spm bio perlis 2009 serta skema

8/9/2019 SPM Bio Perlis 2009 Serta Skema

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74/96 [Lihat sebelah4551/1 SULIT

SCHEME OF PAPER 1 / SKEMA KERTAS I

1. C 26. A

2. B 27. C

3. A 28. C

4. C 29. B

5. C 30. C

6. B 31. C

7. B 32. B

8. A 33. C

9. B 34. D

10. B 35. A

11. B 36. D

12. C 37. C

13. C 38. B

14. D 39. D

15. B 40. C

16. C 41. B

17. B 42. B

18. C 43. B

19. C 44. B

20. A 45. C

21. C 46. B

22. C 47. C

23. A 48. B

24. C 49. C

25. D 50. D


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Skema jawapan Biologi percubaan SPM 2009

www.spm.via.my

1

Structured Question:

No. 1

Item No. SuggestedAnswers: Marks

1 (a) Organelle X : Chloroplast

Organelle Y : Mitocondrion

1

1

(b) Organelle X : Absorbs sunlight for photosynthesis

Organelle Y : Site of aerobic cellular respiration

1

1

(c) i.

ii.

Organelle X : Mesophyll palisad/spongy/guard cells

Organelle Y : Sperm Cell / muscle cell

Organelle X - Photosynthesis cannot be carried out by the cells

Organelle Y Energy cannot be generated by the cells

1

1

1

1

(d) Organelle X absorbs carbon dioxide and releases oxygen while

Organelle Y absorbs oxygen and releases carbon dioxide

Oranelle X use energy to synthesise glucose while Organelle Y breaks

down glucose to produce energy

Organelle X carries out the synthesis/anabolic process whereas

Organelle Y carries out the break down/katabolic process

Organelle X forms organic compound while Organelle Y breaks down

organic compound

(Any two)

Maximum

1

1

1

1

2

(e) - Have cellulose cell wall

- often have a large central vacuole ( Any two )

- have fixed shape

- absence of centrioles

Maximum

TOTAL

1

1

1

1

2

12


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2

No. 2

Item No. SuggestedAnswers: Marks

2 (a)(i)

(ii)

P substrate/ sucrose 3- 2 marks

R enzyme / sucrase 2- 1 mark

S product / glucose / fructose 1- 0 mark

Maximum

1. act specifically

2. can be reused

3. do not destroyed after the reaction (any two)

Maximum

2

1

1

1

2

(b)(i)

(ii)

Lock and key hypothesis

- R/The enzyme/Sucrase will combine with P/substrate (at the

active site) to form a complex enzyme-substrate

- S/the products will leave the active site of R/the enzyme and

R/enzyme remain unchange

1

1

1

(c) i Soften / tenderise the meat 1

(c)ii - Rate of reaction decreases.

- This is because most enzyme protease have been denatured at

higher temperature /more than 60C

1

1

(d) - The enzyme reaction is most effective at a temperature of

30C - 40C

- The low temperature /cold water makes the enzyme inactive.

TOTAL

1

1

12


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3

No. 3

Item No. Suggested answers Marks

3 (a) (i)

(ii)

Euthrophication.

- The waste disposal contains a lot of organic matter /

- Enrichment of nutrient in freshwater

- Very suitable for algae growth.

Maximum

1

1

1

1

2

(b) - The algae layer prevents the penetration of sunlight to the

submerged plants in the pond.

- The algae uses a lot of oxygen for respiration and results in less

oxygen in the pond water. ( Any one )

Maximum

1

1

1

(c) - The decomposers / microorganisms use more oxygen to decompose

the dead plants / organic materials (due to absence of sunlight)

- result in less oxygen / increasing in BOD level

1

1

(d)(i)

(ii)

(iii)

(iv)

To determine / compare the BOD value of water from sample P and Q.

Bacteria

Protozoa / Paramecium / Amoeba

Sample P is - more polluted

- contains less oxygen.

The higher the BOD value, the higher the degree of pollution.

TOTAL

1

1

1

1

1

1

12


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4

No. 4

Item No. Suggested answers Marks

4(a) - Heart / Pump

- Blood vessels

- Blood / medium

1

1

1

(b)(i)

(ii)

P : Vena cava

Q : Pulmonary vein

R : Pulmonary artery

S : Aorta

# Accept any two correct vessels even though P, Q, R or S are not

mentioned.

