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    PERFECT SCORE BIOLOGY 2011

    1

    BAHAGIAN SEKOLAH BERASRAMA PENUH DAN

    SEKOLAH KLUSTER

    KEMENTERIAN PELAJARAN MALAYSIA

    PERFECTSCORE

    BIOLOGY 2011Teachers Module

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    Paper 2 Section A:

    Structural Questions Marks Studentnotes

    1. Diagram 1.1 shows a somatic cell of an insect undergoing meiosis.

    Diagram 1.1(a) Label Q, R and S in Diagram 1.1.

    [3 marks] 3m

    (b) In the space below draw chromosome behaviour during metaphase Iand metaphase II.

    [2 marks]

    2m

    Metaphase IMetaphase II

    Q:Chromosome/chromatid

    R:Centromere

    S: Nuclearmembrane

    Process X

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    (c) Explain how the process X involves in producing variation in organisms.

    P1 - During prophase Meiosis 1, crossing over occurs betweenhomologous chromosomes

    P2 - resulting a new genetic combination.

    P3- It is also producing the exchange of genetic material betweenpaternal chromosome and maternal chromosome // betweenhomologous chromosomes.

    [3 marks]

    Any3=3m

    (d) Diagram 1.2 shows the formation of an ovum.

    (i) What are process M and N?

    M : Meiosis I

    N : meiosis II

    [2 marks]2m

    (ii) Describe the process that occurs if a sperm present at process N.P1: meiosis II completed // ovum form

    P2: (nucleus) ovum is fertilized by sperm nucleus

    P3 : zygote is form

    [2 marks]

    2m

    M

    N

    M

    N

    Primary oocyte (2n)

    Oogonium (2n)

    First polar bodySecondary oocyte

    Second polar body

    Diagram 1.2

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    2. Diagram 2 shows the digested food is being carried from small intestineto the liver and body cell.

    (a)(i) Name process X at the villus.

    Absorption[1 mark]

    1m

    (ii) Explain ONE adaptation of the villus for the process in (a)(i).

    F : Has thin wall//one cell thick wall

    E : Diffusion of nutrient occurs rapidly //

    F: Has network of blood capillary

    E : transport nutrient to body tissue

    1F=1m1E=1m

    Diagram 2

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    F : has lacteal //

    E : to absorb/ transport fatty acid/glycerol to body tissue

    F : numerous

    E : increase total surface area

    [2 marks](b) Vessel P and Q transport digested food from the villi to the liver and body

    cells respectively.

    Name vessel P and vessel Q.

    P :Hepatic portal vein

    Q: Lymphatic vessel[2 marks]

    1m

    1m

    (c) Explain what happens to the excessive amino acids in the liver?

    P1 : Excess amino acid is converted into urea

    P2 : process is called deamination

    [2 marks]

    1m

    1m

    (d) Digested food are used by the body cells for growth, to form complexcompounds or structural components.State how lipids, amino acid and glucose are used in the cell.

    Lipids:L1: is used to build up plasma membrane/phospolipid

    L2 : excess lipid is stored in adipose tissue

    L3 : is used as energy reserve in the body

    Amino acids:

    A1: Amino acids are used in protein synthesis

    A2 : to repair damage tissue

    A3: used in synthesis of enzyme/hormones/antibody

    Glucose:G1 : is used in cellular respiration/ is oxidized to release energy

    G2 : Glucose is stored as adipose tissue.

    [3 marks]

    L=1m

    A=1m

    G=1m

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    (e) Explain what will happen to a person if his liver receives insufficient insulinfrom the pancreas.

    P1 : Blood sugar level increases// Diabetes mellitus

    P2 : Excess glucose cannot be converted to glycogen

    [2 marks]

    1m

    1m

    3. Diagram 3 shows the structure of respiratory system in human.

    Diagram 3

    (a) Based on Diagram 3, explain one adaptation of alveolus for efficientgases exchange.F1 : one cell thickP1 : gas doesnt have far to diffuse //diffuse easily

    F2 : supply with network of blood capillary.P2 : to increase the diffusion // transportation of respiratory gases to/from all the body cells.

    F3 : large surface area // numerous number of alveoliP3 : increase the diffusion of respiratory gases

    F4 : inner surface of alveoli are moistP4 : oxygen dissolve in the film of water

    1F=1m1P=1m

    BronchusP

    Bloodcapillary

    Cells

    Alveolus

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    Any F with correspond P

    (b)(i) Name PTrachea 1m

    (b)(ii)Explain the role of P to prevent dirt and bacteria from entering thealveolus.

    F1 : secrete sticky fluid/mucusP1 : traps dirt / bacteria that are breathed in.

    F2 : cells in P have cilia / tiny hair-like structuresP2 : sweeping the mucus out towards the mouth.

    Any F with correspond P

    1F=1m1P=1m

    (c)(i) On Diagram 3, draw labeled arrow ( ) to show the direction of

    Blood flow (P1)

    Oxygen diffusion (P2)

    Carbon dioxide diffusion (P3)[3 marks]

    Blood flow= arrow from blood capillary to other side of bloodcapillary

    Oxygen diffusion = arrow from alveolus to blood capillary// arrow from blood capillary to cells

    Carbon dioxide diffusion = arrow from blood capillary to alveolus// from cells to blood capillary

    P1=1m

    P2 =1m

    P3 = 1m

    (c)(ii) Explain why the diffusion of oxygen occur at the alveolus.

    F: the partial pressure of oxygen in the air of the alveoli is highercompared to the partial pressure of oxygen in the blood capillary

    P: ( therefore,) oxygen diffuses across the surface of the alveolus tothe blood.

    1m

    1m

    (d) A hard mass of food passing down the oesophagus might indirectlyinterrupt the air supply to lung by pressing on P.Explain how P overcome this problem.

    F : P/trachea is protected (against closure by a series of closelypacked C-shaped) ring of cartilage

    P : cartilage keep the trachea open// prevent from collapse

    1m

    1m

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    Structural Questions Marks Studentnotes

    4. Diagram 4.1 shows a cross section of a leaf..

    Diagram 4.1

    (a)(i) On Diagram 4.1 label the structures P and Q.

    P : Xylem

    Q : Phloem

    [2 marks]

    1m

    1m

    (a)(ii) Explain the stage of cell organization of the leaf .

    F : Organ

    E: made up of ground tissue, epidermal tissue, mesophyll tissueand vascular tissue // consists of various types of tissuescombined together to perform spesific functions.

    [2 marks]

    1m

    1m

    (b)(i) Q are important structure in plant transport system.Explain how structure Q in the leaf help in plant transportation.

    P1: Q / Phloem tissue composed of sieved tubes

    P2: with the end walls of each cell are perforated by pores to form

    sieves plates

    P3:which allow substances to pass from one cell to another.

    [2 marks]

    1m

    1m

    1m

    (b)(ii) Name the process occurs in (b)(i).

    Translocation[1 marks]

    1m

    P: ________________

    Q: _______________

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    Diagram 4.2 shows a longitudinal section of structure Q.

    (b)(iii) On Diagram 4.2 label the structures R and S.[2 marks]

    1m

    1m

    (c) R plays an important role in helping S in the plant transportation.Predict what happen to the plant if structure R is not presence ?

    P1: The plant will be dye

    P2: (without R / companion cell) no energy will be provided to thesieve tube

    P3:hence dissolve organic substances/sucrose/ cannot betransported (from leaves to the storage organ/other part ofplant)

    [2 marks]

    Any 2= 2m

    Diagram 4.3 (a) and 4.3 (b) shows a student removed the ring bark fromthe branch of a woody plant.

    S: Sieve tube cell

    R: Com anion

    Diagram 4.2

    Diagram 4.3(a) Diagram 4.3(b)

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    (d) Predict the effect of removing the ring bark from the branch.Explain your answer.

    P1: The branch will be die

    P2: owing to a lack of organic substances in the parts below the

    ring.

    [2 marks]

    1m

    1m

    5. Diagram 5 shows three types of neurones in the human body.

    (a) Name neurone P and neurone Q.

    Neurone P: Afferent neurone/ sensory neurone.

    Neurone Q: Interneurone

    [2 marks]

    1m

    1m

    (b)(i) Name structure X.

    Synapse[1 mark]

    1m

    (b)(ii) Explain how the transmission of nerve impulse across the X.

    P1 - When an impulses / electrical signals reaches in the axonterminal

    P2 - Stimulates (synaptic) vesicles to move towards (andbind with the presynaptic membrane)

    P3 - The vesicles fuse / release the neurotransmitter/ acetylcoline /example of neurotransmitter into the synapse / X.

    P4 - The neurotransmitter diffuses across the synapse / X

    P5 - This leads to the generation of new impulse/ electrical signalsin which travels along the Q / neurone

    [3 marks]

    Any 3=3m

    Diagram 5

    Q

    R

    X

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    (c) Describe the pathway in the reflex action involved the three neuronsabove.

    P1 - Receptor detect the stimulus and triggers the nerve impulses

    P2 (The nerve impulses) are transmitted along neurone P toneurone Q (in the spinal cord) through/ via synapse.

    P3 (The nerve impulses) are then transmitted from neurone Q to

    neurone R through synapse

    [3 marks]

    Any3 =3m

    (d)

    Based on the above statement describe how endocrine system isinvolved in the fight and flight situation.

