[edu.joshuatly.com] n9 stpm trial 2010 biology [w ans] [7d789a37] (1).pdf

8/18/2019 [edu.joshuatly.com] N9 STPM Trial 2010 Biology [w ans] [7D789A37] (1).pdf

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964/1*This question paper is CONFIDENTIAL until the examination is over. CONFIDENTIAL*

1 What properties of water help it to moderate the changes in temperature?

A Cohesion and adhesion force

B Polarity and hydrogen bonding

C Solvent and high specific heat capacity

D High specific heat capacity and high heat of vaporisation

2 Which of the followings are properties of protein?

I Amphoteric

II Can be denatured by heat over 60 ° C

III Can act as a buffer

A I

B I and II

C II and III

D I, II and III

3 Blood, compact bone and yellow elastic and hyaline cartilages are all classified as

A epithelium tissues

B muscle tissues

C connective tissues

D nerve tissues

4 Which of the following characteristics are true for both mitochondrion and chloroplast?

I Contain enzymes

II Contain DNA

III Able to replicate

IV Present in all eukaryotic cells

A I and II

B I and IIIC I, II and III

D I, II, III and IV

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5 Which of the following statements uses the shortest DNA template.

A A DNA with 75 codons

B An mRNA with 75 codons

C A tRNA with 75 nucleotides

D A polypeptide chain with 75 amino acids

6 The diagram below shows the Lineweaver-Burke plot for an enzyme catalysed reaction.

Which of the following combinations is correct?

Without inhibitor With competitive inhibitor With non-competitive inhibitor A X Y Z

B Z X YC Y Z X

D Z Y X



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10

Based on the diagram above what is the class of enzyme which catalyses reaction I ?

A Isomerase

B Dehydrogenase

C Hydrolase

D Transferase

11 Which of the following is a hydrogen donor in the biosynthesis of purple sulphur bacteriaand green sulphur bacteria ?

A Water

B Hydrogen sulphide

C NADH

D Hydrocarbon compound



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12 Which of the following are the characteristics of saprotrophs ( saprophytes) ?

I No chlorophyll

II Digestive enzymes are secreted outside the body and the soluble products

of digestion are absorbed.

III They are decomposers

IV Many saprotrophs are fungi and bacteria

A I and II

B I, II and III

C II and III

D I, II, III and IV

13 The following graph shows four oxygen dissociation curves for respiratory pigments.

Which curve represents the oxygen dissociation curve for myoglobin ?



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14 The diagram below represents the control of breathing. Which of the following terms best describes this control mechanism ?

A Negative feedback

B Positive feedback

C Reflex arc

D Voluntary control

15 The following diagram shows the hydrostatic pressure (HP), the solute potential of the plasma proteins in the capillary ( Ψs) at the arteriole end and the venule end, and the HPand Ψs in tissue fluid.

Which is true about the diagram ?

At the arteriole end At the venule endA Filtration, filtration value = 3.2kPa Absorption, absorption value = 0.5kPa

B Filtration, filtration value = 2.0kPa Absorption, absorption value = 0.3kPa

C Filtration, filtration value = 1.2kPa Absorption, absorption value = 1.5kPa

D Filtration, filtration value = 2.1kPa Absorption, absorption value = 3.6kPa



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16 Which of the following correctly describes the movement of water across a root?

17 Which of the following is true when a person has defective cell in the islets oflangerhand?

A Glucose concentration is low in the blood

B Glucose concentration is low in the urine

C Glycogen content is high the muscle

D Glycogen content is high in the liver

18 Which of the following processes enable the mammalian kidney to control theabsorptions of water in dry conditions?

A The production of hypotonic urine

B The ability of the tubule cells to detect blood osmotic pressure

C The reabsorptions of water from the glomerulus in the proximal tubule

D An increase in the permeability of the collecting duct through the action of

antidiuretic hormone( ADH)

19 Why do action potential usually propagate only one direction along the axon?

A The ions flow along the axon in one direction only

B The Nodes of Ranvier conduct in one direction only.

C Both sodium and potassium voltage- gated channel open in one direction only

D The refractory period prevents the opening of the voltage- gated Na + channel after

each depolarization stage.

From the root hairs across thecortex

From the cortex across theendodermis

A Apoplast and vacuolar

pathways

Apoplast and vacuolar

pathwaysB Symplast and vacuolar

pathways

Symplast and vacuolar

pathways

C Apoplast, symplast and

vacuolar pathways

Apoplast symplast and

vacuolar pathways

D Apoplast, symplast and

vacuolar pathways

Symplast and vacuolar

pathways



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20 The figure below shows a neuromuscular junction.

The chemical substance release from the regions X and Y are:

X YA Cholinesterase Sodium ions

B Cholinesterase Calcium ionsC Acetylcholine Sodium ions

D Acetylcholine Calcium ions

21 Steroids hormones

A function via second messenger

B after the activity of the genes in cell

C trigger rapid, short-term respond in cell

D bind to receptors on the cell surface membranae

22 Which of the following statement about phytochrome is true?

I phytochrome is a blue green pigment

II phytochrome exist in two form, P r and P fr

III the P r form of phytochrome is the active form

A I

B I and II

C II and III

D I, II and III



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23 Which of the following lymphoid tissues enlarges during childhood but shrinks after puberty?

