PERFECT SCORE 2011 BIOLOGY - Seven Heavenzzat.weebly.com/uploads/1/4/6/7/1467566/perfect-score-bio-2011-ans.pdf · Label Q, R and S in Diagram 1.1. [3 marks] 3m (b) In the space below

PERFECT SCORE BIOLOGY 2011

1

BAHAGIAN SEKOLAH BERASRAMA PENUH DAN

SEKOLAH KLUSTER

KEMENTERIAN PELAJARAN MALAYSIA

PERFECT SCORE BIOLOGY

2011 Teacher’s Module


2

Paper 2 – Section A:

Structural Questions Marks Student notes

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 I and metaphase II.

[2 marks]

2m

Metaphase I Metaphase II

Q: Chromosome/ chromatid

R: Centromere

S: Nuclear

membrane

Process X


3

(c) Explain how the process X involves in producing variation in organisms. P1 - During prophase Meiosis 1, crossing over occurs between

homologous chromosomes P2 - resulting a new genetic combination. P3- It is also producing the exchange of genetic material between paternal chromosome and maternal chromosome // between homologous chromosomes.

[3 marks]

Any 3=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 body (n)

Secondary oocyte (n)

Second polar body (n)

Diagram 1.2


4

2.

Diagram 2 shows the digested food is being carried from small intestine to 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=1m 1E=1m

Diagram 2


5

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 complex compounds 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


6

(e) Explain what will happen to a person if his liver receives insufficient insulin from 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 efficient gases exchange. F1 : one cell thick P1 : gas doesn’t 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 alveoli P3 : increase the diffusion of respiratory gases F4 : inner surface of alveoli are moist P4 : oxygen dissolve in the film of water

1F=1m 1P=1m

Bronchus P

Blood capillary

Cells

Alveolus


7

Any F with correspond P

(b)(i) Name P Trachea

1m

(b)(ii) Explain the role of P to prevent dirt and bacteria from entering the alveolus. F1 : secrete sticky fluid/mucus P1 : traps dirt / bacteria that are breathed in. F2 : cells in P have cilia / tiny hair-like structures P2 : sweeping the mucus out towards the mouth.

Any F with correspond P

1F=1m 1P=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 blood capillary 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 higher

compared to the partial pressure of oxygen in the blood capillary P: ( therefore,) oxygen diffuses across the surface of the alveolus to

the blood.

1m 1m

(d) A hard mass of food passing down the oesophagus might indirectly interrupt 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 closely packed C-shaped) ring of cartilage P : cartilage keep the trachea open// prevent from collapse

1m 1m


8

Structural Questions Marks Student notes

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 tissue and vascular tissue // consists of various types of tissues combined 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: _______________

__________________

____


9

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 the sieve tube P3:hence dissolve organic substances/sucrose/ cannot be transported (from leaves to the storage organ/other part of plant)

[2 marks]

Any 2 = 2m

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

S: Sieve tube cell

_____________

R: Companion

ccecell

____________

Diagram 4.2

Diagram 4.3(a) Diagram 4.3(b)


10

(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 axon

terminal P2 - Stimulates (synaptic) vesicles to move towards (and bind 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 signals

in which travels along the Q / neurone [3 marks]

Any 3= 3m

Diagram 5

P Q

R

X


11

(c) Describe the pathway in the reflex action involved the three neurons above. P1 - Receptor detect the stimulus and triggers the nerve impulses

P2 – (The nerve impulses) are transmitted along neurone P to

neurone 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 is involved 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 of muscle (to run away)

[3 marks]

Any 3= 3m

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


12

6. Diagram 6 .1 shows a part of human brain, kidney and a nephron which involve 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


13

(c) If individual Y eating a very salty food, the adrenal gland will release less hormone Q . What is hormone Q and explain how hormone Q involved in the mechanism to restore the osmotic pressure of the blood back to normal levels. P1: Hormon Q is aldosterone hormone

E1:(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 to collecting 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 poisons or infectious diseases which can lead to kidney failure. Diagram 6.2 shows a haemodialysis machine which can save a kidney patient’s 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


14

7. Diagram 7 shows the changes of four type of hormone which control the menstrual 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 and the 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]

Any 3=3m

Diagram 7

L M

N


15

3(d) If fertilisation occurred, the level of hormones S is maintained and the pregnancy 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 be carried out to overcome this in fertility problem. Discuss the appropriate technique 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 a flower during fertilization.

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

4m

Diagram 8


16

(b)(i) In the space below, draw a section through the ovule, showing all the cells 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 the

triploid zygote to form endosperm [2 marks]

2m

(c)(i) A farmer spraying hormone X on the tomato’s flower to produce mature tomato. 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 develop

into fleshy fruits P2: without pollination and fertilization P3: the process is called partenocarpy P4: where the tomato fruits are seedless

Any 2= 2 m


17

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) DNA ii) Bases iii) 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


18

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 two strands 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


19

Diagram 9.3 shows chromosomes with the alleles for crosses between two varieties of pea plants: yellow and smooth seed variety, with a green and 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 F1

generation.

[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


20

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 blood clotting 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 of heterozygous alleles XHXh

(to produce a male haemophiliac offspring with genotype XhY)

[2 marks]

2m

L K

Q S R

Diagram 10


21

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

Female x Male Phenotype : 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 different genotypes. Explain why? P1 :One of them inherits XH from her father and XH from her mother

P2: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 is the probability of obtaining children that are haemophiliac?. Explain your answer by constructing a genetic diagram for this inheritance. Parental generation :

Phenotype Haemophiliac female x normal male

Genotype XhXh XHY meiosis

Gametes Xh XH Y Fertilisation

Offspring : 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


22

(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 Y chromosomes // male only has one X chromosome

P2: Sex-linked genes are absent in the Y chromosomes//only

presence in X chromosome. P3 : Therefore, the presence of one sex-linked gene in the X

chromosome will affect the male as compared to the female who needs recessive genes to be present in both X chromosomes for her to be affected.

[2 marks]

Any2 =

2m


1


SEKOLAH KECEMERLANGAN




Paper 2

Section B


2

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


3

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 amino

acids

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 and

stored

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


4

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

Salted plum Potato chips Prawn crackers


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


5

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 the factors that cause food spoilage.

