VJC 2021 H2 Bio 9744 P3 Answers

1 Victoria Junior College Biology Department 2021 Prelims H2 Paper 3 – Proposed Answers 1(a) (i) Circle the sex chromosomes in Fig 1.1. [1] Fig 1.1 (ii) Explain what is meant by ‘heterogametic sex”. [2] • The organism where the pair of sex chromosomes are different in shape/size; • Hence there are 2 types of gametes produced – one type containing X chromosome, the other containing Y chromosome; (b) Explain how the behavior of chromosome during meiosis supports Mendel’s law of segregation and independent assortment. [4] [Note: the basis is genes are found on chromosomes.] Law of segregation: • Separation of homologous chromosomes in Anaphase I /meiosis I and sister chromatids in Anaphase II/meiosis II; • results in each gamete containing only 1 allele of each gene; Law of independent assortment: • Arrangement of pairs of homologous c hromosomes across the equator in Metaphase I is random; • Subsequent segregation results in random combination of maternal and paternal chromosomes in the gametes; (c) Describe one way in which translation of the maternal mRNA in the egg can be prevented. [2] [Any 1 below] • Translational repressors/ sequence-specific RNA binding proteins bind to the 5’ UTR of mRNA; • and prevents ribosome binding and formation of translational initiation complex; Or • Binding of a complementary RNA strand to a specific region/critical region of the mRNA; • to block the binding to ribosome;

2 (d) (i) Suggest an advantage of nuclei of the early embryo sharing a common cytoplasm. [1] Any 1 below • Large protein molecules present in the cytoplasm can move directly into the nuclei without having to cross any cell membranes; • Sharing of a common pool of nutrients eg. nucleotides, ATP, enzymes, proteins; • Enable nuclei to respond to the same signal; • AVP; R: allows for nuclear divisions as the 3rd diagram (Fig 1.2) showed that there is no further division. There is only migration of nuclei to the periphery. (ii) Cellularisation occurs when cell membranes are formed around each nucleus to form cells. Explain the importance of cellularisation in the continued development of the embryo. [2] • Compartmentation which allows for setting up of specific environments; • Allowing differentiation/ specialisation of cells via development of tissue specific structures/differential gene expression; • Allows for more cells to be formed as cells undergo cell division/mitosis; • Cells can then differentiate/specialised via expression of different genes/development of tissue specific structures; € Using the information from Fig. 1.3 and 1.4, (i) account for the distribution of bicoid and nanos proteins. [2] • Bicoid and nanos mRNA are only found concentrated at the anterior and posterior ends of the embryo respectively; • When translated, the bicoid and nanos proteins will only be found in these regions respectively; (ii) suggest explanations for the distribution of the hunchback and caudal proteins.[3] • [What is the distribution] Although mRNA concentration of both hunchback and caudal are high and distributed evenly (idea of) throughout the embryo, hunchback is present mainly in the anterior region while caudal is towards the posterior; • Bicoid protein can function as a repressor/ inhibitor for the translation of caudal mRNA; • Nanos can be a repressor for hunchback mRNA; • Very low concentration of nanos + caudal at the anterior end and bicoid + hunchback at the posterior end is due to diffusion; • AVP; (f) Suggest how the bicoid protein can result in the development of specific structures in the anterior region of the embryo. [3] • Bicoid protein act as a transcription factor/ activator; • Bicoid attach to promoter region of specific gene(s) and help recruit RNA polymerase to the promoter;

3 • Results in transcription of specific genes that codes for specifc structures (eg. Mouth) found in the anterior region; A: If reference is made to bicoid being a repressor A: If reference is made to bicoid being a histone acetylase (g) (i) If each of the DNA has undergone 10 rounds of DNA replication, how many DNA molecules are there in one polytene chromosome? [1] • [210 = 1024 x 2(because each polytene chromosome was formed from the replication of one homologous pair of chromosomes) =] 2048 DNA molecules; (ii) Explain why it is unusual for homologous chromosomes to pair up in the salivary gland. [1] • as salivary glands are somatic cells, not involved in the formation of gametes/ pairing of homologous chromosomes only occur in gamete forming cells / during meiosis; (iii) Explain how repeated divisions can result in the banding pattern seen in Fig 1.5. [3] 1. idea that all the sister chromatids are formed as a result of semiconservative DNA replication and hence have identical DNA sequences; 2. Idea of DNA packaging depends on the DNA sequence hence same extent of packaging of DNA for same regions of all sister chromatids; 3. Banding pattern - Darker regions take up more stain as the DNA is more condensed around histones / more tightly packed / heterochromatin OR lighter region takes up less stain as the DNA is less condensed around the histones / loosely packed / euchromatin; (h) Scientists have labelled the banded regions of these giant chromosomes and use them to distinguish between different species of Drosophila. (i) Suggest why different species of Drosophila show different banding patterns for the same polytene chromosome. [2] • Accumulation of independent mutations over time; • Different genes sequences are found / being expressed in different species (Reject different alleles); (ii) Suggest a limitation of using polytene chromosomes for establishing phylogenetic relationship. [1] • Subjectivity in comparing extent of difference in banding pattern to determine relationship; • Small changes/ changes to a few nucleotides may not be reflected as a change in the banding pattern;

4 Question 2 (a) Fig. 2.1 shows an event occurring in a dendritic cell infected with Mycobacterium tuberculosis (indicated with an * in the figure). The arrowheads show this event involving several of the same organelle. B’ is an enlargement of the boxed area. Fig. 2.1 (i) Compare Mycobacterium tuberculosis with the structure labelled A, excluding size differences. [3] 1. Both contain ribosomes 2. Both contain circular DNA; 3. Both are membrane-bound/ consist of their own membrane; M. tuberculosis A (mitochondrion) 4. Number of membrane present/ presence of cristae • Single membrane/ lack cristae • Double membrane / presence of inner membrane which is highly folded to form cristae 5. Presence of (peptidoglycan) cell wall • Yes • No 6. AVP • rod-shaped • bacterial/ prokaryotic cell • spherical shape here • organelle found inside eukaryotic cell (ii) With reference to Fig. 2.1, describe the event that is occurring in the dendritic cell infected with Mycobacterium tuberculosis. [1] 1. Fusion of lysosomes with the (membrane of) phagosome containing the bacterial cells; A

5 (b) Explain what happens to the IgG -coated beads when they are introduced to the macrophages. [3] 1. IgG are antibodies which opsonize the beads; 2. Macrophages have Fc receptors that bind to the antibody constant region; 3. Idea that this promotes / triggers phagocytosis or cause extension of long pseudopodia to engulf the IgG-coated beads; 4. Phagosome containing IgG-coated beads fuses with lysosome to form a phagolysosome/ to release hydrolytic enzymes; (c) With reference to Fig. 2.2, (i) suggest how IgG-coated beads are being used as a control in this experiment. [2] 1. Phagosomes containing IgG-coated beads are successfully delivered to the lysosome causing a drastic drop in pH from 7.25 to 5.0 / rap