A Spleen

B Liver

C Lymp nodesD Thymus gland

24 Which of the following are found in Human Immunodefeciency Virus (HIV)?

I DNA

II RNA

III DNA ligase

IV Reverse transcriptase

A I and II

B I and III

C II and III

D II and IV

25 Which of the following are the similarities between Marchantia and Dryopteris ?I In both plants, movement of antherozoids occur in the presence of water.

II In both plants, sporophyte generations produce spores by meiosis

III In both plants, the sporophyte generation is dominant

IV Both plants can reproduce asexually

A I, and II

B I, II and III

C I, II and IV

D I, II, III and IV



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26 Which of the following is a correct match.

Germ layers Organs or Systems(a) Parthenogenesis (I) The potential to divide and replace damaged

or missing parts.(b) Paedogenesis (II) Larval or pupal forms with the potential to

reproduce before attaining the adult stage.(c) Polyembryony (III) Production of two or more individuals from a

zygote.(d) Regeneration (IV) Production of new individual from an egg

without fertilisation.

Which of the following is correctly paired?

(a) (b) (c) (d)A (II) (IV) (III) (I)

B (II) (II) (I) (III)

C (III) (II) (I) (IV)

D (IV) (I) (III) (II)

27 Which of the following events triggers parturition?

A Oestrogen level drops lower than the progesterone level

B Progesterone level drops lower than the oestrogen level

C Oxytocin level increases

D Prostaglandin level increases

28 Which event occurs during epigeal germination of a seed planted in the soil?

A The epicotyls elongates but the cotyledon remains underground.

B The hypocotyls elongates but the cotyledon remains underground

C The hypocotyls elongates and pushes the cotyledon above the soil surface

D The plumule elongates and pushes the cotyledon above the soil surface



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29 The diagram below shows an allometric growth that occurs in a human.

Which of these statements are correct ?

I Graph P shows the growth of the thymus gland

II Graph Q shows the growth of the brain

III Graph R shows the absolute growth curve of the human body

IV Graph S shows the growth of the reproductive organs

A I, II and III

B I, III and IV

C II, III and IVD I, II, III and IV

30 During pupa formation from the last larval stage in the butterfly, which of thefollowing concentration of hormones occur ?

A Low concentration of ecdysone and juvenile hormone

B High concentration of ecdysone and juvenile hormone

C High concentration of juvenile hormone and low concentration of ecdysone

D High concentration of ecdysone and low concentration of juvenile hormone



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31 In a garden pea, two alleles are present, allele P for the purple flower trait and recessiveallele p for white flower trait. Half of the pollen produced by the plant has the recessiveallele p. Which of the following is most suitable to explain why this has occurred?

A Mutation occurred.

B Mitosis occurred in parent cells.

C Crossing over occurred during prophase I of meiosis.

D Segregation of the two alleles during anaphase I of meiosis.

32 White eye in Drosophila is determined by a recessive sex-linked gene and a dominantallele produces red eye. Which of the following is the correct progeny of a cross

between a heterozygous red eye female with a white eye male.

Female Male Red White Red White

A 1 1 1 1

B - 2 1 1

C 1 1 2 -

D 2 - - 2

33 Which of the following statements about Down’s syndrome are true?

I Trisomy disease

II Caused by non-disjunction during mitosis

III A type of aneuploidy

IV Could be male or female

A I and III only

B I, II and III only

C I, III and IV only

D I, II, III and IV

34 If the original sequence of chromosome is L M N O P Q R S and two chromosomemutation occurs, resulting in the sequence of L N M O P R S, what type of mutation is

it?

A Two translocation

B One translocation and one deletion

C One inversion and one deletion

D Two translocation



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35 In a population of humans in Africa, 1.0% of its individuals suffered from the sickle-cellanemia and died. What is the percentage of the individuals in the population that possesan abnormal haemoglobin, which is heterozygote?

A 1

B 10

C 18

D 90

36 Albinism in man is controlled by a recessive allele a. In a population, the frequency ofalbinism is one for every 40, 000 individuals. What is the frequency of allele a?

A 0.05%

B 0.5%

C 5%

D 50%

37 The diagram below shows the regulatory scheme of lactose operon, where I to IVrepresent the reaction sequence while X, Y and Z represent the genes.

What are the genes represented by X, Y and Z in the above scheme?

X Y Z

A Regulator Operator StructuralB Regulator Structural Operator

C Operator Regulator Structural

D Operator Structural Regulator



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38 If the regulator gene of lactose operon system in Escherichia coli has mutated and isnon-functional, which of the following statements is true?

A β- galactosidase enzyme is synthesized in the presence of lactose.

B β- galactosidase enzyme is synthesized with or without the presence of lactose.

C β- galactosidase is not synthesized in the presence of lactose D β- galactosidase enzyme is not synthesized with or without the presence of lactose

39 The following diagram shows a plasmid having three restriction site tet R and amp R , thegenes for tetracycline and ampicillin resistance respectively.

