NYJC 2023 H2 Biology 9744 P2 QP

9744 / H2 Biology / 02 NANYANG JUNIOR COLLEGE JC 2 Preliminary Examination Higher 2 CANDIDATE NAME CLASS BIOLOGY 9744/02 Paper 2 Structured Questions 14 September 2023 Candidates answer on the Question Paper. No Additional Materials are required. 2 hours READ THESE INSTRUCTIONS FIRST Write your name and CT on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions in the spaces provided on the Question Paper The use of an approved scientific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do no use appropriate units. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 Total This document consists of 24 printed pages. [Turn over

9744 / H2 Biology / 02 2 Answer all the questions. 1 Polysaccharides, such as glycogen, are composed of thousands of monomers. Oligosaccharides are carbohydrates that contain three to ten monomers in their chain. Nystose is one example of an oligosaccharide. The structure of nystose is shown in Fig. 1.1. Fig 1.1 (a) State three differences between the structures of nystose and glycogen, other than the number of monomers in the molecules. 1 2 3 [3] (b) One of the enzymes involved in glycogen synthesis is glycogen synthase. The monomer of the glycogen polymer is α-glucose. (i) Draw the ring form of α-glucose in the space provided. (ii) [2]

9744 / H2 Biology / 02 3 (iii) The gene coding for glycogen synthase is known as GYS1 . Glycogen synthase catalyses the formation of a covalent bond between two α-glucose molecules during glycogen synthesis. Name the type of bond formed. [1] (iv) Glycogen branching enzyme is another enzyme that is required for glycogen synthesis. Suggest why glycogen branching enzyme is needed in addition to glycogen synthase. [1] (c) Table 1.1 shows three functions of cell structures that are involved in the synthesis of glycogen synthase. Complete Table 1.1 by naming the cell structure that carries out the function listed. Table 1.1 function name of cell structure assembles ribosomes for polypeptide synthesis synthesizes ATP to provide a supply of energy for the transcription of GYS1 folds and modifies synthesized polypeptide to produce functioning glycogen synthase [3] [Total: 10]

9744 / H2 Biology / 02 4 2 There are a number of mutations affecting the production of fetal haemoglobin, HbF, and normal adult haemoglobin, HbA. • The HbA allele codes for the normal β-globin polypeptide of haemoglobin. • The HbS allele, caused by a base substitution mutation, codes for an abnormal β - globin polypeptide. The abnormal haemoglobin molecules (HbS) form fibres in low partial pressures of oxygen (pO2). The fibres cause red blood cells to become sickle shaped and the cells can block blood capillaries. Individuals with adult haemoglobin molecules that are all abnormal (HbS) have sickle cell anaemia. This is a painful chronic condition that can be life-threatening. (a) Explain why this mutation causes the HbS to form fibres. [2] (b) Fetal haemoglobin, HbF, is produced by the fetus until just before birth, when adult haemoglobin begins to be made. By the age of six months, adult haemoglobin has replaced most of the HbF. This change occurs when the genes coding for HbF are switched off and the genes coding for adult haemoglobin are switched on. • A base substitution, British-198, causes fetal haemoglobin to continue to be produced. • Normally by the age of six months, the concentration of HbF reduces to less than 1% of total haemoglobin. • With the British-198 mutation, the concentration of HbF may be as high as 20% of total haemoglobin in an adult. • HbF has a higher affinity for oxygen at low p O2 than adult haemoglobin. Individuals who have both sickle cell anaemia and British -198 mutation have reduced symptoms of sickle cell anaemia. Suggest why having the British-198 mutation reduces the symptoms of sickle cell anaemia. [2]

9744 / H2 Biology / 02 5 (c) Gel electrophoresis can be carried out to test individuals for the different versions of haemoglobin: HbA, HbS and HbF. • A buffer with alkaline pH is used to make all haemoglobin molecules negatively charged. • HbS molecules have an additional positive charge compared to HbA. (i) Describe and explain how gel electrophoresis is used to diagnose sickle cell anaemia. [4] (ii) Four individuals had their haemoglobin analysed by gel electrophoresis. One of the individuals was heterozygous for the Hb A and HbS alleles and had a condition known as sickle cell trait (SCT). Some of the results are shown in Fig. 2.1. In Fig. 2.1, lane 1 and lane 5 are complete. Fig. 2.1 Predict the results for the individuals analysed, by adding bands to lanes 2, 3 and 4 on Fig. 2.1. [2] [Total: 10]

9744 / H2 Biology / 02 6 3 (a) The house mouse, Mus musculus, has a diploid number of 40 chromosomes. Fig. 3.1 shows 6 of these chromosomes. Fig. 3.1 Identify one pair of homologous chromosomes on Fig. 3.1 by drawing circles around two chromosomes. [1] (b) Fig. 3.2 shows the banding pattern of chromosome pair 11 of M. musculus. The banding pattern is obtained by staining. Fig. 3.2 (i) Explain why chromosomes, such as those in Fig. 3.2, are described as a homologous pair. [3]

9744 / H2 Biology / 02 7 (ii) State the number of chromosomes that are present in M. musculus spermatozoa. [1] (c) M. musculus produces gametes by meiosis. These gametes are genetically different. There is random fusion of gametes at fertilisation. (i) Explain why meiosis is important in the life cycle of M. musculus, apart from producing genetically different gametes. [2] (ii) Explain how the random fusion of gametes leads to the expression of rare, recessive alleles. [2] [Total: 9]

9744 / H2 Biology / 02 8 4 (a) Cats with black fur or white fur are common in Europe whereas cats with brown fur are less common. A gene, coding for an enzyme involved in pigment production, has two alleles. • The dominant allele, B, results in black fur. • The recessive allele, b, results in brown fur. A second gene can affect fur colour. • The dominant allele, A, prevents pigment production, resulting in a cat with white fur. • The recessive allele, a, has no effect on fur colour. The two genes are on different pairs of autosomes. (i) Use a genetic diagram to show how a cross between two cats, heterozygous at both loci, can produce offspring with three different colours: white, black and brown. State the expected ratio of the different coloured offspring. [4] (ii) Suggest how the presence of allele A prevents pigment production. [3]

9744 / H2 Biology / 02 9 (b) Apart from having different fur colour, a variety of domestic cat does not have a tail. This condition is controlled by a single gene with two alleles. These alleles are • ‘with tail’ • ‘without tail’ Table 4.1 shows the results of four crosses between cats with tails and cats without tails. Each male was crossed with several female