    P0 adrenaline is secreted by adrenal gland

    P1 (adrenaline) cause the heartbeat / breathing rate increase

    P2 more oxygen and glucose are sent/ transported to the tissues

    P3 metabolic rate / cellular respiration increases

    P4 - more energy is produced to contraction and relaxation ofmuscle (to run away)

    [3 marks]

    Any 3=3m

    A dog suddenly barks and chases you. Your heartbeat increasesand your palms become sweaty. You feel so scared and run away.

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    6. Diagram 6 .1 shows a part of human brain, kidney and a nephron whichinvolve in the process of osmoregulation.

    (a) What is the function of kidney in osmoregulation ?

    P1: Kidney regulates salts/solutes and water levels in the blood

    P2: to maintain a constant water potential in the body /regulates the osmotic pressure of the blood

    [1 mark]

    1m

    (b)(i) Individual Y drinks excessive water.What happen to the blood osmotic pressure in his body?

    Osmotic pressure in plasma decreases // water potential increases[ 1 mark ]

    1m

    (b)(ii) Explain how hypothalamus and gland M response to the condition in(b)(i) ?Hypothalamus:

    P1: Osmoreceptor (in hypothalamus) detect the changes /

    less stimulated.Gland M :

    P2:Pituitary gland / gland M is less/ not stimulated/trigger

    P3:Hence less hormone P / ADH secreted

    P4:Less water reabsorbed

    [ 2 marks]

    P1 =1m

    Any P2-P4 = 2m

    Diagram 6.1

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    (c) If individual Y eating a very salty food, the adrenal gland will release lesshormone Q .What is hormone Q and explain how hormone Q involved in themechanism to restore the osmotic pressure of the blood back to normallevels.

    P1: Hormon Q is aldosterone hormoneE1:(Adrenal gland less stimulated) ,

    E2; less aldosterone produced,

    E3: less salt is reabsorbed

    E4:most of it will be secreted through urine

    [ 3 marks ]

    3 any =3m

    (d) Explain what happen to the filtrate that flows from glomerulus tocollecting duct?P 1 : Water /glucose/amino acid is reabsorbed into blood capillary

    P2 : urea is secreted into distal convoluted tubule

    P3 : salt is reabsorb/ secreted depends on osmotic pressure

    [ 2 marks ]

    2 m

    (e) Kidney function may be impaired by excessive blood loss, certain poisonsor infectious diseases which can lead to kidney failure.Diagram 6.2 shows a haemodialysis machine which can save a kidneypatients life.

    (e) Explain how the machine operates.

    P1: blood is pump into semi permeable membrane in the dialysis

    machine

    P2: contains (dialysate) solution

    P3: waste substances /urea diffuse out (from the blood)

    P4: useful substances are not.

    P5: cleaned blood returns to the patient

    [ 3 marks ]

    Any 3=3 m

    Diagram 6.2

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    7. Diagram 7 shows the changes of four type of hormone which control themenstrual cycle and follicle development in the ovaries

    (a) Based on Diagram 7, name the hormone labelled P and R.

    P : Luteinising hormone

    R : Oestrogen hormone[2 marks]

    2m

    (b) Complete the follicle development in boxes L and N in Diagram 7.L : ovum release from graafian follicle/ovulation (diagram)N : size of corpus luteum is smaller than M.(diagram)

    [2 mark]2m

    (c) Based on Diagram 7, explain the relationship between structure M andthe level of hormone S.

    P1 : Structure M /corpus luteum develop after ovulation

    P2 : Structure M secretes Hormone S /progesterone

    P3 : concentration/level of hormone S increase

    P4: when the structure M degenerate, level of hormone S decrease

    [3 marks]

    Any3=3m

    Diagram 7

    L

    M

    N

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    3(d) If fertilisation occurred, the level of hormones S is maintained and thepregnancy is proceed. Explain the importance of hormone S.

    P1 : to thicken the endometrium wall

    P2 : with epithelium tissue/ network of blood capillary//highly

    vascular

    P3 : prepare for implantation of foetus

    [3 marks]

    3m

    (e) If the sperm counts of a husband is too low, artificial insemination can becarried out to overcome this in fertility problem. Discuss the appropriatetechnique should be used.

    P1 : sperm are collected

    P2 : (over a period of time) until the count of sperm will be high

    enough

    P3 : the sperm are injected directly into Fallopian tube

    [2 marks]

    2m

    8. Diagram 8 shows a longitudinal section of the reproductive parts of aflower during fertilization.

    (a). On Diagram 8, name the structure P, Q, R and S.[4 marks]

    4m

    Diagram 8

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    (b)(i) In the space below, draw a section through the ovule, showing all thecells in R.Label the cell involved in the fertilization.

    D= 1m

    L= 2m

    (b)(ii) What is the significance of having two Q structure in the fertilization.

    P1: one cell Q/ male gamete fertilizes an egg cell to form the diploid

    zygotes

    P2: one cell Q/ male gamete fertilizes 2 polar nuclei to form thetriploid zygote to form endosperm

    [2 marks]

    2m

    (c)(i) A farmer spraying hormone X on the tomatos flower to produce maturetomato.What is hormone X?

    Auxin.[1 mark]

    1m

    (c)(ii) Explain the role of hormone X in the production of mature tomato fruits.

    P1: Auxin stimulates the ovaries of the tomato flowers to developinto fleshy fruits

    P2: without pollination and fertilization

    P3: the process is called partenocarpy

    P4: where the tomato fruits are seedless

    Any 2=2 m

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    9. Diagram 9.1 shows the relationship between a cell, chromosome, DNA,genes and bases.

    (a) On Diagram 9.1 , mark and label of the following terms :i) DNAii) Basesiii) Chromosomes

    [3 marks]

    3m

    (b) State the chromosome number of the cell shown in Diagram 9.1.

    Answer : 12 [1 mark] 1m

    (c) What can you deduce about genes based on Diagram 9.1?

    gene consists of a (short) segment of DNA molecule //

    genes carried genetic information in form of sequence of

    nitrogenous base / A, G, T.

    [1 mark]

    1m

    genes

    Diagram 9.1

    Chromosome

    DNA

    Base

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    Diagram 9.2 shows parts of a molecule of DNA.

    (d)(i) Name the basic unit of DNA.

    Answer: Nucleotide[ 1 mark ]

    1m

    (d)(ii) What is K ?

    answer: (Pentose) sugar // Deoxyribose[ 1 mark ]

    1m

    (d)(iii) Complete the Diagram 9.2, to show that DNA molecule consist of twostrands that joined together by hydrogen bond.

    Criteria Correctly

    C1. paired base

    C2. position of polynucleotide ( opposite direction )

    C3. Connection between molecules in polynucleotide

    [ 3 marks]

    C1=1m

    C2=1m

    C3=1m

    Diagram 9.2

    A

    T

    G

    Unit of

    DNA

    K

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    Diagram 9.3 shows chromosomes with the alleles for crosses betweentwo varieties of pea plants: yellow and smooth seed variety, with a greenand wrinkled seed variety. Yellow seed (Y) is dominant over green seed(y) while smooth seed (S) is dominant over wrinkled seed (s).

    (e)(i) Based on Diagram 9.3, indicate the pair of alleles found in the F1generation.

    [1 mark]

    1m

    (e)(ii) Determine the phenotype of the offspring in the F1 generation.

    Answer: Yellow and smooth seed.

    [1 mark] 1m

    Diagram 9.3

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    10. Diagram 10 show the pedigree inheritance of the characteristics,

    haemophilia in a family. The characteristics, haemophilia is controlled by

    a pair of alleles hh that are linked to the sex chromosomes. H is dominant

    to h.

    Key :

    Normal male Haemophiliac male

    Normal female Haemophiliac female

    (a) Explain briefly the characteristics haemophilia.

    (A phenomenon which is) blood fail to clot due to lack of bloodclotting factor.

    [1 marks]1m

    (b)(i) What is the genotype of K?

    Answer:XHXh[1 mark] 1m

    (b)(ii) Explain how to determine the genotype in (b)(i)?

    P1 :K has a male haemophiliac offspring, XhY.

    P2 :K is a normal female, hence she has a pair ofheterozygous alleles XHXh

    (to produce a male haemophiliac offspring with genotype XhY)

    [2 marks]

    2m

    LK

    Q SR

    Diagram 10

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    (c)(i) State the genotypes and gametes for the parent in the third generation.

    Female x MalePhenotype : Normal Normal

    Genotype : XHXh XHY

    Gamete : XH Xh XH Y

    [2 marks]

    2m

    (c)(ii) Q and R are sisters and are normal. They are found to have differentgenotypes. Explain why?

    P1 :One of them inherits X

    H

    from her father and X

    H

    from her motherP2:The other one inherits XH from her father and Xh from her

    P3 :mother. Both are normal but genotypically, one of them is a

    carrier

    [2marks]

    2m

    (c)(iii) If a haemophilic female marries the child S in the third generation, what isthe probability of obtaining children that are haemophiliac?. Explainyour answer by constructing a genetic diagram for this inheritance.