[ 10 marks]

Sample answer:

Concept : Food can be preserved by destroying the microorganism present in the food // by stopping the activities of the microorganism F1: Cooking-.high temperature kill the microorganisms P1: denature the enzyme that cause the breakdown of food F2: Treating food with sugar/salt P2: causes the microorganism to lose water due to osmosis F3: Adding vinegar will reduced the pH P3 that prevent microorganism from growing F4: Fermentation of fruit juices and other food by adding yeast P4: high concentration of alcohol prevent the microorganism from growing

1

1

1

1

1

1

1

1

1


6

F5: Dry under hot sun (meat/fish/fruits) P5: removes water from food – dehydrated F6: Ultravoilets rays P6: kills microorganism F7: Pasteurisation – destroy bacteria which cause tuberculosis and typhoid P7: (technique) -Food is heated to 630C for 30 minutes / 720C for 15 seconds followed by rapid cooling to 10 0C P7.1: (Pasteurisation) retains the natural flavour and nutrients F8: Canning – uses heat sterilization to kill microorganisms and their spores P8 (technique) -.Food is packed in cans, steamed at high temperature and pressure to drive out air P8.1: the vaccum created within the cans prevent growth of microorganism F9: Refrigeration P9: food stored at temperature below 00C prevent growth/germination of microorganism P9.1: food remain fresh for a long period of time Any ten : F + P correctly

1

1

1

1

1

1

1

1

1

1

1

1

1

10

TOTAL MARKS

20


7


3

Diagram 3 shows roots of plants found in mangrove swamp.

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

[10 marks] Sample Answer: F1: Root adaptations increase stability of mangrove trees in the soft 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 of Avicennia adding to the structural stability of the tree. P3: Other species of mangrove trees grow at higher elevations, in drier soils, do not require specialized root structures.

1 1 1 1

Pneumatophore


8

F2: The ability to exclude salts occurs through filtration at the surface of the root. P1: The cell sap is hypertonic to sea water, the water able to diffuse 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-poor sediments through specialized root structures / pneumatophore. P1: these spaces in soil fill with water, containing lower oxygen levels than air.

P2: having well-developed aerial roots or pneumotophores gases exchange

P3 : example Avicennia

P4 : Red mangroves / Rhizophora have prop/stilt roots extending from the trunk and adventitious roots from the branches able to absorb more water and mineral. P5 knee root and buttress root has lenticels that allow air into the roots.

Any 10 points.

1

1 1 1 1 1 1 1 1 1


9


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 that can be absorbed directly by plants and explain how a deficiency of substances K in the soil affect the growth of the plants.

[ 4 marks]

Sample answer:

P1 : (Two form of nitrogenous compound that can be absorbed directly by plants) are nitrate ions and ammonium ions. P2 : substances K is nitrate P3 : Substance K is used in synthesis of protein in plant or animal

1 1 1

Nitrogen in the atmosphere

Substance K

Nitrogen fixation by microorganisms in plant

Nitrites

Ammonium compounds

Nitrogen compounds in

plants

Nitrogen compounds in

animals

Diagram 4


10

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 nitrogen cycle.

[6 marks]

Sample answer:

P1 : Rhizobium sp. (in root nodule of legume plant)// Nostoc sp.//Azotobacter sp. P2 : fix the Nitrogen from atmosphere into nitrate/ substance K. P3 : (Nitrate/K substance) is absorp by roots of plants and converted into protein. P4 : (when the plant /animal die), protein in plant/ animal is decomposed 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]

1 1 1 1 1 1 1 1


11


5.

(a)

Diagram 5 shows the eutrophication process that occurs to a lake due 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 contains nitrates/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 are

supplied with extra nitrate/(phosphate) P4: (they may grow so much) that they completely cover the

water.

1

1

1

1


12

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 eventually die.

P9: Their remains are good source of food bacteria //bacteria

decomposed the dead plant rapidly//bacteria breed rapidly

P10: The large population of bacteria respires using up more

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

organism P12: BOD increased

P13: Those fish which need oxygen have to move other areas or die

Any 10

1

1

1

1

1

1

1

1

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 which provide Nitrate/nutrient for microbe.

P2: Remove harmful bacteria/most of the nutrient which

could course eutrophication before it is released into the rivers.

P3: When sewage has been treated, the water in it can be

used again//sewage treatment enables water to be recycled.

P4: Microorganisms used in sewage treatment.

Any 3

1 1 1 1


13

(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 can easily be leached out of the soil.

4. They release their nutrients gradually (over a long period

of time) giving crops time to absorb them efficiently.

Any 3

1

1

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 impact of 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) eroded soil deposited blocking the flow of water.

P6: The water levels in rivers rise rapidly causing flood to

occur.

Any 4

1 1

1 1

1 1 1

20


14

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 the activities to the mankind and their surroundings. Suggest the ways to overcome this problem.

[12 marks] Sample Answer P1: the problem is green house effect

P2: the activities produce green house gases such as carbon dioxide, methane and nitrogen dioxide.

P3: The gases accumulate and forms a layer at the atmospheric surface

P4: Solar radiation penetrate earth atmosphere and warm the

earth surface.

P5: Part of the heat energy is reflected back by earth surface to 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.

1

1

1

1

1

1

1


15

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 will

increase the rate of photosynthesis or agriculture yield. P2: Increase in earth temperature / global warming will

accelerate evaporation of water and reduce soil humidity.

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

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

P5: Yield of crop / domestic animal reduced

P6: Mass destruction of animal habitat and cause the animal emigration/ 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 using

the 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 and

caring the environment through mass media and environmental campaigns.

P8: planting more tree

Any 2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1


16

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

Diagram 7.1 Describe how the ozone layer becomes thinner. Discuss its effects on humans and the environment and suggest the ways to solve these problems.

[10 marks ]


7 (a)

Sample Answer

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

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

Ultraviolet radiation strikes a CFC molecule

1

1

1

Solar radiation Sinaran suria

Stratosphere Stratosfera

Ozone layer Lapisan ozon

Harmful ultraviolet radiation Sinaran ultra ungu

berbahaya

Troposphere Trofosfera

Earth Bumi


17


cause the chlorine atom to break away

Then the chlorine atom collides with an ozone molecule and combines with an oxygen atom to form chlorine monoxide and oxygen

Then the free atom of oxygen collides with the chlorine monoxide, the two oxygen atoms form a molecule of oxygen

The chlorine atom is released and free to destroy more ozone molecules

The chlorine produced re-enters the cycle

When the ozone layer becomes thinner, more ultraviolet radiation reaches the Earth

The effect of excessive ultraviolet radiation on human

reduction of the body’s immune system

skin cancer

cataract of the eye

Effect on plants

reduction of the rate of growth therefore reducing crop yields

Effect on aquatic organism

death of plankton, reduce food supply to aquatic organism, fisherman’s catch is reduced.

