All Questions: AS & A Level Biology - 9700 Paper 5 2014 Winter Zone 3
Theory
MCQ
01.
Theory 15 Marks
CH1 - CELL STRUCTURE

(a) Many people are intolerant to the disaccharide lactose, which is found in milk. The enzyme lactase is used commercially to catalyse the breakdown of lactose to the monosaccharides glucose and galactose. These sugars taste sweeter and are easier to digest than lactose.

Enzymes can be immobilised in a number of different ways, using different materials.

Fig. 1.1 shows three ways of immobilisation of enzymes.



A student carried out an investigation to compare the activity of the enzyme lactase that had been immobilised in the three different ways shown in Fig. 1.1.

• A solution containing 20 mg cm-3 of lactose was poured through a column containing the immobilised enzyme.
• The solution containing the products was collected and the concentration of glucose measured.

(i) State a null hypothesis that the student could make for this investigation.
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(ii) State the independent variable in this investigation.
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(iii) Suggest how the dependent variable might be measured.
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(b) Outline a procedure the student could use to compare the activity of lactase that has been immobilised in different ways.
Your method should be detailed enough for another person to use.
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(c) Table 1.1 shows the student's results.

[Table_1]

(i) Complete Table 1.1 by calculating the mean total glucose collected for A and C. [1]

(ii) State the other two pieces of information that are required to calculate the rate of hydrolysis of lactose by these immobilised enzymes.
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(d) Table 1.2 shows the results of a number of statistical tests to find out if the differences in the rates of reaction were significant.

[Table_2]

(i) State what is meant by ‘significant at P < 0.05’.
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(ii) The student concluded that immobilising lactase by adsorption onto the surface of glass beads (A) has a greater activity than either of the other ways of immobilising the enzyme. With reference to Table 1.1 and Table 1.2, state the evidence that supports this conclusion.
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02.
Theory 11 Marks
CH1 - CELL STRUCTURE

(a) (i) Explain why the results were converted to cpm per unit area of leaf.
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(ii) State how the mean cpm per unit area of leaf was calculated.
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(b) (i) Fig. 2.2 shows that:
• radioactivity in leaves that were covered is always lower than in the leaves that were not covered
• in both leaves that were covered and leaves that were not covered, radioactivity increased to day 3 and then decreased.
Suggest reasons for these observations.
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(ii) Identify two ways in which the reliability of the results could be improved.
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(c) Radioactivity causes dark spots to appear on X-ray film.
The radioactive phosphate is transported to the leaves in a tissue. Outline a method, using X-ray film, to locate and identify this tissue.
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(d) Another set of plants were exposed to moving air after they were transferred to unlabelled nutrient. Predict, with reasons, the effect on the measurements of radioactivity in leaves that were covered and in leaves that were not covered.
covered leaves ..........................................................
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leaves not covered ..........................................................
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