Maximum

Q and S

1

1

1

1

2

2

(c)(i)

(ii)

(iii)

This is because blood flows through the heart twice in one complete

circulation

- Pulmonary circulation

- Systemic circulation

F 1- prevents the mixing of deoxygenated and oxygenated blood.

E 1- blood of high concentration of oxygen is supplied to the targeted

organs

F 2- the blood pressure is generated by the strong contraction of the left

ventricle

E 2- ensures that blood is been supplied to all parts of the body at an

appropriate level.

F 1 + E 1 = 2

Or F 2 + E 2 = 2

Maximum

TOTAL

1

1

1

1

1

1

1

2

12


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5

No. 5

Item No Suggested Answer Marks

5(a) P : Arch

Q : Whorl

R : Loop

S : Composite

Maximum 2

(b)(i)

(ii)

Genetic factor

- The exchange of genetic materials between chromatids during the

crossing over

- Independent assortment of chromosomes (Any two)

- Random fertilisation

- Mutation

Maximum

1

1

1

1

1

2

(c)(i)

(ii)

- Continuous variation, examples height/weight

- Discontinuous variation, examples - ear lobe/blood group.

Maximum

- Continuous variation influence by genetic factor and environmental

factor while Discontinuous variation influence by genetic factor

- The differences in continuous variation are not distinctive / with

intermediate characters where as the differences in discontinuous

variation in a characters are distinctive / no intermediate characters

- Any acceptable answer

Maximum

2

2

4

1

1

1

All 4 - 2 marks

3 or 2 - 1 mark

1 - 0 mark


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6

(d)

- 2 axes correctly labelled

- correct chart ( bar chart )

TOTAL

1

1

12

# Able to representthe variation shownin the form of barchart


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7

Essay Questions :

No. 6

6 (a) i. Open burning produces smoke, dirt and fine particles in the air.

Haze will form when these substances combine with the water

vapour in the atmosphere.

Total

1

1

2

ii.

iii.

Irritates the lungs, nose and eyes.

Damages the respiratory tracts and lungs.

Causes conjunctivitis, sore throats, influenza, asthma and

bronchitis.

Reduces the light intensity which may cause the decrease in the

rate of photosynthesis

subsequently reduce the yield from the crops.

Fine particles deposit on the leaves and block the gaseous

exchange which may also lower the rate of photosynthesis of the

crops.

Total

Stop open burning

Use well- designed furnaces for the complete burning of fossil fuel.

Total

1

1

1

1

1

1

6

1

1

2


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8

6 (b) F1-The flow of insecticides into the rivers, ponds and lakes may

poison the aquatic organisms and subsequently kill them.

E1-Residents that depend on the river, pond or lakes lost their food /

income sources

E2-Residents that depend on the river, pond or lakes as their water

supply will face with polluted water sources and suffer a lot of

diseases

E3-This will distrupt the food web of the ecosystem.

F2-Deforestation is carried out to develop the new residential area

and improve the infrastructures

E1- this activity may result in :

-soil erosion causes water pollution /

-flash floods cause the damages of properties /

crops and animals (any 3

-landslides cause the damages of properties / answers)

crops and animals

-severe climate changes cause decreasing in crop yield

-loss of biodiversity cause distruption of food chains / food webs

-disruption of carbon cycle leads to global warming and green

house effect

F3-Combustion of fossil fuels from the factories and vehicles releases

sulphur dioxide , nitrogen oxide , carbon dioxide, lead and fine

particles which make a large portion of the air pollutants.

E1-Sulphur dioxide and oxides of nitrogen dissolve in the rain to form

1

1

1

1

1

3

1


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9

acid rain.

E2-Smoke, dirt and fine particles lead to haze formation,

subsequently cause the diseases of eyes, nose and lungs

E3-It also reduces the light intensity which will lower down the rate of

photosynthesis and the yields of crops.

Maximum

1

1

1

10

No. 7

7 (a) Fact 1 : Oogenesis is the process of ovum formation in the ovary.

Explanation 1: It begins in the ovary of a female foetus.

E2: The primordial germ cells divide repeatedly through mitosis to form

diploid oogonia (2n).

E3: Each oogonium grows and develops into a primary oocyte (2n).

E4: - The primary oocyte undergoes meiosis I and completes meiosis I

at puberty and

- forms two haploid cells;- a secondary oocyte and polar body.

F2: During ovulation, the secondary oocyte is released from ovary.

F3: When fertilisation occurs, the secondary oocyte undergoes meiosis

II and forms an ovum (n) and a polar body (n).

E5: - The first polar body also undergoes meiosis II

- to form another two haploid polar bodies.