Steps to overcome this problem

Reduce or stop using CFC or chlorine-based products

Replace CFC with HCFC

Use wrapping papers instead of polystyrene boxes

Patch up the holes in the ozone layer by firing frozen ozone balls into the atmosphere

1

1

1

1

1

1

1

1

1

1

1

1

1

1

Max 5 Max 3 Max 2 TOTAL 10marks


18

7 (b) Diagram 7.2 shows a phenomenon X that occurs from air pollution. Describe the

formation and the effects of the phenomenon on agriculture and aquatic ecosystem.

[10 marks]


9(b)

Able to name the phenomenom X

Sample Answer

F1 : X is acid rain

P1 : combustion of fossil fuels in power station/factories/domestic boilers

P2 : produce sulphur dioxide

P3 : and oxide of nitrogen

P4 : both gases combine with water vapour

P5 : form sulphuric acid and nitric acid

P6 : fall to the Earth with pH less than 5.0

1 1 1 1 1 1 1

Max 6

Diagram 7.2


19

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

1 1

1

1

4

Total 10 marks


20

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 doesn’t experience any response to hot object.


21


8 (a)

Sample Answer

• When an impulses arrives in the axon terminal

• Stimulates (synaptic) vesicles to move towards and bind with the presynaptic membrane • The vesicles fuse / release the neurotransmitter into the synapse • The neurotransmitter molecules across the synapse to the dendrite of another neurone • Stimulated to trigger a new impulses which travels along the neurone

1

1

1

1

1

Max 4

(b)

SampleAnswer

F1 - P is afferent neurone which transmits nerve

impulse from the receptors to the interneurone. P1 - If P damaged, impulse from receptor cannot be transfered to the interneurone. P2 - (As a result), individual A cannot feel any pain P2 - R is efferent neurone which transmits nerve impulse from interneurone to the effector P1 - If R damaged, impulse from interneurone cannot be transfered to the effector P2 - (As a result), individual A cannot withdraw the finger // pull the hand away from the pointed needle

1

1

1

1

1

1

Max 6


22

9

Mr. Q is married to Mrs. Q for more than 10 years but did not 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 they decided not to have any more kids. Explain how reproduction technologies able to help these two families.

[10 marks]

Sample Answer F1: Mr Q have problem with infertility, that is low sperm 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 further development.

or F2: artificial insemination P3: sperms are collected until the number of sperms are enough P4: sperms are injected into the fallopian tube of Mrs Q P5: during ovulation

Any 5 points – 5 marks

1

1

1

1

1

1

1

1

1

1


23

Sample Answer

F3: For Mr V family the problem is to control the birth rate/ stop pregnancy P1: Mrs. V have high risk if pregnant due to high blood pressure and heart problem P2: use contraceptive pills, to stop ovum development P3: use condom during copulation, prevent sperm from reaching uterus P4: Tubal ligation or tubectomy – the fallopian tube is tied/cut P5 : blocking the ovum from entering the uterus/ Prevent sperm from reaching the ovum

Any 5 points – 5 marks

1

1

1

1

1

1


24

No Essay Questions Mark Student notes

10.

The variation of ABO blood group determined by three different alleles, but an individual carry only two of the three allele. With schematic diagram , explain the possibilities of the blood group and genotypes of the offspring if the father’s blood group is A and the mother’s blood group is AB.

[10 marks] Sample Answer

Schematic diagram:

1

1

1

1

1

1

1

1

1

1

Father Mother X Parent :

IB

Phenotype F1: Blood group A

Blood group B

Parent

genotype : X IA I

A

I

A I

B

Parent

genotype :

Meiosis

Gamete : IA

IA

Fertilisation

Genotype F1: IA I

A

IA I

B

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

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

IB

IB I

o

Parent

genotype : X IA I

o

IA I

B

Parent

genotype :

Meiosis

Gamete : IA

IA

Fertilisation

Genotype F1: IA I

A

IA I

B

Io

IA I

o


25

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 of offspring 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 of offspring is 50% blood group A , 25% is blood group B and 25% blood group B //refer to schematic diagram

[Total : 10 marks]

10


26


11. (a).

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

Diagram 11.1 Diagram 11.2 Based on the biology knowledge, identify the variation and explain 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 species P4: Both type of variation are caused by genetic factor Differences: P5: Height is continuous variation while fingerprints is

discontinuous variation P6: Graf distribution of continuous variation shows a

normal distribution while Graf distribution of discontinuous variation shows a discrete distribution.

P7: The characters of continuous variation are quantitative

/ can be measured and graded from one extreme to the other while the characters of discontinuous variation are qualitative / cannot be measured and graded from one extreme to the other.

1 1 1 1

1

1 1


27

P8: Continuous variation exhibits a spectrum of

phenotypes with intermediate character while discontinuous variation exhibits a few distinctive phenotypes with no intermediate character.

P9: Continuous variation influenced by environmental

factors while Discontinuous variation is not influenced by environmental factors.

P10: In continuous variation two or more genes control the

same character while In discontinuous variation single genes determines the differences in the traits of the character.

P11: In continuous variation the phenotype is usually

controlled by many pairs of alleles while in discontinuous variation the phenotype is usually controlled by a pair of alleles.

Any 10

1 1 1 1

(b).

Diagram 11.3 shows the variants P, Q and R of a species of fish.

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

1 1 1


28

between the non-sister chromatid. P4: the exchange of materials between the chromatids

results in new combination of genes P5: By independent assortment P6: during metaphase I of meiosis, homologous

chromosomes arrange themselves randomly at the equator

P7: the random arrangement and separation of each

homologous 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 that

can fertilise any of the ova which also has unique combination of genes.

P12: the fertilisation of gametes produced zygote/offspring

which has various of variation. F2: by environmental factors. P13: environmental factors that cause variation included

abiotic factors P14: such as light intensity / temperature / water / humidity

/ nutrients / soil fertility P15: these factor affect the growth rate of the organism.

