2013 a-Level Answers P3 Ans Edited

7/24/2019 2013 a-Level Answers P3 Ans Edited

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2013 A-Levels H2 Paper 3 Answers

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(a) State two other ways in which RFLP can be used as a biological tool. [2]

1. To detect differences in alleles responsible for genetic diseases;

2. To construct DNA fingerprinting, which can be used for tracing evolution

lineage, determine paternity of a child or in forensics; @ 1m

(b) (i) EcoRI BamHI and EcoRI

Number of restriction sites 2 3

Size of fragments / kb 2, 3, 10 1, 2, 3, 9

(ii)

2 9 1 3

(c) Outline why gel electrophoresis separates DNA fragments. [4]

1. DNA is separated on the basis of their rate of movement through a gel in an

electrical field;

2. Ref. to agarose gel acting as molecular sieve;

3. DNA being negatively charged will move from the negative to the positive

electrode;

4. The longer the DNA fragment, the slower it travels to the positive end / vice

versa;

@ 1m

(d) Outline the process of DNA hybridization that allows the RFLP pattern for a particular

gene to be visualized. [5]

1. Carry out Southern blotting and transfer the sample DNA onto a nitrocellulose

membrane;

2. Add alkaline solution to denature double stranded DNA into single strands;

3. Introduce single stranded, radioactive gene probe, which will anneal to the

gene of interest via complementary base pairing (through the formation of

hydrogen bonds);

4. Place an X-ray film onto the nitrocellulose membrane and expose it to the

radioactivity by probe during development using autoradiography equipment.5. Dark bands formed indicate the position of the DNA fragments containing the

particular gene sequence;

@ 1m

[Total: 15]


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2

(a) (i) Cytosine is a pyrimidine. Name one other pyrimidine. [1]

Thymine; @1m

(ii) Suggest how methylation of cytosine nucleotides prevents the DNA of a prokaryote

from being cut by its own restriction enzymes. [2]

1. Restriction enzymes binds to specific short sequences of DNA and cleaves

DNA;

2. DNA methyltransferase adds methyl groups only to cytosine molecules,

resulting in a change in 3D conformation of the DNA molecule;

3. Hence, DNA no longer complementary to the active site of the restriction

enzyme and hence not cleaved; @1m, max 2

(b) (i) The restriction enzyme, Bbe1, cuts at the same restriction site but produces a GCGC-

3 overhang. On the diagram below, use arrows to indicate the cut sites. [1]

5 GGCGCC3

CCGCGG

(ii) Compare the ways in which recombinant DNA could be produced from DNA digested

with Bbe1 and DNA digested with Sfo1. [4]

Sfo1 Bbe1

1. Produced blunt ends and hence linker

DNA has to be added to the cut ends of

DNA

Produced sticky ends and hence addition of

linker DNA to cut DNA is not required;

2. Attached linker DNA has to be cut byanother restriction enzyme to produce

sticky ends before recombinant DNA can

be produced

Only one enzyme / XmaI is required toproduce recombinant DNA;

3. This process to produce recombinant

DNA takes longer and requires more

energy

Less time and energy required when XmaI is

used in the production of recombinant DNA

4. Both enzymes require the addition of DNA ligase to form recombinant DNA

(c) Suggest why the scientist made this prediction. [2]

1. BspL I can recognize the same sequence that make up the restriction site of

Sfo1which can make 20 cuts in the standard DNA;2. Since N can be any of the four DNA nucleotides, BspL I can make more than 20

cuts; @1m

[Total: 10m]


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(a) (i) Explain why the gene has at least 200000 base pairs, but the protein only has 1480

amino acids. [3]

1. Gene contains introns and other non-coding DNA regions (e.g. promoter

which are not transcribed onto mRNA);

2. mRNA has fewer base pairs than the gene as introns are removed during

RNA splicing;

3. In cases where alternative splicing takes place, not all the exons present

may be joined together to form the protein; @ 1m

4. Gene contains regions that are transcribed to 5UTR and 3UTRwhich are

not translated. (5 UTR are the ribosomal binding site while 3UTR contains

polyadenylation sequence. @ 1m

(ii) Describe the effects of cystic fibrosis on a person and how these are linked to the

reduction in removal of chloride ions. [3]

