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 CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2 STUDENT NAME CLASS 2T INDEX NUMBER BIOLOGY 9648/03 Applications Paper and Planning Question MONDAY 02 SEPTEMBER 2013 Paper 3 2 hours Additional Materials: Answer Paper READ THESE INSTRUCTIONS FIRST Write your index number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] 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, fasten all work securely together. The number of marks is given in brackets [ ] at the end of each question or part of the question.  This document consists of 24 printed pages. [Turn over For Examiner’s Use 1 [15] 2 [10] 3 [15] Planning 4 [12] Free-response 5a [7] 5b [6] 5c [7] TOTAL 72

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    CATHOLIC JUNIOR COLLEGEJC2 PRELIMINARY EXAMINATIONSHigher 2

    STUDENTNAME

    CLASS 2TINDEXNUMBER

    BIOLOGY 9648/03Applications Paper and Planning Question MONDAY 02 SEPTEMBER 2013Paper 3 2 hours

    Additional Materials: Answer Paper

    READ THESE INSTRUCTIONS FIRST

    Write your index number and name on all the work you hand in.Write in dark blue or black pen on both sides of the paper.[PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED]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 allquestions.

    At the end of the examination, fasten all work securely together.The number of marks is given in brackets [ ] at the end of each question or part of the question.

    This document consists of24 printed pages.

    [Turn over

    For Examiners Use

    1 [15]

    2 [10]

    3 [15]

    Planning

    4 [12]

    Free-response

    5a [7]

    5b [6]

    5c [7]

    TOTAL72

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    2

    Answer allquestions.

    1 (a) Explain what is meant by a restriction enzyme. [2]

    [L1]Any two points:1. A restriction enzyme is an endonuclease which detects palindromic restriction site which is

    4-6 base pairs in length.

    2. Restriction enzymes bind to the restriction site, i.e. a specific sequence of bases of DNA andhydrolyse the phosphodiester bonds within polynucleotide strandsof the DNA molecule.3. They hydrolyse covalent phosphodiester bond of both DNA strands, often in a staggered way

    creating single-stranded ends called sticky ends. Some restriction enzymes make a simplecut across both strands at a single point, forming bluntor flush ends.

    (b) Fig. 1.1 illustrates part of a DNA sequence that is recognised by a restriction enzyme.Complete the sequence by filling up the blanks in Fig. 1.1. [1]

    Fig. 1.1

    [L2]

    (c) What do the 5and 3ends in Fig. 1.1 represent? [2]

    [L2]Any two points1. The 5 end represents a phosphate group linked to the sugar-phosphate backbone viacarbon

    5 of deoxyribose sugar.2. The 3 end represents thehydroxyl group linked to the sugar-phosphate backbone viacarbon

    3 of the deoxyribose sugar.

    3. Both strands of a DNA molecule are antiparallel. As seen in Fig. 1.1, the top strand of aDNA molecule runs from 5 to 3directionand the complementary bottom strand runsfrom 3 to 5 direction.

    5 A G C _ _ _ 3

    3 _ _ _ _ _ _ 5

    5 A G C G C T 3

    3 T C G C G A 5

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    Fig. 1.2 illustrates a typical plasmid used in genetic engineering.

    Fig. 1.2

    (d) The plasmid shown in Fig. 1.2 lacks two features that are essential for genetic engineering.Add these two features to the plasmid in Fig. 1.2 and label them. [2]

    [L1]1. Label ori(origin of replication)

    [L2]2. Label Multiple Cloning Site / any restriction site with a relevant example within either of theantibiotic resistance genes.

    (e) Explain the significance of one of the features added in part (d). [1]

    [L2]1. Presence of origin of replication within the plasmids allow the recombinant plasmids to

    multiply independently from the host DNA and make many copies of the insertedgene/gene of interest.

    2. Multiple cloning sites/ Restriction site with a relevant example. This serves as a convenientsite into which foreign DNA may be inserted.

    Ampicillin resistancegene

    Tetracycline resistancegene

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    A dot blot is a simplified version of Southern Blotting. It can be employed to test for the presence of genemutations. In this molecular technique, undigested DNA is applied directly onto a membrane through avacuum. These dots containing DNA will be hybridised to a sequence-specific oligonucleotide probe andbe subjected to autoradiography. Dots which successfully hybridise to a probe will be seen as a dark doton the autoradiogram.

