2011 ri prelim bio h2 p3 answers

16
RAFFLES INSTITUTION Higher 2 BIOLOGY 9648/03 Paper 3 20 September 2011 2 hours Additional materials: Answer Booklet/Paper READ THESE INSTRUCTIONS FIRST Write your index number, CT group and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. At the end of the examination, hand in your essay SEPARATELY. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 11 printed pages. CIVICS GROUP 1 1 S0 3 CANDIDATE NAME INDEX NUMBER Raffles Institution Internal Examination For Examiner’s Use 1 /13 2 /14 3 /13 4 /12 5 /20 Total /72

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Page 1: 2011 RI Prelim BIO H2 P3 Answers

RAFFLES INSTITUTION Higher 2

BIOLOGY 9648/03 Paper 3 20 September 2011

2 hours Additional materials: Answer Booklet/Paper

READ THESE INSTRUCTIONS FIRST Write your index number, CT group and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. At the end of the examination, hand in your essay SEPARATELY. The number of marks is given in brackets [ ] at the end of each question or part question.

This document consists of 11 printed pages.

CIVICS GROUP

1 1 S0 3

CANDIDATE

NAME

INDEX NUMBER

Raffles Institution Internal Examination

For Examiner’s Use

1 /13

2 /14

3 /13

4 /12

5 /20

Total /72

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Answer all questions.

1 Fig. 1.1 below illustrates how a human protein, YS56, can be produced using bacteria.

Fig 1.1

Human cell

producing YS56

protein

DNA from

human cell

YS56 gene cut out

of DNA

YS56 gene spliced

into plasmid

Bacterium makes

YS56 protein

Bacterium

Plasmid from

bacteria

Plasmid is cut

open

Hybrid plasmid is

put back into

bacteria

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(a) Restriction enzymes are used to cut the YS56 gene out from human DNA. Explain the mechanism of action of restriction enzymes. [3]

Restriction enzyme recognizes and binds to a specific restriction site on DNA; which is a sequence of 4-6 nucleotide base pairs that are palindromic; Specificity due to complementary shape between restriction site and active site of

enzyme. Enzyme hydrolyses phosphodiester bond on both strands of DNA between specific

bases.

(b) State the enzyme that is used to splice the YS56 gene into the plasmid, and describe the reaction it catalyses. [2]

DNA ligase;

formation of phosphodiester bonds between adjacent nucleotides;

(c) Explain how the recombinant plasmid can be “put back” into the bacteria. [2]

bacterial cells made competent (i.e able to take up DNA) by addition of calcium ions / CaCl2 solution;

brief heat shock treatment to create transient pores in bacterial cell membrane to allow entry/uptake of DNA; [Reject: electroporation – this is less common for bacteria]

(d) Explain two considerations that must be fulfilled to ensure proper expression of the gene in the bacterium. [4]

To get full marks, must have 2 points together with 2 elaboration

Prokaryotic promoter

to allow for binding RNA polymerase and sigma factor for transcription initiation; [idea of prokaryote must be there];

Gene inserted in correct orientation downstream of promoter / promoter is upstream of gene

ensure correct strand of DNA transcribed;

Introns should be absent / can use cDNA as insert

prokaryotes cannot remove/splice intron

(e) Suggest the advantage of using yeast cells instead of bacteria as a host for the production of human proteins. [2]

[Must have first point]

Yeast is also a euk but bac is prok;

Allows post-translational modifications [eg. glycosylation] for proper folding/function of protein;

Allows splicing of introns in euk to produce functional protein

[Total: 13]

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2 Restriction fragment length polymorphism, RFLP, refers to differences in fragment length generated by restriction enzymes. These differences are used to detect variations in genomes. The technique is widely used in molecular biology.

(a) Fig 2.1 below shows an autosomal dominant disease segregating in a family. Genomic DNA from each family member was assayed for RFLP for EcoRI sites, on Southern blots using probes derived from the gene. The results for all individuals, except those for the foetus, are shown below the pedigree chart. The sizes of the DNA fragments obtained are as indicated. I-1 II-1 II-2 II-3 II-4 I-2 10 kb 6 kb 4 kb

Fig 2.1

(i) Sketch a suitable diagram to illustrate the region of genomic DNA recognised by the probe in individual I-1. Show the positions of EcoRI sites and the distance between them in kb. [1]

All or nothing

[Note: There should be 2 alleles since each individual is a diploid. Since the probe detects both alleles, diagram should include both allele 1 and 2.]