    Parental generation :

    Phenotype Haemophiliac female x normal male

    Genotype XhXh XHYmeiosis

    Gametes Xh XH Y

    FertilisationOffspring :Genotype XHXh XhY

    Phenotype: Normal (female) Haemophiliac (male)

    Probability : 50% normal and 50% haemophiliac //

    Phenotypic ratio : 1 normal female : 1 haemohilia male

    [4 marks]

    1m

    1m

    1m

    1m

    1m

    Max 4m

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    (d) Sex-linked trait such as haemophilia and colour-blindness are usually

    associated with males. Explain why?

    P1 :Males are determined by the presence of the X and Ychromosomes // male only has one X chromosome

    P2: Sex-linked genes are absent in the Y chromosomes//onlypresence in X chromosome.

    P3 : Therefore, the presence of one sex-linked gene in the Xchromosome will affect the male as compared to thefemale who needs recessive genes to be present in bothX chromosomes for her to be affected.

    [2 marks]

    Any2 =2m

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    BAHAGIAN SEKOLAH BERASRAMA PENUH DAN

    SEKOLAH KECEMERLANGAN

    KEMENTERIAN PELAJARAN MALAYSIA

    PERFECTSCORE

    BIOLOGY 2011Teachers Module

    Paper 2Section B

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    Paper 2 Section B:

    1. Diagram 1.1 show the human digestive system and Diagram 1.2 show the structure

    inside organ P.

    No Essay Questions Marks Student notes

    1(a)(i)

    Explain the process of absorption of glucose and amino acids in

    organ P.

    [4 marks]

    Sample answer

    F1: involve process diffusion and active transport

    P1: from the lumen into epithelial cell by facilitated

    diffusion

    P2: across the epithelial lining by active transport

    P3: both are absorb into blood capillaries

    1

    1

    1

    1 4

    1(a)(ii)

    Explain three structural adaptation of organ P for effective

    absorption of food

    [6 marks]

    Diagram 1.1

    Diagram 1.2

    Organ R

    Organ P

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    Sample answer

    F1: the largest section of alimentary canal

    P1: increases the surface area of absorption

    F2: Inner surface has numerous number of villi

    P2: Form brush border to increase the surface area of

    absorption

    F3: Epithelial lining is only one cell thick

    P3: Increases the rate of diffusion process

    1

    1

    1

    1

    1

    1

    6

    1(b) Describe the process of assimilation in organ R.

    [10 marks]

    Sample answer

    P1: organ R is a liver

    P2: act as a checkpoint // control the amount of

    nutrients released into blood circulatory

    P3: involve in assimilation of amino acids and glucose

    P4: (organ R) synthesizes plasma protein from aminoacids

    P5: converted amino acids into glucose when a short

    supply of glucose/glycogen

    P6: broken down/ convert excess amino acids through

    deamination

    P7: to form urea as waste products

    P8: glucose is used for respiration

    P9: excess glucose is converted into glycogen andstored

    P10: if full, excess glucose is converted into lipids

    P11: glycogen is converted back into glucose if needed

    1

    1

    1

    1

    1

    1

    1

    1

    1

    1

    1

    Any 10 10

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    2. Diagram shows various processed food on a supermarket shelf.

    Salted plum Potato chips Prawn crackers

    No Essay Questions Marks Student notes

    2(a) Based on your biology knowledge,

    Explain the good and the bad of food processing on human being.

    [10 marks]

    Sample answer

    Good(G) Explanation(P)

    G1 ; to preserve food P1: Avoid wastage of

    food/prevent food

    spoilage/can be stored(for

    future use)

    G2: to increase its

    commercial value/uses of

    food additives

    P2: improve the

    taste/appearance/texture of

    food/to preserve the

    freshness

    G3:to diversify the uses of

    food substances

    P3: to increase the variety of

    product//any example

    Max ; 5 marks

    1,

    1

    1,

    1

    1,

    1

    5

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    Sample answer

    Bad(B) Explanation(P)

    B1 ; uses food additive P4:give long term side

    effect/examples//reduce the

    nutrient/vitamin in the food.

    B2: too much sugar P5: increases the risk of

    diabetes

    B3: foof colouring/yellow

    dye/tetrazine

    P6: causes allergy reaction

    B4: too much salt P7:increase the risk of high

    blood pressure

    B5: Sodium nitrate P8:causes nausea/athma(to

    certain people)

    Any 3B with respective P

    Max 5 marks

    1,1

    1,1

    1,1

    1,

    1

    1,

    1

    5

    (b)

    Explain the food processing methods which are related to thefactors that cause food spoilage.

    [ 10 marks]

    Sample answer:

    Concept : Food can be preserved by destroying themicroorganism present in the food //by stopping the activities of the microorganism

    F1: Cooking-.high temperature kill the microorganismsP1: denature the enzyme that cause the breakdown of food

    F2: Treating food with sugar/saltP2: causes the microorganism to lose water due to osmosis

    F3: Adding vinegar will reduced the pHP3 that prevent microorganism from growing

    F4: Fermentation of fruit juices and other food by addingyeast

    P4: high concentration of alcohol prevent themicroorganism from growing

    1

    1

    1

    1

    1

    11

    1

    1

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    F5: Dry under hot sun (meat/fish/fruits)P5: removes water from food dehydrated

    F6: Ultravoilets raysP6: kills microorganism

    F7: Pasteurisation destroy bacteria which causetuberculosis and typhoid

    P7: (technique) -Food is heated to 630C for 30 minutes /720C for 15 seconds followed by rapid coolingto 100C

    P7.1: (Pasteurisation) retains the natural flavour andnutrients

    F8: Canning uses heat sterilization to kill microorganismsand their spores

    P8 (technique) -.Food is packed in cans, steamed at hightemperature and pressure to drive out air

    P8.1: the vaccum created within the cans prevent growth ofmicroorganism

    F9: RefrigerationP9: food stored at temperature below 00C prevent

    growth/germination of microorganismP9.1: food remain fresh for a long period of time

    Any ten : F + P correctly

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    10

    TOTAL MARKS20

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    No Essay Questions Marks Studentnotes

    3 Diagram 3 shows roots of plants found in mangrove swamp.

    Discuss how these roots are adapted for stability, salt tolerance andless oxygen of water logged mangrove swamp soil.

    [10 marks]

    Sample Answer:

    F1: Root adaptations increase stability of mangrove trees in thesoft sediments along shorelines.

    P1: Prop roots descending from the trunk and branches,providing a stable support system.

    P2: Shallow wide spreading roots, surrounds the trunks ofAvicennia adding to the structural stability of the tree.

    P3: Other species of mangrove trees grow at higher elevations, indrier soils, do not require specialized root structures.

    1

    1

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    1

    Pneumatophore

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    F2: The ability to exclude salts occurs through filtration at thesurface of the root.

    P1: The cell sap is hypertonic to sea water, the water able todiffuse into the root cell by osmosis

    P2: Salt is removed through hidatodes located on each leaf.

    F3: Mangrove trees are adapted for survival in oxygen-poorsediments through specialized root structures /pneumatophore.

    P1: these spaces in soil fill with water, containing lower oxygenlevels than air.

    P2: having well-developed aerial roots or pneumotophoresgases exchange

    P3 : example Avicennia

    P4 : Red mangroves / Rhizophora have prop/stilt roots extendingfrom the trunk and adventitious roots from the branches ableto absorb more water and mineral.

    P5 knee root and buttress root has lenticels that allow air into theroots.

    Any 10 points.

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    No Essay Questions Marks Studentnotes

    4. Diagram 4 shows part of a nitrogen cycle .

    (a)(i) The atmospheric nitrogen cannot be absorbed directly by plants.Based on Diagram 4, state two form of nitrogenous compounds thatcan be absorbed directly by plants and explain how a deficiency ofsubstances K in the soil affect the growth of the plants.

    [ 4 marks]

    Sample answer:

    P1 : (Two form of nitrogenous compound that can be absorbeddirectly by plants) are nitrate ions and ammonium ions.

    P2 : substances K is nitrate

    P3 : Substance K is used in synthesis of protein in plant oranimal

    1

    1

    1

    Nitrogen in theatmosphere

    Substance K

    Nitrogen fixation bymicroorganisms in plant

    Nitrites

    Ammoniumcompounds

    Nitrogencompounds in

    plants

    Nitrogencompounds in

    animals

    Diagram 4

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    P4 : (deficiency of substance K), less synthesis of protein

    P5 : plant growth is retarded/slow/ stunted

    [ 4 marks]

    1

    1

    4

    (ii) Based on Diagram 4, explain role of the microorganism in nitrogencycle.

    [6 marks]

    Sample answer:

    P1 :Rhizobium sp. (in root nodule of legume plant)// Nostocsp.//Azotobacter sp.

    P2 : fix the Nitrogen from atmosphere into nitrate/ substance K.

    P3 : (Nitrate/K substance) is absorp by roots of plants andconverted into protein.

    P4 : (when the plant /animal die), protein in plant/ animal isdecomposed by decomposer/fungi

    P5 : into ammonium compound

    P6 : Nitrosomonas sp. converts ammonium compound into

    nitrite

    P7 : Nitrobacter sp. Convert nitrite into nitrate/substance K

    P8: Denitrifying bacteria convert nitrate back into Nitrogen

    [Max : 6 marks]

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    No Essay Questions Marks Student notes5.

    (a)

    Diagram 5 shows the eutrophication process that occurs to a lakedue to the human activities.

