2012 a Level H2 Biology P3 Ans

2012 A level H2 Biology Paper 3 Question 1 (a) 1. A collection/ set of cloned DNA fragments; 2. of the entire genome of an organism; 3. genome is cleaved by a specific restriction enzyme; 4. restriction fragments are inserted into plasmids before introducing into bacteria; (b) 1. Reverse transcriptase; uses mRNA as a template to synthesise cDNA; 2. RNA-ase; to degrade the RNA template; 3. DNA polymerase; synthesizes a complementary strand using the cDNA as template to produce double stranded DNA; (c) 1. cDNA library is produced using mRNA which are transcribed at a particular time in a cell; 2. gene expression might be different at different time; 3. different mRNA might be present; 4. depending on the function and need of the cell; 5. genomic library is produced from the entire set of genome in the cell; 6. which remains the same all the time; (d) (i) 1. 2. 3. 4.

Annealing of primers; to DNA template; by forming hydrogen bonds; between complementary base pairs;

(ii) Free deoxyribonucleoside triphosphate (dNTPs); Unused primers; Taq polymerase; Double stranded DNA molecule; Double stranded DNA molecule with regions annealed to DNA primers; Double stranded DNA molecule with Taq polymerase bound to it;

(e) 1. 2. 3. 4.

A - Denaturation of double stranded DNA; breaking of hydrogen bonds between complementary base pairs; B - Binding of single stranded radioactive probes; to complementary region on a single stranded DNA;

2012 A level H2 Biology Question 2 (a) Multipotent (b) 1. 2. 3. 4.

cells from donor trachea might be incompatible to the recipient;; recognized by recipient immune system as foreign; Elicit immune response; resulting in tissue rejection/ destroying the donor trachea;;

(c) 1. stem cells are undifferentiated; 2. when treated with chemicals, trigger the cells to divide indefinitely and differentiate;; 3. into tracheal cells; 4. causing certain genes to be turned on/ turned off; (d) 1. 2. 3. 4. 5. 6.

Presence of telomerase in stem cells; Lengthen the telomeres at the ends of chromosomes; No telomerase in other cells; DNA daughter strand shortened after every round of mitotic division; when telomeres shortened to a particular length triggers cell death; unable to continue to divide mitotically;

Question 3 (a) (i) 1. Plasmids have multiple restriction sites; 2. that can be recognized and cleaved by different restriction enzymes; 3. DNA fragment isolated from any species by restriction digestion; 4. can be inserted into the plasmid if cut with the same restriction enzyme; (ii) 1. Allows plasmid to replicate independently; 2. forming multiple copies within a bacterium;; (b) (i) Complementary base pairing;; (ii) 1. 2. 3. 4. 5. 6. 7.

Plasmid reanneals; forming back the original plasmid; Gene of interest anneals at both ends; forming a circular DNA; due to complementary base pairing; between sticky ends; CTTC and GAAG;

1. 2. 3. 4.

Treat bacteria with CaCl2; To neutralize the negative charge on the DNA and phospholipid bilayer; Place the mixture of bacteria and recombinant plasmids on ice; Perform heat shock treatment at 42oC for 30s;

(c)

2012 A level H2 Biology 5. creates temperature imbalance which sweeps the plasmid into the bacteria; (d) Grow bacteria in medium containing antibiotic; Non-transformed bacteria would die; Transformed bacteria with reannealed plasmid or recombinant plasmid would survive; Conduct replica plating; Grow transformed bacteria in another antibiotic; Transformed bacteria with recombinant plasmid would die; due the antibiotic resistance gene being disrupted by insertion of gene of interest; (e) Plasmid Q;; Plasmid contain both PstI and FokI recognition sites for insertion of genes S and T respectively; Plasmid Q has medium copy number more efficient than plasmid P; Question 4 Aim Effect of concentration of sucrose solution and the rate of respiration in yeast cell. Reaction Sucrose  Glucose + Fructose C6H12O6 + 6O2  6CO2 + 6H2O CO2 + Ba(OH)2  BaCO3 (s) + H2O Background information 1. Sucrose can by hydrolysed into glucose and fructose. 2. Yeast cell undergoes aerobic respiration using glucose. 3. Carbon dioxide is released as by-product of oxidative decarboxylation during link reaction and Krebs cycle. 4. Reactions are catalyzed by respiratory enzymes Relate method of measurement to dependent variable. 5. Barium hydroxide can react with the CO2 to produce barium carbonate 6. The remaining barium hydroxide can be neutralized with hydrochloric acid 7. when all the remaining barium hydroxide is completely neutralised, 8. end point of the reaction can be determined using indicator phenolphthalein 9. as it turns from pink to colourless 10. amount of CO2 released is inversely proportional to the amount of HCl required to neutralize the remaining barium hydroxide 11. In the absence of CO2, 5 cm3 of HCl is required to completely neutralize the 10cm3 of barium hydroxide 12. 1 cm3 of HCl is equivalent to 2.2mg of CO2 13. mass of CO2 can be calculated using the formula below: 14. (5 - Volume of HCl used to neutralise Barium hydroxide) x 2.2mg 15. rate of respiration= mass of CO2 produced/ fixed duration of experiment Expected trend: 16. As the concentration of sucrose increases, the mass of CO2 produced in a fixed duration increases, thus higher rate of reaction.

