2023 TMJC A Level Paper 2 (A)

1 2023 ‘A’ Level Examinations H2/9744 Biology Paper 2 General comments by Cambridge examiners The candidates’ responses were generally of a high standard and displayed very good subject knowledge. Most candidates were able to accurately interpret graphical information and apply their knowledge to novel situations. Candidates did not always use correct scientific terminology and there was a tendency for responses to include irrelevant details that did not address the requirements of the question. Explanations were sometimes not fully developed.

2 QUESTION 1 Fig. 1.1 shows the effect of increasing the external concentration of glucose or fatty acids on the rate of uptake of these molecules into a cell. Fig. 1.1 (a) With reference to Fig. 1.1, explain the effect of increasing the external concentration of glucose on the rate of uptake of glucose into a cell. [3] 1. [describe] Rate of uptake into the cell via facilitated diffusi on increases as external concentration increases. 2. [explain] Due to increase in glucose concentration gradient. 3. [describe] Rate no longer increases beyond a certain external concentration. 4. [explain] The glucose transporters are all saturated with glucose. Examiners’ Comments Many responses clearly explained the relationship between glucose concentration and glucose uptake. Weak responses often did not consider the specific mechanism by which glucose molecules cross cell surface membranes or recognise the role of specific membrane components in this process. Some described the relationship shown in the graph rather than trying to explain it. (b) Explain why there is no uptake of glucose into the cell at X even though glucose is present outside the cell. [1] 1. The concentration gradient is not steep enough for glucose to diffuse into the cell / the concentration of glucose outside and inside the cell are at equilibrium. 2. The affinity of glucose transporter is not high enough to bind to glucose at such low concentrations. Examiners’ Comments Most candidates were able to give a correct explanation. Some responses incorrectly referred to a lag time or a lack of ATP.

3 (c) One suggested mechanism for the transport of fatty acids into cells is shown in Fig. 1.2. This mechanism is referred to as a flip-flop mechanism. Fig. 1.2

4 With reference to Fig. 1.2, explain the flip-flop mechanism for the transport of fatty acid molecules across the cell surface membrane, including the role of hydrogen ions. [5] 1. Fatty acid contains a long hydrophobic hydrocarbon and a carboxylate group / COO− 2. Due to the charged COO− group, fatty acid cannot interact with the hydrophobic fatty acid tail of the membrane to be directly transported. 3. The hydrocarbon of the fatty acid first inserts into the outer phospholipid layer of the membrane, with the COO− interacting with the phosphate group and aqueous environment. 4. Protons (H+) then reacts with COO− group to neutralize the charge (─COOH) 5. This allows the fatty acid to traverse through the hydrophobic fatty acid core of the membrane to be inserted into the inner phospholipid layer of the membrane. 6. The proton then dissociates from the fatty acid to restore its COO− group. 7. The fatty acid is then released into the cytosol. Examiners’ Comments Strong responses made effective use of scientific terminology to explain the alignment of fatty acid molecules in the phospholipid bilayer and the involvement of hydrogen ions in the flip -flop mechanism. Some candidates were unclear about the distinction between phospholipids, triglycerides and fatty acids. Weaker responses often referred to a phosphate head or glycerol on the fatty acid molecule. [Total: 9]

5 QUESTION 2 Hydrogen peroxide is a harmful by-product of many metabolic processes in cells and is quickly broken down into oxygen and water by the enzyme catalase. Fig. 2.1 shows the energy changes during the breakdown of hydrogen peroxide into water and oxygen with, and without, catalase. Fig. 2.1 (a) Explain how the change in activation energy shown in Fig. 2.1 affects the rate of a catalase - controlled reaction. [2] 1. Catalase lowers the activation energy to speed up the rate of reaction. 2. As lesser energy is needed for the substrate to overcome to reach transition state. Note: the x-axis “progress of reaction” is NOT time. It refers to the state of the reaction: substrate to the unstable intermediates and eventually to products. Examiners’ Comments Most candidates interpreted the graph correctly and recognised the effect of enzymes on activation energy. Strong responses provided an explanation for the effect of changes in activation energy on the rate of reaction. Weak responses often simply quoted figures from the graph without addressing the question.

6 (b) The effect of storage temperature on the activity of catalase was investigated over a period of 1.6 hours. The results are shown in Fig. 2.2. Fig. 2.2 (i) With reference to Fig. 2.2, state what can be concluded from the results of the investigation. [3] 1. The higher the storage temperature, the faster the relative activity of catalase falls. 2. As the storage time increases, the relative activity of catalase continues to decrease. 3. However, the relative activity of catalase does not fall by the same magnitude with every 5C increase / decrease in relative activity is the most drastic when the storage temperature is increased from 40C to 45C, e.g. @0.6h, from 0.7 to 0.2. 4. At 50C, relative catalase activity reaches 0 by 1.2h, the fastest drop among the storage temperatures. [Accept: 1.1h to 1.4h] Examiners’ Comments Many candidates found this a challenging question and simply listed data from the graph rather than drawing conclusions. Several did not appreciate that the question was about storage temperature of the enzyme and answered the question in terms of the effect of temperature on enzyme activity.

7 (ii) Explain the effect of storage temperature on the activity of catalase, as shown in Fig. 2.2. [5] 1. The higher the temperature, the greater the kinetic energy of catalase 2. Higher rate of breakage of hydrogen bonds and ionic bonds between R-groups 3. Disrupts the 3D conformation / tertiary structure of catalase 4. Leads to denaturation of catalase 5. Change in shape of the active site 6. Unable to bind to substrate to form enzyme-substrate complex, (hence decrease in relative activity as temperature increases). Examiners’ Comments This was generally well answered, with many candidates expressing themselves clearly and showing a good understanding of the reasons why different storage temperatures affect catalase activity. Weaker responses often omitted to consider relevant details such as how storage temperature affected the tertiary structure of the enzyme. Some provided descriptions of the graph rather than developing an explanation. [Total: 10]

8 QUESTION 3 Bacteria can take up DNA fragments from the environment and incorporate them into their genomes. This process is known as transformation. As a result of transformation, bacteria may acquire new genetic traits such as resistance to antibiotics. The DNA fragments usually come from dead bacteria lysing (splitti ng open) and releasing their contents into the surrounding area. Fig. 3.1 shows the main features of environmental DNA uptake by bacteria. Fig. 3.1 Fig. 3.2 represents an enlargement of the region labelled 'uptake of DNA' in Fig. 3.1. It shows in detail how environmental DNA enters the recipient cell. Fig. 3.2

9 (a) Name the parts of Fig. 3.2 labelled B, C and D. [3] B peptidoglycan cell wall C cytosol / cytoplasm D transport protein / channel pro