Any 10

1 1 1 1 1 1 1 1

1 1 1 1 1 1

20


29


12 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 and

land 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 in the 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

1 1 1 1 1 1 1 1 1 1 1


1


SEKOLAH KECEMERLANGAN




PAPER 3

QUESTION 1


2

Question 1 :

No. Questions Marks Student notes

1 A group of students carried out an experiment to study the effect of the concentration of

glucose 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


3

Percentage concentration of glucose / %

Final height of coloured liquid in the manometer after 10 minutes /cm

10

15

20

3

5

8


4


(a) Complete Table 1.2 by recording the height of coloured liquid 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 final

height of coloured liquid after 10 min is 8 cm

(ii) State the inference which corresponds to the observation in 1(b(i). 1. Low activity of yeast in lower concentration of

glucose, less carbon dioxide is released

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

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

Variable

Method to handle the variable

Manipulated variable: The concentration of glucose

Use different concentration of nutrients/glucose

Responding variable: Height of coloured liquid// The rate of yeast activity

Record the height of coloured liquid by using a metre rule // Calculate rate of yeast respiration using formula: = height of coloured liquid time

Controlled variable : Volume of yeast suspension /mass of yeast/volume of glucose/pH/light intensity/temperature/time taken

Fix the volume of 100cm3 of yeast suspension /the mass of 4 g of yeast /pH5 /light intensity at distance of 50cm /temperature at room temperature/time taken for 10 minutes

3

3

3


5

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

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

Percentage concentration of glucose

Height of coloured liquid

The rate of the activity of yeast

Percentage concentration of glucose (%)


(cm)

The rate of the activity 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 between the rate of the activity of yeast and the concentration of glucose. Explain your answer. When the concentration of glucose increases/decreases, the rate of yeast activity increases/decreases, more substrate for yeast to use for energy production, more yeast reproduced.

(f) Based on the experiment, define anaerobic respiration in yeast operationally. An anaerobic respiration is when yeast using glucose to produce gas that causes the rising of liquid in manometer tube and the process is affected by concentration of glucose

3

3

3

3

3


6

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

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

(h) The following list is part of the apparatus and material used in this experiment. Complete Table 1.3 by matching each variable with the apparatus and material used in the experiment.

Variables

Apparatus

Material

Manipulated

Measuring cylinder

Glucose

Responding

Coloured liquid

Metre ruler

Controlled

electronic balance

Yeast

Yeast, metre rule, coloured liquid, electronic balance, glucose solution, measuring cylinder

3

3


7

Question 2 :


2 Lemna minor is a species of free-floating aquatic plants from the duckweed family Lemnaceae. The plants grow mainly by vegetative reproduction: two daughter plants bud off from the adult plant. An experiment is carried out to investigate the effect of abiotic factor such as pH on Lemna sp. growth. Experiment is done under controlled conditions: 12 hours a day light exposure and using the same Knop’s solution. Petri dish is filled with 20 ml Knop’s solution with different pH value and 5 Lemna sp. each. The Knop’s solution is treated by adding acid or alkali to achieve the pH value needed. ** Knop’s 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 .

pH value

Petri dish

Number of Lemna sp.

3

4

Lemna minor

Petri dish

Knop’s solution


8

5

5

7

8

9

11

11

5


9

13

1

Table 1.1


(a) State the number of Lemna sp. in the spaces provided in Table 1.1

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

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

pH ( Manipulated Variable)

Number of Lemna sp (Responding Variable) Sample answers: 1. At pH 2 (Knop solution), the number of Lemna sp is 4 2. At pH 8 (Knop solution), the number of Lemna sp is 11 3. At pH 12 ( Knop solution), the number of Lemna sp is

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

grow is less than at pH 2/4/6/8/10 5. 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 observations in 1(b)(i). Able to make one logical inference for each observation based on the criteria

suitable abiotic factor

Favourable for Lemna sp growth Sample answers:

3

3

3


10

1. Strong acidic condition is not favorable for Lemna growth.

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

3. Strong alkaline is not favorable for Lemna growth. 4. Strong alkaline condition is the least favorable for

Lemna growth compare with other conditions. 5. Neutral/Slight alkaline condition is the best/moss

favorable 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 in the experiment and how the variables are operated.

Variables How the variables are operated

Manipulated:

pH

Add/Use acid or alkali to the Knop solution to get different pH condition// Use pH solution: pH2, pH4, pH6, pH8, pH10,pH12 // change/alter the medium condition

Responding:

Number of Lemna sp

Count and record the number of Lemna sp. plants after 7 days.

Fixed:

Light exposure /

Volume of Knop solution

Fix 12 hours light exposure every day /

Maintain the volume at 20ml

(d) State the hypothesis for this experiment.

Able to state a hypothesis to show a relationship between the manipulated variable and responding variable and the hypothesis can be validated, based on 3 criteria:

manipulated variable

responding variable

relationship Sample answer :

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

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

3

3


11

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 the experiment. Your table should contain the following title.

pH of water

Number of Lemna sp.

Able to draw and fill a table with all columns and rows labeled with complete unit Sample answers

pH of water Number of Lemna sp

2 4

4 5

6 8

8 11

10 5

12 1

(e) (ii) Plot a graph showing the number of Lemna sp against the pH in 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 relationship between the Lemna sp growth and the condition of the medium.

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

P1- Alkali, acidic or neutral (abiotic factor)

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

1. In the acidic medium the Lemna sp. growth is less, and increase when the medium become neutral but decrease when in alkali condition.

2 Lemna sp. grow very well in neutral medium and less growth rate in alkali or acidic medium

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

3

3

3


12

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

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 Lemna sp in the pond. Explain your answer.

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 Lemna sp, P2- Because water is contaminated with soap/detergent contain alkali, P3- Which is not suitable/favourable for Lemna to grow

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

Able to classify all 4 pairs of the abiotic and biotic factors in ecosystem Sample 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


13

Question 3:


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 transparent quadrat on the East-facing surface of the tree trunk. He counted the number of squares that contained half or more than half of the areas covered by the Lichen. Square with less than half 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 1 cm2.

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


14

South

35 cm2

North

45 cm2

West

52 cm2

Table 1

10 cm

10 cm

10 cm


15

a) Count the total surface area of Lichen for each quadrat and record the 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 Lichen is 60 cm2 (RV). Observation 2: At the surface facing south (MV), the total surface area of Lichen is 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 Lichen because it receives more light intensity, so higher rate of photosynthesis. Inference from observation 2: At the south aspect is least suitable for the growth of Lichen because it receives less light intensity, so lower rate of photosynthesis.

(c) Complete Table 2 based on this experiment.

Variable Method to handle the variable

Manipulated variable Direction facing on the tree trunk //

Use different direction on the tree trunk such as east, north, south and west.

Responding variable Total surface area coverage by Lichen

Count and record the total surface area coverage by lichen by using the quadrat.

