TJC 2023 JC1 H1 EOY ANS

1 JC1 2023 H1 PROMOTIONAL EXAMINATION ANSWERS SECTION A: MCQ 1 A 6 A 11 D 2 B 7 D 12 B 3 D 8 A 13 B 4 A 9 D 14 D 5 C 10 B 15 B 1 Fig.1.1 shows a cell releasing insulin in response to glucose uptake. The uptake of glucose through Glucose Transporter 2 (GLUT2) causes a rise in ATP levels in the cell. This closes the potassium channel and in turn causes calcium channel to open. The influx of calcium ions leads to the release of insulin. Fig. 1.1 (a) (i) With reference to Fig. 1.1., describe two differences in the way insulin and glucose are transported across the cell surface membrane. [2] Any 2 1. Transport of insulin involves the secretory vesicles , but the transport of glucose involves transport / carrier proteins. 2. Insulin is released when the secretory vesicles fuse with the cell surface membrane, but glucose enters the cell when they bind to [QF] GLUT2 which undergoes conformational change to facilitate the transport of glucose. 3. Glucose is transported down a concentration gradient but not insulin. 4. Transport of insulin requires energy from ATP (R: active transport) but not the transport of glucose.

Feedback / comment 1. Candidates had difficulty recalling the different types of transport across the membrane. While most knew that glucose is transported via facilitated diffusion, most did not realize from the Fig. that insulin is released / secreted via exocytosis (i.e., involving secretory vesicles). 2. Transport protein could either be channel protein or carrier protein. Note that glucose is transported into the cell via carrier protein and not channel protein. (ii) Explain how the membrane proteins in Fig. 1.1 are able to transport the substances across the cell surface membrane. [2] 1. The membrane proteins are transmembrane proteins 2. with hydrophilic channel to shield polar molecules (e.g., glucose) and charged ions (e.g., calcium ions) [1] Note: If no reference to glucose / potassium ions / calcium ions = ½ mark 3. from hydrophobic core of phospholipid bilayer / cell surface membrane. This helps to transport them across the membrane. Extra point: 4. Potassium channels have binding sites for ATP to bind to such that they are closed and prevent K+ ions from leaving the cell. This will help calcium channels to open. Feedback / comment 1. Candidates were mostly unable to relate to the property of membranes and the properties of substances that are transported across the membrane. 2. Candidates should use the information presented in the figure to answer (i.e., QF the types of substances in Fig. 1.1). (b) Suggest with a reason how a secretory cell will differ in terms of organelles from a non-secretory cell. [1] Any one 1a) There will be more mitochondria. 1b) Reason: To produce more ATP for the synthesis of proteins / movement of vesicles / exocytosis [any 1 example] 2a) There will be more rough endoplasmic reticulum. 2b) Reason: Attachment of more ribosomes on its surface to synthesise more proteins. 3a) There will be more Golgi apparatus. (Note: Uppercase G for Golgi). 3b) Reason: So that m ore proteins can be modified, sorted and packaged for secretion. Feedback / comment 1. Transport proteins are not organelles. 2. Reason was often lacking in the answers. Candidates are reminded to answer both parts of the question.

3 Fig. 1.2 shows an organelle in a eukaryotic cell. Fig. 1.2 (c) With reference to Fig. 1.2; explain why the organelle cannot be a lysosome. [1] Any one 1. Nucleus is a double-membrane bound organelle while lysosome is a single membrane- bound organelle. 2. Presence of a prominent nucleolus in the nucleus which is absent in a lysosome. 3. Presence of nuclear pores which are absent in the lysosome. Feedback / comment Most candidates were able to identify the organelle as nucleus, and answer in terms of the structural difference, as seen in Fig. 1.2, between nucleus vs lysosome. [Total: 6] 2 (a) Fig. 2.1 shows the photomicrograph during one stage of mitosis occurring in a root tip cell of a diploid flowering plant. The diploid number is 14. Fig. 2.1

Complete Table 2.1 to show the number of chromosomes found in the cell at the end of a given stage of the cell cycle. [2] Table 2.1 Feedback / comment 1. More than half of the candidates were able to gain at least one mark. Some students who drew the chromosomes were able to gain full credit. stage no. of chromosomes per cell G1 14 S phase 14 cytokinesis 14 Interphase: G1 No. of chromosomes per cell = 2 (i.e., 2 unduplicated chromosomes.) [End of] Interphase: S No. of chromosomes per cell = 2 (i.e., 2 duplicated chromosomes). Prophase No. of chromosomes per cell = 2 [End of] Cytokinesis No. of chromosomes per cell = 2 For STQ 2a No. of chromosomes per cell = 14 (given) For STQ 2a No. of chromosomes per cell = 14 For STQ 2a No. of chromosomes per cell = 14 Metaphase No. of chromosomes per cell = 2 Anaphase No. of chromosomes per cell = 4 Note: Telophase marks the end of mitosis. Telophase: No. of chromosomes per cell = 4 No. of chromosomes per nucleus = 2

5 (b) Outline one difference between a pair of homologous chromosomes and sister chromatids of a chromosome. [1] Feature A pair of homologous chromosomes Sister chromatids 1) Genetic similarity Not genetically identical to each other [1/2] Genetically identical to each other before crossing over [1/2] Explanation for point 1 A diploid individual inherits one homologous chromosome from his father and one homologous chromosome from his mother. Hence, the pair of homologous chromosomes are genetically different / not identical. On the other hand, a pair of sister chromatids (of a duplicated chromosome) is the result of DNA replication. Hence, they are genetically identical. 2) Physical connection / attachment Not connected to each other Connected to each other at the centromere 3) Anaphase I Separate Do not separate 4) Alleles at corresponding gene loci. Same number and type of genes buthave different alleles for the same genes. Max ½ mark if “may” is missing from answer. Same number and type of genes and have same alleles for the same genes. Explanation: It is possible for the homologous chromosomes to have the same alleles for the same genes. In Fig 17 of Cell cycle notes (page 19), the pair of homologous chromosomes have different alleles for gene A/a but the same alleles for gene B/b. Explanation for point 4 For the pair of homologous chromosomes, the corresponding alleles of one gene may be identical (i.e., homozygous for the particular gene) or they may be different (i.e., heterozygous for the particular gene). On the other hand, a pair of sister chromatids will always have identical alleles for every gene because they are the result of DNA replication. 5) Crossing over Crossing over occurs during prophase I. Max ½ mark if “prophase I” is missing. No crossing over. Feedback / comment

1. Mixed performance for this question. A few candidates have misconceptions regarding point 1 and 4. Point 1 and 4 are related to each other. (c) (i) Describe the characteristics of embryonic stem cells. [2] Any four 1. Capable of dividing and renewing themselves. 2. Pluripotent 3. and can differentiate into almost any cell type to form any organ or type of cell. 4. Are unspecialized. 5. are able to give rise to specialized cells. 6. They are not totipotent but are multipotent. Feedback / comment 1. Candidates had difficulty recalling the characteristics of embryonic stem cells and no candidates were able to gain full credit. (ii) State the challenges of using embryonic stem (ES)