(ii) Which RFLP fragment did the mother (I-1) contribute to the diseased children? [1] 10 kb fragment

(iii) What is the probability of the foetus having the disease? Explain your answer. [2] 50% or ½ probability

Mother’s genotype is heterozygote; father’s genotype is homozygous recessive so chance of receiving dominant allele ½ [Accept Punnett Square with defined alleles]

Genotype of foetus is either 10 kb (disease allele), 6 kb-4 kb OR 6 kb - 4kb, 6 kb – 4 kb

Foetus Normal Disease

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(b) A different study was carried out on another family. Fig. 2.2 below shows a Southern blot with an RFLP detected by using a probe for a sequence located on a sex chromosome. The variant exhibits two forms, one at 8 kb and one at 6 kb. The mother, father and daughter are normal.

Mother Father Daughter Son 1 Son 2

Fig 2.2

(i) Which son is abnormal? [1] Son 1

(ii) Explain how you arrived at your answer. [3] Son 1 is a case of XXY/ Klinefelter’s syndrome

8 kb band found on an X chromosome, as it is same as mother’s 8 kb fragment

6 kb band found on X chromosome from father, as it is same as father’s 6 kb fragment. [Y chromosome is not shown because the probe detects only the X chromosome]

(c) Before the genome of an organism is sequenced, preliminary mapping is usually first carried out. Shown below is a chromosome map for the X chromosome of the fruit fly Drosophila melanogaster.

(i) What is a chromosome map? [1] A diagram showing position / order /sequence where known genes occur along a

chromosome and their relative distance apart Reject: show linkage [doesn’t explain]

8 kb

6 kb

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(ii) Explain how restriction fragment length polymorphism, RFLP, can be used with linkage mapping to construct such a chromosome map. [4]

Genes are linked to RFLPs / can be detected as RFLPs / represented by RFLPs

Parents with known genotype at two RFLP loci are crossed to find recombinant frequency

Recombination frequency (%) of linked RFLPs indicates relative distance between 2 RFLP loci

1 % = 1 centiMorgan or If % recombination between 2 RFLPs on a chromosome is high, they are far apart / if % is low, they are close together.

Give formula for calculating recombination frequency

(iii) Suggest one limitation scientists would have faced if they used linkage mapping techniques to map the human genome. [1]

Linkage mapping requires crossing over to obtain recombinants - unethical for human beings

Linkage mapping requires short generation time in the studied organism - human beings’s generation time is too long

Linkage mapping involve observable phenotypes - human genome mapping aims to find all phenotypes, including non-observable phenotypes

Linkage mapping involve heritable diseases - human genome mapping aims to find all types of diseases including non-heritable

Any reasonable limitations

[Total: 14]

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3 Round-Up ReadyTM soybeans are genetically modified to be resistant towards Round-UpTM herbicide, a glyphosate-based herbicide that normally inhibits an important enzyme in the aromatic amino acid synthesis pathway in plants. This allows farmers to freely apply Round-UpTM to destroy weeds without affecting the Round-Up ReadyTM soybeans. Round-Up ReadyTM soybeans may be produced by first transfecting soybean callus cells with the resistance gene from Agrobacterium.

(a) To obtain a callus culture, suitable explants must first be chosen. Suggest a suitable explant and explain your choice. [2]

a Root tips/ shoots tips/ cambium/ bud/ hypocotyls b Contains meristematic tissue which are fast-growing / tend to be disease-free

(b) Outline the steps in obtaining a callus culture from an explant. [3]

a Surface-sterilize explants by immersing in a solution of sodium hypochlorite/bleach/chlorox

b Cut explants into small fragments c Transfer to culture vessels containing nutrient agar media [reject: agarose] d with intermediate auxin: cytokinin for inducing callus formation e Seal and incubate in sterile containers