1. CF patient has thick and sticky mucus which clogs up the airways of the

lungs, pancreatic duct, bile duct and the reproductive ducts;

2. Bacterial trapped in the thick mucus lead to higher chance of infection;

3. Reduction in removal of Cl ions causes them to accumulate in the cells,

trapping water within the cell, and attracting water (from the external mucus

layer) into the epithelial cells;

3. Less water in the external mucus layer results in the formation of thick

mucus; @ 1m

(iii) Suggest, with reasons, why other mutations of the CFTRgene vary in the extent to

which they cause symptoms of cystic f ibrosis. [3]

1. Other mutations may include frameshift mutations involving deletion or

insertion of a number of nucleotides (not divisible by 3) on the CFTRgene;

2. Causes the reading of the codons on mRNA after the mutation to code for

different amino acids, resulting in a non-functional CFTR

3. Substitution of nucleotides on the CFTRgene;

4. May change a codon on mRNA, resulting in a different amino acid (with

different chemical property) and hence a non-functional CFTR is

synthesized;

5. Frameshift mutation will lead to large change in protein channel 3D

conformation hence severe symptom while substitution a smaller change in

protein 3D conformation;

(b) (i) Outline how liposomes can be used as part of a gene therapy treatment for cystic

fibrosis. [3]

1. The normal CFTR gene is cloned into a plasmid, which then is placed intoliposomes;

2. They are sprayed into the nose and mouth of CF patients as an aerosol spray;

3. The liposomes then fuse with the cell membrane of tracheal (epithelial) cells,releasing (the plasmid containing) the normal CF gene into the cytoplasm ofcells;

4. The introduced (normal) CF gene then enters the nucleus, is transcribed into


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mRNA and translated into the normal CFTR protein;

@ 1m, max 3

(ii) Describe and explain the factors that prevent this method of gene therapy for cystic

fibrosis from becoming an effective treatment. [3]

1. Low efficiency of transfer;2. due to non-specificity/inefficient cell-specific binding/low cellular uptake ofDNA by target cells;

3. degradation of gene by lysosomal enzymes as well as cytosolic nucleases;;

4. Risk of losing delivered gene due to gene not incorporated into target cells

genome;

5. epithelial cells are also constantly being shed; @ 1m, max 3

2013 SPA

Mark Scheme

1. Theoretical consideration or rationale of the plan to justify the practical procedure,including (1m)

How would the independent variable ([copper sulfate]) affect the dependent

variable (amount of leakage of pigments from beetroots cells)?

(a) Red pigment molecules in vacuoles of beetroot cells are hydrophilic and are not ableto pass through hydrophobic core of phospholipid layer of cell membrane.

(b) Copper sulphate denatures tertiary structure of proteins embedded on cell surfacemembrane by breaking ionic and hydrogen bonds. Cell surface membrane no longselectively permeable. Red pigment diffuse out of cells into the solution to give thesolution a red colour.

How would the dependent variable be measured?

(c) Membrane permeability can be measured by measuring the leakage of pigment frombeetroot tissue under different Cu2So4 concentration and measuring the absorbanceof the solution at red colour wavelength using a colorimeter.

(d) The lowest concentration of Cu2So4 that resulted in the steepest increase inabsorbance at red colour wavelength is the lowest concentration that has an effect on

the leakage of pigment.

Correct use of technical and scientific terms :(1m)

a) hydrophobic core of phospholipid bilayer

b) selectively permeable


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3 IndependentVariable

At least 5 concentrationof Cu2So4solution withregular interval rangingfrom 0.06, 0.12, 0.18, 0.24, 0.3 mol dm-3 (not inclusive of 0.0 mol dm-3 asthis is the control).

Dilution table to make up different concentrations of sucrose solution.Volume of distilled water, volume of Cu2So4and concentration of Cu2So4must be indicated (check total volume is the same for all different

concentrations of Cu2So4)4 DependentVariable

% absorbance at red colour which indicates permeability of membrane

measured using a colourimeter.

