2021 Bio H2 MCQ Explanations

1. N21Q1 B chloroplast mitochondrion ribosome phosphorylation ✓ oxidation ✓ translation ✓ polymerisation ✓ transcription ✓ replication  transcription ✓ reduction ✓ polymerisation ✓ translation ✓ phosphorylation ✓ transcription  Chloroplast – it contains circular DNA and 70S ribosomes: transcription and translation can occur. The synthesis of mRNA and polypeptides are also polymerisation reactions. In photosynthesis, ATP is synthesised in the light - dependent reactions. This involves phosphorylation of ADP to form ATP. Mitochondrion – it contains circular DNA and 70S ribosomes: transcription and translation (good to note this even though it is not in the question) can occur. Aerobic respiration involves the mitochondrion where glucose is oxidised. There are also several reactions that are oxidation by dehydrogenation in the Krebs cycle. Passing of electrons down the electron transport chain involves a series of redox reactions: the electron donor is oxidised, whereas the electron acceptor is reduced. (Note that electron transport chains are also in the chloroplast even though this is not in the question.) Ribosome – it is the site of translation and the process involves the polymerisation of amino acids to form polypeptides; 2. N21Q2 D According to endosymbiosis, mitochondrion was once a bacterial-like single cell organism. Thus it shows similarity with bacteria such as circular DNA, 70S ribosomes and reproduction by binary fission. The ability to synthesise proteins is not unique to bacteria / prokaryotes. 3. N21Q3 D Consider the features and select those that are unique to viruses: 1: genome of circular double -stranded DNA – this is a feature of some (not all) viruses and is also a feature of bacteria 2: enclosed within a protein coat – this describes non-enveloped viruses 3: layer of lipids – this describes the viral envelope, but is also true of any other cell that has a phospholipid bilayer as its cell surface membrane. 4: replicates only in the cytoplasm of a eukaryotic cell – this describes an obligate parasite which is a feature of viruses 4. N21Q4 A Only bacteriophage has no phospholipid while influenza virus and HIV are enveloped virus and Mycobacterium tuberculosis is a bacteria with plasma membrane. 5. N21Q5 D The difference between -glucose and - glucose is the position of the OH group attached to carbon -1. For -glucose, the OH group is below the plane of the ring (on the opposite side of the ring as carbon -6), while for -glucose, the OH group is above the plane of the ring (on the same side of the ring as carbon-6). GCE A Level H2 Biology 9744 Biology November 2021

Options A and B are incorrect as the OH groups cannot be replaced. Option D describes carbon -1 correctly as carbon -1 is attached to O in the ring and the O of the OH group, so it is the carbon that forms bonds with 2 oxygen atoms. 6. N21Q6 D For this type of questions, look for things that are inconsistent/incoherent in the options given: Option A – Water moves by facilitated diffusion down an electrochemical gradient is incorrect. The correct term should be water potential gradient. Electrochemical gradient is used for charged particles. Option B – H+ moves by active transport down a diffusion gradient is incorrect. Diffusion is passive transport and not active transport. Option – K+ description is incorrect (same as option B). Water moves across the membrane by osmosis, down a water potential gradient. This is a passive process and should not require an ATP-powered pump. 7. N21Q7 C Based on the context of the question, the enzyme is able to fluoresce when illuminated with UV light. Changes in shape are able to the change the intensity or the wavelength of the fluorescence. To interpret the effect of adding serine and the effect of adding serine + indole to the enzyme – compare the graphs with the graph of enzyme only. For enzyme + serine, the peak fluorescence obtained is higher, showing that serine binds to the enzyme to change the conformation of the enzyme. This supports the idea that an enzyme-serine complex is formed. For enzyme + serine + indole, the peak fluorescence obtained is lower, showing that serine + indole binds to the enzyme to change the conformation of the enzyme. The shape is different from that of enzyme + serine, hence the peak fluorescence is also different. This supports the idea that an enzyme -serine- indole complex is formed. For both graphs, the wavelength at which the peak fluorescence occurs is unchanged, so statement 2 is incorrect. 8. N21Q8 B The question context says that hormones from the pituitary gland stimulate cells in the mammary glands so the cells become able to secrete milk. Stem cells are unspecialised, so they must first differentiate to become cells that can secrete hormones. 9. N21Q9 A The events occur in this order: virus recognises host cell by (statement 3) hemagglutinin in the viral envelope binding to sialic acid receptors on host cell. Then the virus is taken into the cell by endocytosis. Hence, (statement 1) the virus becomes enclosed in an endosome. The endosome then fuses with a lysosome, resulting in (statement 4) a low pH in the endosome. The low pH causes a conformational change in hemagglutinin (statement 5) that changes its binding specificity, such that it brings together the viral envelope and the endosomal membrane. This causes (statement 2) the viral envelope to fuse with the endosomal membrane. 10. N21Q10 A This question is on transduction. In stage 2, the 1 st bacterial host is killed, suggesting that this is lytic cycle. New phages are released, including defective phages that have enclosed parts of the bacterial DNA in its capsid head. In stage 3, new phages infect another culture of bacteria. At the end of stage 4, there is no description that the bacteria is killed (unlike in stage 2), suggesting that this is the lysogenic cycle. The final result for type 1 bacteriophage produced bacteria showing a range of different characteristics sugge sting that different DNA sequences are

introduced. This describes generalised transduction where random sequences from the 1st bacterial host is introduced into the 2nd bacterial host. In contrast, for type 2 bacteriophage, all bacteria show the same characteristics, suggested that the same DNA sequence is introduced. This matches with specialised transduction where specific segments of bacterial chromosome adjacent to the prophage may be wrongly excised and packaged into the defective phage, and transferred to the 2nd bacterial host. 11. N21Q11 C For this type of question, consider the options to choose the best one that matches the figure. A: Incorrect as the DNA also has sequences like the promoter, as suggested by the binding of RNA polymerase. B: Incorrect as RNA polymerase (which is a protein) continues to bind to the DNA as it transcribes through the entire section (both coding and non-coding). C: Correct. Grey shaded regions are removed from the pre-mRNA showing that they are introns that are found between coding regions. D: Incorrect as the grey regions (non-coding regions corresponding to introns) appear in the pre-mRNA, showing that they are transcribed. Introns are later removed from the pre-mRNA during post-transcriptional modification. Note that for this question, the term intron is used to refer to the non-coding region on the DNA as well as the non-coding region on the pre-mRNA. 12. N21Q12 C The lack of a nuclear envelope means that transcription and translation cannot be compartmentalised. So as soon as the mRNA is made during transcription in the nucleoid region, the 70S ribosomes can attach to the mRNA for translation purpose. 13. N21Q13 A This is the correct order of events: (bullet pt 3) gel electrophoresis – to separate DNA fragments according to size (bullet pt 4) DNA denatured