(c) State two advantages of using callus cells in producing genetically modified soybeans. [2] a Modified callus cells may be sub-cultured to obtain high multiplication rates

a2 / high reproductive capacity or high fecundity – get many plants within short time a3/ genetic uniform – desirable traits can be pass down the progeny a4/ idea of controlled microenvironment vs natural environment for plant growth - with examples a5/ reasons for getting disease-free plants – benefits like high yield and robust, healthy plants b Undifferentiated modified callus cells may be induced to form plantlets or embryoids c Callus cells can be easily screened for special characteristics d Easy to obtain protoplasts to insert foreign DNA

(d) Explain how cells are screened to check which ones have successfully taken up the resistance gene. [2]

[0 marks if there is any reference to screening of bacterial cells such as replica plating]

Spray plant cells with glyphosate / grow in media with glyphosate / Round-Up on callus [Reject: tetracycline, ampicillin]

Only cells which have successfully taken up resistance gene will survive (ORA)

(e) The desired callus cells can then be induced to form plantlets. State the ratios of plants growth regulators required for this. [2]

High cytokinin : auxin concentration ratio for shoot formation [ORA]

High auxin : cytokinin concentration ratio for roots formation [ORA]

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(f) Despite offering benefits to farmers and consumers in terms of better yield, various concerns have also been raised regarding the use of Round-Up ReadyTM soybeans. State one concern and suggest a way of overcoming it. [2]

Resistance genes may be passed onto weed relatives through cross-pollination/ horizontal gene transfer

[Reject: gene transfer]

Create a large buffer zone / isolation distance by planting unrelated crops OR

Intense use of same herbicide will eventually select for resistant weeds, for which Round-UpTM will no longer be effective

Alternate use of Round-UpTM with other types of herbicides OR

Forced dependence of farmers on Round-Up ReadyTM soybeans as they are not allowed to replant seeds from their crops (WTTE)

Allow farmers to replant seeds OR

Allows excessive use of Round-UpTM herbicide, causing runoffs which are toxic to fish and amphibians

Promote responsible use of herbicide among farmers OR Any logical concern without solution, max 1

[Total: 13]

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Planning question

4 Plants carry out photosynthesis under sunlight or white light that is make up of different wavelengths. Different wavelengths may affect the rate of photosynthesis to different extent. An action spectrum shows the effect of the wavelength of light on the rate of photosynthesis. Design an experiment, using a suitable aquatic plant, e.g. Elodea sp. to investigate: The effect of different wavelengths of light on the rate of photosynthesis of an aquatic plant. Your planning must be based on the assumption that you have been provided with the following equipment and materials which you may choose to use:

Different coloured filters of known wavelengths and maximum transmission

Large beaker

Filter funnel

Delivery tube

Gas syringe

Stopwatch

Thermometer

Lamp

Sodium bicarbonate

Light meter

[Total: 12]

Your plan should have a clear and helpful structure to include:

a description of the method used including the scientific reasoning behind the method,

an explanation of the dependent and independent variables involved,

relevant, clearly labeled diagrams,

how you will record your results and ensure they are as accurate and reliable as possible,

proposed layout of results tables and graphs with clear headings and labels,

the correct use of technical and scientific terms,

relevant risks and precautions taken.

Answer 1. Aim: To investigate effect of different wavelengths of light on rate of photosynthesis 2. Theory: 1. In light reactions, photoactivation occurs where light energy is absorbed by chlorophyll

pigments of photosystem II and photosystem I 2. Photosystems contain light harvesting complexes that contain pigments e.g.

chlorophylls a, b and carotenoids that absorb light of different wavelengths 3. Photosystem I absorbs wavelength of 700 nm while photosystem II absorbs wavelength

at 680 nm that excite electrons to higher energy levels 4. Photolysis occurs where water is split to release oxygen and transfer electron to

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reduced photosystem II 5. Action spectrum will show that red light and blue light will result in higher rate of

photosynthesis which can be determined by measuring rate of oxygen produced 3. Procedure (PANCR): (a) Pilot Test

Conduct a pilot experiment to determine suitability of apparatus, suitable coloured filters each with its range of wavelength, optimum conditions and amount of materials used.