F4: All three polar bodies will eventually degenerate.

Maximum

1

1

1

1

1

1

1

1

1

1

1

10

7 (b) i.

ii.

F1-After ovulation, if the ovum is not fertilised, the ovum will break

followed by the thick endometrium

E1-The excess blood and the endometrium tissue together with the

unfertilised ovum will leave the uterus through the vagina

E2-This process is called menstruation

E3-It usually last from three to seven days.

Maximum

F1-After ovulation, if the ovum is fertilised, the endometrium continues

to thicken

E1-due to the increasing progesterone level produced by the corpus

luteum.

E2-The zygote continues to develop until it is implanted into the

thickened endometrium.

E3-It then develops into an embryo, then foetus and born as baby.

Maximum

1

1

1

1

3

1

1

1

1

3


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10

7(c) Functions of placenta:

Allows dissolved food substances (glucose, amino acid, mineral

salts) and oxygen to diffuse from the mothers blood into foetal

blood.

Allows metabolic waste products (urea, and carbon dioxide) to

diffuse from the foetal blood into the mothers blood.

Allows antibodies from the mothers blood to diffuse into the foetal

blood to protect the foetus from diseases.

Produces progesterone and oestrogen to maintain the thickness of

the endometrium so that the embryo is allowed to attach firmly in

the uterus.

Maximum

1

1

1

1

4


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11

No. 8

8 (a) i.

Maximum

1

4

(a) ii. Less enzyme erepsin / peptidase /maltase / sucrase / lactase /

lipase is produced.

The rate of food digestion is slow.

Glucose/ galactose/ fructose /amino acids / glycerol and fatty

acid / vitamin / mineral ions are less / slow to absorb from the villi

into the blood capillary.

The rate of absorption of digestive food is slow.

The patient becomes tired easily because less glucose is

absorbed into the blood / body.

Patient faces the problem of late healing of wound.

Because less new cells are formed due to less absorption of

amino acid / fatty acids and glycerol.

Maximum

1

1

1

1

1

1

1

6

Drawing neat and tidy

Label - 4-5 - 3 marks

- 2-3 - 2 marks

- 1 - 1 mark


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12

8(b) Cow and rabbit are herbivores while humans are omnivores.

The main food source for the cow and the rabbit is plant while the

food sources for human are plants and animals.

In the digestive systems of the cow and rabbit, there are

symbiotic bacteria or protozoa whereas there is none in the

human digestive system.

Cellulase is produced by the symbiotic bacteria or protozoa in the

digestive system of cow and rabbit, no enzyme cellulase is

produced in human.

Thus cellulose is hydrolysed or broken down by the enzyme

cellulase in the digestive system of the cow and the rabbit while

this form of hydrolysis does not occur in the human digestive

tract.

Glucose is produced from cellulose in the digestive system of the

cow and the rabbit. In human, fibre, an equivalent of cellulose is

not digested but taken to prevent constipation.

Cow has 4 stomach chambers while human and rabbit have only

1 stomach chamber each.

Only the cow regurgitates the foods that have entered the

reticulum / stomach into the mouth to be rechewed and then into

the other stomach chambers to be hydrolysed.

Rabbit has a large caecum which both human and cow do not.

The rabbit eats again its soft faeces to be redigested, while

human and cow do not.

Maximum

1

1

1

1

1

1

1

1

1

1

10


87/96


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13

No. 9

9(a) The parents genotypes are RR and rr in which RR represents

homozygous for round seed and rr represents homozygous for

wrinkled seed.

Through meiosis , gametes produced are R from parent RR and r

from parent rr.

Fertilisation produces offspring with genotype Rr.

All offsprings have round seeds due to the presence of the

dominant allele R.

Maximum

1

1

1

1

4

9 (b) Let R represents allele for red flowers / r represents allele for

white flowers. T represents allele for tall plant / t represents allele

for short plant.

Parents :

Phenotype: Tall plant Short plant

red flower X white flower

Genotype: TTRR ttrr

Gametes TR tr

Offspring TtRr

Genotype

Phenotype: Tall plant red flowers

1

1

1

1

1

1

6


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Maximum

9(c) Diagram 9.2 is a cross between a homozygous dominant for

purple flower plant and a homozygous recessive plants for white

flower plant.

