RI N22 H2 P2 Answers

2022 H2 Paper 2 Section A Answer all questions 1(a) Outline the main features of the cell theory. [3] 1. The cell is the smallest unit of life ; 2. All known living organisms are made up of cells ; 3. All cells come from pre-existing cells ; (b) (i) Complete Table 1.1 to show the name and function of the structures labelled A, B, C and D in Fig. 1.1. [4] label name function A mitochondrion Site of aerobic respiration* where ATP* synthesis occurs; B nucleus Contains DNA where transcription* occurs to produce mRNA*; C Golgi* body/apparatus Modify lipids and proteins by glycosylation* to form glycolipids and glycolipids respectively; To sort and package proteins into different vesicles and target the proteins to different parts of the cell or for secretion out of cell; D Rough endoplamic reticulum Translation* of mRNA to protein/ protein synthesis by bound ribosomes*; Glycosylation* of proteins; Transportation of proteins in transport vesicles to the Golgi apparatus; (ii) Compare the process of asexual reproduction in yeast with binary fission in bacteria. [2] 1. Similarity – both result in 2 genetically identical* daughter cells; 2. Difference – budding involves the evagination* of the cell surface membrane to form a daughter cell but binary fission involves the invagination* of the cell surface membrane to separate the 2 daughter cells; [Total: 9]

2 (a) With reference to Fig. 2.1, describe the effect of increasing lopinavir concentration on the inhibition of HIV protease. [3] 1. As the concentration of lopinvar increased, the percentage inhibition increased; 2. From 0.0 to 0.4 µmoldm-3 lopinavir concentration, there is a gradual increase in percentage inhibition, from 0.0 to 8.0%; 3. From 0.4 to 1.6 µmoldm-3 lopinavir concentration,there is a steep increase in percentage inhibition from 8.0 to 97.0%; 4. From 1.6 to 2.4 µmoldm-3 lopinavir concentration, the pecentage inhibition starts to plateau from about 97.0 to 99.0%; (b) Explain how lopinavir inhibits HIV protease. [4] 1. Lopinavir has a 3D conformation that is complementary in shape and charge to the active site* of HIV protease; 2. and lopinavir has a similar shape as the substrate/polyproteins; 3. Lopinavir competes with the substrate/polyproteins for the active site of HIV protease and prevents the substrate from binding to active site and hence reduces the rate of reaction; 4. The binding of the inhibitor to the active site is not permanent/reversible; 5. However, at high substrate concentration, the effect of the inhibitor can be overcome and Vmax can be reached; (c) Translation of viral RNA in cells infected with the human immunodeficiency virus (HIV) results in large polypeptides known as polyproteins. HIV protease cuts each of these viral polyproteins into smaller molecules. Suggest how inhibition of HIV protease may limit the ability of HIV to reproduce. [3] 1. If HIV protease activity is inhibited, then viral polyproteins cannot be cleaved; 2. Hence proteins such as reverse transcriptase, integrase, protease and capsid proteins cannot be synthesised; 3. Fully functional virions cannot be produced and the ability for HIV to reproduce will be limited; [Total: 10] 3 (a) Name the structures labelled P, Q and R in Fig. 3.1. [3] P enhancer Q promoter R structural gene/ coding region

(b) Compare transcription in prokaryotes and eukaryotes. [4] Feature Prokaryotes Eukaryotes Binding of RNA polymerase RNA polymerase binds directly to the promoter* via the sigma factor; RNA polymerase is recruited by general transcription factors* to bind to the promoter* forming the transcription initiation complex*; Genes transcribed Transcription of several genes in an operon controlled by one promoter; Transcription of one gene controlled by one promoter; mRNA formed Polycistronic* mRNA is formed; Monocistronic* mRNA is formed; Upregulation of transcription In lac operon, cAMP binds to catabolic activator protein* (CAP). Active CAP binds to CAP binding site* in the promoter region and increase affinity of RNA polymerase* binding to the promoter*, thus increasing the frequency of transcription; Activator* binds to enhancer* causing bending of spacer DNA and promoting the assembly of the transcription initiation complex which will increase frequency of transcription; Downregulation of transcription Binding of repressor* to operator* prevents the RNA polymerase binding to the promoter and hence preventing transcription. E.g. lac repressor bind to operator in lac operon/ trp repressor bind to operator in trp operon; Binding of repressor* to silencer* will prevent the assembly of the transcription initiation complex which will decrease frequency of transcription; (c) Explain how the introduction of mRNA into the cytoplasm of adult cells can result in the production of iPSCs and why these iPSCs are footprint free. [4] How introduction of synthetic mRNAs can produce iPSCs: 1. Translation* of the synthetic mRNAs will produce proteins which are needed to 2. switch off certain genes in specialised adult cells allowing them to de-differentiate; 3. and switch on expression of certain specific stem cell genes allowing the adult somatic cells to achieve pluripotency; Why are these iPSCs footprint-free: 4. The synthetic mRNAs do not enter the nucleus and will not intergrate with the genomic DNA – they do not modify the genome of the adult cells; 5. The synthetic mRNAs are degraded after a while and will not have a trace inside the cells; [Total: 11]

4(a) Explain how the end replication problem arises during DNA synthesis and describe the consequences of this problem. [5] 1. Occurs during replication of linear DNA. 2. Each round of DNA replication will result in the shortening of daughter molecules at the 5’ ends of the telomeres. 3. because RNA primers* at the 5’end is removed, but DNA polymerase is unable to replace with DNA; (idea of end replication problem) 4. result in overhangs at the 3’ end of the telomeres. 5. However, since telomeres are non-coding, this ensures that vital genetic information/genes are not lost / eroded with each round of replication; 6. If the telomeres shorten to critical length, cell will be signalled to stop dividing; (b) (i) With reference to Fig. 4.1, describe the role of the short piece of RNA that forms part of the telomerase complex. [3] 1. 5 nucleotides of the telomerase RNA anneals and forms complementary base pairs* with the single-stranded overhang at 3’ end of the telomere; 2. which aligns the telomerase reverse transcriptase with respect to the DNA; 3. The telomerase RNA then serves as the template* for formation of a complementary DNA sequence; 4. whereby adenine* pairs with uracil*, thymine* with adenine*, cytosine* with guanine*, and guanine* with cytosine*; 5. Resulting in tandem repeat sequences; (ii) With reference to Fig. 4.1, describe the activity of the reverse transcriptase part of the telomerase complex. [3] 1. Telomerase has an active site* that is complementary in conformation and charge* to a specific* telomeric DNA sequence; 2. Using telomerase RNA as a template*, telomerase reverse transcriptase forms a complementary DNA* sequence through complementary base pairing*; (whereby adenine base pair with uracil, thymine with adenine, cytosine with guanine, and guanine with cytosine) 3. Catalyzes the formation of phosphodiester bonds* between deoxyribonuceotides; 4. translocation in the 5’  3’ direction to produce a series of tandem repeats of GGTTAG, thus elongating the telomere/DNA at the 3’ overhang; [Total: 11] 5(a) (i) a loss of function mutation of the tumour suppressor gene p53. [2] 1. loss-of-function mutation in both alleles/two copies of p53 results in a non- functional gene product/ gene product/p53 p