Download - Perfect Score Bio 2011 Ans
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PERFECT SCORE BIOLOGY 2011
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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
11
11
1
1
1
1
1
1
1
11
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
1
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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1
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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
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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
(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
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
11
1
1 4
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:
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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
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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
3
3
3
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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
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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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78
9
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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.
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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
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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.
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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.
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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
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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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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