Let P represents the allele for purple flowers / p represents the

allele for white flowers

Parents :

Phenotype: Purple flowers White flowers

Genotype PP pp

Gametes:

Offspring

Genotype:

Phenotype:

( any 5 )

Maximum

1

1

1

1

1

1

1

5

P p

Pp

Purple flowers


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15

Diagram 9.3 is a cross between a heterozygous purple flower

plant and a homozygous recessive white flower plant.

Let P represents the allele for purple flowers / p represent the

allele for white flowers

Parents :

Phenotype: Purple flowers x White flowers

Genotype: Pp pp

Gametes:

Offspring:

Genotype:

Pp pp

Phenotype: Purple flowers White flowers

Maximum

Total (Maximum)

1

1

1

1

1

1

1

5

10

ppP


90/96

Skema jawapan biologi

Peperiksaan Percubaan Biologi SPM 2008

Panitia Biologi Negeri Perlis www.spm.via.my

1

MARKING SCHEME : PAPER THREE TRIAL BIOLOGY 2008Question 1 : 1(a)

Score Explanation

3

Able to record all readings of lengths of air column correctly.Initial = 6.4 cmP = 5.7 cmQ = 6.0 cm

R = 6.2 cm2 Able to record any three lengths.1 Able to record any two lengths.0 No response or wrong response

1(b)(i)Score Explanation

3

Able to state two correct observations based on following criteria.C1 levels of vigorous activityC2 The lengths of air column.

Sample Answer:(either 2):

1. For activity level P the length of air column after treatment with KOH is 5.7 cm2 . For activity level R the length of air column after treatment with KOH is 6.2 cm

2 Able to state one correct observation and one inaccurate response.1 Able to state one correct observation or two inaccurate response or idea.0 No response or wrong response (response like hypothesis)

1(b) (ii)Score Explanation

3

Able to state two reasonable inferences for the observation.

Sample answer:1. More carbon dioxide is absorbed by KOH because activity P is more vigorous.2. Less carbon dioxide is absorbed by KOH because activity R is less vigorous.

2 Able to state one correct inference and one inaccurate inference.1 Able to state one correct inference or two inaccurate inference or idea.0 No response or wrong response (inference like hypothesis)


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2

1(c)Score Explanation

3

Able to state all the variables and the method to handle variable correctly() for each variable and methodManipulated Variable: Levels of vigorous activity ()Method to handle: The athlete is asked to perform different levels of vigorous

activity ()

Responding Variable: The length of air column after treatment with KOH/ thepercentage of carbon dioxide released. ()

Method to handle: Measure and record the lengths of air column usingruler. / calculate the percentage of carbon dioxide releasedusing formulae:

Percentage of carbon dioxide = change in length of air column X 100%Initial length of air column ()

Controlled variable : duration for athlete to perform the activities/ the sameathlete. ()

Method to handle: fix the duration for each activity/ ask the same athlete performall activities. ()

Able to get 6 2 Able to get 45 1 Able to get 23 0 No response or wrong response

1(d)Score Explanation

3

Able to state the hypothesis correctly based on the following criteria:

V1 State the level of vigorous activity.V2 State the length of air column / percentage of carbon dioxide.R - State the relationship between V1 and V2.

As the level of vigorous activity increases, the length of air column decreases. //As the level of vigorous activity increases the percentage of carbon dioxideincreases.

2 Able to state the hypothesis but less accurate.1 Able to state the idea of the hypothesis0 No response or wrong response


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3

1(e)(i)Score Explanation

3

Able to construct a table and record the result of the experiment with thefollowing criteria: State all three levels of vigorous activities. () Transfer all data correctly. () Calculate the percentage of carbon dioxide with unit. (%) ()

If without unit (x).

- Title with correct unit

Tajuk dengan unit yang betul

- Initial length and final length of air column

Panjang awal dan panjang akhir turus udara

- Change in the length of column

Perubahan panjang turus udara

- Percentage of carbon dioxide releasedPeratus karbon dioksida yang dibebaskan

Levels ofvigorousactivity

Length of aircolumn / cm

Change in thelength of aircolumn / cm

Percentage of carbondioxide released / %

Initial FinalP 6.4 5.7 0.7 10.94/ 11.0Q 6.4 6.0 0.4 6.25/ 6.3R 6.4 6.2 0.2 3.13/ 3.1

2 Able to construct a table and record any two criteria

1 Able to construct a table and record any one criteria0 No response or wrong response


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4

1(e)(ii)Score Explanation

3

Able to draw a bar chart of percentage of carbon dioxide released against the levelof vigorous activity.Axes (A) both axis are labeled with units, uniform scales, independent

variable on horizontal axis. ()Point (P) All points are correctly plotted. ()Shape (S) All bars are correctly drawn ().

