RI Nov 16 H2 P3 ans

2016 H2 Biology P2 9744/03 1 (a) Describe the natural functions of these restriction enzymes. [2] 1. Used as a defense mechanism by bacteria against bacteriophage by cutting up foreign DNA, hence restricts multiplication of viruses; 2. Enzyme that recognizes and bind a specific 4-6 base pair DNA sequence called a restriction site* as its active site is complementary to the DNA sequence; 3. Enzyme cuts/breaks phosphodiester bonds* on specific positions on both DNA strands; (b) (ii) Use the information in Table 1.1 to com pare the effect of pH on immobilised P1 nuclease and free P1 nuclease. [4] 1. Compare peak + quote data The optimum pH for immobilized P1 nuclease was higher at pH 6.0 whereas free P1 nuclease peaked at pH 5.5; or The maximum nuclease activity for immobilized P1 nuclease was higher at 68 arbitrary units(a.u.), whereas maximum nuclease activity for free P1 nuclease was 65 arbitrary units. 2. Immobilised P1 nuclease activity was lower than free P1 nuclease from pH 5 to pH5.5. + quote data from either pH e.g. immoblised P1 nuclease acti vity was 49 a.u. whereas free P1 nuclease was 56 a.u. at pH 5.0, or immoblised p1 nuclease activity was 7 a.u. lower than free P1 nuclease at pH 5.0 3. Immobilised P1 nuclease activity was higher than free P1 nuclease from pH6.0 to pH7.0. + quote data from any 1 pH e.g. immoblised P1 nuclease acti vity was 68 a.u. whereas free P1 nuclease was 48 a.u. at pH 6.0, or immoblised p1 nuclease activity was 20 a.u. higher than free P1 nuclease at pH 6.0); 4. Both had similar trend of increasing in activity from pH4.5 to pH5.5 then decreasing from pH6.0 to pH7.0 + quote data (ii) With reference to your knowledge of enzyme structure and the information in (b), explain the higher activity of immob ilised P1 nuclease, compared to free P1 nuclease, at temperatures above 30 °C, as shown in Table 1.2. [3] 1. At higher temperatures, intramolecular vibrations increase, which results in the breaking of bonds that determine the conformation of the enzyme in the free P1 nuclease. 2. Resulting in the denaturation of the enzyme due to a change in conformation of the enzyme active site, decreasing the ra te of enzyme activity of free P1 nuclease 3. Binding of P1 nuclease to matrix, st abilize the enzyme’s structure at higher temperatures, reducing intramolecular vibrations, prevents breaking of bonds, and retaining the conformation of enzyme active site. 4. Activity of free P1 nuclease starts to drop at 60°C which is lower than that of 70°C in immobolised P1 nuclease. (ii) Suggest why P1 nuclease was needed in the production of this particular recombinant DNA molecule. [3] 1. P1 nuclease was needed to break down the different sticky ends produced by the digestion of HindIII and EcoRI. 2. As it is able to break down single- stranded DNA, leaving the double stranded DNA fragment intact.

3. Scientist will then be able to use the double stranded DNA fragment to synthesize complementary stick ends/add ligase to join the desired DNA fragments directly to form a recombinant DNA molecule. 2 (a) From your knowledge of collagen structure, explain why the length of any exon is divisible by 9. [3] 1. Collagen has a repeating amino acid sequen ce of glycine-X-Y/three amino acids; 2. Where each amino acid is deterimined by a codon of three nucleotides on the mRNA, 3. Thus, three of such codons will be 9 nucleotides in the exon. 3 (a) The genetic disease X-linked SCID (severe combined immunodeficiency) can be treated by bone marrow transplants. This process can be carried out very early in a child's life. (i) The bone marrow contains blood stem cells that are described as being multipotent. State what is meant by multipotent. [1] 1. Multipotent stem cell has ability to divide* and differentiate* to form limited range of cell types / differentiate into red and white blood cells.; (ii) Describe the characteristics that identify all stem cells. [2] 1. A stem cell is an undifferentiated/unspecialized* cell capable of undergoing proliferation* and self-renewal*; 2. and retains potential to differentiate* to produce specialized cells upon receiving appropriate molecular signals*; (c) (i) Suggest how the mitochondrial genome can code for about 40 genes when it is only four times the size of the SCID gene. [2] 1. As mitochondrial genome are similar to prokaryotic genome, they are much smaller, as it has lesser non-coding sequences compared to genes in the chromosomes. 2. SCID genes are large due to introns* which are sequences that are spliced out during post-transcriptional modifications. = 4 Main theory  Effectiveness of active compounds in garlic solution against Staphylococcus can be determined by placing garlic solutions of different concentrations onto wells on prepared 100 mm diameter agar plates, with a lawn of Staphylococcus sp.*.  Garlic solution will diffuse through agar and inhibit growth of bacteria. Measurable quantity  This results in a clear zone around the well that has no bacteria growth. This is known as zone of inhibition and its diameter is a measure of effectiveness of garlic solution against bacteria. Trend  The higher the concentration of garlic, the more it will diffuse across agar, hence a zone of inhibition with a larger diameter due to more bacteria dying.  Most effective concentration of garlic solution is lowest concentration of garlic solution that gives largest zone of inhibition. Further increase in concentration of garlic solution does not increase diameter anymore. Independent variable  20% garlic solution* diluted at 4 concentrations using syringes and sterile distilled water* should be prepared first, under aseptic techniques: Concentration of garlic solution (%) Volume of stock garlic solution (ml) Volume of distilled water* (ml) 20 10.0 0.0 16 8.0 2.0 12 6.0 4.0 8 4.0 6.0 4 2.0 8.0

Dependent variable  Growth of bacteria, as indicated by diameter of zone of clearance (mm) measured using a mm ruler. Controlled variables 1) Volume of garlic solution poured into wells on agar plate  (how) using same cork borer eg: 10 mm diameter, puncture 6 wells of same size on given Petri dish with agar  (why) wells need to be same size so that constant volume of garlic solution will be placed in the wells as amount of active compounds in garlic solution affects bacterial growth 2) Temperature where bacteria is incubated  (how) incubator set at 37 C for 12 hours  (why) temperature needs to be kept constant as bacteria growth is affected by temperature Replicates  (how) 3 replicates are the 3 different petri dishes, each with 5 wells with different concentrations of garlic solution  (why) check for anomalous results Repeats  (how) repeat entire experiment 2 more times  (why) check for reproducibility Negative control  (what) set up control experiment by filling up well with sterile distilled water on bacteria lawn. All other variables and procedures should remain constant.  (why) control shows no clear zone, where bacteria growth is not inhibited. Bacterial growth in other wells is due to presence of garlic solution. Annotated diagram Results Concentration of garlic solution (%) Diamete r of zone of inhibition (mm) Average diameter of zone of inhibition (mm) Replicate 1 Replicate 2 Replicate 3 20 16 12 8 4 100 mm diameter agar plates containing lawn of Staphylococcus bacteria 4% garlic solution 8% garlic solution 12% garlic solution 20% garlic solution Zone of inhibition measured in mm. This is measure after incubating agar plates 37C for 12 hours. Fig. 1 - Set up of experiment to find 16% garlic solution Zone of inhibition measured in mm. This is measure after incubating a