    Based on the Diagram 5, explain what is meant by`eutrophication`

    [10 marks]

    Sample answer

    P1 : Farmers use fertilizers that usually containsnitrates/phosphate

    P2: Fertilizer/animal waste/silage which contain

    nitrate/phosphate may washed out in water when it rains/leaching/run into the lake.

    P3: Algae/green plant in the lake grow faster (when they aresupplied with extra nitrate/(phosphate)

    P4: (they may grow so much) that they completely cover thewater.

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    P5: block out the light for plants growing beneath them.

    P6: Photosynthesis rate reduced

    P7: Dissolve oxygen also reduced

    P8: Plant on the top of water and beneath water eventuallydie.

    P9: Their remains are good source of food bacteria //bacteriadecomposed the dead plant rapidly//bacteria breedrapidly

    P10: The large population of bacteria respires using up moreoxygen

    P11: so there is very little oxygen left for other livingorganism

    P12: BOD increased

    P13: Those fish which needoxygen have to move other areasor die

    Any 10

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    1

    (b) Explain how each of the following can reduce water pollution:

    (i) Treating sewage(ii) Using organic fertilizers rather than inorganic ones.

    [6 marks]

    Sample answer:

    (i) Treating sewage

    P1: The sewage contains harmful bacteria /substance whichprovide Nitrate/nutrient for microbe.

    P2: Remove harmful bacteria/most of the nutrient whichcould course eutrophication before it is released into therivers.

    P3: When sewage has been treated, the water in it can beused again//sewage treatment enables water to berecycled.

    P4: Microorganisms used in sewage treatment.

    Any 3

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    (ii) Using organic fertilizers rather than inorganic

    Sample answer

    1. Example of organic fertilizers : Manure

    2. Example of inorganic fertilizer : Ammonium nitrate

    3. Organic fertilizers do not contain many nitrates(which caneasily be leached out of the soil.

    4. They release their nutrients gradually (over a long periodof time) giving crops time to absorb them efficiently.

    Any 3

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    1

    1

    (c) Explain how deforestation of rainforest can cause flash flood.

    [4 marks]

    Sample answer

    F: deforestation can cause soil erosion

    P1 : The leafy canopy trees protect the soil from the impactof falling rain.

    P2: The roots of the trees hold soil and water

    P3: (With the trees removed) the soil is exposed directly to

    the rain//water runoff becomes intense.

    P4: Topsoil/fertile layer, get washed away during heavy rain.

    P5: (heavy rainwater flows down hillside to river with) erodedsoil deposited blocking the flow of water.

    P6: The water levels in rivers rise rapidly causing flood tooccur.

    Any 4

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    20

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    No Essay Questions Marks Student notes

    6(a) Diagram 6.1 shows the mankind activities.

    Based on your knowledge in biology, explain the effects of theactivities to the mankind and their surroundings. Suggest theways to overcome this problem.

    [12 marks]

    Sample Answer

    P1: the problem is green house effectP2: the activities produce green house gases such as

    carbon dioxide, methane and nitrogen dioxide.

    P3: The gases accumulate and forms a layer at theatmospheric surface

    P4: Solar radiation penetrate earth atmosphere and warm theearth surface.

    P5: Part of the heat energy is reflected back by earth surfaceto the atmosphere in the form of infrared radiation.

    P6: Heat energy that is reflected back is trapped by

    greenhouse gases.

    P7: Higher concentration of greenhouse gases on the

    atmosphere cause more reflected energy being trapped.

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    P8: This will increase the earth temperature and can cause

    global warming.

    Any 5

    The Effect:

    P1: Increase of carbon dioxide and temperature of earth willincrease the rate of photosynthesis or agriculture yield.

    P2: Increase in earth temperature / global warming willaccelerate evaporation of water and reduce soilhumidity.

    P3: Climate change / changes in wind direction / change thedistribution of rainfall / drought /flood

    P4: Melting of ice in north and south poles increase the sealevel and cause flooding of low level areas.

    P5: Yield of crop / domestic animal reduced

    P6: Mass destruction of animal habitat and cause the animalemigration/ reduces of animal population.

    Any 5

    Ways to overcome:Use of technology such as :

    P1: less the emission of CO2 by the motor vehicles by usingthe unleaded petroleum.

    P2: using the filter on the chimney to prevent harmful gases

    P3: car pool/ use public transporti

    P4: less open burning

    P5: less the using of CFC and change to HCFC

    P6: Using catalytic converter in the car exhaust

    P7: educate the public on the importance of protecting andcaring the environment through mass media andenvironmental campaigns.

    P8: planting more treeAny 2

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    7(a) . Diagram 7.1 shows the ozone layer in atmosphere that protects earth from ultraviolet raysfrom the sun.

    Diagram 7.1

    Describe how the ozone layer becomes thinner. Discuss its effects on humans and theenvironment and suggest the ways to solve these problems.

    [10 marks ]

    No Essay Questions Marks Studentnotes

    7 (a)Sample Answer

    Thinning of the ozone layer is due to thewidespread use of CFC

    It is used in aerosol, industrial solvents,electronics and Freon in air conditioners

    Ultraviolet radiation strikes a CFC molecule

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    1

    Solar radiationSinaran suria

    StratosphereStratosfera

    Ozone layerLa isan ozon

    Harmfulultraviolet radiationSinaran ultra ungu

    berbahaya

    TroposphereTrofosfera

    EarthBumi

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    No Essay Questions Marks Studentnotes

    cause the chlorine atom to break away

    Then the chlorine atom collides with an ozonemolecule and combines with an oxygen atomto form chlorine monoxide and oxygen

    Then the free atom of oxygen collides with thechlorine monoxide, the two oxygen atomsform a molecule of oxygen

    The chlorine atom is released and free todestroy more ozone molecules

    The chlorine produced re-enters the cycle

    When the ozone layer becomes thinner, moreultraviolet radiation reaches the Earth

    The effect of excessive ultraviolet radiation onhuman

    reduction of the bodys immune system

    skin cancer

    cataract of the eye

    Effect on plants

    reduction of the rate of growth thereforereducing crop yields

    Effect on aquatic organism

    death of plankton, reduce food supply toaquatic organism, fishermans catch isreduced.

    Steps to overcome this problem

    Reduce or stop using CFC or chlorine-based

    products Replace CFC with HCFC

    Use wrapping papers instead of polystyreneboxes

    Patch up the holes in the ozone layer by firingfrozen ozone balls into the atmosphere

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    Max 5

    Max 3

    Max 2

    TOTAL10marks

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    7 (b) Diagram 7.2 shows a phenomenon X that occurs from air pollution. Describe theformation and the effects of the phenomenon on agriculture and aquatic ecosystem.

    [10 marks]

    No Essay Questions Marks Studentnotes

    9(b) Able to name the phenomenom X

    Sample Answer

    F1 : X is acid rain

    P1 : combustion of fossil fuels in powerstation/factories/domestic boilers

    P2 : produce sulphur dioxide

    P3 : and oxide of nitrogenP4 : both gases combine with water vapour

    P5 : form sulphuric acid and nitric acid

    P6 : fall to the Earth with pH less than 5.0

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    11 Max 6

    Diagram 7.2

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    Effects:

    On agriculture

    P1 : soil become acidic// leaching of minerals

    P2 : not suitable for culativation/grow of crops

    On aquatic ecosystem

    P1 : accumulation of insoluble aluminium ion in

    water sources// increase acidity in the

    ecosystem

    P2 : kill aquatic organisms

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    Total10 marks

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    8. Diagram 8 shows three types of neurone in individual A.

    Diagram 8

    a) Describe the process X in Diagram 8[4 marks]

    b) Explain the above situation.

    [6 marks]

    X

    Neurone P

    Neurone Q

    Neurone R

    After an accident , individual A doesnt experienceany response to hot object.

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    No Essay Questions Marks Student notes

    8 (a)

    Sample Answer

    When an impulses arrives in the axon terminal

    Stimulates (synaptic) vesicles to move towards andbind with the presynaptic membrane

    The vesicles fuse / release the neurotransmitter intothe synapse

    The neurotransmitter molecules across the synapseto the dendrite of another neurone

    Stimulated to trigger a new impulses which travelsalong the neurone

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    Max 4

    (b)

    SampleAnswer

    F1 - P is afferent neurone which transmits nerveimpulse from the receptors to theinterneurone.

    P1 - If P damaged, impulse from receptor cannotbe transfered to the interneurone.

    P2 - (As a result), individual A cannot feel any pain

    P2 - R is efferent neurone which transmits nerveimpulse from interneurone to the effector

    P1 - If R damaged, impulse from interneuronecannot be transfered to the effector

    P2 - (As a result), individual A cannot withdraw thefinger // pull the hand away from the pointed

    needle

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    Max 6

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    9

    Mr. Q is married to Mrs. Q for more than 10 years but didnot have any child due to low sperm count in Mr. Q.Mr. and Mrs. V have 6 children in 12 years of marriage.Mrs. V has high blood pressure and heart problem, so theydecided not to have any more kids.

    Explain how reproduction technologies able to help thesetwo families.

