Nov 13 H2P3 ans print

Raffles Institution Nov 2013 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 Nov 2013 H2 Bio Paper 3 . N13P3Q1 (a) State two other ways in which RFLP can be used as a biological tool. [2] RFLP can be used in the analysis of 1. detection of disease e.g. sickle cell anaemia; 2. DNA fingerprinting* in forensics / paternity testing (b) (i) Complete Table 1.1 to show the number of restriction sites and the size of the fragments resulting from digestion of each 15kb DNA molecule, for the three samples. [2] Table 1.1 BamHI EcoRI BamHI and EcoRI Number of restriction sites 1 2 3 Size of fragments / kb 4,11 2,3,10 1,2,3,9 (ii) Using the information from Fig. 1.1 and Table 1.1, it is possible to map the 15kb length of DNA for BamHI and EcoRI. The restriction map for BamHI is shown in Fig. 1.2. Complete the restriction map in Fig. 1.3 for both BamHI and EcoRI, by adding the positions of the EcoRI restriction sites. Indicate the size, in kb, of each fragment. [2] Teacher’s comments: The 2 answers below cannot get the 10kb fragment, hence incorrect. (c) Outline why gel electrophoresis separates DNA fragments. [4] 1. Negatively-charged DNA*; 2. migrates towards the positive electrode/anode when subjected to an electric field / current; 3. Fragments migrate through agarose gel matrix, made up of a meshwork of polysaccharides ; which impedes movement of longer fragments more than shorter fragments; 4. Longer fragments migrate slower compared to shorter fragments;

Raffles Institution Nov 2013 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 (d) Outline the process of DNA hybridization that allows the RFLP pattern for a parti cular gene to be visualized [5] To detect the RFLP pattern of a gene, after the DNA from is digested with restriction enzyme and the digested DNA separate using gel electrophoresis 1. ds DNA is denatured / made single -stranded and by alkaline / NaOH solution and transferred to a nitrocellulose membrane; exactly the same position as they were in the gel 2. The nitrocellulose membrane incubated with a radioactive single stranded DNA probe *, that is complementary in sequence* to part of the target sequence / gene. 3. DNA fragments containing this part of the target sequence will hybridise to the probe by complementary base-pairing*; 4. After hybridisation, membrane is washed to remove any unhybridised probes. 5. Using Autoradiography/X-ray film* over the membrane , the banding pattern can be visualised. (The radioactivity of the bound probes exposes the film to form an image corresponding to the bands that have base-paired to the probe.) N13P3Q2 2 (a) (i) Cytosine is a pyrimidine. Name the other pyrimidine. [1] Thymine* (ii) Suggest how methylation of cytosine nucleotides prevents the DNA of a prokaryote from being cut by its own restriction enzymes. [2] 1. Methylation of cytosine in prokaryotic DNA changes the conformation of the DNA at the restriction site * such that it is no longer complementary in shape and charge * to the restriction enzyme’s active site*. 2. Hence, the restriction enzyme will be unable to recognise and cleave prokaryotes’ own DNA. (b) (OUT OF SYLLABUS) (c) Another restriction enzyme, BspLI, has the same restriction site Where N can be any nucleotide. Using Table 2.1, a scientist predicted that BspLI would cut at more than 20 sites in the standard DNA. Suggest why the scientist made this prediction. [2] 1. There are 4 possible nucleotides (adenosine triphosphate/ Guanosine triphosphate / cytosine triphosphate / thymine triphosphate), 2. Since BspLl restriction site is less specific than of Sfol, there will be more than 20 cut sites; N13P3Q3 (a) (i) Explain why the gene has at least 200 000 base pairs, but the protein only has 1480 amino acids. [3] 1. Besides the 4 440 bases which make up exons, code for the 1480 amino acids; 2. there are non-coding DNA such as introns; 3. e.g. promoter, stop codon, UTR (reject control elements such as enhancers/ silencers); (ii) (OUT OF SYLLABUS)

Raffles Institution Nov 2013 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 (iii) Suggest, with reasons, why other mutations of the CFTR gene vary in the extent to which they cause symptoms of cystic fibrosis. [3] 1. Single base substitutions which lead to coding of another amino acid with similar R -group properties will cause l ittle effect on symptoms of cys tic fibrosis as it can still fol d to similar conformation as original protein; 2. Mutations such as single base insertion or single base deletion within exons 3. can lead to frameshift mutation leading to an incorrect sequence of amino acid formed with different R -groups which cannot fold to form a functional protein leading to serious complications. 4. Base substitutions which lead to a premature stop codon can lead to a truncated protein which is non-functional will lead to have severe symptoms of cystic fibrosis. (b) (OUT OF SYLLABUS) N13P3Q4 Planning Question 1. Aim : To investigate the lowest concentration of copper sulfate solution that has an effect on the leakage of pigment from beetroot cells. 2. Theory (Main theory) [Max 1m] [T1] Betacyanin is found in beetroot cells is prevented from leaking out of cells by membranes (vacuole & plasma membrane) which are made up of phospholipid bilayer embedded with proteins [T2] The betacyanin are too large to pass through the transient pores in the phospholipid bilayer /The betacyanin are charged/polar and are unable to pass through the non-polar hydrophobic hydrocarbon core of the phospholipid bilayer. [T3] Copper ions in the copper sulphate causes denaturation of the membrane proteins embedded in the phospholipid bilayer. The loss of 3D conformation of the membrane proteins increases the permeability of the membrane. Accept when make reference to copper ions disrupting ionic interactions in tertiary structures. (Measurable quantity)[ 1] [T4] the degree of permeability of the membrane will be reflec ted by the amount of betacyanin pigment leaking out of cells. The permeability is determined by measuring the time taken for the colour of the bathing solution to match the colour standard. (Predicted trend) [1] [T5] Increasing concentration of copper sulfate increases the membrane permeability which results in a shorter time taken for the colour of the bathing solution to match the colour standard. [T6] Calculate rate of leaking and extrapolating lowest concentration that has an effect from the graph 3. Procedure (PAN CR) a) [P] Pilot test* Conduct a pilot experiment to determine suitability of apparatus , suitability of range of independent variable (e.g. concentration of copper sulfate solutions) , optimum conditions , amount of materials used (number of pieces of beetroot cylinders, volume of distilled water) [1]

Raffles Institution Nov 2013 (H2 Biology) Paper 3 (for 9744 syllabus) 2017 b) Annotated diagram c) Variables [DV] Dependent variable AND description of how it is measured : Time taken/s for solution to match the colour standard [2] Comment: Dependent variable  Rate of leakage of pigment/ s-1 How is this measured  1/time taken [IV] Independent variable (at least 5 concentrations at equal interval) AND description of how it is controlled : 0.2%, 0.4%, 0.6%, 0.8% and 1.0% copper sulfate solution [2] [CV] Controlled variable and description of how it is cont