RI N23 P2 ans

2023 Nov P2 answers Question 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. As external concentration of glucose increases, the rate of glucose uptake into the cell increases steeply at first, but the increase slows down and eventually reaches maximum rate at a plateau. 2. As glucose is polar*, and hence hydrophilic, glucose molecules would be repelled by the hydrophobic core of the phospholipid bilayer. 3. The entry of glucose requires transmembrane carrier proteins, but at high external glucose concentration, the carrier proteins become saturated and become a limiting factor and hence the rate of uptake plateaus. 4. As the external concentration increases, the concentration gradient is steeper and hence the rate of uptake by facilitated diffusion increases. (b) Explain why there is no uptake of glucose into the cell at X even though glucose is present outside the cell.[1] 1. There is no diffusion gradient between the cytosol and the solution outside the cell as the concentration of glucose in the cell and outside the cell is the same. (c) 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. Positively charged H+ ions surround the negatively charged carboxyl groups to reduce the negative charge and allow the fatty acid molecule to be embedded on the top half of the phospholipid bilayer; 2. The fatty acid molecule is orientated in such a way that the hydrophobic non-polar hydrocarbon tail is embedded between the hydrophobic non-polar hydrocarbon tails of the phospholipid in the top half of the phospholipid bilayer; 3. And the negatively charged carboxylic acid group is between the charged phosphate head of the phospholipid molecules; 4. Positively charged H+ ions then bind to the negatively charged carboxylic acid group to neutralise the negative charge so that the non-polar fatty acid molecule can flip within the hydrophobic core of the phospholipid bilayer without being repelled; 5. Once the non-polar fatty acid molecule is embedded at the lower half of the phospholipid bilayer, the H+ ions dissociate from the carboxyl group that now orientates outwards towards the internal cytosolic side of the cell; [Total: 9]

Question2 (a) Explain how the change in activation energy shown in Fig. 2.1 affects the rate of a catalase- controlled reaction.[2] 1. Enzyme lowers activation energy* by forming 3 different enzyme-substrate complexes required for chemical reaction to take place, from 75kJ mol-1 without enzyme to 28 kJ mol- 1 with enzyme; 2. Increases number of substrate molecules with the required energy to cross the activation energy* barrier so