MI 2023 J2 H2 BI Prelim P1 ANS

1 2023 End-of-Year Exam Pre-university 3 Biology Higher 2 9744/01 Paper 1 Multiple Choice 21 September 2023 1 hour Additional Materials: Optical Answer Sheet READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, Adm No. and class on all the papers you hand in. There are thirty questions in 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. Each correct answer will score one mark. A mark will not be deducted for wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.

2 1 B 11 B 21 B 2 D 12 A 22 A 3 A 13 B 23 B 4 B 14 C 24 C 5 D 15 A 25 A 6 A 16 D 26 D 7 C 17 B 27 C 8 C 18 B 28 D 9 C 19 B 29 A 10 B 20 C 30 B

3 1 A part of a cell is shown in the electron micrograph below. Which of the following correctly describes the labelled organelles? membrane-bound contains nucleic acids found in both animal and plant cells A 1, 4 1 2, 3, 4 B 1, 4 1, 2 1, 2, 4 C 1, 2, 3 2, 3 3, 4 D 1, 2, 4 1 1, 2, 3, 4

4 2 It is possible for a bacterium to synthesise a eukaryotic protein. This involves introducing a eukaryotic gene into the bacterial DNA. The eukaryotic gene is then expressed into protein by the bacterium. Which of the following explains why a bacterial cell can produce a eukaryotic protein but cannot produce a eukaryotic glycoprotein? A Bacteria do not have rough endoplasmic reticulum. B Bacteria do not have nuclear envelope. C Bacteria do not have mitochondria. D Bacteria do not have Golgi bodies. 3 Which of the following correctly describes starch, glycogen and cellulose? starch glycogen cellulose A Can be easily hydrolysed to release α-glucose monomers Angle of α-(1,4)-bonds and CH2OH side chains results in helical chains Made up of long linear chains due to 180o rotation of alternate β- glucose B Insoluble due to the –OH groups projecting into the interior of helices Made up of α-glucose monomers with (1,4)- glycosidic bonds resulting in highly branched chains Cellulose chains are organised into microfibrils and macrofibrils C Highly branched amylose due to α-(1,6)-glycosidic bonds Highly compacted molecule that serves as a good energy storage in animal cells Provides structural support and prevents cells from bursting when turgid D Highly compacted molecule that serves as a good energy storage in plant cells Made up of β-glucose units with (1,4)-glycosidic bonds resulting in highly branched chains High tensile strength is due to the accumulative strength of the covalent cross linkages between cellulose chains

5 4 The table compares three molecules, X, Y and Z, which contain the elements carbon, hydrogen and oxygen only. The percentage of carbon, hydrogen and oxygen atoms in each molecule are shown below. Which row correctly identifies the three molecules? molecule % carbon % hydrogen % oxygen X 25.0 50.0 25.0 Y 28.5 47.7 23.8 Z 34.6 61.6 3.8 molecule X Y Z A monosaccharide disaccharide polysaccharide B monosaccharide polysaccharide triglyceride C polysaccharide triglyceride monosaccharide D triglyceride monosaccharide polysaccharide

6 5 The activity of an enzyme, E, is measured under different conditions. The same amount and concentration of enzyme was used in each set up. Set up Activity of E (A.U.) 10% substrate, 30⁰C 5 10% substrate, 40⁰C 10 10% substrate, 50⁰C 13 10% substrate, 1% inhibitor, 40⁰C 7 20% substrate, 1% inhibitor, 40⁰C 6 Which row correctly shows a statement that can be concluded and the supporting evidence? Statement Supporting evidence A The optimum temperature of the enzyme is 50⁰C. It is the temperature with the highest enzyme activity. B There is end-product inhibition. At higher substrate concentration and at the same temperature, the enzyme activity is lower. C The inhibitor binds to the active site of the enzyme. An increase in substrate concentration did not overcome the inhibition. D The inhibitor changes the active site of the enzyme. Enzyme activity did not increase at higher substrate concentration.

7 6 The figure below shows different mechanisms of transport across membrane, labelled A to E. Which of the following correctly categorises the mechanisms based on their features? A passive process active process A B, C, D, E B transport polar molecules transport non-polar molecules A, B, C B, D, E C transport molecules down concentration gradient transport molecules against concentration gradient A, C B, C, D, E D transport large molecules transport small molecules A, B, C A, D, E

8 7 The bacterium Escherichia coli, or E. coli, divides once every 50 minutes. E. coli were grown on a medium containing only heavy nitrogen, 15N, until all of the bacterial DNA contained heavy nitrogen. This was counted as sample time of 0 minutes. Some of the bacteria were moved from a heavy nitrogen medium and cultured in a medium with only light nitrogen, 14N. After every generation, some bacteria were collected, and their DNA was extracted and centrifuged. The diagram shows the possible positions (upper, middle, lower) of the bands of DNA. The actual position of bands in the first sample is shown. What proportion of the DNA of the sample taken at 150 minutes will be at the upper position? A 25% B 50% C 75% D 100%

9 8 A population of bacteria is exposed to the antibiotic penicillin. Most of the bacteria die. However, after a few generations, bacteria that were found to survive were found to have an allele coding for an enzyme that breaks down penicillin. Which could explain how some of these bacterial cells could have acquired this allele? 1 A mutation during DNA replication. 2 Transcription error that leads to a different mRNA being formed. 3 Ribosomes failing to associate with mature mRNA during translation. A 1, 2, and 3 B 1 and 3 only C 1 only D 2 and 3 only 9 If DNA is damaged, checkpoints in the cell cycle can either trigger DNA repair, allowing the cell to progress through the cell cycle or, if this cannot be carried out, divert the process to programmed cell death, also known as apoptosis. Damaged DNA can b e repaired using proteins like p53 and Chk1. About half of all cancer cells have non-functional p53 proteins. An inhibitor for Chk1 protein has been developed as a treatment for cancer patients to improve tumour shrinkage during radiation treatment. How would this Chk1 inhibitor benefit these patients? A Chk1 genes would be damaged and unable to repair DNA. B Fewer healthy cells would have damaged DNA. C More cells with non-functional p53 protein would undergo apoptosis. D The radiation treatment would kill all the tumour cells.

10 10 A bacteria colony produces a normal protein with a known amino acid sequence. The bacteria colony was treated with the same chemical mutagen twice. This gave rise to two mutant strains of bacteria where each had a single nucleotide change at a particular mRNA codon resulting in a change of amino acid as shown in the diagram. The mRNA codons for some amino acids are shown in the table. alanine (ala) leucine (leu) threonine (thr) valine (val) GCU GCC GCA GCG UUU UUC UUA UUG ACU ACC ACA ACG GUU GUC GUA GUG Assuming that both treatments resulted in a single nucleotide change each, which diagram correctly shows the codons that were translated into alanine, methionine, threonine and valine? A B C D ala met val thr first treatment second treatment GCU AUG GUA ACU GCG AUG G