    (f) Describe two key differences between the process of obtaining a dot blot and a Southern blot. [2]

    [L3]Any two points1. For the dot plot,the entire undigested DNA is being used, whereas for Southern Blot, DNA

    digested with restriction enzymes are being used.2. For the dot plot, DNA can be transferred directly onto a membrane, whereas for Southern Blot,

    DNA fragments must go through gel electrophoresis first before transference.3. For the dot plot, transfer of DNA is through a vacuum, whereas for Southern Blot, transfer

    of DNA is through capillary action.Also accept: if student only stated the difference in dot plot without a point-by-pointcomparison.

    Cystic fibrosis (CF) has been linked to several mutations. Two probes have been synthesized to

    hybridise with a mutated sequence associated with CF as well as the corresponding normal sequence.The dot blots of nine patients clinically diagnosed with CF are shown in Fig. 1.3.

    Probe for normal sequence

    Probe for mutated sequence

    CF patient

    Adapted from: http://www.glowm.com/resources/glowm/cd/pages/v5/ch076/framesets/006f.html

    Fig. 1.3Dot plot of 9 CF patients.

    (g) State one mutation linked to cystic fibrosis. [1]

    [L1]1. delta F508OR1. Deletion of codon 508 of the CFTR gene.

    (i) Account for the presence of two dots in individuals 1, 7 and 9. [2]

    [L3]1. Human cells are diploid/ have homologous chromosomes.2. These individuals are heterozygous for the sequence, i.e. they have a copy of the normal

    sequenceas well as a copy of the mutated sequence.

    (j) With reference to Fig. 1.3, explain why the mutation associated with CF is most likely to berecessive. [2]

    [L2]

    1. Most CF patients/ 5 out of 9patients/ CF patients 2, 3, 4, 5 and 8have only one dotshown,2. indicating that 2 copies of the CFTR mutated allele is neededfor the manifestation of cysticfibrosis.

    [Total: 15]

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    2 A patient was found to have a defective gene that results in the failure to produce an important chemicalsubstance, chemical substance X.

    Three experiments were carried out to compare three different techniques for introducing chemicalsubstance X into such patients:

    Experiment 1:The chemical substance X was injected directly into the patient.

    Experiment 2: The gene coding for chemical substance X was packaged into a viral vector that wasmodified to render it harmless. The viral vector containing the gene was injected into thepatient.

    Experiment 3:The gene coding for chemical substance X was placed into a protein capsule, which wasthen inserted into the tissues of the patient.

    Fig. 2.1 shows the results from these experiments.

    Experiment 2

    Experiment 3

    Experiment 1Direct injection of chemical

    substance X

    Experiment 2

    Injection of viruses carryinggene coding for substance X

    Experiment 3Injection of protein capsulecontaining gene coding for

    substance X

    Fig. 2.1

    (a) (i) Describe the results of the three experiments. [3]

    [L2]1. For Experiment 1, direct injection of substances results in a rapid/ sudden/ sharpincrease / risein the concentration of the substance in the tissues, followed by a rapiddecrease/ short time in the harmful concentration range / effective concentrate range.

    2. For Experiment 2, injection of viruses carrying the gene for the substance / viral gene vectorshows a slowerincrease / riseof the concentrationof the substance in the tissues; theconcentration decreases gradually/ remains in the effective/harmful range for a longerperiod of time

    3. For Experiment 3, using protein capsule containing the gene, results in the slowestincrease / risein the concentrationof the substance; the substance remainsat the

    effective concentration rangeover the longestperiod of time / (accept: idea of)

    Additional Note: Both experiments 1 and 2 results in having the substance at concentrationsthat can cause harmful side effects but not for experiment 3.

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    (ii) Using the information in the graph, suggest two advantages and one disadvantage of theprotein capsule method, as compared to the other methods.

    [L3]Advantage: [2]1. The chemical substance never reaches / does not reach harmfullevels / concentration

    that may cause possible harmful side effectsto occur, hence less likelyto cause / littleornodamage / harmto the organism.

    2. The chemical substance remains longerin the effective concentration range, hence canprovide longer relief from the symptoms / needs less frequent treatment.

    Disadvantage: [1]3. The concentration of substance in tissues increases / appears and reaches effective range

    veryslowly, hence treatment takes a longer timeto take effect / to be become effective.

    Duchenne muscular dystrophy (DMD) is a common lethal X-linked human genetic disease caused bythe absence of the protein dystrophin in muscle fibres.