    [10 marks]

    Sample Answer

    F1: Mr Q have problem with infertility, that is lowsperm count

    P1: not enough sperm/ less sperm produce by Mr Q/less chance for the sperm to reach fallopian tube

    F2: technology applied : in vitro fertilization

    P3: sperm and egg are taken from Mr. and Mrs. Q

    P5: fertilize in petri dish/test tube

    P6: embryo is inserted into Mrs Q uterus for furtherdevelopment.

    or

    F2: artificial insemination

    P3: sperms are collected until the number of spermsare enough

    P4: sperms are injected into the fallopian tube of MrsQ

    P5: during ovulation

    Any 5 points 5 marks

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    Sample Answer

    F3: For Mr V family the problem is to control the birthrate/ stop pregnancy

    P1: Mrs. V have high risk if pregnant due to highblood pressure and heart problem

    P2: use contraceptive pills, to stop ovum development

    P3: use condom during copulation, prevent spermfrom reaching uterus

    P4: Tubal ligation or tubectomy the fallopian tube istied/cut

    P5 : blocking the ovum from entering the uterus/Prevent sperm from reaching the ovum

    Any 5 points 5 marks

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    No Essay Questions Mark Studentnotes

    10. The variation of ABO blood group determined by three differentalleles, but an individual carry only two of the three allele.

    With schematic diagram , explain the possibilities of the blood groupand genotypes of the offspring if the fathers blood group is A and themothers blood group is AB.

    [10 marks]Sample Answer

    Schematic diagram:

    1

    1

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    1

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    1

    1

    1

    1

    1

    Father MotherXParent :

    IB

    Phenotype F1: Blood group A Blood group B

    Parent

    genotype : XIA IA IA IB

    Parent

    genotype :

    Meiosis

    Gamete :IA

    IA

    Fertilisation

    Genotype F1: IA

    IA

    IA

    IB

    Phenotypic Ratio: 1 blood group A : 1 blood group B

    Phenotypic Ratio: 2 blood group A : 1 blood group A : 1 bloodgroup AB

    IB

    IB Io

    Parentgenotype : XI

    AIo

    IA

    IB

    Parent

    genotype :

    Meiosis

    Gamete :IA

    IA

    Fertilisation

    Genotype F1: IA

    IA I

    AIB

    Io

    IA

    Io

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    Explain :

    P1 : Allele IA and IB are codominant.

    P2: Father has 2 possibilities of genotype

    P3 : (either) IA IA //homozygous dominant or IA Io//heterozygous

    P4 : (if genotype of father is IA IA ), possibility of blood group ofoffspring is 50% blood group A and 50% is blood group

    B//refer to schematic diagram

    P5 : (if genotype of father is IA Io), possibility of blood group ofoffspring is 50% blood group A , 25% is blood group B and

    25% blood group B //refer to schematic diagram

    [Total : 10 marks] 10

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    No Essay Questions Marks Student notes11.(a).

    Diagram 11.1 shows a group of boys with different height andDiagram 11.2 shows the various types of fingerprints.

    Diagram 11.1 Diagram 11.2

    Based on the biology knowledge, identify the variation andexplain the similarities and differences in Diagram 11.1 and

    Diagram 11.2. [10 marks]

    Able to:(i) Identify the continuous variation and discontinuous variation.

    (ii) Explain the similarity and the contrast of continuous variation

    and discontinuous variation.

    Sample answer:

    P1: Diagram 11.1 (height) is continuous variation

    P2: Diagram 11.2 (fingerprints) is discontinuous variation

    Similarities:

    P3: Both create varieties in the population of speciesP4: Both type of variation are caused by genetic factor

    Differences:

    P5: Height is continuous variation while fingerprints isdiscontinuous variation

    P6: Graf distribution of continuous variation shows anormal distribution while Graf distribution ofdiscontinuous variation shows a discrete distribution.

    P7: The characters of continuous variation are quantitative/ can be measured and graded from one extreme to theother while the characters of discontinuous variationare qualitative / cannot be measured and graded fromone extreme to the other.

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    P8: Continuous variation exhibits a spectrum ofphenotypes with intermediate character whilediscontinuous variation exhibits a few distinctivephenotypes with no intermediate character.

    P9: Continuous variation influenced by environmentalfactors while Discontinuous variation is not influencedby environmental factors.

    P10: In continuous variation two or more genes control thesame character while In discontinuous variation singlegenes determines the differences in the traits of thecharacter.

    P11: In continuous variation the phenotype is usuallycontrolled by many pairs of alleles while in

    discontinuous variation the phenotype is usuallycontrolled by a pair of alleles.

    Any 10

    1

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    1

    1

    (b). Diagram 11.3 shows the variants P, Q and R of a species offish.

    Describe how the variation occurs in the species of fish.

    [10 marks]

    Sample Answer

    F1: Variation occurs because of genetic factors

    P1: By crossing over

    P2: during prophase I of meiosis

    P3: when two homologous chromosomes are intertwine

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    between the non-sister chromatid.

    P4: the exchange of materials between the chromatidsresults in new combination of genes

    P5: By independent assortment

    P6: during metaphase I of meiosis, homologouschromosomes arrange themselves randomly at theequator

    P7: the random arrangement and separation of eachhomologous pair is independent of one another

    P8: and result various genetic combination in the gametes.

    P9: By random fertilisation

    P10: the fertilisation of sperm and ovum occurs randomly

    P11: each gamete has unique combination of genes thatcan fertilise any of the ova which also has uniquecombination of genes.

    P12: the fertilisation of gametes produced zygote/offspringwhich has various of variation.

    F2: by environmental factors.

    P13: environmental factors that cause variation includedabiotic factors

    P14: such as light intensity / temperature / water / humidity/ nutrients / soil fertility

    P15: these factor affect the growth rate of the organism.

    Any 10

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    No Essay Questions Marks Student notes12 Diagram 12.1 shows a mangrove swamp forest and

    Diagram 12.2 shows the same area 50 years later.

    Diagram 12.1 Diagram 12.2

    Discuss the impact of the exploitation on the ecosystem.[10 marks]

    Sample answer:P1: swampy area is change to densely populated / town /

    commercial area

    P2: the change requires activities such as deforestation andland reclaimation

    P3: more and more buildings/ glass buildings built in the are

    P4: could be the factors for air / thermal / noise pollution

    P5: and greenhouse effect as well as heat island

    P6: lost of vast quantity of flora and fauna / biodiversity inthe area

    P7: less water catchment area / less of reproductive area

    P8: landslide and soil erosion

    P9: which frequent flash flood and muddy flood

    P10: water pollution in the nearby river

    P11: which kill most of the aquatic organisms

    Any 10

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    BAHAGIAN SEKOLAH BERASRAMA PENUH DAN

    SEKOLAH KECEMERLANGAN

    KEMENTERIAN PELAJARAN MALAYSIA

    PERFECTSCORE

    BIOLOGY 2011Teachers Module

    PAPER 3QUESTION 1

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    Question 1 :

    No. Questions Marks Studentnotes

    1 A group of students carried out an experiment to study the effect of the concentration ofglucose on the activity of yeast . Diagram 1.1 shows the method used by the students.

    The initial height of the coloured liquid in the manometer is shown in Diagram 1.2.

    The experiment was repeated using different concentrations of glucose. Table 1.1 shows the

    results of the experiment after 10 minutes.

    Diagram 1.1

    DIAGRAM 1.2

    rubber tubing

    Manometer with

    coloured liquid

    Initial height of

    coloured liquid

    Boiling tube containing yeast

    suspension

    Glass tube

    clip

    Rubber stopper

    Initial height of

    coloured liquid :

    1 cm

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    Percentage concentration ofglucose / %

    Final height of coloured liquid in themanometer after 10 minutes /cm

    10

    15

    20

    3

    5

    8

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    No. Questions Marks Studentnotes

    (a)Complete Table 1.2 by recording the height of colouredliquid in the manometer after 10 minutes

    (b) (i) Based on Table 1.1, state two observations .1. At 10% concentration of glucose ,the final

    height of coloured liquid after 10 min is 3 cm

    2. At 20% concentration of glucose , the finalheight of coloured liquid after 10 min is 8 cm

    (ii) State the inference which corresponds to the observation in1(b(i).

    1. Low activity of yeast in lower concentration ofglucose, less carbon dioxide is released

    2. High activity of yeast in high concentration ofglucose, more carbon dioxide is released

    (c) Complete Table 1.2 for the three variables based on theexperiment.

    Variable Method to handle the variable

    Manipulated variable:

    The concentration ofglucose

    Use different concentration ofnutrients/glucose

    Responding variable:

    Height of coloured liquid//The rate of yeast activity

    Record the height of colouredliquid by using a metre rule //Calculate rate of yeastrespiration using formula:= height of coloured liquid

    timeControlled variable :

    Volume of yeastsuspension /mass ofyeast/volume ofglucose/pH/lightintensity/temperature/timetaken

    Fix the volume of 100cm3ofyeast suspension /the mass of4 g of yeast /pH5 /lightintensity at distance of 50cm

    /temperature at roomtemperature/time taken for 10minutes

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    (d) State the hypothesis for the experiment.The higher/ lower the concentration of glucose, thehigher / lower the rate of yeast activity

    (e) (i) Based on Table 1.1, construct a table and record the resultsof the experiment which includes the following aspects:

    Percentage concentration of glucose

    Height of coloured liquid

    The rate of the activity of yeast

    Percentageconcentrationof glucose (%)

    Height ofcoloured liquid

    (cm)

    The rate of theactivity of yeast

    (cm/min)

    10 3 0.3

    15 5 0.5

    20 8 0.8

    Table 1.1

    (e) (ii) Draw a graph of the rate of the activity of yeast against the

    concentration of glucose

    (iii)

    Based on the graph in 1(e)(ii), state the relationship betweenthe rate of the activity of yeast and the concentration ofglucose. Explain your answer.

