2026 EJC JC2 H2 Biology Prelims P3 (A)

©EJC 2026 9477/03/J2H2PRELIM/2026 [Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examinations 2026 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME ANSWERS CIVICS GROUP 2 5 - REGISTRATION NUMBER H2 Biology Paper 3 Long Structured and Free-Response Questions 9477/03 16 September 2026 2 hours Additional Materials: 4-page Answer Booklet READ THESE INSTRUCTIONS FIRST Write your name, civics group and registration number on all the work you hand in. Write your answers in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue, or correction fluid/tape. Section A Answer all questions on the Question Paper. Section B Answer one question on the 4-page Answer Booklet provided. Write your answer to each part of the question on a fresh sheet of paper. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, ensure that you submit both the Question Paper and Answer Booklet. This document consists of 21 printed pages. For Examiner’s Use Section A 1 / 30 2 / 13 3 / 12 Section B 4 OR 5 / 20 Total 75

2 ©EJC 2026 9477/03/J2H2PRELIM/2026 Section A Answer all questions on the Question Paper. 1 A species of leaf -eating beetle, Coleoptera eunoia, feeds on a specific plant that produces a highly hydrophobic, non-polar toxic metabolite. A resistant population of C. eunoia produces a water-soluble, globular enzyme called Eunoiase in their gut, which effectively binds to and degrades this toxin. Eunoiase is coded for by the eun gene. There are no disulfide bonds in Eunoiase. (a) Describe how the structure of the Eunoiase enzyme is maintained in the aqueous environment of the beetle's gut. [3] 1. To remain soluble in an aqueous environment, the polypeptide folds such that most non- polar hydrophobic R-groups face away from water, 2. while polar hydrophilic R-groups face outwards towards the aqueous exterior; 3. The structure is maintained by hydrogen bonds, ionic bonds and hydrophobic interactions formed between R-groups 4. The structure is maintained by weak hydrogen bonds formed between polar R-groups; 5. It is further stabilised by ionic bonds between oppositely charged R-groups; 6. Also maintained by weak hydrophobic interactions formed between non-polar R - groups; (b) (i) The eun gene is expressed at high levels in the gut, but not in the beetle's eyes. It is thought that DNA methylation might have occurred in the eye cells. Explain how DNA methylation prevents the expression of the eun gene in the eyes. [3] 1. DNA methylation involves the chemical attachment of methyl groups by the enzyme DNA methyltransferase, to cytosine bases in the DNA; 2. This recruits transcriptional repressors / histone deacetylases / repressive chromatin remodeling complexes; 3. This causes the DNA to coil more tightly around histone proteins, forming highly condensed heterochromatin; 4. This prevents RNA polymerase and general transcription factors from accessing and binding to the promoter, preventing the formation of the transcription initiation complex; (ii) Multiple variants of the Eunoiase enzyme were found in various parts of the beetle’s body. Researchers found that their polypeptide lengths were different, despite being coded by the same eun gene. Explain the mechanism that resulted in the multiple variants of Eunoiase enzyme. [3] 1. During post-transcriptional modification, alternative splicing occurs to the primary RNA transcript / pre-mRNA transcribed from the eun gene; 2. where all introns are removed but different combinations of exons are joined together; (R: RNA splicing) 3. This produces different lengths of mature mRNA from the same gene, which is then translated into polypeptides of different lengths.

3 ©EJC 2026 9477/03/J2H2PRELIM/2026 [Turn over (c) A susceptible population of beetles possesses a mutant allele (eun-m). This allele arose from a deletion of 100 base pairs within the coding sequence of the eun gene. Fig. 1.1 shows the toxin degradation activity of the wild-type (normal) Eunoiase and mutant Eunoiase over time. Explain the difference in toxin degradation activity of the two enzymes as shown in Fig. 1.1. [4] 1. The mass of toxin degraded by wild -type Eunoiase reaches 50 µg at 60/80 minutes while the mass of toxin degraded by mutant Eunoiase remains at 0 µg; 2. The deletion of 100 base pairs (which is not a multiple of 3) causes a frameshift mutation / shifts the reading frame of the mRNA sequence downstream of the mutation OR introduces a premature stop codon (nonsense mutation); 3. This results in a completely different sequence of amino acids being encoded OR a truncated polypeptide; 4. Thus changing the 3D conformation of the active site, making the mutant Eunoiase enzyme non-functional; Wild-type Eunoiase Mutant Eunoiase Fig. 1.1 Time / min Wild-type (normal) Mutant Mass of toxin degraded / µg

4 ©EJC 2026 9477/03/J2H2PRELIM/2026 (d) Researchers isolated genomic DNA from a wild-type beetle and a homozygous mutant beetle. They used Polymerase Chain Reaction (PCR) to amplify the eun gene before separating the DNA fragments via agarose gel electrophoresis. (i) Fig. 1.2 shows the gel slab under UV light after gel electrophoresis. A molecular ladder was added to the well in Lane 1. Lanes 2 and 3 contained the PCR products from the wild-type allele and Lane 4 contained the PCR product from the mutant allele (eun-m). Draw on Fig 1.2 to indicate the band for the PCR product of the eun-m allele. [1] Band should be same thickness as bands in Lanes 2 and 3 and just below the position of the 3kb band since 2900 bp (2.9 kb) fragment would move slightly further than the 3kb fragment. (ii) Explain the position of the band you have drawn in (i). [1] 1. The mutant DNA fragment is shorter / 2.9 kb long. The agarose gel will impede the movement of the mutant DNA fragment less and will therefore migrate faster towards the positive electrode (anode) / from the well / down the gel compared to the slightly longer wild-type fragment; Lane 1 Lane 2 Lane 3 Lane 4 4kb 3kb 2kb 3.5kb 2.5kb Fig. 1.2

5 ©EJC 2026 9477/03/J2H2PRELIM/2026 [Turn over (e) Colour in the beetles are determined by two genes on different autosomes. A pure -breeding brown beetle mated with a pure-breeding albino (white) beetle. The F1 offspring were all brown in colour. When two F1 offspring were mated, the resultant F2 offspring were found to be in the following numbers: 198 brown, 66 tan and 88 albino. Using the symbols B and b to represent type of colour pigment, as well as E and e to represent pigment deposition, draw a genetic diagram to show the cross between the F1 beetles and the resulting F2 phenotypic ratio. [4] F1 phenotype brown beetle X brown beetle F1 genotype BbEe X BbEe Meiosis Gametes Random fertilisation Male gametes Female gametes BBEE brown beetle BBEe brown beetle BbEE brown beetle BbEe brown beetle BBEe brown beetle BBee albino beetle BbEe brown beetle Bbee albino beetle BbEE brown beetle BbEe brown beetle bbEE tan beetle bbEe tan beetle BbEe brown beetle Bbee albino beetle bbEe tan beetle bbee albino beetle F2 genotypic ratio 9 B_E_ : 3 bbE_