Constant variable

3

3

3


16

Quadrat size Type of organism Sampling time

Fix the size of quadrat at 10 cm X 10 cm. Fix the organism use in the experiment that is Lichen Sampling experiment is carried out at same time

Table 2

d) State the hypothesis for this experiment: 1. The total surface area of Lichen on the tree trunk (RV) is

higher (R) when the light intensity is high (MV). 2. When the Lichen is facing east (MV), the total surface area

covered by Lichen/population of Lichen (RV) is increase (R). 3. The higher the light intensity (MV), the higher (R) the total

surface area covered by Lichen / the higher the population of Lichen (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 by Lichen (cm2)

East 60

South 35

West 52

North 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 the relationship between the population of Lichen against the directions facing on the tree. The population of Lichen is represented by the total surface area covered in the quadrat.

3

3

3


17

(f)

Based on the graph in 1 (e)(ii), explain the relationship between the population 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 by Lichen is higher at east direction which receives high light intensity.

2. Population of Lichen / The total surface area covered by Lichen is low at south direction which receives low light intensity.

3. Population of Lichen / The total surface area covered by Lichen is higher at east direction than at the south direction because Lichen at east direction receives high light intensity so rate of photosynthesis is higher.

(g) State the operational definition for population distribution of Lichen. P1 – Total surface area covered by Lichen P2 – Size of quadrat P3 – Abiotic factor that influence the population distribution Sample answer:

3

3


18

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

(h) Lightning strike the tree and cause the tree to fall. The Lichen under study 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 Lichen after a month. Explain your prediction. P1: Prediction of total surface area of Lichen P2: Effect of light intensity P3: Effect on the Lichen Sample answer: Size of total surface area covered by lichen is increase / more than 60 cm2 because Lichen receive more sunlight / light intensity, so more photosynthesis by Lichen and more growth to Lichen.

(i) The following is a list of biotic and abiotic factors. Classify these factors in the Table 3.

Abiotic factors Biotic factors

pH of water Humidity

Temperature

Pigeon orchid Bird

Elodea sp

pH of water, pigeon orchid, humidity, bird, temperature, Elodea sp.

3

3

3


19

Question 4 :


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 these three 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 blue solution. 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 added with methylene blue

solution

Sample Q Sample R

10 minutes

23 minutes

42 minutes


20


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

colourless in the boxes provided in Table 1.


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

o Sample o Time taken o 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 colourless

for sample R is 42 minutes 3. Time taken for methylene blue to become colourless

for sample Q is 23 minutes 4. Time taken for sample P is 10 minutes that is shorter

than time taken for sample R that is 42 minutes to become colourless

(ii) State the inferences which corresponds to the observations in 1(b)(i). Able to make one logical inference for each observation based on the criteria

o Sample o Oxygen concentration o 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/ less time taken

2. Oxygen concentration in sample R is high, the methylene blue become colourless slow/ longer time taken 3. Oxygen concentration in sample P is lower than

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

3

3

3


21


Variables How the variables are operated

Manipulated:

Water sample

Water sample is collected from three different locations.

Responding variable

Time taken to decolourise methylene blue

Time taken for methylene blue to become colourless is recorded by using a stopwatch.

Fixed variable

Metlhylene blue concentration / volume/ volume of water sample

0.1% of Methylene blue is used for all experiments/ 1 ml volume/ 200 ml of water sample.

Table 2


Able to state a hypothesis to show a relationship between the manipulated 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


22

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

o Table draw with labeled column. o Sample o 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 for methylene blue solution become colourless against water samples.

Able to draw a bar chart base on criteria:

o Correct chart o Axis with correct scale o Correct value

(g)

What is the relationship between time taken, oxygen concentration and BOD value of water in this experiment? Able to state clearly and accurately the relationship between:

o time taken o oxygen content o BOD value

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

become colourless, less oxygen in the water and BOD 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 according to the criteria:

o Amount of oxygen in the water sample o used by microorganisms o shown by time taken

Sample answer:

3

3

3


23

2. BOD is amount of oxygen in the water sample that used by microorganisms and can be shown by time taken of methylene blue to become colourless.

(i) This experiment is repeated by using water sample from chicken farm areas. Predict the time taken for methelyne blue to become colourless. Able to predict the result accurately.

o Expected time o Compare to which o Reason

Sample answer:

The time taken for methylene blue to become colourless is 5 minutes, less than water sample P, because chicken farm water can be contaminated with chicken faeces/ or any other answer.

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

polluted.

Able to arrange the 3 level of polluted water Sample answer:

Types of water Polluted

P Most

Q Moderate

R Least

Most polluted least polluted

P Q R

3

3

3


24

Question 5 :


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

35oC.

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

Diagram 1


25

Temperature oC

Time taken for air bubble to move a distance of 10 cm (min)

First reading

Second reading Average Reading

20

25

32 28

41

30.0

39 40.0


26

Temperature Suhu oC

Time taken for air bubble to move a distance of 10 cm (min)

First reading

Second reading

Average Reading

30

35


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

of 10 cm and average reading in Table 1.


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 20 20.0

11 9 10.0

3


27

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 of

water lost from the leaf is higher/lower. So the rate of

transpiration is higher/lower when the temperature is

higher/lower


Variable Method to handle the variable

Manipulated Variable

Temperature

Place the

apparatus/potometer at

different temperature / 20 oC,

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


28


Able to make a hypothesis based on the following aspects

P1 : MV- Temperature

P2 : RV – rate of transpiration

H : 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

Temperature

Suhu oC

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 mark

3

3

3


29

(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 the

plant 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 aspects

P1 : 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


30

(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 kutikel

air movement pergerakan angin

stomata stomata

light intensity keamatan cahaya

3


31

1


1. Diagram 2 shows three types of fruits.

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

You can use the common chemicals and science apparatus that can be found in the laboratory. The planning of your experiment must 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 correctly that included criteria:

Manipulated variable

Responding variable

Relation in question form and question mark (?) Sample Answer 1. 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 [Rajah 2]

Papaya [Betik]

Orange [Oren]

Water melon [Tembikai]

2

Hypothesis: Able to write a suitable hypothesis correctly base on the 3 criteria:

Manipulated variable

Responding variable

Relationship of the variables

Sample Answer 1. 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 Answer

Manipulated variable : type of fruit juice

Responding variable : percentage of vitamin C

Fixed variable : concentration of DCPIP / volume of DCPIP / concentration of ascorbic 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/orange

juice/papaya juice Apparatus : 1. Beakers (A) 2. Measuring cylinder 3. Syringe with needle 4. 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% DCPIP solution

3

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

4. Place the syringe needle into the DCPIP solution and release the ascorbic acid drop by drop into the DCPIP solution in A1

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

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

7. Repeat step 3 – 6 for A2 and A3 and calculate the average volume.

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 juice and orange juice that discolourised the DCPIP in the table and calculate the average volume

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

Results: Able to draw a complete table to record the relevant data base on the 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

4

Sample Answer

Type of juices Volume of Juice to decolourise 1

ml DCPIP (cm3)

Percentage of ascorbic acid injuices (%) // Concentration of

vitamin C in juice (mg/cm3)

0.1 % Ascorbic

Acid

Papaya juice

Orange juice

Watermelon juice


2. A group of students did a study on the function of enzymes as an

organic catalyst that can regulate and increase the rate biochemical

reactions in the cell. Enzymes are very sensitive to a change in

temperature and functions efficiently at an optimum temperature.