    Two possible treatments have been proposed to correct the deficiency of dystrophin:

    (1) Transplant normal cells from genetically compatible donors into the patient;(2) Gene therapy.

    Gene therapy involves inserting the gene coding for dystrophin into an adenovirus and injecting therecombinant adenovirus into the patients muscle.

    (b) Gene therapy using the recombinant adenoviruses did not provide a permanent cure for DMD. Thetrials showed that the transgene expression declined after about 2 weeks and was negligible afteronly 4 weeks. Explain this phenomenon. [2]

    [L2]1. Transfected / modified muscle cellsdifferentiated and non-dividing cells, are replaced

    when worn-out / die. The therapeutic gene that the transfected muscle cells carry is hence lostwhen these muscle cells die.

    2. Transgene/ gene coding fordystrophin fails / does notintegrateinto the genome/ remainsextrachromosomal, affecting the extent andeffectivenessof the gene expression.

    (c) Suggest a suitable vector for delivery of genes designed to kill cancer cells. Explain your choice. [2]

    [L3]1. using retroviruses / modified retroviruses (accept: modified HIV)

    Reject: adenovirus since it can infect all kind of cells)

    Any one of the following:2. Can only infect dividing cells, thus can target actively dividing cancer cells3. Cannot targetnon-dividingcells / cells that have stopped dividing, hencewill not damage or

    harmto surrounding normal cells

    Reject: retrovirus ability to integrate gene into genome, leading to long-term stable geneexpression this is out of context

    [Total: 10]

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    3 (a) Genetic engineering has been used to improve the quality and yield of many crop plants.One method of genetic engineering involves modifying Ti plasmid from a bacterium, Agrobacteriumtumefaciens, and transferring the modified Ti plasmid containing the transgene into crop plants. Thismethod has been used for the production of Btcorn.

    (i) State one other method that can be used for the genetic engineering of crop plants. [1][L1]1. Gene gun.

    (ii) The Ti plasmid is usually modified to contain an antibiotic resistance gene (e.g. ampicillinresistance gene).

    Explainthe purpose of this modification. [2]

    [L2]1. The antibiotic resistance gene acts as a selection marker.2. To identify the Agrobacterium tumefaciens colonies that has been successfully

    transformedwith recombinant Ti plasmid containing the transgene/foreign gene.

    The successfully genetically engineered crop plant cells may be stimulated to grow into plantlets

    using tissue culture.

    (iii) Describe how a plantlet may be produced from cells in tissue culture. [2]

    [L1]1. Plant cells may be induced to grow into a callus by adding specific ratioof plant growth

    regulatory factorssuch as auxin and cytokinin and addition of nutrients;2. The callus is stimulated to develop shoots and roots by altering the concentration of

    auxin and cytokininsuch that it is at the specific ratio for shoot and root growth;

    (iv) Describe two advantages of using tissue culture to clone genetically modified plants. [2]

    [L3]Any two of the following:1. Once a new gene / transgene is successfullyintroduced / transformedinto a plant cell,

    the transformed/ modified plant cellcan be grown into a whole plantand then clonedto produce many genetically identicalplants, all containing thattransgene(GM plants)

    2. Allows forrapid productionof large numbers / mass productionof GM plants from justone or a few stockplants

    3. Production of virus-free / disease-free plantsby selecting only meristematic tissueofthe GM plant / explants containing meristematic tissuefor propagation;

    4. Allows all year round productionof GM plants / can be produced at any time of the year /not affected by seasons higher productivity / annual yield

    5. Reliability since quality control can be achieved by standardising growingconditionssuch that batch after batch of standard GM plants can be produced.6. Land area needed for initial growth of these GM plants through tissue culture is much

    less, hence saving valuable arable space7. Prevents the loss of transgene due to sexual reproduction / through gamete formation;

    preservation of transgene within the entire crop8. Prevents GM plants from being completely wiped out by changing environmental factors;

    as the plants are always available as stored in the cold storage;9. AVP

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    (b) A number of different crop plants have been genetically engineered to express a gene for aninsecticidal toxin (Bttoxin) from a bacterium, Bacillus thuringiensis.

    The Bt toxin binds to cell membranes of the cells in the insect gut. The receptor for the toxin hasbeen identified as a membrane-anchored protein encoded by the cadherin gene. Bttoxin forms ionchannels that cause an efflux of potassium ions from the insect gut cells, leading to cell lysis.