2 Graph with any two criteria.1 Graph with any one criteria.0 No response or wrong response.

1(f)Score Explanation

3

Able to explain the relationship between the level of vigorous activity and thepercentage of carbon dioxide released correctly.

When the level of vigorous activity increases the percentage of carbon dioxideincreases due to the increase in the rate of respiration.

2 Able to explain briefly the relationship between the level of vigorous activity andthe percentage of carbon dioxide released

1Able to explain the idea the relationship between the level of vigorous activity andthe percentage of carbon dioxide released

0 No response or wrong response


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5

1(g)Score Explanation

3

Able to state the definition of exhaled air correctly, based on the following criteria.

C1 the content of carbon dioxideC2 the effect on length of air column after treatment with KOH.C3 influence by vigorous activity

Exhaled air is air that contains carbon dioxide that will cause the length of aircolumn to decrease after treatment with KOH solution

2 Able to state the definition of transpiration based one of the two criteria.1 Able to state the idea of exhaled air.0 No response or wrong response

1(h)Score Explanation

3

Able to predict correctly and explain the prediction based on the following item:

C1 the length of air column.

C2 duration of activityC3 rate of respiration

The length of air column is less than 5.7 cm because the longer the time taken toperform the activity the more carbon dioxide is released due to the increase in therate of respiration.

2 Able to predict based on any two criteria.1 Able to predict based on any one criteria.0 No response or wrong response

1(i)Score Explanation

3

Able to classify the levels of vigorous activity and the rate of respiration.

Levels of vigorous activity

Tahap kecergasan aktiviti

Rate of respiration

Kadar respirasi

P High

Q Medium

R Low

Able to classify all the levels of vigorous activity and the rate of respiration

correctly.

2Able to classify two of the levels of vigorous activity correctly.

1Able to classify one of the levels of vigorous activity correctly.

0 No response or wrong response


95/96




6

Question 2 :

Aspect Sample Answer Remarks

Aim/objective To investigate the effect of different quantities of water intake onurine output.

Problemstatement

KB061201

Do different quantities of water intake affect the volume of urineoutput?

3 marks

HypothesisThe more the volume of water intake, the larger the volume of urineoutput.

3 marks

Variables Manipulative variable: quantity of water intakeResponding variable: quantity of urine outputFixed variable: types of drink

Only two correct variablesOnly one correct variable

Apparatusand materials

KB061205

Materials : mineral water

Apparatus : drinking cup, containers for collecting urine andmeasuring cylinders.

All present3marks1 materialsand 2 app2marks

1 mark

Techniqueused

Measure and record the volume of urine collected with a measuringcylinder. B

Procedure

KB061204

1. Four students of same gender, age and size are selectedas the respondents.- KP,KF

2. The respondents are not allowed to consume any food ordrinks 3 hours before the experiment.- KP,KF

3. The respondents are instructed to empty their bladdersbefore the beginning of the experiment.- KP

4. During the experiment the respondents are asked to drink100ml, 300ml, 500ml and 700ml respectively.- KMV

5. A stop watch is started immediately after consuming thewater.- KP

6. At the interval of half an hour, until two hours , therespondents will empty their bladder and collect the urinesample.- RRV

7. Measuring cylinders are used to measure the volume ofurine collected.- KRV8. After each sampling the urine sample are discarded into the

toilet bowl.- KP9. The results are tabulated.- KP10. Make sure all urine collected is measured.- KC

KP Step 1,2,3,5,8,9 (any 4)KMV Step 4

KRV Step 6,7KF Step 1,2KC Step 10

8 - 9P3 m

6 - 7P2 m

3 - 5P1 m

3 marks


96/96

Skema jawapan biologi 7

Presentationof data

Volume of water intake (ml)

Volume ofurine

produced(ml)

100 300 500 700

Total volume

(ml)

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

Volume of water intake

Volume of urine released

The units in ml or cm3

Sample Answer

B

Conclusion The larger the volume of water intake, the larger the volume ofurine output.Hypothesis accepted.

PlanningKB061203

Able to state correctly8 9 aspects (correct) - 3 marks6 7 aspects (correct) - 2 marks3 5 aspects (corect) - 1 mark< 3 - 0 mark

3 marks

Report Able to state correctly:

presentation of data and

technique.

2 correct - 2 marks1 correct - 1 mark

2 marks

spm bio perlis 2009 serta skema

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