(b) Well-Annotated diagram (c) Numbered steps in procedure (DICE) 1. Set up as shown 2. Cut out a 5 cm long section from stem of a Elodea with 6 leaves with scapel. Cut two

more sections as replicates, to identify anomalous values. 3. Place each in a boiling tube containing 40 cm3 of 2 g/dm3 NaHCO3 solution, as

source of CO2. 4. Place tube into a water bath at 20oC; maintain this temperature by adding warm/cool

water, monitor with thermometer 5. Use a delivery tube to connect boiling tube to a frictionless 5 cm3 gas syringe. 6. Place a blue coloured filter between lamp and beaker 7. Ensure there is no other light source by switching off all other lights/ carry experiment

out in a dark room 8. Fix distance of lamp to beaker [eg 20 cm]

OR use a light meter to measure a light intensity of 100 lux, adjust distance of lamp

9. Allow plant to acclimatize to wavelength and temperature of water bath for 3 minutes till rate of bubbling has stabilized and record initial value on syringe

10. Collect gas produced over 5 minutes and record final value on syringe. 11. Calculate dependent variable, which is rate of oxygen produced by dividing volume

of oxygen by time 12. Repeat steps 1-11 under:

delivery tube

thermometer

frictionless gas syringe

retort stand to hold

gas syringe upright

and boiling tube

Elodea

water bath with

maintained by

adding warm and

cool water

boiling tube containing sodium

hydrogen carbonate

Colour filter

Lamp

Light meter placed in

line with plant

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other coloured filters [indigo, purple, green, yellow, orange and red] OR filters which give suitable wavelength [in range of 400 to 700 nm (independent variables)]

Repeat the whole experiment twice to check for reproducibility.

(d) Controls Set up control experiment with a new plant specimen. All conditions remain the same EXCEPT there is no light source. This shows plant photosynthesize in white light with wavelengths between 400 to 700 nm (purpose).

4. Data recording and processing: Table showing rate of photosynthesis at different wavelengths

Colour of light *

Wavelength /nm *

Volume of O2 given off / cm3

Rate of O2 produced

/ cm3 min-1

Replicate 1

Replicate 2

Replicate 3

Average

Violet 400

Indigo 445

Blue 475

Green 510

Yellow 570

Orange 590

Red 650

* IMPORTANT : EITHER colour of light column OR wavelength column.

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* IMPORTANT : EITHER X axis showing colour OR showing wavelength 5. Risks & Precautions 1. Avoid touching hot lamp to prevent burns. 2. When using scalpel, cut Elodea in direction away from hand to avoid getting cuts. Mark scheme

Wavelength/nm

Ra

te o

f p

hoto

synth

esis

cm

3 m

in-1

Action spectrum

RJC©

400 500 600 700

B G R Colour

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1.

Theory:

In light reactions, photoactivation occurs where light energy is absorbed by chlorophyll pigments of photosystem II and photosystem I, excite electrons to higher energy levels

2. Theory: Measurement rate

Photolysis occurs where water is split to release oxygen and transfer electron to reduced photosystem II, rate of photosynthesis is measured by rate of oxygen produced

3. Theory: Predicted results

Action spectrum will show that red light and blue light will result in higher rate of photosynthesis

4. Pilot test

5. Well annotated diagram showing clearly how oxygen produced can be measured

6. Coherence of procedure

7. How different values of independent variable can be created

Use of coloured filters to get different wavelengths e.g. red, blue filter

8. State other conditions to be kept constant and elaborate how constant conditions can be achieved (Any one)

fixed volume and concentration of sodium bicarbonate

fixed time for experiment

fixed light intensity

fixed temperature

9. Details of how acclimatization of plant to wavelength is achieved, and purpose of acclimatization e.g. acclimatization duration for photosynthetic rate to stablise

10. Details on how to obtain readings of measurable quantity – oxygen produced e.g. gas syringe/measuring cylinder etc OR Well annotated diagram showing clearly how the oxygen produced can be measured

11. Calculation of dependent variable – rate of oxygen produce

12. 5 values of independent variable

Wavelength

400, 445, 475, 510, 570, 590 & 650 nm (any 5 inclusive of 445 and 650 nm) OR

Colour

Violet, indigo, blue, green, yellow, orange, red (any 5 including red and blue)

13. Precaution for accuracy: No other light source

14. Details + purpose of control experiment:

Control experiment where plant is not illuminated and all other conditions remain the same.