    When the concentration of glucose increases/decreases,the rate of yeast activity increases/decreases, moresubstrate for yeast to use for energy production, moreyeast reproduced.

    (f) Based on the experiment, define anaerobic respiration inyeast operationally.

    An anaerobic respiration is when yeast using glucose toproduce gas that causes the rising of liquid inmanometer tube and the process is affected byconcentration of glucose

    3

    3

    3

    3

    3

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    (g) The experiment is repeated by using 1 ml of 0.1 mol dm-3 ofsodium hydroxide solution is added into the boiling tube.Predict the manometer reading after 10 minutes. Explainyour prediction.

    1 cm, not increase, sodium hydroxide is alkali, themedium is not suitable for yeast.

    (h)The following list is part of the apparatus and material usedin this experiment.

    Complete Table 1.3 by matching each variable with the

    apparatus and material used in the experiment.Variables Apparatus Material

    Manipulated Measuringcylinder

    Glucose

    Responding Coloured liquid Metre ruler

    Controlled electronic balance Yeast

    Yeast, metre rule, coloured liquid, electronicbalance, glucose solution, measuring cylinder

    3

    3

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    Question 2 :

    No. Questions Marks Student notes

    2 Lemna minor is a species of free-floating aquatic plants from the duckweed familyLemnaceae. The plants grow mainly by vegetative reproduction: two daughter plantsbud off from the adult plant.

    An experiment is carried out to investigate the effect of abiotic factor such as pH onLemna sp. growth. Experiment is done under controlled conditions: 12 hours a daylight exposure and using the same Knops solution.Petri dish is filled with 20 ml Knops solution with different pH value and 5 Lemnasp.each.The Knops solution is treated by adding acid or alkali to achieve the pH value needed.

    ** Knops solution is a solution which contains essential nutrient for plants growth.

    Figure 1

    After 7 days, the observation is made and the result shown in Table 1.1.

    pHvalue

    Petri dish Number of Lemnasp.

    3

    4

    Lemna minor

    Petri dish

    Knops solution

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    5

    5

    78

    9

    11

    11

    5

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    131

    Table 1.1

    No. Questions Marks Student notes

    (a) State thenumber of Lemnasp. in the spacesprovided in Table 1.1

    (b) (i) Based on Table 1, state two different observations .

    Able to state any two observations correctly according to 2criteria:

    pH ( Manipulated Variable)

    Number of Lemna sp (Responding Variable)

    Sample answers:1. At pH 2 (Knop solution), the number of Lemna sp is 42. At pH 8 (Knop solution), the number of Lemna sp is 113. At pH 12 ( Knop solution), the number of Lemna sp is

    14. At pH 12 (Knop solution), the number of Lemna sp

    grow is less than at pH 2/4/6/8/105. At pH 8 (Knop solution), the number of Lemna sp is

    more than at pH2/4/6/10/12

    *1,2 &3 is a horizontal observation

    *4 & 5 is a vertical observation

    (ii) State the inferences which corresponds to the observationsin 1(b)(i).

    Able to makeone logical inferencefor each observationbased on the criteria

    suitable abiotic factor

    Favourable for Lemna sp growthSample answers:

    3

    3

    3

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    1. Strong acidic condition is not favorable for Lemnagrowth.

    2. Weak/slight alkaline // neutral condition is mostfavorable for Lemna growth.

    3. Strong alkaline is not favorable for Lemna growth.

    4. Strong alkaline condition is the least favorable forLemna growth compare with other conditions.

    5. Neutral/Slight alkaline condition is the best/mossfavorable condition for Lemna growth.

    *1,2 &3 is a horizontal inference*4 & 5 is a vertical inference

    (c) Complete Table 1.4 to show the variables involved inthe experiment and how the variables are operated.

    Variables How the variables are operated

    Manipulated:

    pH Add/Use acid or alkali to theKnop solution to get differentpH condition// Use pHsolution: pH2, pH4, pH6, pH8,pH10,pH12 // change/alter themedium condition

    Responding:

    Number of Lemna sp

    Count and record the number

    of Lemna sp. plants after 7days.

    Fixed:

    Light exposure /

    Volume of Knopsolution

    Fix 12 hours light exposureevery day /

    Maintain the volume at 20ml

    (d) State the hypothesis for this experiment.

    Able to state a hypothesis to show a relationship between themanipulated variable and responding variable and the

    hypothesis can be validated, based on 3 criteria: manipulated variable

    responding variable

    relationshipSample answer :

    1. In low pH, number of Lemna sp is less than in ahigher pH.

    2. The higher pH the higher number of Lemna sp.3. In a neutral condition the number of Lemna sp.

    3

    3

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    plants is the highest /the most.4. The more alkali the medium is the less number of

    Lemna sp.

    (e) (i) Construct a table and record the results of theexperiment.

    Your table should contain the following title. pH of water

    Number of Lemnasp.

    Able to draw and fill a table with all columns and rowslabeled with complete unitSample answers

    pH of water Number of Lemnasp2 44 56 8

    8 1110 512 1

    (e) (ii) Plot a graph showing the number of Lemnasp against the pHin the graph below

    Able to plot a graph with 3 criteria:

    A(axis): correct title with unit and uniform scale

    P (point) : transferred correctly

    S (Shape): able to joint all points, smooth graph, bell

    shape.

    (iii) Referring to the graph in (e) (ii), describe the relationshipbetween the Lemnasp growth and the condition of themedium.

    Able to state clearly and accurately the relationship betweenthe condition of medium and Lemna growth based on thecriteria:

    P1- Alkali, acidic or neutral (abiotic factor)

    P2- Lemna sp. growthSample answer:(Associates each of the condition with the Lemna growth)

    1. In the acidic medium the Lemna sp. growth isless, and increase when the medium becomeneutral but decrease when in alkali condition.

    2 Lemnasp. grow very well in neutral medium andless growth rate in alkali or acidic medium

    (f) Based on the experiment, define operationally the abioticfactor in an ecosystem.

    3

    3

    3

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    Able to explain the abiotic factor operationally base on 3criteria:

    Lemna sp (organism)

    affected (growth)

    pH of medium (abiotic factor in ecosystem)

    Sample answer:1. Abiotic factor is pH of the medium that affect the

    Lemna sp growth in an ecosystem.

    (g) The effluent from laundry shop flows into a pond nearby,predict the population of Lemnasp in the pond. Explain youranswer.

    Able to predict the result accurately base on 2 criteria.

    Expected population of Lemna sp

    The reason of the answer Not suitable for growth

    Sample answer:P1- No Lemna sp found/ very small population of Lemnasp,P2- Because water is contaminated with soap/detergentcontain alkali,P3- Which is not suitable/favourable for Lemna to grow

    (h) Classify the biotic and abiotic factors from the listprovided below.

    Able to classify all 4 pairs of the abiotic and biotic factors inecosystemSample answer

    Abiotic factors Biotic factors

    Humidity Decomposer

    Light intensity Parasite

    Soil texture Symbiotic organism

    Topography invertebrates

    3

    3

    3

    Humidity, light intensity, decomposer,

    parasites, symbiotic organism, soil

    texture, invertebrates, topography

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    Question 3:

    No. Questions Marks Studentnotes

    1. A group of students conducted an experiment to study the effect of light intensity on the

    population distribution of Lichen on the tree trunk. He placed a 10 cm x 10 cm transparentquadrat on the East-facing surface of the tree trunk. He counted the number of squares thatcontained half or more than half of the areas covered by the Lichen. Square with less thanhalf of the covered areas were not included.The procedures were repeated for the surfaces that face the direction of North (N), south (S)and west (W).

    Figure 1 shows how a quadrat is placed on the tree trunk. Each small square represent 1cm2.

    Figure 1

    Table 1 shows the areas covered by the Lichen on the different surface of the tree trunk.Direction/position of

    surface

    Total surface area covered by

    Lichen

    East

    60 cm2

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    South

    35 cm2

    North

    45 cm2

    West

    52 cm2

    Table 1

    10 cm

    10 cm

    10 cm

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    a)Count the total surface area of Lichen for each quadrat and recordthe answer in the spaces provided in Table 1.

    b) (i) State two different observation based on the diagram in Table 1.

    Observation 1:

    At the surface facing east (MV), the total surface area of Lichenis 60 cm2 (RV).

    Observation 2:

    At the surface facing south (MV), the total surface area of Lichenis 35 cm2 (RV).

    (ii) State the inferences from the observation in 1 (b) (i).

    Inference from observation 1:

    At the east aspect is most suitable for the growth of Lichenbecause it receives more light intensity, so higher rate ofphotosynthesis.

    Inference from observation 2:

    At the south aspect is least suitable for the growth of Lichenbecause it receives less light intensity, so lower rate ofphotosynthesis.

    (c) Complete Table 2 based on this experiment.

