2026 NYJC H2 Biology Prelim Paper 1 MS

9477 / H2 Biology / 01 NANYANG JUNIOR COLLEGE JC 2 Preliminary Examination Higher 2 CANDIDATE NAME ANSWERS CLASS BIOLOGY 9477/01 Paper 1 Multiple Choice 23 September 2026 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name and CT on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. Calculators may be used. This document consists of 24 printed pages. [Turn over

9477 / H2 Biology / 01 2 1 The diameter of living cells varies considerably. The diameter of a typical eukaryotic cell is 1.5 X 101 µm. The diameter of a typical prokaryotic cell is 7.5 x 102 nm. Using these measurements, what is the maximum number of each cell type which could fit along a line 1 cm long? number of white blood cells number of Streptococcus cells A 6.7 x 104 1.3 x 102 B 6.7 x 103 1.3 x 105 C 6.7 x 102 1.3 x 104 D 6.7 x 101 1.3 x 103 2 Which roles of the cell surface membrane result from the properties of the phospholipids? CCCCC

9477 / H2 Biology / 01 3 3 The diagrams show the results of an investigation into the composition of different mixtures of amino acids. Each mixture of amino acids was separated using chromatography. Each chromatogram was then turned through 90° and the amino acids separated again by electrophoresis. Which diagram shows an amino acid mixture in which the solubility of some of the amino acids is the same but the charge on those particular amino acids is different? BBBBBBBBBBBBBB

9477 / H2 Biology / 01 4 4 The diagram shows a phospholipid molecule divided into three regions, X, Y and Z. R, in region X, represents a range of possible chemical groups. Regions X, Y and Z affect the properties of cell surface membranes in different ways. Which row shows the effect of each region on the properties of a cell surface membrane? increases permeability of hydrophobic region repels polar molecules attracts water molecules A X X Y and Z B Y Y and Z X C Y and Z X Y and Z D Z Y and Z X

9477 / H2 Biology / 01 5 5 A scientist investigated the rate of breakdown of hydrogen peroxide. Four experiments were carried out using different mixtures. • substrate only • substrate + enzyme • substrate + enzyme + competitive inhibitor • substrate + enzyme + non-competitive inhibitor The results are sketched in the graph. Which row shows the correct lines for two of the experimental mixtures? substrate only substrate + enzyme + competitive inhibitor A 1 2 B 4 2 C 1 3 D 4 3

9477 / H2 Biology / 01 6 6 The diagram shows different molecules in a solution. Which statement could explain what happens when some of the molecules are mixed together? A Molecule P forms an enzyme– substrate complex with the non- competitive inhibitor molecule Q. B Molecule Q binds to molecule P, increasing the activation energy. C Molecules R and S bind to the active site of molecule P. D Molecules S and R are the products of the breakdown of molecule P. 7 Some of the features of different types of stem cells X, Y and Z are listed below. 1 They are able to develop into all the cell types of the body to form a whole organism. 2 They can develop into a wide range of different types of cell. 3 They can give rise to cells of the three primary layers. 4 They can only develop into a limited range of cell types. 5 They are unspecialised cells found in differentiated tissue. Which of the following options correctly shows the features of the different stem cells? X (inner cell mass) Y (16- cell embryo) Z (umbilical cord blood) A 1,2 and 3 3 and 4 2 and 5 B 2 and 3 1, 2 and 3 4, 5 C 1, 2 and 3 2, 5 1, 3 D 2 and 3 4, 5 2, 3

9477 / H2 Biology / 01 7 8 The table shows the percentage of the four nucleotide bases, A, C, G and T, in DNA taken from yeast and from two different tissue samples of an individual rat. source of DNA percentage of nucleotide bases A C G T yeast 31.3 17.1 18.7 32.9 rat tissue 1 28.6 21.5 21.4 28.5 rat tissue 2 28.2 22.2 21.8 27.8 Which of these explanations could help to account for these percentages? 1 50% of nucleotide bases are either C or T because base pairs are formed either by hydrogen bonding between two purines or by hydrogen bonding between two pyrimidines. 2 The percentage of each nucleotide base in the two rat tissues is slightly different because different genes are expressed in different tissues. 3 The percentages of the four nucleotide bases in rats and yeast are different because they have different proteins that are coded for by genes with their own specific base sequences. A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 3 only 9 Which statement concerning the expression of a new allele that is formed as a result of a gene mutation in a diploid organism is correct? A If the gene mutation introduces a chain termination (stop) sequence, then RNA polymerase cannot bind for transcription and the new allele will not be expressed. B A new allele that is dominant will not always be expressed in every cell of a diploid organism. C A new allele cannot be expressed unless the mutation is present in gametes and the new allele will then only be expressed in the offspring or in subsequent generations. D The new allele is more likely to be expressed in the phenotype than the original allele.

9477 / H2 Biology / 01 8 10 In Drosophila, three of the genes involved in the embryological development are named bicoid, caudal and hunchback. Bicoid protein affects the expression of caudal and hunchback genes. To determine the effect of bicoid protein on the other two genes, the levels of biocoid, caudal and hunchback proteins in a developing Drosophila embryo were determined. The results are shown in the diagram below. Which of the following statement best describes the possible mechanisms by which the bicoid protein might affect the expression of the hunchback and caudal genes? Effect on hunchback gene Effect on caudal gene A Bicoid protein binds to hunchback gene enhancer sequence and increases transcription of hunchback gene. Bicoid protein binds to caudal gene enhancer sequence and increases the transcription of caudal gene. B Bicoid protein binds to hunchback gene silencer sequence and decreases transcription of hunchback gene. Bicoid protein binds to caudal gene enhancer sequence and increases transcription of caudal gene. C Bicoid protein binds to hunchback gene enhancer sequence and increases transcription of hunchback gene. Bicoid protein binds to caudal mRNA and prevents translation of caudal mRNA. D Bicoid protein binds to hunchback mRNA and prevents translation of hunchback mRNA. Bicoid protein binds to caudal gene silencer sequence and decreases transcription of caudal gene.

9477 / H2 Biology / 01 9 11 The data below shows the results of electrophoresis of PCR fragments amplified using primers for the site that has been shown to be altered in Huntington's disease. The inherited mutation in the Huntington's disease gene abnormally repeats the nucleotide sequence CAG from 36 up to more than 120 times of that. The male parent, shown as individual 2, had the onset of Huntington's disease when he was