15. How to obtain 3 replicates for each wavelength using different sprigs of same length (R: continue with 2 other duration with same plant)

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AND repeating experiment 2 more times

16. Correct table with headings and units

17. Graph

x-axis: rate of photosynthesis (cm3/min)

y-axis: wavelength (nm)

show expected results

18. 1 relevant/specific risk and precaution

Free-response question

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Write your answer to this question on the separate answer paper provided.

Your answer:

should be illustrated by large, clearly labeled diagrams, where appropriate;

must be in continuous prose, where appropriate;

must be set out in sections (a), (b) etc, as indicated in the question.

5 (a) Describe the viral-mediated approach to gene therapy to treat SCIDs and explain why the ex vivo approach was adopted. [8]

Max 6 marks a. SCIDs is a heritable genetic disease affecting T and B lymphocytes where patients

suffer from immunodeficiency; b. Gene affected is a loss-of-function mutation in gamma chain gene in X chromosome or

adenosine deaminase; c. Retroviral vector used e.g. lentivirus; d. Remove haematopoietic stem cells from bone marrow of patient e. because they have ability to self-renew and proliferate so that treatment is permanent; f. Infect stem cells in vitro with virus carrying normal functional allele [Reject: gene]; g. Viral RNA is converted into cDNA by reverse transcriptase which is then integrated into

host cell genome using integrase; h. Reintroduction of transfected stem cells back to patient’s bone marrow;

Max 3 marks i. Ex vivo approach used because it allows specific screening of target haematopoietic

stem cells; j. Allows monitoring of gene expression before reintroduction as success rate is very low; k. Safer because can monitor for cancer cells because retroviruses cause insertional

mutagenesis which can knock out tumour suppressor genes or activate proto-oncogenes;

l. You don’t introduce retrovirus directly into patient – already immunocompromised;

(b) Discuss the factors which prevent gene therapy from becoming an effective treatment. [6] a. Therapy is typically short-lived because of difficulty in integrating normal, functional

allele into host cell genome b. natural death of treated cells means that new cells need to be constantly transformed; c. Many diseases are a result of multiple genes and/or multiple factors (e.g. Alzheimer’s

disease, diabetes etc) OR as such will be difficult to treat with gene therapy which currently treats only single gene disorders;

d. Some diseases such as thalassemia require a precise level of expression of the normal gene, which gene therapy cannot achieve [with reason]

e. For viral mediated gene therapy, problems associated with the viral vector include immune response against the vector such as allergies, inflammatory and toxicity responses [give one example];

f. Cannot find a suitable viral vector due to viruses’ host specificity g. Viruses may recover its ability to cause disease once in host; h. If virus vector triggers the immune response the first time, the immune system will

mount a stronger and faster response the second time rendering the therapy ineffective subsequently;

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i. Insertional mutagenesis caused by retroviruses may cause inactivation of tumour suppressor genes and/or activation of proto-oncogenes, leading potentially to cancer;

Examiners report (Confidential): Responses to this part of the question varied in quality. Those candidates who scored well covered a range of ideas and expressed each idea clearly and succinctly. Ethical and economic ideas were not considered to be appropriate to this question and so were not credited. Some candidates discussed those to the exclusion of any other aspects.

(c) Describe the unique features of stem cells. Explain the differences between stem cells and

cancer cells. [6]

Features: (3 marks to come from here) a. Unspecialised/undifferentiated cells with no tissue specific structures/do not carry out

any specific function; b. Can undergo mitosis/divide indefinitely and undergo self-renewal to renew pool of stem

cells; c. They have the ability to differentiate to produce specialized cells upon receiving

appropriate molecular signals; Differences: (next 3 marks can come from here) d. Unlike stem cells, cancer cells do not differentiate and is unresponsive to molecular

signals; e. Cancer cells divide indefinitely while stem cell division is determined by molecular

signals [that either stimulates cell division or stop it altogether]; f. Cancer cells experience no contact inhibition and is invasive while stem cells

experience contact inhibition; g. Cancer cells metastasize (dislodge from original tumour and form secondary tumours)

while stem cells remain in tissue of origin; h. AVP of comparison;

[Total: 20]