    The cotton bollworm, Helicoverpa armigera, is one of the most serious insect pests of cotton in Asia,

    Australia, and Africa. A major deletion of the cadherin gene was associated with a high level of Bttoxin resistance in H. armigera. Three alleles (r1, r2, and r3) of the cadherin gene in the resistantstrain have been identified: r1 allele lacks 24-bp

    r2 allele has a 202-bp deletion

    r3 allele has a 126-bp deletion

    (i) Explain how deletions of the cadherin gene could confer resistance to Bt toxin in H. armigera. [4][L2]1. Deletion of the cadherin gene results in the removalof corresponding mRNA codons;2. Leading to missing / removalof certain amino acidsin the polypeptide chain / a

    truncated / shorterpolypeptide chain3. Affecting the overall foldingof polypeptide chain / changes in protein / receptorprotein

    conformation4. Bttoxin unable to bindto cell membrane, unable to exert its effect, hence prevent / no

    cell lysis

    A survival test was conducted on larvae of different genotypes to assess their resistance to Bttoxin.

    Frequency

    ofsurvival

    Cadherin genotype of larvae

    Legend:

    s - s allele coding for normal

    cadherin gener1 - r1 allele lacks 24-bpr2 - r2 allele has a 202-bp deletionr3 - r3 allele has a 126-bp deletion

    Fig. 3.1

    (ii) Suggestwhylarvae of genotype ss was included in this test. [1]

    [L3]1. It acts as a negative controlto ensure that any survival of the larvae in Bt cotton is due

    to the effect of r1, r2 or r3 alleles and not due to any other factors.

    (iii) Describe the effects of the genotypes r1r1, r1r3 and r3r3 on the resistance of larvae to Bt toxin. [3][L2]1. r1r1 increased the resistance of larvae by a frequency of 0.38as compared to the ss

    genotype;2. r1r3 conferred the greatest increasein the resistance of larvae, by conferring an increase

    in resistance by a frequency of 0.56as compared to the ss genotype;3. r3r3 was the least effectiveand increased the resistance of larvae only by a frequency

    of 0.8as compared to the ss genotype.[Total: 15]

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    4 Planning question

    You are required to investigate the rate of photosynthesison a certain species of algae,Scenedesmus Chlorella.

    Algae are tiny and difficult to work with directly in water. Hence, sodium alginate is usually used to

    trap large numbers of algal cells into agar beadsso that the cells are all contained in one piece.

    Source: Copyright Science & Plants for Schools: www.saps.org.uk

    Fig. 4.1

    Hydrogen carbonate indicatorcan be used to measure the rate of photosynthesis as it is verysensitive to changes in carbon dioxide level. The indicator is red when the amount of CO2inthe solution is in equilibrium with atmospheric air, becomes orange to yellow withincreased CO2in the solutionand changes from red through magenta to deep purple asCO2is removed from the solution. This is represented in the diagram below.

    PURPLE MAGENTA RED ORANGE YELLOW

    Carbon dioxideconcentration

    0.04% CO2Atmospheric air

    Low High

    Source: Copyright Science & Plants for Schools: www.saps.org.uk

    Fig. 4.2

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    The colour changes in hydrogen carbonate indicator can be measured objectively using a

    colorimeter. A colorimeter measures the absorbance of the solution in the cuvette. Thecuvette holds the working solution.

    0.20A Absorbancereading

    Cuvetteholder

    Cuvette

    Source: Copyright Science & Plants for Schools: www.saps.org.uk

    Fig. 4.2

    Different colours absorb different wavelengths of light to a different extent. The darker thecolour, the greater the absorbance value (A)will be. The known absorbance values (A) ofthe following colours of the hydrogen carbonate indicator are shown in the table below.

    Table 4: Table of known absorbance values

    Color of Hydrogen Carbonate Indicator Absorbance of Colorimeter/ AYellow 0.10Orange 0.20Red 0.30Magenta 0.50

    Purple 0.70A = Absorbance units

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    Using all the information above, you are required to plan, but not carry out, an investigation into:

    The effect of light intensity on the rate of photosynthesis in the algae,Scenedesmus Chlorella

    Your planning must be based on the assumption that you have been provided with the following

    equipment and materials, which you must use:

    150Scenedesmus Chlorellaalgal beads

    100 cm3of hydrogen carbonate indicator solution containing 387 parts per million (ppm) CO2, which isat equilibrium with the 0.04% atmospheric CO2.