    Variable Method to handle the variableManipulatedvariable

    Directionfacing on thetree trunk //

    Use different direction on the tree trunk sucheast, north, south and west.

    Responding

    variable

    Total surfacearea coverageby Lichen

    Count and record the total surface areacoverage by lichen by using the quadrat.

    Constantvariable

    3

    3

    3

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    Quadrat size

    Type oforganism

    Sampling time

    Fix the size of quadrat at 10 cm X 10 cm.

    Fix the organism use in the experiment thatis Lichen

    Sampling experiment is carried out at sametime

    Table 2

    d) State the hypothesis for this experiment:

    1. The total surface area of Lichen on the tree trunk (RV) ishigher (R) when the light intensity is high (MV).

    2. When the Lichen is facing east (MV), the total surface areacovered by Lichen/population of Lichen (RV) is increase (R).

    3. The higher the light intensity (MV), the higher (R) the totalsurface area covered by Lichen / the higher the population ofLichen (RV).

    e) (i) Construct a table and record all data collected in this experiment.Your table should have the following aspect:

    Title with correct unit

    Position of direction

    Total surface area covered by Lichen

    Position of direction Total surface area covered byLichen (cm2)

    East 60South 35West 52North 45

    (ii) Use the graph paper provided to answer this question.Using the data in 1 (e) (i), draw a bar chart graph to show therelationship between the population of Lichen against the directionsfacing on the tree.

    The population of Lichen is represented by the total surface areacovered in the quadrat.

    3

    3

    3

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    (f) Based on the graph in 1 (e)(ii), explain the relationship between thepopulation distribution of Lichen and the light intensity.

    P1 Population of Lichen / Total surface area covered by Lichen

    P2 Position direction of quadrat

    P3 Degree of light intensity

    Sample answer:

    1. Population of Lichen / The total surface area covered byLichen is higher at east direction which receives highlight intensity.

    2. Population of Lichen / The total surface area covered byLichen is low at south direction which receives low lightintensity.

    3. Population of Lichen / The total surface area covered byLichen is higher at east direction than at the southdirection because Lichen at east direction receives highlight intensity so rate of photosynthesis is higher.

    (g) State the operational definition for population distribution of Lichen.

    P1 Total surface area covered by LichenP2 Size of quadratP3 Abiotic factor that influence the population distribution

    Sample answer:

    3

    3

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    1. Population distribution is defined as total surface areacovered by Lichen (P1) within the quadrat size of 10 cm x 10cm at different direction of compass (P2) which influence bythe light intensity (P3).

    (h) Lightning strike the tree and cause the tree to fall. The Lichen understudy is then exposed to direct sunlight from 7.00 a.m. to 6.00 p.m.daily.Predict what will happen to the total surface area covered by Lichenafter a month.

    Explain your prediction.

    P1: Prediction of total surface area of LichenP2: Effect of light intensityP3: Effect on the Lichen

    Sample answer:Size of total surface area covered by lichen is increase / morethan 60 cm2 because Lichen receive more sunlight / lightintensity, so more photosynthesis by Lichen and more growth toLichen.

    (i) The following is a list of biotic and abiotic factors.

    Classify these factors in the Table 3.

    Abiotic factors Biotic factorspH of water

    HumidityTemperature

    Pigeon orchidBird

    Elodeasp

    pH of water, pigeon orchid, humidity, bird,temperature, Elodeasp.

    3

    3

    3

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    Question 4 :

    No. Questions Marks Student notes

    4 An experiment was carried out to investigate the water pollution level or BOD in three different

    locations from a suspected polluted Rivers. Three water samples are collected from thesethree locations and labelled as P, Q and R as in Diagram 1.200 ml of each sample is put in a reagent bottle and added with 1 ml of 0.1% methylene bluesolution. All the bottles are kept in dark cupboard.Observations are made every minute to see the changes in the methylene blue colour.

    Diagram 1

    Table 1 shows the results of this experiment.

    Water sample P Q R

    Time taken for

    methylene

    blue solution

    become

    colourless

    Table 1

    Sample P

    Each sample is addedwith methylene blue

    solution

    Sample Q Sample R

    10 minutes 23 minutes 42 minutes

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    No. Questions Marks Student notes

    (a) Record the time taken for methylene blue solution become

    colourless in the boxes provided in Table 1.

    (b) (i) Based on Table 1, state two different observations .

    Able to stateany twoobservations correctly according to thecriteria:

    o Sampleo Time takeno Become colourless

    Sample answers:1. Time taken for methylene blue to become colourless

    for sample P is 10 minutes.

    2. Time taken for methylene blue to become colourlessfor sample R is 42 minutes

    3. Time taken for methylene blue to become colourlessfor sample Q is 23 minutes

    4. Time taken for sample P is 10 minutes that is shorterthan time taken for sample R that is 42 minutes tobecome colourless

    (ii) State the inferences which corresponds to the observations in1(b)(i).Able to make one logical inference for each observationbased on the criteria

    o Sampleo Oxygen concentrationo Duration of time for methylene blue to become

    colourless

    Sample answers:1. In sample P, oxygen concentration is low, the

    methylene blue become colourless very fast/ lesstime taken

    2. Oxygen concentration in sample R is high, themethylene blue become colourless slow/ longer

    time taken3. Oxygen concentration in sample P is lower than

    oxygen concentration in sample R, the time taken formethylene blue to become colourless is shorter.

    3

    3

    3

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    (c) Complete Table 2 based on this experiment.

    Variables How the variables areoperated

    Manipulated:

    Water sample Water sample is collectedfrom three differentlocations.

    Responding variable

    Time taken todecolourise methyleneblue

    Time taken for methyleneblue to becomecolourless is recorded byusing a stopwatch.

    Fixed variable

    Metlhylene blueconcentration / volume/volume of watersample

    0.1% of Methylene blue isused for all experiments/1 ml volume/ 200 ml ofwater sample.

    Table 2

    (d) State the hypothesis for this experiment.

    Able to state a hypothesis to show a relationship between themanipulated variable and responding variable and the

    hypothesis can be validated, base on 3 criteria: manipulated variable

    responding variable

    relationship

    Sample answer :1. The most polluted water has shortest time for

    methylene blue to become colourless.2. Sample water P is the most polluted has shortest time

    for methylene blue to become colourless.3. Sample water R is less polluted compare to water

    samples P and Q, has longest time for methylene blue

    to become colourless,

    (e) (i) Construct a table and record all the data collected in this

    experiment based on the following criteria:

    Water sample

    Time taken

    3

    3

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    Able to tabulate a table and fill in data accurately base onthree criteria:

    o Table draw with labeled column.o Sampleo Time taken with unit.

    Sample answers :

    Water Sample Time taken ( minutes)

    P 10

    Q 23

    R 42

    (f) Based on the data in 1(e) draw a bar chart of time taken formethylene blue solution become colourless against watersamples.

    Able to draw a bar chart base on criteria:

    o Correct charto Axis with correct scaleo Correct value

    (g)What is the relationship between time taken, oxygenconcentration and BOD value of water in this experiment?Able to state clearly and accurately the relationship between:

    o time takeno oxygen contento BOD value

    Sample answer:1. The shorter time taken for methylene blue to

    become colourless, less oxygen in the water andBOD value is high.

    (h) Based on the result of this experiment, state the operational

    definition for BOD

    Able to explain BOD base on experiment correctly accordingto the criteria:

    o Amount of oxygen in the water sampleo used by microorganismso shown by time taken

    Sample answer:

    3

    3

    3

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    2. BOD is amount of oxygen in the water sample thatused by microorganisms and can be shown bytime taken of methylene blue to becomecolourless.

    (i) This experiment is repeated by using water sample fromchicken farm areas. Predict the time taken for methelyne blueto become colourless.

    Able to predict the result accurately.o Expected timeo Compare to whicho Reason

    Sample answer:

    The time taken for methylene blue to become colourlessis 5 minutes, less than water sample P, because chicken

    farm water can be contaminated with chicken faeces/ orany other answer.

    (j) Arrange the water samples from the most polluted to the least

    polluted.

    Able to arrange the 3 level of polluted waterSample answer:

    Types of water Polluted

    P Most

    Q Moderate

    R Least

    Most polluted least polluted

    P Q R

    3

    3

    3

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    Question 5 :

    No. Questions Marks Student notes

    5 Transpiration is the evaporation of water from a plant to the surroundings. The rate of transpiration

    is affected by environmental factors such as temperature.

    A group of students carried out an experiment to study the effect of temperature on the rate of

    transpiration. Diagram 1 shows the set up of the apparatus. An air bubble was trapped in the

    capillary tube. The apparatus was placed in an air-conditioned room at 20oC.

    The time taken for the air bubble to move a distance of 10 cm was recorded. The experiment was

    repeated for a second time to get average readings.

    The experiment is repeated by placing the apparatus at three more different temperatures: an air-

    conditioned room at 25oC , an air-conditioned room at 30oC and in a non air-conditioned room at

    35

    o

    C.

    Table 1 shows the reading of stopwatch for air bubble to move a distance of 10 cm at differenttemperature

    Diagram 1

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    TemperatureoC

    Time taken for air bubble to move a distance of 10 cm (min)First reading Second reading Average

    Reading

    20

    25

    3228

    41

    30.0

    39 40.0

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    TemperatureSuhuoC

    Time taken for air bubble to move a distance of 10 cm (min)First reading Second reading Average

    Reading

    30

    35

    No. Questions Marks Student notes

    (a) Record the time taken for the air bubbles to move a distance

    of 10 cm and average reading in Table 1.

