2026 VJC H2 Biology Prelim P2 Ans

2026 H2 Biology Prelim Paper 2 Answer 1 Catalase is a ubiquitous enzyme found in nearly all living organisms that are exposed to oxygen. It catalyses the decomposition of hydrogen peroxide (H2O2) into water and oxygen. Fig. 1.1 shows the results of an investigation into the effect of pH on the initial rate of reaction of catalase. Fig. 1.1 (a) With reference to the structure of enzymes, explain the shape of the curve in Fig. 1.1. [2] • As pH moves away from optimum, concentration of H+ and OH- ions changes; • Disrupt hydrogen / ionic bonds holding tertiary structure of enzyme; • Enzyme denatured, fewer enzyme-substrate (ES) complexes form; (b) Table 1.1 shows the effect of increasing substrate concentration on the initial rate of reaction of a fixed concentration of catalase at its optimum pH and temperature. Table 1.1 substrate concentration (H2O2 / mmol dm-3) initial rate of reaction (O₂ produced / cm3 s⁻¹) 0.5 0.5 1.0 1.0 2.0 2.0 5.0 4.0 10.0 5.0 20.0 5.0

(i) Account for the relationship between substrate concentration and the initial rate of reaction. [3] • From 0.5 to 2.0mmoldm-3, the rate increases proportionally; • At 10.0mmoldm-3 and above, the rate plateaus at 5.0cm3s-1; • Active sites are saturated with substrate; (ii) A student hypothesised that the addition of a heavy metal ion, such as lead ( Pb2+), would decrease the initial rate of reaction of catalase at all substrate concentrations. Describe how the student could modify the investigation in Table 1.1 to test this hypothesis. [3] • Repeat experiment with addition of lead ions; • Add a control without lead; • Compare rates – rate with lead ions should be lower;

2 Lipids have a variety of important functions in cells. Phospholipids are major components of cell surface membranes, while triglycerides are stored in lipid droplets as an energy reserve. The structure and properties of lipids influence membrane structure, membrane function and the movement of substances across cell surface membranes. Linoleic acid is an example of fatty acid found in some phospholipids. Phospholipids are major components of the cell surface membrane. Fig. 2.1 shows a molecule of linoleic acid. Fig. 2.1 At the start of the cold season, the proportion of phospholipids containing linoleic acid in the cell surface membrane of chickpea, Cicer arietinum, increases. Plants that do not make this change are less likely to survive at low temperatures. (a) Explain how an increase in the proportion of phospholipids containing linoleic acid helps chickpea cells survive at low temperatures. [3] • Unsaturated fatty acid; • C=C double bonds forming kinks; • Prevent phospholipids from packing too close;

(b) Hepatocytes (liver cells) are involved in lipid metabolism and store triglycerides in lipid droplets. Fig. 2.2 is a diagram of a hepatocyte. Fig. 2.2 The triglycerides stored in a lipid droplet are surrounded by a single layer (monolayer) of phospholipids. Explain why phospholipids, rather than triglycerides, are used for the outer monolayer of the lipid droplet. [3] • Phospholipids are amphipathic; • Hydrophilic phosphate heads face cytoplasm, hydrophobic fatty acid tails face triglycerides; • Triglycerides lack phosphate heads to interact with cytoplasm; (c) Carbon dioxide (CO₂) is transported across the cell surface membrane of hepatocytes by a different mechanism from glucose. Name the transport mechanism for CO₂ and glucose and explain why these substances are transported across the membrane by different mechanisms. [3] • CO₂ by simple diffusion, glucose by facilitated diffusion; • Hydrophobic core of phospholipid bilayer consists of non-polar fatty acid tails; • CO₂ is non-polar and can diffuse directly through phospholipid bilayer; • Glucose is polar and requires specific transport protein to move across cell surface membrane;

3 Gentamicin is an antibiotic used to treat severe bacterial infections in children. Some children have a mutation in the MT-RNR1 gene. If gentamicin is given to children with this genetic mutation, it can cause deafness. Before gentamicin can be given to a child with a severe bacterial infection, polymerase chain reaction (PCR) and electrophoresis are used to test whether the child has this mutation. If the mutation is found, a different antibiotic must be given. (a) Describe and explain the role of Taq polymerase in PCR. [2] • Taq polymerase synthesises DNA molecules during PCR; • Adds DNA nucleotides to elongate DNA strand in 5′  3′ direction; • Catalyses formation of phosphodiester bonds between adjacent DNA nucleotides; (b) Taq polymerase synthesises new DNA strands during PCR. Explain how the antiparallel arrangement of DNA determines how new DNA strands are synthesised. [2] • Taq polymerase elongates DNA strands in 5′  3′ direction; • So two DNA strands elongated in opposite directions as template DNA strands are in opposite directions; (c) PCR with primers specific to the MT -RNR1 gene is used to amplify DNA from the child that is being tested. The primers are designed so that the amplified product of the normal allele of MT-RNR1 is longer than the amplified product of the mutant allele. Gel electrophoresis is used to separate the PCR products. Fig. 3.1 shows the results of gel electrophoresis after using this method of PCR on DNA samples collected from three children. Fig. 3.1

State which of the children in Fig. 3.1 cannot receive the antibiotic gentamicin to treat a severe bacterial infection. [1] • Child 2 and child 3; (d) Explain how gel electrophoresis produces the pattern of results shown in Fig. 3.1 from the PCR products of the MT-RNR1 gene. [4] • Gel electrophoresis separates DNA fragments by size; • DNA fragments are negatively charged and move towards positive electrode; • Smaller DNA fragments move more quickly and travel further; • Normal MT-RNR1 allele produces longer PCR product that moves a shorter distance, while mutant MT-RNR1 allele produces shorter PCR product that moves a longer distance;

4 The Arctic krill (Euphausia superba) utilises bioluminescence produced by luciferase enzyme (encoded by LUC1 gene) for schooling behaviour and counter-illumination against predators. The eukaryotic LUC1 gene consists of 5 exons interspersed by 4 introns. Researchers isolated a mutant strain of E. superba, strain A, and did a sequence comparison with the wild-type strain. Fig. 4.1 shows the pre- mRNA sequence at the Exon 2 – Intron 2 junction for the wild- type and strain A LUC1 gene. wild-type pre-mRNA : 5' - ... UAC - GCA - CAG | GUU - AAG - UUG ... - 3' strain A pre-mRNA : 5' - ... UAC - GCA - CAG | AUU - AAG - UUG ... - 3' (Note: The symbol | denotes the boundary between Exon 2 and Intron 2.) Fig. 4.1 (a) With reference to Fig. 4.1, explain what is meant by the term gene mutation. [2] • Change in DNA sequence of a gene; • Cytosine in template strand changed to thymine; (b) Outline the processes that have to occur before pre-mRNA can be used for translation. [3] • Addition of cap to 5’ end of the mRNA by mRNA guanyltranferase; • Excision of introns and splicing of exons by spliceosome; • Addition of poly-A tail to 3’ end of the mRNA by poly (A) polymerase; (c) With reference to Fig. 4.1, explain the impact of the gene mutation on strain A. [3] • Removal of splice site; • Intron 2 is not removed from the mature mRNA; • Translated into non-functional luciferase; (d) In a different strain B, sequencing revealed that the entire coding sequence of the LUC1 gene was identical to the wild-type. However, no LUC1 mRNA or protein could be found inside the cells. Suggest a specific region of the DNA where a mutation occurred in strain B and explain how it results in this outcome. [1] • Promoter – RNA polymerase can