Design an experiment to study the effect of different temperatures

on the activity of salivary amylase on starch.

The planning of your experiment must include the following

aspects:

Statement of identified problem

Hypothesis

Statement of variables

List of materials and apparatus

Experimental procedure

Presentation of data

[17 marks]

Sample Answer:

Problem Statement:

What is the effect of different temperatures on the activity of

salivary amylase on starch?

Variables:

Manipulated variables: Temperature.

Responding variables: Rate of reaction.

Controlled variable: pH /enzyme concentration/ substrate

concentration.

5

Hypothesis:

The rate of reaction of salivary amylase on starch increases

when the temperature increase until it reaches the optimum

temperature.

Apparatus:

Test tube, a dropper, a stopwatch, beakers, a thermometer, a

white tile, a Bunsen burner, a tripod stand, test-tube rack, a

wire gauze.

Material:

1 % starch solution, salivary amylase solution, ice cubes,

distilled water, iodine test solution.

Procedure:

1. Collect 5 ml of saliva and dilute it with 5 ml of distilled water

and labeled test tube X.

2. Put 1 ml saliva in each test tube labeled P, Q, R, S and T.

3. Pour 5 ml of 1 % starch solution into each test tube labeled

P, Q, R, S and T. .

4. Immerse test tube P and X, in water bath where the

temperature is fixed at 5C and leaves it for 10 minutes to

maintain the temperature.

5. Prepare a piece of white tile and drop iodine solution on it.

6. After 5 minutes pour 2ml of saliva from test tube X to the

starch solution in test tube P. Start the stop watch.

7. Take out the mixture and drop into iodine solution on white

tile.

8. Repeat the iodine test for the mixture from test tube P at

intervals of 1 minute using stopwatch.

9. Record the time taken for the mixture from P to change in

colour of iodine solution until it does not change.

10. Repeat steps 4 to 8 for test tubes Q, R, S and T with water

temperatures fixed at 28ºC, 37ºC, 45ºC and 60ºC respectively.

11. The results are recorded in a table.

Presentation of data:

Test tube P Q R S T

Temperature(ºC) 5 28 37 45 60

Time taken for the iodine to change from yellowish brown to blue black (minute)

Rate of reaction (1/minute)

6


3. Human needs energy to maintain the body temperature at 37ºC

and to carry out daily activities.

The energy is gain from oxidation of food in body cell respiration.

Energy value is measured in Joule per gram unit.

Base on the information; design a laboratory experiment to

investigate the energy value in three types of food samples.

Your experimental design should include:

Problem statement

Variables

Hypothesis

Material and apparatus

Procedures


[17 marks]

Problem statement:

Which of the food samples contains higher energy value?

Variables:

Manipulated: type of foods

Responding: energy value// final water temperature.

Control : volume of water.

Statement of hypothesis:

Cashew nut has highest energy value compare to peanut and

dried bread

List of materials:

Dried bread, peanut, cashew nut // any suitable foods, distilled

water, plasticine.

List of apparatus:

Boiling tube, pin, matches, Bunsen burner, electronic balance,

retort stand and thermometer.

Experimental procedure:

1. Weight bread, peanut and cashew nut by electronic balance and record.

2. Fill in boiling tube with 20ml of distilled water and clamp to a retort stand.

3. Record the initial temperature of distilled water

7

4. Use pin to hold the food 5. Burn the food and place it under the boiling tube 6. Place the windshield to prevent heat lost to the

surrounding 7. Record the final temperature of distilled water by

thermometer. 8. Repeat the experiment by using different type of food//

cashew nut, bread. 9. All data is tabulated.

Presentation of data:

Type of food

Mass of food (g)

Initial temperature

(0C)

Final temperature

(0C)

Energy value (Joule/g)

Bread

Peanut

Cashew nut


4. The activity of microorganisms is affected by abiotic component in habitat. Based on the above statement, plan an experiment to study the effect of light intensity on the activity of yeast. The planning of your experiment must include the following aspect:

Problem statement

Hypothesis

Variables

List of apparatus and material

Experimental procedure or method


[17 marks]

Sample Answer: Problem statement: Does the light intensity affect the activity of yeast? What is the effect of light intensity on the activity of yeast? Hypothesis: The higher the light intensity, the lower the activity of yeast.

8

Variables:

1. Manipulated : light intensity 2. Responding : Height of coloured liquid in manometer 3. Constant : Volume of yeast suspension/ temperature.

Apparatus and material: *Yeast suspension, glucose solution, distilled water // coloured liquid (manometer) Boiling tube, test tube, glass tube, clip, retort stand, rubber stopper, (manometer tube), *lamp , rubber tubing. Procedure: 1. Set-up an apparatus as shown in the diagram.

2. The boiling tube is filled with 1 ml yeast suspension and 5

ml glucose solution.

3. The boiling tube is allowed to stand in water bath at room

temperature.

4. A light source is set up. Place the apparatus 50 cm from

the light source.

5. After 10 minutes, observe the height of coloured liquid in

manometer.

6. Record the height of coloured liquid in table provided.

7. Repeat an experiment with different distance that are 40, 30cm, 20 cm and 10 cm.

8. Make sure that all the connection of apparatus is tied

tightly.

9

Data:

Distance from light sources

(cm)


(cm) 10 20 30 40 50


5. During vigorous exercise such as running, more sweat but less urine is produced by an individual. Design an experiment to study the effect of time of vigorous exercise on the volume of urine produce by a group of students. They are given the same amount of water to drink before the exercise. Your plan must include the following aspects:

Problem statement

Statement of hypothesis

Variables

List Material and Apparatus

Experimental Procedure

How data communicated [17 marks]

Problem statement: Does time of vigorous exercise affect the volume of urine produce?