    (b) (i) Based on Table 1, state two different observations .

    1. When temperature is 20oC, the average time taken for

    air bubble to move a distance of 10 cm is 40 minutes

    2. When temperature is 35oC , the average time taken for

    air bubble to move a distance of 10 cm is 10 minutes.

    20 2020.0

    11 9 10.0

    3

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    3. When temperature is 20oC ,the average time taken

    for air bubble to move a distance of 10 cm is

    longer than the average time taken when

    temperature is 35oC

    (ii) State the inferences which corresponds to the observations

    in 1(b)(i).

    1. (When temperature is low) , the amount of water lost

    from the leaf is low(P1). So the rate of transpiration is

    low (P2)

    2. (When temperature is high) , the amount of water lost

    from the leaf is high(p1). So the rate of transpiration is

    high (P2)

    3. When the temperature is higher/lower, the amount ofwater lost from the leaf is higher/lower. So the rate of

    transpiration is higher/lower when the temperature is

    higher/lower

    (c) Complete Table 2 based on this experiment.

    Variable Method to handle the variable

    Manipulated Variable

    Temperature Place the

    apparatus/potometer at

    different temperature / 20

    o

    C,25 oC, 30 oC and 35 oC

    Responding Variable

    1.Rate of transpiration

    2. Time taken for air

    bubble to move a

    distance of 10 cm

    1.Calculate and record the

    rate of transpiration by using

    formula : Distance / time

    2. Record the time taken for

    air bubble to move a distance

    of 10 cm by using stopwatch

    Constant Variable

    1.Type of plant

    2.Distance travelled by

    air bubble

    1.Use the same plant

    2.Fix the distance travelled by

    air bubble at 10cm

    3

    3

    3

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    (d) State the hypothesis for this experiment.

    Able to make a hypothesis based on the following aspects

    P1 : MV- Temperature

    P2 : RV rate of transpirationH : Relationship (Higher. Higher)

    Sample Answer:

    1. The higher the temperature, the higher the rate of

    transpiration//vice versa

    (e) (i) Construct a table and record all the data collected in this

    experiment.

    Your table should have the following aspects:

    - Temperature

    - Average time taken for air bubbles to move a

    distance of 10 cm .- Rate of transpiration

    Rate of transpiration = Distance

    Time

    Able to construct a table based on the following aspects

    1. Title with correct unit - 1 mark

    2. Data - 1 mark

    3. Rate of transpiration - 1 mark

    Sample Answer

    TemperatureSuhuoC

    Average time taken for air by

    air bubble to move a

    distance of 10 cm (min)

    Rate of

    transpiration

    cm/min

    20 40.0 0.25

    25 30.0 0.33

    30 20.0 0.5

    35 10.0 1.0

    (e) (ii) Using the data in 1(e)(i), draw the graph of the rate of

    transpiration against the temperature

    Able to draw the graph correctly

    Axes : Uniform scales on both horizontal and vertical axis

    with correct unit 1 mark

    Points : All points plotted correctly - 1 mark

    Curve : smooth without touching the axes - 1 mark3

    3

    3

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    (f) Based on the graph in 1(e)(ii), explain the relationship

    between the rate of transpiration and temperature.

    Able to explain the relationship between the rate of

    transpiration and temperature based on the following

    aspects.

    P1 State the relationship

    P2 kinetic energy of water

    P3 evaporation

    Sample Answer

    When the temperature increases, the rate of

    transpiration increases. When the temperature

    increases, kinetic energy of water molecules (in the leaf)

    increases, causes the rate of evaporation increase.

    (g) Based on the result of this experiment, state the operational

    definition for process of transpiration.

    Able to define operationally the process of transpiration

    based on the following aspects:

    P1 water loss from plant at different places

    P2 Air bubble in capillary tube move at 10 cm

    P3 The rate of transpiration is influenced by temperature

    Sample Answer

    Transpiration is a process where water is lost from theplant when it is placed at different temperature which

    causes the air bubble in capillary tube move a distance

    of 10 cm. The rate of transpiration is influenced by the

    temperature.

    (h) If the surface of the leaves of a plant at temperature of 35 oC

    are covered with vaselin, predict the time taken for air

    bubble to move a distance of 10 cm. Explain your prediction.

    Able to predict the outcome of the experiment based on the

    following aspectsP1 : Correct prediction

    P2 : Effect

    P3 : Reason

    Sample Answer

    Time taken for air to move a distance of 10 cm is more

    than 10 minutes. Rate of transpiration decreases

    because vaselin covered the stomata/stomata closed

    3

    3

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    (i)The following list is a factor that affecting transpiration.

    Classify the factors into two group in Table 3.

    Environmental factor Morphology factors

    1. Relative humidity

    2. Air movement

    3. Light intensity

    1. Cuticle

    2. Stomata

    Table 3

    Relative humidity Kelembapan relatif

    cuticle kutikelair movement pergerakan angin

    stomata stomata

    light intensity keamatan cahaya

    3

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    1

    No. Questions Marks Student notes1. Diagram 2 shows three types of fruits.

    Plan a laboratory experiment to investigate the percentage ofvitamin C content in each fruit. DCPIP (dichlorophenolindophenol)0.1% solution is used to test the presence of vitamin C in the fruitjuices.

    You can use the common chemicals and science apparatus thatcan be found in the laboratory. The planning of your experimentmust include the following aspects:

    Problem statement

    Hypothesis

    Variables Apparatus and materials

    Procedures

    How data is communicated

    [17 marks]

    Problem statement:

    Able to state the problem statement of the experiment correctlythat included criteria:

    Manipulated variable

    Responding variable

    Relation in question form and question mark (?)

    Sample Answer1. What is the percentage / concentration of vitamin C in

    watermelon, orange and papaya?2. Which fruit juice has the highest percentage /

    concentration vitamin C?3. Does the percentage/concentration of vitamin C in

    watermelon, orange and papaya are same?4. Does orange juice contain higher percentage /

    concentration vitamin C than papaya and water melon?

    Diagram 2

    Papaya[Betik]

    Orange[Oren]

    Water melon[Tembikai]

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    Hypothesis:

    Able to write a suitable hypothesis correctly base on the 3 criteria: Manipulated variable Responding variable

    Relationship of the variables

    Sample Answer1. Orange juice has the highest percentage / concentration

    of vitamin C compare to other fruits.2. Watermelon has the lowest content of vitamin C than

    orange juice and papaya juice.

    Variables:Able to identify all the three variables correctly

    Sample AnswerManipulated variable : type of fruit juice

    Responding variable : percentage of vitamin C

    Fixed variable : concentration of DCPIP / volumeof DCPIP / concentration ofascorbic acid

    Material and Apparatus:

    Able to state material and apparatus:Compulsory to use in : MV, RV and FV

    Materials : 1. DCPIP solution(M) 2. 0.1 % Ascorbic Acid

    3. Fruit juices / watermelon juice/orangejuice/papaya juice

    Apparatus : 1. Beakers(A) 2. Measuring cylinder

    3. Syringe with needle4. Specimen tube with cap

    Procedures:

    Able to write five procedures P1. P2, P3, P4 and P5 correctly.

    P1 : Steps to set up the apparatus ( at least three P1)P2 : Steps to handle the fixed variable ( one P2)P3 : Steps to handle the manipulated variable (one P3)P4 : Steps to record the responding variable (one P4)P5 : Precautionary steps / steps taken to get accurate results /

    readings (one P5)

    1. Three specimen tubes are labeled as A1, A2 and A3.

    2. Filled each specimen tubes with 1 ml of 0.1% DCPIPsolution

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    3. Use a syringe to take 10 ml of 0.1 % ascorbic acid

    4. Place the syringe needle into the DCPIP solution andrelease the ascorbic acid drop by drop into the DCPIPsolution in A1

    5. Observe the change of DCPIP colour and stop releasingthe ascorbic acid when the DCPIP become colourless

    6. Record the volume of ascorbic acid used to dicolourisedthe DCPIP using syringe.

    7. Repeat step 3 6 for A2 and A3 and calculate the averagevolume.

    8. Repeat the step 2 7 by using fruit juices to replace the

    0.1 % ascorbic acid.

    9. Do not shake the bottle to prevent from DCPIP is oxidized.

    10. Record the volume of watermelon juice, papaya juiceand orange juice that discolourised the DCPIP in thetable and calculate the average volume

    11. Calculate the percentage/concentration of vitamin C ineach of the fruit juice using the formula below:

    Results:

    Able to draw a complete table to record the relevant data base onthe 3 criteria:

    Type of juices

    Juice volume (ml //cm3) Percentage of ascorbic acid in juices (%)

    Percentage of vitamin C = volume of 0.1% ascorbic acid X 0.1 %

    in fruit juice volume of fruit juice

    Concentration of vit. C = volume of 0.1% ascorbic acid X 1.0mgcm-1

    in fruit juice volume of fruit juice

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    Sample Answer

    Type of juices Volume of Juiceto decolourise 1

    ml DCPIP(cm3)

    Per