2024 EJC Prelim Biology P3 (A)

©EJC 2024 9744/03/J2H2PRELIM/2024 [Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examinations 2024 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME ANSWERS CIVICS GROUP 2 3 - REGISTRATION NUMBER H2 Biology Paper 3 Long Structured and Free-response Questions 9744/03 16 September 2024 2 hours Additional Materials: 12-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 12-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 14 printed pages and 2 blank pages. For Examiner’s Use Section A 1 2 3 Section B 4 OR 5 Total 75

2 ©EJC 2024 9744/03/J2H2PRELIM/2024 Section A Answer all questions on the Question Paper. 1 The development of a mouse from a fertilised egg into an adult is regu lated by variations in DNA methylation. Fig. 1.1 shows the developmental stages of a mouse with corresponding levels of DNA methylation. R, S and T represent the zygote, blastocyst and embryo respectively. Fig. 1.1 (a) (i) Compare the features of a cell derived from the zygote with that from the inner cell mass of the blastocyst. [3] 3m = 2 similarities + 1 difference OR 1 similarity + 2 differences [Similarities] 1. Both cells are unspecialized and undifferentiated; (any one) 2. Both cells can be differentiated into specialized cells; 3. Both cells are capable of self-renewal and extensive proliferation; [Differences] 4. Cell derived from the zygote are totipotent while cell derived from inner cell mass of blastocyst is pluripotent; (A: former higher potency than latter); 5. Cell derived from the zygote has higher level of methylation while cell derived from inner cell mass of blastocyst has lower level of methylation; R: Cell derived from the zygote is a zygotic stem cell while cell derived from the inner cell mass of the blastocyst is an embryonic stem cell [Identity; not a feature difference] developmental stage zygote blastocyst embryo egg cell zygote morula blastocyst embryo adult relative level of DNA methylation R T S sperm cell

3 ©EJC 2024 9744/03/J2H2PRELIM/2024 [Turn over (ii) Explain how changes to DNA methylation from R to S bring about differentiation. [4] 1. Relative level of DNA methylation decreases from R to S 2. DNA methylation involves addition of methyl groups to CpG islands / selected cytosine nucleotides in promoters of totipotency genes, catalysed by DNA methyltransferase; 3. It prevents transcription of genes necessary for maintaining totipotency (OWTTE) 4. Blocking binding of (general) transcription factors and hence, preventing the assembly of transcription initiation complex at promoter and/or 5. Recruiting chromatin remodeling complexes / histone deacetylases (at least 1) to condense chromatin / cause DNA to be more tightly wound around histones (iii) At different developmental stages of the mouse, the control of the telomerase gene expression is crucial. Suggest if the telomerase gene in cells is likely to be methylated from T to an adult mouse. [1] Not likely to be methylated because telomerase is required even in adult stem cells OR Likely to be methylated because telomerase is inactivated in terminally differentiated / specialized cells. (iv) Active telomerase can be found in some cell types in a mouse. Using a named example, explain the role of telomerase. [3] 1. Telomerase elongate / extend telomeres that would otherwise be shortened after each cell division (idea of compensating for end-replication problem), to allow for maintenance of telomere length. 2. and allows for stem cell / cancer cell (reference to specific cell) to continue dividing indefinitely; 3. Stem cell: maintain pool of stem cells / differentiate to form more specialized cells over time OR Cancer cell: supports sustained tumour growth / accumulate genetic mutations over time

4 ©EJC 2024 9744/03/J2H2PRELIM/2024 In an experiment, chromatin from various tissues were isolated and treated with DNase, an enzyme that degrades naked double-stranded DNA. After digestion, the enzyme was removed. Any remaining intact DNA was extracted and mixed with radioactively label led DNA probes specific for certain genes, under conditions that favoured nucleic acid hybridisation. The levels of binding of the labelled DNA probes were measured. Some of the resu lts are shown in Table 1.1 below. Table 1.1 Sample Tissue source of chromatin Gene radioactive DNA probe is specific to Percentage binding of radioactive DNA probe 1 Skeletal muscles Myosin gene 25 2 Pancreas Myosin gene 91 3 Skeletal muscles Chymotrypsin gene 93 (b) (i) Outline the steps between extraction of intact DNA and mixing with radioactively labelled DNA probes for nucleic acid hybridization. [3] 1. Gel electrophoresis is carried out to separate DNA fragments by molecular size; 2. Alkaline solution to denature double-stranded DNA into single-stranded DNA; 3. Single-stranded DNA are transferred from agarose gel slab to nitrocellulose membrane (A: nylon membrane); (ii) State two similarities between the probes used in the experiment and primers used in Polymerase Chain Reaction (PCR). [2] 1. Both are complementary in sequence to target gene (A: undergo complementary base-pairing with target gene); 2. Both are single-stranded polynucleotide chains; 3. Both are made of DNA; 4. AVP e.g. both are artificially synthesized; (iii) Suggest how DNA in the chromatin are protected from digestion by DNase. [2] 1. Negatively-charged DNA are wound around positively-charged histone octamers (A: histones) to form nucleosomes, held together by electrostatic interactions (A: ionic bonds); 2. DNase can only degrade naked double-stranded DNA and are not able to access and digest the DNA when they are wound around histone octamers; (iv) With reference to Table 1.1, explain which gene is more actively transcribed in skeletal muscle cells. [2] 1. Myosin gene; 2. The percentage binding of radioactive DNA probe to myosin gene is 25%, which is lower than 93% for chymotrypsin gene; (A: difference) (v) Explain one type of protein modification that supports your answer to (b)(iv). [2] 1. Histone acetylation (OR histone de-methylation); 2. Myosin gene is less tightly wound around histones; (A: chromatin decondense) 3. Hence, its promoter is more accessible to transcription factors / RNA polymerase; Dystrophin is a cytoskeletal protein found in human muscles. The gene which encodes

5 ©EJC 2024 9744/03/J2H2PRELIM/2024 [Turn over dystrophin has 79 exons. Mutations in this gene leads to Duchenne muscular dystrophy (DMD), which causes progressive muscle impairment in children. One of the most common mutations in the dystrophin gene, which occurs in exon 44, results in a non-functional protein. The asterisks (*) in Fig. 1.3 show the positions where de letions were detected in exon 44 and the reading frame is indicated by the boxes. To treat DMD, scientists modified the dystrophin gene by removing exon 44, producing a partially functional protein. Fig. 1.3 (c) Explain how