5+6

ConstantVariables(max 2)

a. Constant thickness and diameter of the beetroot discs.b. Constant no. of discs used for each concentration.c. Constant volume of Cu2So4to be used (max 5cm

3)d. Specify period of time for the expt (max 30min)e. Specify temperature of experiment (e.g. 300C)

7 Equilibration Equilibrate beetroot and Cu2So4 separately by placing beetroot (in isotonicsolution) and Cu2So4 solution in separate tubes. Place both tubes inthermostatically controlled water bathat specific temp(temperature of

experiment)] for fixed amt. of time (1-5min).8 Control Replace Cu2So4 solution with equivalent volume of distilled water

(e.g. 5cm3) and subject to same experimental conditionsPurpose of control is to show that the leakage of pigments from thebeetroots are due to presence of Cu2So4 that disrupt membrane proteins.

9 Results Measure absorbance (%) at specific wavelength of 550nm red colour.

Table of results with clear headings and units (concentration of Cu2So4 / %and Absorbance/%)

10 Accuracy Repeat experiment twice to get replicates readingsRepeat entire experiment 2 more times for repeats readings

11 Data Analysis

& Graph

Calculation of average absorbance

Plot graphs of absorbance (a.u) (y-axis) against concentration of sucrose(%) (X-axis)Sketch expected trend.

12 LabelledDiagram

Labeled diagram showing the following apparatus:

a) beaker with test-tubes with Cu2So4 and beetroot discs

b) white card used as background

13 Use ofapparatus

a. Use of colorimeter to measure absorbance (a.u) at a specificwavelength 550nm.b. Use of Bunsen burner to prepare water bath

c. Use of white tile / card to judge / observe the red colouration in the test

tubes / intensity of red coloration.

Safety considerations:

a) When using scalpel, cut the beetroot away from the hand to avoid getting cut.

b) Use insulating gloves when handling hot water to prevent scalding.

c) Handle Bunsen burner with care to prevent burnts.


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Propose procedure

Explanation of theory

(a) Red pigment molecules in vacuoles of beetroot cells are hydrophilic and are not ableto pass through hydrophobic core of phospholipid layer of cell membrane.

(b) Copper sulphate denatures/disrupt 3-D structure or tertiary structure of proteinsembedded on cell surface membrane and cell surface membrane no long selectivelypermeable. Red pigment diffuse out of cells into the solution to give the solution a redcolour.

(c) Membrane permeability can be measured by measuring the leakage of pigment frombeetroot tissue under different Cu2So4 concentration and observing the colour of thesolution against the white card.

(d) The lowest concentration of Cu2So4 that resulted in the appearance of red colourationin the solution is the lowest concentration that has an effect on the leakage ofpigment.

Sample procedure

1. Prepare 5 different concentrations of Cu2So4using the dilution table below.

2. Boil 100cm3of water using the Bunsen burner*. This is for the set-up of 300C waterbath.

3. Set up a 300C water bath by mixing the hot water and cold tap water and monitor thetemperature using the thermometer*

4. Label 5 test-tubes 0.06, 0.12, 0.18, 0.24, 0.30, for the 5 independent variables of0.06%, 0.12%, 0.18%, 0.24%, 0.30% Cu2So4respectively.5. Using a cork borer*(diameter 1cm), cut out cylinders of beetroot*. Cut 40 pieces of 2

mm thick discs from the cylinder using scalpel*.6. Rinse the discs in isotonic solution until the water is colourless and gently dry the

beetroot with toilet rolls.7. Place 3 pieces of washed beetroot discs in each test tube.8. Add 6 cm3of 0.06% Cu2So4*into the test tube labelled 0.06.9. Place the test tubes and the test tube containing the beetroot discs into the 300C water

bath to equilibrate for 1min.10. After 1min, pour the 6 cm3of 0.06% Cu2So4 into the 0.06 tubes and start the stopwatch.11. Conduct the expt for 30 min.