2012 A level H2 Biology 17. This is because as concentration of sucrose increase 18. the frequency of successful collision between the active site of respiratory enzymes and substrates increases, 19. more enzyme-substrate complexes formed per unit time and 20. more CO2 formed per unit time 21. rate of respiration increases until saturation point. 22. Rate of respiration will remain at maximum when all the enzyme active sites are occupied by substrates Experiment variables 23. Independent variable – concentration of sucrose 24. Range: 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm -3 of sucrose Dependent variable 25. mass of CO2 produced, based on the volume of HCl required to neutralize the remaining barium hydroxide Controlled variables Controlled variables 26. Time taken for reaction 27. Volume of yeast suspension

Quantity 1 min 5.0 cm3

28. Concentration of yeast suspension

1%

29. Volume of sucrose solution

10.0 cm3

30. Volume of barium hydroxide 31. Concentration of barium hydroxide 32. Concentration of HCl 33. Volume of phenolpthalein 34. Temperature

10 cm3 0.025 moldm-3 0.1 moldm-3 0.5 cm3 37°C

Experimental Procedures 1. Conduct pilot experiment to determine suitability of apparatus, optimum conditions and amount of material used. 2. You are given 1.0moldm-3 sucrose stock solution. 3. Label plastic vials 0.2, 0.4, 0.6, 0.8 and 1.0. Dilute the sucrose stock solution, using labeled plastic vials, to give 30 cm3 of each concentrations: 0.2 moldm3 , 0.4 moldm-3, 0.6 moldm-3, 0.8 moldm-3, 1.0 moldm-3 Dilution table of sucrose solution Concentration Volume of Volume of of diluted stock sucrose/ water/ cm3 sucrose/ cm3 -3 moldm 0.2 2.0 8.0 0.4 4.0 6.0 0.6 6.0 4.0 0.8 8.0 2.0 1.0 10.0 0.0

2012 A level H2 Biology 4. Label test tubes 0.2, 0.4, 0.6, 0.8 and 1.0. Aliquot 10cm3 of each concentration of the diluted sucrose from the plastic vials into respective test tubes. 5. Place test tubes containing diluted sucrose solution and yeast suspension in a water bath maintained at 37°C for 1 minute. 6. Add 10cm3 of barium hydroxide into a test tube using syringe 7. Add 5 cm3 of active yeast suspension to the test tube containing 0.2 moldm-3 sucrose solution and mix well. 8. Cover the test tube with rubber bung connected to delivery tube, place delivery tube into the test tube containing barium hydroxide immediately and start the stopwatch. 9. After 1 minute, remove the text tube containing barium hydroxide, add 0.5 cm3 of phenolphthalein 10. Add HCl to the test tube using syringe, 0.5 cm3 at a time. 11. Record the volume of HCl required for phenolphthalein to turn from pink to colourless. 12. Repeat steps 6 to 11 for each concentration of diluted sucrose. 13. Repeat procedures two more times to obtain 2 more readings for each concentration. 14. Repeat the entire experiment two more times. 15. Carry out statistical test to determine whether there is any significant difference between the means. Negative Control 35. A negative control is subjected to the same factors as that for the experiment, except that the 10 cm3 of sucrose is replaced with 10 cm3 of distilled water. 36. It is expected that 5 cm3 of HCl will be required to neutralize the solution as no CO2 is produced. 37. This proves that it is indeed the presence of sucrose and not any other factors, causes the production of CO2. Data manipulation 38. Calculate the mass of CO2 produced using the formula below: (5 – volume of HCl required) x 2.2mg 39. The mass of CO2 produced by each mixture is converted to rate (R) by the following equation: R = mass of CO2 produced/ 1min 40. Table showing the effect of concentration of sucrose solution on the rate of respiration Concentration Volume of HCl Mass of CO2 Rate of sucrose required, V/cm3 produced/mg respiration solution,c in yeast _ /moldm-3 cell, R/mg V1 V2 V3 V min-1 0.2 0.4 0.6 0.8 1.0