    7 cuvettes

    Colorimeter with a digital display to 2 decimal places and a cuvette holder.

    7 glass tubes with screw-on-lids. The walls of these tubes are transparent.

    A 150 W bench lampwhich emits very little heat.

    5 cm3syringe

    Stop watch.

    Ruler.

    Your plan should:

    have a clear and helpful structure such that the method proposed can be repeated by anyonereading it,

    be illustrated by relevant diagram(s) to show, for example, the arrangement of the apparatusused,

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

    include full details and explanations of the procedures that you would adopt to ensure that the

    results obtained were as quantitative, precise and reliable as possible.

    [Total: 12]

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    ANSWERS

    Planning question

    MARKS TIMEPLAN 6

    AIM 0 0.5

    HYPO 0 1

    INTRO 2 3min

    VARIABLES 2 3

    APPARATUS 0 0.5

    PROCEDURE 5-7 10.5

    SAFETY 1 1.5

    RESULTS 1 2.5

    INTERPRETATION 1 1.5Correct Use/

    Scientific Reasoning1

    MAX 12 30

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    AIM

    AIM MARKS TIME0 0.5 min

    APPROACH

    AIM can be derived directly from the Question Paper:

    you are required to plan, but not carry out, an investigation into the effect of light intensity onthe rate of photosynthesis in a certain species of algae, Scenedesmus Chlorella.

    This experiment aims to investigate the effect of light intensity on the rate ofphotosynthesis in a certain species of algae, Scenedesmus Chlorella.

    HYPOTHESIS

    HYPO MARKS TIME0 1 min

    APPROACH

    State your STANDwith reference to the AIMS.

    State the General Hypothesis.

    State the Hypothesis specific for experiment.

    [General Hypothesis]

    Increase in light intensity increases the rate of photosynthesis in ScenedesmusChlorella.Note:The rate of photosynthesis will continue to increase until the light saturation pointis reached.After the light saturation point, light is either no longer the limiting factor.

    [Hypothesis specific for experiment]

    The closer the distance of the Scenedesmus Chlorellaalgal beads to the

    bench lamp, thehigher the absorbance valuebeing shown on the colorimeter,

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    INTRODUCTION[1 - 4 MAX 2m]

    INTRO MARKS TIME2 3min

    APPROACH

    Give THEORETICAL EXPLANATIONSto SUPPORT HYPOTHESESmade.

    LIGHT

    =>CO2intake

    => Hydrogen Carbonate turns REDMAGENTA PURPLE

    =>Absorbance Value on Colorimeter

    => RATE

    NO NEEDto give additional information that is UNRELATED to the hypothesis.e.g. The types of chlorophyll pigments present in the plant or the theory of chemiosmosis.

    LIGHT

    =>CO2intake

    => Hydrogen Carbonate turns REDMAGENTA PURPLE

    =>Absorbance Value on Colorimeter

    => RATE

    1] As light intensity is increased, more light is absorbed by the photosynthetic pigmentslocated at the thylakoid membranes of the chloroplasts.2] This would increase the rate at which the electrons are being excited from the primaryreaction centres/ primary pigment moleculeof the photosystems, resulting in a faster rate ofproduction of ATP and NADPH, which are needed in the Calvin Cycle.3] This makes a greater amount ofATP and NADPHavailablefor CO2fixation by RuBP in theCalvin Cycle.4]This leads to a higher rate of absorptionof CO2 from the hydrogen carbonate indicator,

    leading to a higher absorbance value on the colorimeter.

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    VARIABLES[5 - 9 MAX 2m]

    VARIABLES MARKS TIME2 3 min

    APPROACH

    Independent variable: A variable that can be changed by the investigator [i.e. light intensity].*Clearly state HOW you would varyit and at least 5independent variablesat fixed intervalsmust be given.

    Dependent variable: The variable being investigated in this experiment [i.e. rate ofphotosynthesis].

    *Clearly state HOW it is being measured.

    Controlled variables: Ensure that all OTHER variables are controlledand

    *Clearly suggest HOW you would control these variables.

    Negative Control: Ensures that any change in the dependent variable is due to the independentvariable

    and NOT any other variable.

    Positive Control: Ensures that the reagents and apparatus used in the experiment are in workingcondition.

    PRIORITYgoes to Independent Variable, Dependent Variable and Controlled Variable.

