Nov 14 H2P3 ans print

Raffles Institution Nov 2014 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 Nov 2014 H2 Bio Paper 3 N14P3Q1 1(a) (i) Name one genetic disease which has been treated with stem cell transplantation. [1] Bone marrow haematopoietic stem cells transplants from normal healthy bone marrow donors to leukaemia patients Or Neural stem cell transplant for Parkinson’s disease/multiple sclerosis by introducing adult neural stem cells into damaged tissue. (ii) Explain why stem cell is suitable for this purpose. [3] 1. Adult stem cell is multipotent that differentiates* into the respective specialized* cell type, thus restoring function of damaged or diseased tissue. 2. Self-renewing* nature of stem cells ensures that transplanted stem cells constantly replicate* in the patient to maintain a constant pool of stem cells 3. As the ‘healthy’ stem cell carries the normal and functional allele thus they can produce normal levels of functional protein and be used to treat genetic diseases b(i) Using the letter R, label Fig. 1.1 to identify a feature that allows the virus to bind to cells. [1] (ii) With reference to your knowledge of retroviruses, explain how expression of an inserted gene (transgene) is brought about following infection of host cells with the lentiviral vector. [3] 1. Once inside the host cell, retroviruses create double stranded DNA copies of their RNA genomes via reverse transcriptase*. 2. Viral genome together with the transgene are integrated randomly into host chromosomes via integrase. 3. Transgene will undergo transcription and translation by host enzymes to produce normal, functional protein. (iii) (OUT OF SYLLABUS) (c) (OUT OF SYLLABUS) [Total : 12] N14P3Q2 (OUT OF SYLLABUS) Note: R can be any one of the 2 types of glycoproteins, one arrow is sufficient.

Raffles Institution Nov 2014 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 N14P3Q3 (a) Describe the limitations of PCR. [3] Feature Limitations Taq polymerase lacks 3’ to 5’ proofreading ability Errors occurring early in the PCR reaction will get compounded with each replication cycle and all daughter molecules resulting from this early error will be exponentially affected. Synthesis of PCR primers depends on sequence information from target region Success of PCR requires knowledge of sequences flanking target region to be amplified. If the flanking sequences of a gene of interest are unknown, no proper primers can be synthesised to amplify the target DNA sequence. If primers are designed incorrectly, no amplification occurs / wrong DNA fragment(s) may be amplified. Limit to size of DNA fragment to be amplified DNA fragments to be amplified are limited to about 3 kb 1. Further increase in length of target sequence decreases efficiency of amplification. This is because the polymerase tends to ‘fall off’ DNA template before chain extension is complete. Exponential amplification of contaminant DNA It is possible to contaminate a fresh PCR reaction with minute amounts of contaminant DNA due to poor laboratory skills. Such unwanted DNA sequences may be amplified to significant amounts, alongside the target DNA sequences. (b) (i) Describe the role of the buffer solution in the gel electrophoresis protocol. [2] 1. Buffers contain ions which allows conduction of electric current 2. Thus allowing the negatively charged DNA molecules to move from the negative electrode to the positive electrode (b) (ii) Describe the role of the loading or tracking dye in the gel electrophoresis protocol. [3] 1. Contains glycerol which makes the DNA sample denser than buffer so that DNA sample can sink to the bottom of the well. 2. DNA is invisible, so the dyes colour the DNA sample showing if it has been loaded correctly into well. 3. 2 coloured dyes act as visual markers to show the progress of migration of DNA fragments in the gel. 4. One dye typically (moves at speed corresponding to 100bp DNA fragment which) runs ahead of sample and another (moves at speed of 1100bp DNA fragment which) runs after sample.

Raffles Institution Nov 2014 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 (c) (i) Outline the process of genetic fingerprinting using RFLP that could be used to test this seized ivory. [4] 1. Genomic DNA is extracted from the soft tissue/dried blood and cut with same restriction enzyme* to obtain different-sized DNA fragments. 2. DNA is separated according to size in gel electrophoresis where negatively- charged DNA* migrates towards the positive electrode/anode when subjected to an electric field / current; 3. Meshwork of agarose fibres impedes movement of longer fragments more than shorter fragments resulting in smallest fragments moving furthes t/largest fragments least far from well 4. ds DNA is denatured / made single -stranded and by alkaline / NaOH solution and transferred to a nitrocellulose membrane 5. Carry out Southern blotting/nucleic acid hybridisation by incubating membrane with single strand radioactive probe* which will hybridise with DNA fragment through complementary base pairing 6. Using autoradiography/X-ray film* over the membrane, the banding pattern can be visualised. (ii) Explain how the genetic fingerprints of the seized ivory could be used to confirm that it originated from elephants in Malawi. [4] 1. Genetic fingerprint is due to different alleles/markers producing different bands in gel resulting in the unique banding pattern in individuals 2. Different bands arise due to polymorphic nature of DNA in different individuals, there will be variations in number and location of restriction sites and number of tandemly repeated nucleotide sequence among individuals. 3. Genetic fingerprint of animals that provided ivory can be compared against fingerprint of elephants from Malawi to see how closely related they are. 4. If fingerprint pattern is similar to pattern to that of Malawi elephants, then ivory haul from Malawi. N14P3Q4 4 Planning [Total: 12] Suggested answer scheme: Part 1: Aim To investigate effect of temperature and pH on rate of sucrase activity of 2 enzymes P and Q. Part 2: Theory (Main Theory): [T1: 1 mark for any 2 points from 1 to 5] 1. Active site of sucrase has a specific conformation , complementary in shape and charge to substrate, sucrose. 2. Conformation of active site, hence rate of sucrase activity, is affected by temperature and pH. 3. Excess [H +] or [OH -] ions disrupt ionic, hydrogen bonds , which determine tertiary/quarternary structure of enzyme hence its any conformation. 4. Increasing temperature (up to denaturation) increases kinetic energy of molecules, increasing rate of effective collisions between enzymes and substrate to form enzyme-substrate complex. 5. At optimum temperature and pH of each enzyme, rate of production of reducing sugars is maximum. (Measurable variable):

Raffles Institution Nov 2014 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 6. Mass of brick red precipitate formed when products (glucose, fructose) obtained from hydrolysis of substrate (sucrose) are tested with Benedict’s solution. [T2: 1 mark] Dependent variable : rate of sucrase activity (P and Q) as indicated by rate of brick red precipitate formed Indepe