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Official June 2026 AQA AS BIOLOGY 7401/1 Paper 1 Merged Question Paper + Mark Scheme (54pages)

AQA PAPERS AND MARK SCHEME Feb 17, 2025
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Answer all questions in the spaces provided. box 0 1 . 1 Figure 1 shows a diagram of the human gas exchange system. Figure 1 Structure B is muscle. The volume of the thorax is increased when structure B contracts. Name the structures labelled A and B. [2 marks] A B 3 *03* Turn over ► IB/M/Jun24/7401/1 Do not write outside the 0 1 box . 2 Figure 2 is a diagram of an alveolus and a blood vessel associated with the alveolus. The direction of the movement of substances is shown by the arrows labelled C, D, E, F and G. Figure 2 Complete Table 1 by writing the correct letter, C, D, E, F or G, in the box next to each statement. You may use each letter once, more than once or not at all. [2 marks] Table 1 Box Statement represents the diffusion of oxygen represents the transport of a gas mixture represents blood moving at the highest pressure 4 *04* IB/M/Jun24/7401/1 Do not write outside the 0 1 box . 3 Describe and explain one adaptation of a single alveolus for gas exchange. [2 marks] 6 5 *05* Turn over ► IB/M/Jun24/7401/1 Do not write outside the 0 2 box . 1 Table 2 contains information on the digestion of two biological molecules. Complete Table 2. [4 marks] Table 2 Biological molecule Enzyme Name of bond hydrolysed Product of digestion Starch Maltose Dipeptide Dipeptidase A student investigated starch digestion by mixing starch with a solution of the enzyme used to digest starch. The student did a biochemical test for protein when starch digestion was completed. 0 2 . 2 Describe a biochemical test to show the presence of protein. [2 marks] Question 2 continues on the next page 6 *06* IB/M/Jun24/7401/1 Do not write outside the 0 2 box . 3 The student’s test for protein was positive. Explain why. [2 marks] 0 2 . 4 An enzyme’s turnover number (kcat) is the number of substrate molecules converted into product molecules by one enzyme molecule in 1 second. It is determined using this equation. kcat = Maximum rate of enzyme-controlled reaction / μmol dm–3 s–1 Enzyme concentration / μmol dm–3 A scientist investigated the action of a protease enzyme. The scientist prepared a reaction mixture with a protease concentration of 0.0118 µmol dm–3 . The kcat for the protease is 110 substrate molecules per second. Use this information and the formula to calculate the maximum rate of the protease-controlled reaction. Give your answer to 3 significant figures. Show your working. [2 marks] Answer μmol dm–3 s–1 10 7 *07* Turn over ► IB/M/Jun24/7401/1 Do not write outside the 0 3 box . 1 Describe how viruses are replicated. [4 marks] Question 3 continues on the next page 8 *08* IB/M/Jun24/7401/1 Do not write outside the 0 3 box . 2 A scientist investigated the effect of a virus species on a bacterial culture. The scientist: • mixed a diluted sample of the virus with a bacterial culture in liquid agar • poured the liquid agar into a Petri dish and let the agar set • incubated the virus-bacterial culture at 30 °C for 24 hours. Figure 3 shows the appearance of the Petri dish after 24 hours.


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Official June 2026 AQA AS BIOLOGY 7401/1 Paper 1 Merged Question Paper + Mark Scheme (54pages)

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