    5] Independent Variable:Light intensity,Varied by the distance of the bench lamp from the algal beads 5cm, 10cm, 15cm, 20cm,25cm.

    6] Dependent Variable:Rateof photosynthesis,Measuredby the absorbance value shown on the colorimeter.

    7] Controlled Variables:a] Number of agar beads used,b] Duration of experiment,c] Concentration of CO2 in hydrogen carbonate solution, controlled by ensuring that thehydrogen carbonate indicator is red at the start of each experiment.d] Temperatureof the surroundings kept constant by holding experiment in a thermostaticallyregulated room and through use of bench light that emits very little heat.

    8] Negative control:Set up the glass tube placed in a dark box/ wrapped in black paper.This is to ensure that the change in absorbance values in the colorimeteris due to light

    and not other environmental factors.ORThis is to ensure that the change in colour of the hydrogen carbonate indicatoris due tolight and not other environmental factors.

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    9] Positive control:Set up glass tube that has 0cm distance between the bench lamp and the glass tubecontaining the algal beads.This simulates 100% light reaching the algal beads.This is to ensure that all reagents and apparatus used in the experiment is in workingcondition.

    APPARATUS

    APPARATUS MARKS TIME0 0.5 min

    APPROACH

    DO NOTREWRITEall apparatus given in question.

    You cannot request for a larger quantity of a given apparatus.

    As given in question, with addition of a dark box/ black opaque paper with 0% lighttransmission.

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    PROCEDURE[11 - 18 MAX 6m]

    PROCEDURE MARKS TIME5-7 10.5 min

    APPROACH

    ALL apparatusshould be mentioned/ used.

    QUANTITYused should be stated.

    PURPOSEof the step should be stated [when applicable].

    REPLICATES VS REPEATS

    REPLICATES - additional measurements of the dependent variable made to CALCULATEAVERAGES.

    REPEATS- additional runs of the experiment made to ensure results are REPRODUCIBLE.

    Basic Experimental Procedure [11 - 21 MAX 5m]

    10] Label 7 glass tubes and 7 cuvettes A, B, C, D, E, positive control and negative control.

    11] Transfer 10 agar beads of Scenedesmus Chlorellato all 7 glass tubes.

    12] Using a 5 cm3 syringe, transfer 3 cm3 of hydrogen carbonate indicator to all 7 glass

    tubes.

    13] Darken the laboratoryto ensure that there is no external light source influencing the rateof photosynthesis.

    14] Allow 2 minutes for the algal beads to equilibrate to the experimental conditions [e.g.hydrogen carbonate solution].

    15] Using a ruler, position the glass tubes at their respective distances from the bench lamp, inthe table shown below.

    GlassTube

    A B C D E Positivecontrol

    Negativecontrol

    Distanceof Tubefrombenchlamp/cm

    5 10 15 20 25 0 0

    GlassTube

    Transparent Transparent Transparent Transparent Transparent Transparent Placed indark box/Fitted with

    blackopaquepaper

    Cuvette A B C D E Positivecontrol

    Negativecontrol

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    16] Turn on the bench lamp and start the stop watch.

    17] Turn off the bench lampwhen the time reaches 5 minutes.

    18] Fill the cuvettes with the indicatorfrom their respective glass tubes, and record theirrespective absorbance values from the colorimeter.

    Labelled Diagram [19 MAX 1m]

    19]

    1 105 15 20 25 30

    PositiveControl

    NegativeControl

    A C

    B D

    E

    150Wbenchlamp

    GlassTubes

    hydrogencarbonatesolution

    Scenedesmus

    Chlorella

    algal beads

    Ruler/cm

    Modified from Source: Copyright Science & Plants for Schools: www.saps.org.uk

    Ensure Reliability of Experiment [20 - 21 MAX 1m]20] Obtain 3 replicates, replenishing the set-up with 3 cm3of fresh hydrogen carbonate indicator

    solution between each replicate.Calculate the average absorbance reading for each glasstube.

    21] Repeat steps 10- 19 [the entire experiment] using another 10 agar beads to ensurereproducible results.

    22]Plot a graph showing the relationship between light intensity and rate of photosynthesis.

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    SAFETY[23 MAX 1m]

    SAFETY MARKS TIME1 1.5 min

    APPROACH

    Risk and Precaution must be stated.DO NOT write blindly Read question carefully.

