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 the reaction can proceed faster. (b) (i) With reference to Fig. 2.2, state what can be concluded from the results of the investigation.[3] 1. Lower temperature of 35ºC is the most ideal for storage compared to higher temperature (40ºC to 50ºC); 2. Catalase stored at 35ºC retained 72% activity after 1.6 hours while higher temperatures had lower activity (quote from any other temperature); 3. Catalase lost its activity the fastest when stored at higher temperatures (quote relative activity or reference to gradient from 0.0 to 0.2 hours); 4. Within 1.6 hours, storage at temperatures at or above 45ºC resulted in loss or almost complete loss of activity, but 35 ºC and 40 ºC still retained 50% to 70% activity; 5. AVP; Activity decreased when stored at all temperatures, indicating that storage is not idea for catalase; (c) Explain the effect of storage temperature on the activity of catalase, as shown in Fig. 2.2. [5] 1. the higher the storage temperature, the greater the amount of thermal agitation / intramolecular vibrations; 2. at temperatures beyond the optimum temperature of an enzyme , weak interactions such as hydrogen bonds, ionic bonds and hydrophobic interactions between R groups will break;

3. the 3D conformation will be lost, resulting in denaturation* 4. the active site* of catalase is no longer complementary in shape* and charge to the substrate and the activity decreases; 5. enzymes exposed to increased temperatures for longer durations will have lower activity; [Total: 10] Question 3 (a) Name the parts of Fig. 3.2 labelled B, C and D. [3] B : peptidoglycan cell wall C : cytoplasm D : transport channel protein (b) With reference to Fig. 3.2, suggest how environmental DNA fragments enter the bacterial cell. [3] 1. Double-stranded DNA passes through the peptidoglycan cell wall and binds to protein A; 2. which is attached to a membrane bound transmembrane channel protein at the outer surface of the phospholipid bilayer of the bacterial cell membrane; 3. The hydrogen bonds between the 2 strands of the DNA are broken and only a single DNA strand enters the cytosol via the channel protein; (c) Two other ways in which DNA can enter bacterial cells are transduction and conjugation. Describe how DNA enters bacterial cells by transduction and conjugation. [4] transduction General transduction 1. A phage* infects a bacterium*, injecting its viral genome into the host cell The bacterial DNA is degraded into small fragments, one of which may be randomly packaged into a capsid* head during the spontaneous assembly* of new viruses; 2. Upon cell lysis, the defective phage will infect another bacterium and inject bacterial DNA from the previous host cell into the new bacterium; 3. The foreign bacterial DNA can replace the homologous region in the recipient cell’s chromosome if crossing over/homologous recombination * takes place, possibly allowing the expression of a different allele from the previous host; OR Specialised transduction 4. A temperate phage* infects a bacterium*, injecting its viral genome into the host cell The viral DNA is integrated into bacterial chromosome forming a prophage* which may be improperly excised to include adjacent segment of bacterial DNA during an induction* event; 5. Bacterial DNA may be packaged into a capsid head during the spontaneous assembly* of new viruses; 6. Upon cell lysis, the defective phage will infect another bacterium and inject bacterial DNA from the previous host cell into the new bacterium;

7. The foreign bacterial DNA can replace the homologous region in the recipient cell’s chromosome if crossing over/homologous recombination * takes place, possibly allowing the expression of a different allele from the previous host; Conjugation 8. Sex pilus* of F+ bacterial cell* makes contact with a F- bacterial cell* and retracts to bring the F- cell closer so a mating bridge* is formed between the 2 cells; 9. One of the 2 strands of the plasmid DNA in F+ cell is nicked and transferred from the F+ cell to the F- cell through mating bridge (via rolling circle mechanism) as the other DNA strand is used as a template for elongation; 10. The single strand F plasmid DNA circularises in F- cell and is used as a template* to synthesise a complementary strand for a double-stranded F plasmid DNA resulting in F+ cell; [Total: 10] Question 4 (a) (i) Use the equation to calculate the melting temperature at which this primer DNA sequence separates from template DNA: TCGACTTCCTCGMCC Tm = 64.9 + [41 x (2 + 7-16.4) / (3 + 4 + 2 + 7)] = 64.9 – 18.96 = 45.9 ºC