12. At the end of 30min, place the white card against the test-tubes to determine if theCu2So4 solution has turned red.

Concentration of

Cu2So4 / %

Volume of stock

solution / cm3

Volume of H2O /

cm3

Total volume /

cm3

0.06 3.0 12.0 15.0

0.12 6.0 9.0 15.0

0.18 9.0 6.0 15.0

0.24 12.0 3.0 15.0

0.30 15.0 0.0 15.0


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13. Decant the liquid from tube and pour into a cuvette.14. Fill a second cuvette with Cu2So4.15. Place a filter that results in light of 550nm into the colorimeter and use the second

cuvette with Cu2So4to adjust the reading to zero absorbance.16. Measure the % absorbance of the test solution using colorimeter.17. Repeat the whole experiment from step 8 to 16 twice for reading of 2 more replicates

18. Repeat the whole experiment from step 8 to 17 for 4 other concentrations of Cu2So4.19. To set up the control, repeat steps 8 to 16 using 6 cm3 of distilled water instead ofCu2So4. Calibrate the colorimeter using a cuvette with distilled water instead. Subjectthe control tubes to the same experimental conditions. This shows that without Cu2So4there will be no leakage of red pigment from the beetroot cells.

20. Repeat the entire experiment two more times.21. Record the results in a table.

Concentration

of Cu2So4/%

Absorbance/%

Try 1 Try 2 Try 3

Average

0.06

0.12

0.18

0.24

0.30

0.00 (Control)

22. Plot a graph showing intensity of colour of extract obtained from incubation at differentCu2So4.

Absorbance/ %

Cu2So4concentration/%


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Safety precautions

a) When using scalpel, cut the beetroot away from the hand to avoid getting cut.

b) Use insulating gloves when handling hot water to prevent scalding.

c) Handle Bunsen burner with care to preven burnts.

5 (a) Describe features of zygotic stem cells and embryonic stem cells that distinguish them

from each other. [5]

1. Zygotic stem cells are totipotent;2. Ref. to the potential to differentiate into any cell type including the extra-

embryonic membranes;3. For example, the zygote and the first 16 cells produced by mitotic division of the

zygote;

4. Embryonic stem cells are multipotent;5. Ref. to the potential to differentiate into any cell type except those of the extraembryonic membranes;

6. For example, inner cell mass of the blastocyst;7. Zygotic stem cells have unique gene expression to produce different sets of

proteins and enzymes to maintain its totipotency, which is different from that ofembryonic stem cells;

@ 1m, max 5

(b) Describe the features of blood stem cells and explain their normal functions. [8]

1. Blood stem cells are undifferentiated with no tissue specific structures nor

specialised functions;2. They are able to divide and renew themselves for long periods of time via mitotic

division;3. They have unique differential gene expression to produce proteins required to

maintain their undifferentiated state;4. Their gene coding for telomerase is expressed;5. To maintain the length of telomere and prevent replicative senescence;6. They are multipotent;7. They can differentiate into all the specialised blood cell types under the

presence of appropriate chemical signals;8. E.g. red blood cells, white blood cells (neutrophils, basophils, macrophages, T

and B lymphocytes), platelets; (at least 3 examples)

9. To replace the blood cells lost through normal tear and wear or when fightinginfection; @1m, max 8

(c) Therapeutic genes can be introduced into stem cells. Discuss why the genes used are

more likely to be obtained from a cDNA library, than a genomic DNA library. [7]

1 cDNA is much shorter in length compared to genomic DNA;2 Thus, a wider range of vectors can be used to carry the gene e.g. retroviral

vector;3 More genes are available to be carried by the vector to treat more genetic

disorders (Since certain genes are larger in size);4 Ref. to increase efficiency of gene delivery since it is more compacted in

size;5 (Intact) cDNA is derived from the reverse transcription of mRNA and does


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not have introns;6 To minimize expression problem associated with the regulatory sequences

found in genomic DNA / unintended effects on the expression of adjacentnon-target genes;

7 Ref. to genomic DNA may not be intact as it may be cut by restrictionenzyme into fragments and may be difficult to ligate the fragments together

in order;8 Ref. to mutations in genomic DNA, which may disrupt the normal structureand function of target proteins may not be detected;

9 Which is not usually encountered when mRNA from a normal, healthy cell isused and hence less problem when screening for cells with functional targetproteins;@ 1m

@ 1m, max 7

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2013 a-Level Answers P3 Ans Edited