    23]*Risk:Glass bottles are fragile and glass bits can cause injury when bottles are broken.*Precaution: Handle glassware carefully by placing them away from the edge of theworkbenchRationale:So as not to break them and accidentally injure oneself.

    Reject: Care not to touch hot lamp near the bulb so as not to get scalded.

    RESULTS [24 25 MAX 2m]

    RESULTS MARKS TIME1 2.5 min

    APPROACH

    Table should contain INDEPENDENT VARIABLE [Light] and DEPENDENT VARIABLE[Rate of P/S]

    VALUES/TRENDS MUST BE SHOWN to SUPPORT HYPOTHESIS.

    24] Table showing relationship between distance from lamp/cm and absorbance reading/ A

    TubeDistance frombench lamp/

    cm

    Absorbance reading from colorimeter/ AAveragereading /

    AReading 1 Reading 2 Reading 3PositiveControl

    0 0.70 0.71 0.69 0.70

    A 5 0.60 0.61 0.59 0.60B 10 0.50 0.51 0.49 0.50C 15 0.40 0.41 0.39 0.40D 20 0.30 0.31 0.29 0.30E 25 0.20 0.21 0.19 0.20

    NegativeControl

    NIL 0.10 0.11 0.09 0.10

    Reject: No values in the table.Reject: If average was not calculated.

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    25] Graph showing relationship between distance from lamp/cm and absorbance reading/ A

    Distance from

    bench lamp/ cm

    Absorbance

    reading/ A

    *PRIORITY: TREND must be shown.*May include values from table but do not spend too much time scaling the points.Also accept: Straight line

    INTERPRETATION [26MAX 1m]

    INTERPRETATION MARKS TIME1 1.5 min

    APPROACH

    Reference MUST be made to RESULTS.i.e. As shown by the graph, ..

    SUPPORTHypothesis.

    26]As shown by the graph, the smaller the distance between the bench lamp andthe algal beads, the higher the absorbance reading.

    This supports the hypothesis that an increase in light intensity increases the

    rate of photosynthesis.27] Correct use of scientific terms/ correct scientific reasoning.

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    Free-response question

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

    Your answer:

    should be illustrated by large, clearly labelled 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 unique features of stem cells and explain the normal functions of stem cells in a livingorganism. [7]

    [L1]Planning outline:Qn KeyWd

    Level ExampleCharacteristic StemCells

    Function Living Org

    Ans KeyWd

    1] Growth ofentire

    organism

    1]Embryonic

    stem cells

    2] Unspecialised Are pluripotent/totipotent;

    To become3] Can divide and

    grow indefinitelycreate organsand

    specialised layers4] able to

    DifferentiateGive rise to 3 germs

    layers: endoderm,mesoderm andectoderm;

    5]Replenishmentdue to wear andtear

    Adult SC5]

    Hematopoetic stem

    cells/bloodstem cells

    6] Unspecialised Are pluripotent;

    7] Can divide andgrow indefinitely

    to replenish numbers

    Neural /Epidermal/Epithelial.

    8] able toDifferentiate

    Can differentiate into allblood cell types;

    To form granulocytes/monocytes/erythrocytes/megakaryocytes

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    (b) Explain what is meant by restriction fragment length polymorphism (RFLP) and how it can bedetected. [6]

    [L2]What is meant by RFLP[max 2m]

    1. Restriction fragment length polymorphism refers to inherited genetic differences in a singlelocusfound within a population.

    2. These differences may be due to mutations such as substitution, addition or deletions; ordue to differences in the number of variable number of tandem repeats (VNTR) or shorttandem repeats (STR).

    3. These genetic differences in nucleotide sequences lead to a difference in the number orlocation of restriction sites between individuals, which will leads to a variation in the lengthand/or number of restriction fragments formedwhen the same specific restriction enzymeis usedon different individuals.

    How it can be detected[max 4m]RFLP between individuals can be detected by RFLP Analysis, which consists of restriction digest,gel electrophoresis and Southern Blotting (DNA transfer, nucleic acid hybridisation andautoradiography).

    4. The same specific restriction enzyme is first used to hydrolyse the genomes of differentindividuals within a population, resulting in the production of different restriction fragmentslengths and/or numbers;

    5. The digested DNA samples are then separated by their restriction fragment lengths byagarose gel electrophoresis;

    6. Restriction fragments are then transferred from the agarose gel to the nitrocellulosemembrane by capillary action;

    7. The single stranded DNA fragments will then undergo nucleic acid hybridisation, where theDNA fragments on the nitrocellulose membrane are hybridized with specific radioactivelabelled probesthat are complementary to the polymorphic locus. The excess probes arewashed away.