Variables: Manipulated: Time of vigorous exercise. Responding: Volume of urine produced. Control : volume of drinking water

Statement of hypothesis: The volume of urine produce decreases when the time of vigorous exercise increases.

List of materials: (5) students, drinking water List of apparatus: Stopwatch, measuring cylinder, beaker, cup.

10

Experimental procedure:

1. 5 boys at same age, gender and same body weight are selected.

2. They are asking to empty their bladder before the experiment.

3. They are given 1 litre of drinking water to drink. 4. Each of the boy is asked to do exercise as follow:

Boy A – no exercise is done Boy B – run on the spot for 5 minutes Boy C – run on the spot for 10 minutes Boy D – run on the spot for 15 minutes Boy E – run on the spot for 20 minutes

5. After the exercise they were asked to rest for a while, and then the urine produce are collected, measured and record by using measuring cylinder.

6. The data collected are recorded in a table.

Presentation of data: Data is presented in a table with the right units for - Time of exercise - Volume of urine produced

Time of exercise (minutes) Volume of urine produced

(ml)

0

5

10

15

20


6. The rate of photosynthesis is influenced by different environment factors. Base on the above situation; plan a laboratory experiment to determine the effect of light intensity on the rate of photosynthesis. The planning of your experiment must include the following aspects:

Statement of identified problem

Variables

Statement of hypothesis

List of materials and apparatus

Experimental procedure


[17 marks]

11

Problem statement: 1. Does light intensity affect the rate of photosynthesis? 2. What is the effect of different light intensity on the rate of

photosynthesis?

Variables: Manipulated: light intensity/distance of Hydrilla sp. to sources

of light. Responding: Number of gas bubbles that are release in 1

minute/ rate of photosynthesis. Control : Concentration of carbon dioxide/temperature of water.

Statement of hypothesis: The rate of photosynthesis increase when the light intensity is increase.

List of materials: Hydrilla plant, 0.3 % sodium hydrogen bicarbonate, plasticine, List of apparatus: 60 W electric bulb, 500 ml beaker, a glass funnel, test tube, stop watch, razor blade, thermometer, meter ruler

Experimental procedure: 1. The apparatus setup as diagram above. 2. The temperature of water in beaker is maintained at 28oC. 3. A few strands of Hydrilla sp. is chosen and the stem end is

cut obliquely with a sharp razor blade under water to avoid bubbles in the xylem.

4. The strands of Hydrilla sp. S is placed inside a glass filter funnel.

5. The funnel is placed upside down in a 500 ml beaker. 6. The beaker is filled with 400 ml of 0.3 % sodium

bicarbonate. 7. The beaker is placed at a distance of 50 cm from the 60 W

bulb as a light source. 8. The number of gas bubbles released in one minute are

counted and recorded in a table. This step is repeated twice.

9. Step 7 is repeated by placing the apparatus at distance 40 cm, 30 cm, 20 cm and 10 cm from the light sources.

10. The results are recorded in a table. 11. The graph of the rate of photosynthesis against the light

source is plotted.

Presentation of data: Data is presented in a table with the right units for - Distance of light sources. - Number of gas bubbles. - The rate of photosynthesis (number of bubbles/time)

12

Distance of light source (cm)

Number of gas bubbles

The rate of photosynthesis (number of bubbles/time)

50

40

30

20

10


7. Rats can be found in urban and rural area. The population size of rats in these places is different. Based on the above statement, plan experiment to estimate the number of rats in urban and rural area. The planning of your experiment must include the following aspects:

Problem statement

Hypothesis

Variables

List of apparatus and material

Experimental procedure or method


[17 marks]

Sample Answer:

Problem statement:

What is the population size of rats in urban and rural area? Hypothesis: The population size of rat in urban area is higher than in rural area. Variables: Manipulated: area

Responding: size of population

Fix: type of organism

Apparatus & material: Rat, cage trap , paint

13

Procedure: 1. Habitat of rat in urban area is selected. The area is fixed as

a research area.

2. Place 10 cage traps at strategic area.

3. A number of rats are caught (assume as X number). 4. The backs of rats are marked with white paint.

5. Make sure that the marked is small / permanent. 6. Release all the rats in 1st capture to their habitat / original

place.

7. After 3-7 days, a second capture is carried out at random in the same habitat to the 1st capture. (A second capture is assumed as Y number).

8. The number that is marked is counted (assume as Z number)

9. The size of population is calculated by using the formula

below:

Number of 1st capture x Number of 2nd capture Number of marked in second capture

10. The experiment is repeated in rural area. Data: Area Urban Rural Number of rats in 1st capture Number of rat in 2nd capture Number of marked rat in the second capture


8. Transpiration is the lost of water vapour from plants, especially from the leaves. Transpiration occurs 90 % through the stomata. The amount of water lost depends on its size, surrounding light intensity, temperature, humidity and air movement.

14

Base on the information; design an experiment to be conducted in the laboratory to investigate the effect of number of leaves on the rate of transpiration in hibiscus plant. The planning of your experiment must include the following aspects:

Problem statement

Hypothesis

Variables

Apparatus and materials

Procedures

How data is communicated

[17 marks]

Problem Statement:

1. What is the relationship between the number of leaves and the rate of transpiration in hibiscus plant?

Hypothesis: 1. As the number of leaves increase the rate of transpiration

is increase. 2. The higher the number of leaves the higher the rate of

transpiration

Variables:

Manipulated variable : number of leaves

Responding variable : distance travelled by air bubble // the rate of transpiration

Fixed variable : hibiscus / type of plant // light intensity // surrounding temperature

15

Material and Apparatus: Materials : 1. Hibiscus shoot/plant (M) 2. Water 3. Vaseline Apparatus : 1. Ruler (A) 2. Capillary tube + rubber tubing // potometer 3. Stopwatch 4. Beakers 5. Basin 6. Sharp knife / cutter 7. String / marker 8. Tissue paper

Procedure: 1. Obtain a hibiscus shoot and immediately immerse in the

water 2. By using the sharp knife, cut 4 cm of the hibiscus stem

under water. 3. Fill in the capillary tube with attached rubber tubing /

photometer with water. 4. Fix the stem of the hibiscus shoot into the rubber tubing /

photometer. 5. Make sure no air bubble trapped. 6. Immerse the capillary tube / photometer in a beaker of

water. 7. Count the number of leaves to 5 leaves.

