2012 a Level H2 Biology P3 Ans

8/11/2019 2012 a Level H2 Biology P3 Ans

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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 particulartime 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. Annealing of primers;2. to DNA template;

3. by forming hydrogen bonds;4. 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. A- Denaturation of double stranded DNA;2. breaking of hydrogen bonds between complementary base pairs;3. B- Binding of single stranded radioactive probes;4. to complementary region on a single stranded DNA;


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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 wouldsurvive;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 Trespectively;Plasmid Q has medium copy number more efficient than plasmid P;

Question 4

AimEffect of concentration of sucrose solution and the rate of respiration in yeast cell.

ReactionSucroseGlucose + FructoseC6H12O6+ 6O26CO2+ 6H2OCO2+ Ba(OH)2BaCO3 (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 CO2to produce barium carbonate6. The remaining barium hydroxide can be neutralized with hydrochloric acid7. 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 colourless10. amount of CO2released is inversely proportional to the amount of HCl

required to neutralize the remaining barium hydroxide11. In the absence of CO2, 5 cm

3of HCl is required to completely neutralize the10cm3 of barium hydroxide

12. 1 cm3of HCl is equivalent to 2.2mg of CO213. mass of CO2 can be calculated using the formula below:14. (5 - Volume of HCl used to neutralise Barium hydroxide) x 2.2mg15. rate of respiration= mass of CO2 produced/ fixed duration of experiment

Expected trend:16. As the concentration of sucrose increases, the mass of CO

2produced in a

fixed duration increases, thus higher rate of reaction.


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17. This is because as concentration of sucrose increase18. the frequency of successful collision between the active site of respiratory

enzymes and substrates increases,19. more enzyme-substrate complexes formed per unit time and20. more CO2formed per unit time21. 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 variables23. Independent variableconcentration of sucrose24. Range: 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm-3of sucrose

Dependent variable25. mass of CO2produced, based on the volume of HCl required to neutralize the

remaining barium hydroxide

Controlled variablesControlled variables Quantity

26. Time taken for reaction 1 min

27. Volume of yeast suspension 5.0 cm

28. Concentration of yeastsuspension

1%

29. Volume of sucrose solution 10.0 cm3

30. Volume of barium hydroxide 10 cm3

31. Concentration of barium

hydroxide

0.025 moldm-3

32. Concentration of HCl 0.1 moldm-3

33. Volume of phenolpthalein 0.5 cm3

34. Temperature 37C

Experimental Procedures1. Conduct pilot experiment to determine suitability of apparatus, optimum

conditions and amount of material used.2. You are given 1.0moldm-3sucrose 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 cm

3

of each concentrations: 0.2 moldm

-

3, 0.4 moldm-3, 0.6 moldm-3, 0.8 moldm-3, 1.0 moldm-3

Dilution table of sucrose solution

Concentrationof dilutedsucrose/moldm-3

Volume ofstock sucrose/

cm3

Volume ofwater/ cm3

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


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4. Label test tubes 0.2, 0.4, 0.6, 0.8 and 1.0. Aliquot 10cm3of eachconcentration of the diluted sucrose from the plastic vials into respective testtubes.

5. Place test tubes containing diluted sucrose solution and yeast suspension ina water bath maintained at 37C for 1 minute.

6. Add 10cm3of barium hydroxide into a test tube using syringe7. Add 5 cm3of 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 andstart the stopwatch.

9. After 1 minute, remove the text tube containing barium hydroxide, add 0.5cm3of phenolphthalein

10. Add HCl to the test tube using syringe, 0.5 cm3at 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 eachconcentration.14. Repeat the entire experiment two more times.

15. Carry out statistical test to determine whether there is any significantdifference between the means.

Negative Control35. 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 cm3of distilled water.36. It is expected that 5 cm3of 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 otherfactors, causes the production of CO2.

Data manipulation38. Calculate the mass of CO2 produced using the formula below:

(5volume of HCl required) x 2.2mg39. The mass of CO2produced by each mixture is converted to rate (R) by the

following equation: R = mass of CO2 produced/ 1min40.

Table showing the effect of concentration of sucrose solution on the rate ofrespiration

Concentration

of sucrosesolution,c/moldm-3

Volume of HCl

required, V/cm3Mass of CO2produced/mg

Rate

respirationin yeast

cell, R/mgmin-1V1 V2 V3

_V

0.2

0.4

0.6

0.8

1.0


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Graph showing the effect of concentration of sucrose solution on the rate ofrespiration

Safety precaution

41. Phenolphthalein indicator is toxic. Wear gloves and goggles to avoid contactwith skin and eyes.

42. Barium hydroxide and hydrochloric acid are corrosive. Wear gloves andgoggles to avoid contact with skin and eyes.

43. AVP i.e. Use a piece of rag to hold the beaker containing hot water (whenpreparing water bath) to prevent burns / scalds.

Question 55(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 meristematiccells;

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

S3. Cells are sterilized using dilutesodium hypochlorite;

R3. to kill bacteria and fungus/ preventcontamination;

S4. Cells are grown in culture vessels;

S5. on nutrient agar containing nutrientsand plant growth regulators;S6. to form callus;

R4. To stimulate cells to divide by

mitosis;

S7. Cut callus into several pieces;S8. Place the pieces into separateculture vessel;

R5. To increase the number of callus;R6. by subculturing;

S9. Different levels of auxin and cytokininare added;S10. Differentiated tissues will grow intoa plantlet;

R7. To induce cell differentiation;R8. Low level of cytokinin and high levelof auxin triggers formation of rootsgrowth;R9. High level of cytokinin and low levelauxin triggers shoot growth;

S11. Transfer of plantlets to soil; R10. Acclimatisation before transferringto the field;

Concentration ofsucrose/moldm

-3

Rate ofrespiration,R/ mgmin-1


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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;;

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

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 systemmay be passed to E. colifound in the gut, making them resistant toantibiotics, potential impact to human health;;

GMO may be carried by wind to other places and establish themselves asweeds and pose difficulties to be eradicated;;

Cross-pollination between GMO and wild relatives, spreading herbicide-resistance to weeds, resulting in superweeds, unable to eradicate byherbicides, outcompete with the crop plants and took over agricultural field;;

GMOs outcompete the local native species and thereby upset the balance inthe natural ecosystem and species/ loss of biodiversity;;

Violation of the natural organism's intrinsic values with the mixing genesamong 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 theenvironment;


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Likely to outcompete the original species when they occupy the same

ecological niche, establish themselves as new species;

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;

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2012 a Level H2 Biology P3 Ans