitive inhibitor.

(e)[2]

Enzyme inhibitors can also inhibit carrier proteins in cell surface membranes. Explain why carrier proteins are required in cell surface membranes.

(f)[3]

Describe three roles of cell surface membranes, excluding the transport of substances into and out of cells.

Worked solution & mark scheme

This 16-mark question has a full step-by-step worked solution and mark scheme. One marking point: Activity rises to a maximum/plateau ($V_{max}$)

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