Wipe dry the leaves with tissue paper.

8. Leave the setup for 5 minutes for the plant to adapt with the environment

9. Lift the capillary tube from the water to trap a column of air bubble // trap an air bubble in the capillary tube.

10. Tie a string on the capillary tube to mark the initial position of the air bubble.

11. Start the stopwatch

12. After 5 minute tie another string to mark final position of the air bubble.

13. Repeat step 12 to get another reading.

14. Measure both distances by using a ruler. Calculate the average distance travelled by the air bubble in 5 minute. Record in a table.

15. By using the same plant, Repeat steps 7 to 13 by

removing 1-2 leave each time. 16. Calculate the rate of transpiration:

Rate of transpiration = distance traveledl by air bubble (cm)

Time taken (minutes)

16

Data:

Number of leaves

Distance travelled by air bubble in 5 minutes (cm)

Rate of transpiration

(cm minutes-1)

1 2 Average 5 3 1


9. Competition is an interaction between organisms which live together in a habitat and compete for the same resources that are in limited supply. The competition between individuals of the same species is called an intraspecific competition. A farmer doesn’t realized his mango trees do not produced high quantity of mangoes because the mango trees are planted too close to each other. Based on the above information and situation, design a laboratory experiment to show to the farmer on how the distance between the plants can affects the growth rate of a named plant. The planning of your experiment must include the following aspects:

Problem statement

Hypothesis

Variables

List of apparatus and materials

Experimental procedures


[17 markah]

Problem statement: What is / Does the distance of seedlings affects the growth rate of maize plants?

17

Hypothesis: 1. The longer/shorter the distance of seedlings, the higher/lower the growth rate of plants 2. The longer/shorter the distance of seedlings, the higher/lower the heights of seedlings 3. The longer / shorter the distance of seedlings the heavier / lighter the mass of seedlings 4. The longer /shorter the distance of seedlings, the more / lesser number of leaves.

Variables: P1- manipulated variable: Distance of seedlings P2-responding variable: The growth rate of plants (maize /paddy / any suitable plants) / the height of seedlings / mass of seedlings/ numbers of leaves P3-fix variable:

Type of seedlings / types of soil/ amount of water/ light intensity / time taken

Procedures: 1- Three planting trays are prepared and filled with 3 kg of garden soil in each tray. 2- The trays are labeled as A, B and C with waterproof paint . 3- 30 numbers of maize seeds are planted in tray A at a distance of 10 cm intervals, 30 numbers of maize seeds in tray B at a distance of 5 cm intervals and 30 numbers of maize seeds in tray C at a distance of 2 cm intervals as shown below (not in correct scale). 2cm

γ-10cm- γ γ γ-5cm-γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ A B C

4- Each tray is watered daily with the same amount of water for

10 days 5- After 10 days, 10 maize seedlings are picked randomly from

tray A and the root of seedlings

-10cm

-

18

are washed under running water 6- The height of maize seedlings are then measured by using metre rule. The average height are calculated by using formula = the total height of seedlings/cm 10 The growth rate is calculated by using formula = the average height of seedlings/cm time taken / day 7- Step 5-6 are repeated for seedlings from tray B and C. The average height and the growth rate of seedlings in tray B and C are measured and calculated separately. 8- The result are recorded in a table.

Material and Apparatus: MATERIALS (M): Maize seeds/ Paddy seeds /any suitable seeds Water Garden soil APPARATUS (A): Metre rule Tray / Basin / Container Waterproof paint /marker pen Spade * Beam / electronic / compression balance mass of * Oven seedlings

Data:

The height of seedlings / cm The distance

of seedlings/

cm(Tray)

1 2 3 4 5 6 7 8 9 10 Average

heights of

seedlings/

cm

The growth

rate of plants

cm/day

10(A)

5(B)

2(C)

No. Questions Marks Student

notes 10. Villages P, Q and R are situated along the Galas River as in Diagram

10.1. The village folks depend heavily on the river to earn a living. The

river provides them with transport, water for cooking, drinking, washing,

etc. They also catch fish from the river.

Lately the three villages’ folks are complaining about the lower catch

}

19

from the river. They attribute this problem to a rubber factory built two

years ago at the upstream of the river. Village P, Q and R are 10km,

20km and 30km away from the rubber factory respectively.

Diagram 10.1

As an Environment Department officer, you are sent to the site to make

a thorough investigation into the matter. Your task is to investigate the

effect of the distance of the rubber factory and the villages on the

levels of water pollution.

Your investigation must include the following aspect:

o Problem statement

o Hypothesis

o Variables

o List of apparatus and materials

o Experimental procedure or method

o Presentation of data

[17 marks]

Problem Statement:

1. What is the effect of (different) distance between the rubber

factory and the village on the level of water pollution?

2. Does the (different) distance between the rubber factory and

the village affect the level of water pollution?

Hypothesis:

The longer /further the distance between the rubber factory and the villages (P1), the lower the level of (river) water pollution. (vice versa)

Variables:

Manipulated : The distance between the rubber factory and the

Rubber

factory

Village P

Village Q Village R

20

villages.

Responding : The level of water pollution // The time taken for the methylene blue solution to be decolourised.

Fixed: volume of water

Material and Apparatus:

Materials:

Water samples, methylene blue solution.

Apparatus:

Reagent bottle, stop-watch, syringe, cupboard

Procedure:

1. One water sample is taken from the river near the villages P, Q and R using dark bottles.

2. Using a measuring cylinder, 100ml of water sample from village P is transferred into the reagent bottle P.

3. Using a syringe, 1ml of 1% methylene blue solution is slowly injected, drop by drop into the water sample P.

4. The tip of the syringe is used to stir the solution slowly.

5. The bottle is capped and placed in a cupboard.

6. Do not shake the reagent bottle.

7. The stop watch / clock is started.

8. The time taken for the methylene blue to decolourised is taken (using stopwatch) and recorded down.

9. Steps 3 to 11 are repeated, replacing the water sample from village P with that of villages Q and R.

Data:

Water

sample

Time taken for methylene

blue solution to turn

colourless /decolourise (min)

The level of

water

pollution

P

Q

R

Documents

PERFECT SCORE 2011 BIOLOGY - Seven Heavenzzat.weebly.com/uploads/1/4/6/7/1467566/perfect-score-bio-2011-ans.pdf · Label Q, R and S in Diagram 1.1. [3 marks] 3m (b) In the space below