2012 A level H2 Biology

Graph showing the effect of concentration of sucrose solution on the rate of respiration Rate of respiration, -1 R/ mgmin

Concentration of -3 sucrose/moldm

Safety precaution 41. Phenolphthalein indicator is toxic. Wear gloves and goggles to avoid contact with skin and eyes. 42. Barium hydroxide and hydrochloric acid are corrosive. Wear gloves and goggles to avoid contact with skin and eyes. 43. AVP i.e. Use a piece of rag to hold the beaker containing hot water (when preparing water bath) to prevent burns / scalds. Question 5 5(a) Steps for plant tissue culture

Scientific reason

S1. Explants (i.e. shoot tips and root tips) are excised from the plant to be cloned; S2 Explant must contain meristematic cells; S3. Cells are sterilized using dilute sodium hypochlorite;

R1. Meristematic cells are totipotent; R2. Can regenerate into a whole new plant;

S4. Cells are grown in culture vessels; S5. on nutrient agar containing nutrients and plant growth regulators; S6. to form callus; S7. Cut callus into several pieces; S8. Place the pieces into separate culture vessel; S9. Different levels of auxin and cytokinin are added; S10. Differentiated tissues will grow into a plantlet;

R4. To stimulate cells to divide by mitosis;

S11. Transfer of plantlets to soil;

R3. to kill bacteria and fungus/ prevent contamination;

R5. To increase the number of callus; R6. by subculturing; R7. To induce cell differentiation; R8. Low level of cytokinin and high level of auxin triggers formation of roots growth; R9. High level of cytokinin and low level auxin triggers shoot growth; R10. Acclimatisation before transferring to the field;

2012 A level H2 Biology 5(b) Positive aspects [max 2]  Inclusion of pest-resistance genes, reduced use of pesticides - chemical-free plant crops for consumers;; - lower production cost for farmers since there will not be a need to purchase pesticides;; 

Inclusion of genes to increase the nutritional value of crop plants - nutritionally enhanced food can overcome the problem of malnutrition in developing countries;;

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Inclusion of gene to increase crop yield - reduce the problem of food shortage;;

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Inclusion of delayed ripening gene - greater commercial value for crop plant to be freight over longer distances to consumers;;

Negative aspects [max 4]         

Threat to human safety as there is a potential of transfer of allergens;; Antibiotic resistance gene if not properly broken down by the digestive system may be passed to E. coli found in the gut, making them resistant to antibiotics, potential impact to human health;; GMO may be carried by wind to other places and establish themselves as weeds and pose difficulties to be eradicated;; Cross-pollination between GMO and wild relatives, spreading herbicideresistance to weeds, resulting in “superweeds”, unable to eradicate by herbicides, outcompete with the crop plants and took over agricultural field;; GMOs outcompete the local native species and thereby upset the balance in the natural ecosystem and species/ loss of biodiversity;; Violation of the natural organism's intrinsic values with the mixing genes among species may evoke strong responses from naturalists;; Objections from specific religious groups with strict dietary restrictions;; Patenting of GMOs;; Labelling of GM food is not mandatory in some countries;;

5(c) Considered as new species  Biological concept defines species as a group of organisms capable of interbreeding and producing fertile offspring;  Creation of GMO may result in organism being sterile due to polyploidy;  Unable to interbreed with original population of 2n species to produce fertile offspring;   

Ecological niche concept defines species as a group of organisms sharing the same ecological niche; No two species can share the same ecological niche; GMO are modified to be able to withstand the unfavourable condition in the environment;

2012 A level H2 Biology 

Likely to outcompete the original species when they occupy the same ecological niche, establish themselves as new species;

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Based on morphological concept, a group of organisms should share a unique set of structural species; Genes from another organism are introduced into GMO; giving desired traits, different phenotype from original species;

   

According to the phylogenetic concept a group of organisms bound by a unique ancestry and share one or more derived character; Mixing of genes of more than one species, no evolutionary relationship of GMO with other species;

Not considered new species  GMO may still exist as a diploid organism, capable of producing fertile offspring with the original species;  Little/no changes to structural morphology, still resemble original species;