    8. The membrane is laid over a photographic film and autoradiography is then used tovisualize the relative positions of the DNA fragments bound to their probes.

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    (c) Using a named example, explain how RFLP analysis can be used for disease detection. [7]

    [L3]

    Detecting Sickle Cell Anaemia using DdeI Restriction Site

    Note: Candidates may choose to illustratepoints using labeled drawings.

    1 For Sickle cell anemia, base pair substitution of the normal beta-globingene (HbA) from CTT to CAT results in a harmful recessive allele, HbS. Themutation results in the loss of a DdeI restriction site.

    2 Restriction digest of HbAusing DdeIwould result in 2restriction fragments, 175kb and 201 kb. Restrictiondigest of HbS would resultin only 1 restrictionfragment of 376kb.

    376kb

    A]

    HbA

    HbS

    DdeI

    175kb 201kb

    CHR. 11

    CHR. 11

    3 Gel electrophoresis is used to separate the restriction fragments based on their

    lengths. Negatively charged DNA will move towards the anode,with shorterfragments moving further.

    4 The separated restriction fragments are being denatured by an alkalinesolution and the single stranded DNA molecules are transferred to anitrocellulose membrane via capillary action. Single stranded DNAmolecules cross-linkto nitrocellulose membrane.

    5 Nitrocellulose membrane issoaked in radioactive

    probes that arecomplementary to 175kbfragment of HbA and376kb fragment atchromosome 11.Note: Also accept anyprobe positions that allowfor differentiation betweenHbA and HbS.

    376kb

    A]

    HbA

    HbS

    DdeI

    175kb 201kb

    CHR. 11

    CHR. 11

    Radioactiveprobe

    Radioactiveprobe

    6 Autoradiography exposes the radioactive bands.7 Homozygous normal

    (HbAHbA) will have 1 thickband at 175kb.

    Carriers (HbAHbS) will have1 band at 175kb and 1band at 376kb.;

    Homozygous recessive(HbSHbS) will have 1 thickband at 376kb.

    Carrier

    Homo

    .

    reces

    sive

    Homo

    .

    Norm

    al

    175kb

    376kb

    8 By comparing the symptomless carrier with the bands of the 3 known

    genotypes, one can identify his genotype.

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    ORDetecting Sickle Cell Anaemia using MstII Restriction Site

    Note: Candidates may choose to illustratepoints using labeled drawings.

    1 For Sickle cell anemia, base pair substitution of the normal beta-globingene (HbA) from CTT to CAT results in a harmful recessive allele, HbS. The

    mutation results in the loss of a MstII restriction site.2 Restriction digest of HbAusing MstIIwould result in 2restriction fragments,1150 kb and 200 kb.Restriction digest of HbSwould result in only 1restriction fragment of1350kb.

    B]

    HbA

    HbS

    MstII1150 200

    1350

    CHR. 11

    CHR. 11

    3 Gel electrophoresis is used to separate the restriction fragments based on their

    lengths. Negatively charged DNA will move towards the anode,with shorterfragments moving further.

    4 The separated restriction fragments are being denatured by an alkalinesolution and the single stranded DNA molecules are transferred to anitrocellulose membrane via capillary action. Single stranded DNAmolecules cross-linkto nitrocellulose membrane.

    5 Nitrocellulose membrane issoaked in radioactiveprobes that arecomplementary to 200kbfragment of HbA and

    1350kb fragment atchromosome 11.Note: Also accept anyprobe positions that allowfor differentiation betweenHbA and HbS.

    B]

    HbA

    HbS

    MstII1150 200

    1350

    CHR. 11

    CHR. 11

    Radioactiveprobe

    Radioactiveprobe

    6 Autoradiography exposes the radioactive bands.7 Homozygous normal

    (HbAHbA) will have 1 thickband at 200kb.

    Carriers (HbAHbS) will have

    1 band at 200kb and 1band at 1350kb.;

    Homozygous recessive(HbSHbS) will have 1 thickband at 1350kb.

    Carrier

    Homo.

    recessive

    Homo.

    Normal

    175kb

    1350kb

    8 By comparing the symptomless carrier with the bands of the 3 knowngenotypes, one can identify his genotype.

    [Total: 20]END OF PAPER