2025 SAJC H2Bio PE P2 (A)

1 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations Civics Group Index Number Name (use BLOCK LETTERS) ST. ANDREW’S JUNIOR COLLEGE 2025 JC2 PRELIMINARY EXAMINATIONS H2 BIOLOGY 9744/2 Paper 2 Wednesday 3rd September 2025 2 hours Materials: Question Paper READ THESE INSTRUCTIONS FIRST Write your name, civics group and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagram, graph or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Write your answers in the spaces provided on the question paper. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiners’ Use 1 /10 2 /12 3 /10 4 /10 5 /10 6 /6 7 /9 8 /10 9 /8 10 /10 11 /5 Total /100 This document consists of x printed pages and 0 blank page. [Turn over H2

2 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations QUESTION 1 Fig. 1.1 shows the structure of the dipeptide consisting of cysteine and tryptophan. Fig.1.1 Picture credits : https://linkinghub.elsevier.com/retrieve/pii/S0039602810003869 (a) Complete the diagram to show the hydrolysis of the dipeptide. ……………………………………………………………………………………………[2]

3 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations 1 Water used in the hydrolysis reaction ; 2 Cysteine amino acid with a - COOH group ; Tryptophan amino acid with a - NH2 group ; (No labels are needed) In bacteria like Escherichia coli, tryptophan is incorporated into polypeptides during protein synthesis. The availability of tryptophan within the cell is tightly regulated by the trp operon which is known as a repressible operon. Fig. 1.2 summarises the structure and control of the trp operon. 1. + H2O + Cysteine Tryptophan 2. 3.

4 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations Fig. 1.2 Picture modified from 9700/W22/42 (b) Suggest and explain the advantage of having repressible operons in prokaryotes. …………………………………………………………………………………………....[2] 1 [Suggest] Essential enzymes / proteins are continuously produced ; 2 [Explain] End product inhibition occurs when enzymes / proteins are synthesized until product / tryptophan concentrations are high ; to prevent wastage of resources; A number of mutations have been found in the trp operon. One of these mutations results in a mutant operator (OC) (c) Predict and explain the likely effect of the O C mutation on the synthesis of tryptophan in E. coli. …………………………………………………………………………………………....[3] [Predict] 1 no repression of tryptophan synthesis / tryptophan synthesized continuously (even when in abundance); [Explain] 2 trp repressor (which is synthesized in inactive form) is no long able to (recognise and) bind to OC / mutated operator ; 3 RNA polymerase can bind to the promoter, transcription of the 5 trp structural genes, Trp E, Trp D, Trp C, Trp B, Trp A, occurs ; 4 Proteins / enzymes encoded by structural genes catalyse steps in tryptophan biosynthetic pathway;

5 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations (d) Describe three differences in the structure and organisation of prokaryotic and eukaryotic genomes. ……………………………………………………………………………………...........….[3] Bacterial genome Eukaryotic genome Points specified in syllabus 1. Genome size / Amount of DNA Smaller genomes, 0.6 to 10Mb / less total DNA per cell [for general info only: Mb – Megabase is the unit length for DNA fragments equal to 1 million nucleotides]. Larger genomes, 10 Mb – 100,000 Mb / more total DNA per cell, about 1000 times more DNA. 2. Gene length Shorter gene sequences / more compact genetic organisation Longer gene sequences / presence of more intragenic (within genes) spaces (e.g. introns) 3. Chromosome structure Circular DNA molecule which is closed covalently Linear DNA molecule with 2 ends 4. Packing of DNA Prokaryotic DNA is not complexed with histones; (DNA is not packaged into nucleosomes. DNA is supercoiled and later folded and condensed via non-histone proteins.) Eukaryotic DNA is complexed with histones and other proteins to form chromatin; (DNA is coiled around histone octamer core and subsequently further packed into higher order chromatin structure.) 5. Introns No introns within genes. (Coding sequence proceeds from start to finish without interruption by introns) Presence of introns within genes. (Introns account for the main difference in average length between human and prokaryotic genes) Points not specified in syllabus 6. Chromosome number Single chromosome / Haploid Many chromosomes / Diploid or polyploid 7. Presence and absence of operons Presence of operons, where two or more genes may be expressed and regulated as a unit Absence of operons. 8. Repetitive sequences Few repetitive DNA sequences. Many repetitive DNA sequences

6 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations 9. Coding and non- coding DNA Most of DNA are coding sequences (codes for protein, tRNA, or rRNA). Most of DNA are non-coding. 10. Origins of replication One origin of replication present Many origins of replication present 11. Presence of extrachromosomal DNA Independent small, double stranded, circular DNA called plasmids Circular, double-stranded DNA in mitochondria / chloroplasts. 12. Telomeres Absent Present [Total: 10]

7 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations QUESTION 2 (a) In chickens, feather color is either white, black or speckled. The alleles for black feathers and white feathers are denoted by “ C B ” and “ C W ” respectively. A cross between a white chicken and a black chicken gave rise to a speckled chicken. White Black Speckled Fig. 2.1 (i) Using the information provided, explain the appearance of the speckled chicken. ………………………………………………………………………...…………………………[2] 1. The black / CB and white / CW alleles for feather color are codominant. 2. Both alleles of the gene for feather color, C B and C W, are equally expressed in the phenotype of the heterozygote (resulting in speckled phenotype). [Reject: incomplete dominance] (ii) Using the symbols provided, draw a genetic diagram to show the results of a sibling mating between two speckled chickens. ………………………………………………………………………...…………………………[3] F1 phenotype: Speckled chicken X Speckled chicken F1 genotype: CBCW X CBCW F1 gametes: CB CW X CB CW F2 genotype: CBCB , CBCW , CWCW F2 phenotype: black chicken, speckled chicken, white chicken F2 phenotypic ratio: 1 : 2 : 1 ;1m for gametes ;1m for genotypes ;1m for phenotype and phenotypic ratio

8 St Andrew’s Junior College 2025 9744/02/Preliminary Examinations [No ecf for wrong F1 genotypes] [Penalise 1m if correct symbols are not used] (b) Drosophila melanogaster, commonly known as the fruit fly, has been a cornerstone of genetic research for over a century. The practical benefits of using Drosophila are numerous and contribute significantly to its widespread adoption in laboratories. (i) Suggest why fruit flies are good experimental organisms for carrying out crosses in genetic research. ………………………………………………………………………………………………..[1] [Any 1] 1. Fruit