What will be obtained if the plasmid is treated with Bam HI?

A A single linear fragment is obtained.

B Two linear fragments will be obtained.

C Six linear fragments will be obtained.

D There will be no change to the structure of the plasmid



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40 The diagram below shows DNA cloning using a bacterial plasmid.

Why is it necessary to carry out screening?I There are bacteria that are not transformed.

II There are bacteria that do not undergo amplification.

III There are bacteria that are transformed with non-recombinant plasmids.

A I and II

B I and III

C II and III

D I, II and III

41 The following identification key is prepared for five types of animals

1 Animals with back bone………………………………… 2Animals without back bone…………………………….. 3

2 Body with dry scales…………………………………… PBody without scales……………………………………. Q

3 A pair of antennae……………………………………….. 4More than two pairs of antennae……………………….. R

4 Three pairs of mouthparts……………………………… SA pairs of jaws…………………………………………. T

What are animals P, Q, R, S and T?P Q R S T

A Snake Frog Cockroach Prawn Centipede

B Frog Snake Cockroach Centipede Prawn

C Snake Frog Centipede Prawn Cockroach

D Snake Frog Prawn Cockroach Centipede



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42 The table below shows the taxonomic hierarchy from kingdom to genus for threedifferent organisms.

Taxonomichierarchy

Examples of organisms

Kingdom Plantae Animalia AnimaliaPhylum W Annelida ChordataClass Dicotyledoneae X MammalianOrder Ranales Terricolae YFamily Ranunculacae Lumbricidae HominidaeGenus Ranunculus Lumbricus Homo

What do X, Y and Z represent?

W X YA Angiospermophyta Anthozoa Echinodermata

B Bryophyte Nematoda MolluscaC Coniferophyta Cestoda Cnidaria

D Angiospermophyta Oligochaeta Primates

43 The table below shows four plant phyla and their examples.

Phylum ExampleI FungiII BryophytaIII Filicinophyta

IV Angiospermophyta

a) Helianthusb) Dryopterisc) Marchantia

d) Mucor

Which of the following is the correct match for the above four phyla and theirexamples?

I II III IVA (a) (b) (c) (d)

B (c) (d) (a) (b)

C (d) (a) (b) (c)

D (d) (c) (b) (a)



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44 The following features are characteristics of a group of plants

Rhizoid is unicellular They are ancestors of terrestrial plants Sporophyte stages is dominant Spores are contained within a sporangium Haploid spores germinate into gametophyte

These plants belong to the phylum

A Algae

B Bryophyta

C Coniferophyta

D Angiospermophyta

45 Which of the following could cause phenotype variation in organisms that have thesame genotype?

A Mutation

B Exposure to different environment

C Continous variation within the species

D Crossing over in gametogenesis

46 The best definition of soecies involves the

A interbreeding capability

B morphological similarities

C habitat similarities

D physiological similarities

47 Which of the following statements are true about a climax community?

I The plant community is at dynamic equilibrium

II The plant community has the highest species diversity

III The plant community has achieved maturity and stability

IV The plant community is formed at the end of a succession process

A I, II, and III

B I, II and IV

C II, III and IV

D I, II, III and IV



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48 Both an aphid and a caterpillar are living on the same leaf. The caterpillar feeds on theleaf, while the aphid sucks the fluid from the phloem of the leaf. Each species occupies

A the same niche in the same habitat

B the same niche in different habitats

C different niches in the same habitatD different niches in different habitats

49 Which population growth curve will be produced if environmental resistance is absent?

A Sigmoid curve

B S-shaped curve

C Linear curve

D Exponential curve

50 In capture-mark-release-recapture method done to estimate the population of rats in anoil palm plantation, 600 rats were caught in the first catch and 650 in the second catch.350 of the 650 in the second catch did not have markings. What is the rat population sizein the oil palm plantation?

A 323

B 379

C 1114

D 1300

END OF QUESTION PAPER



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2

Section A [40 marks ] Answer all questions in this section.

1 The diagram below represents a longitudinal section through part of a striated muscle.

(a)(i) What is the term used to describe the basic unit of muscle contraction shown above?

…………………………………………………………………………………………[1 mark]

(ii) Identify the parts labelled A,B and C.

A: ………..……………………………………………………………………………..

B:…………………………………………………………………………………..……

C:……..………………………………………………………………………..………..[3 marks ]

(b) The diagram shows the A band, the I band and the H zone. Which of the structures:

(i) contain actin but no myosin

………………………………………………………………………………………...[1 mark]

(ii) shortens when the muscle contracts

………………………………………………………………………………………....[1 mark]



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(c) Describe the role played by ATP and Ca 2+ in muscle contraction

(i) ATP:

……..……………………………………………………………………………………

…….………………..…………………………………………………………………..

………………………………………………………………………………………….[2 marks]

(ii) calcium ions:

……………………………………………………………………………………......

….…………………………………………………………………………………….

………….……………………………………………………………………………..[2 marks]

2 Graph M shows the growth curve of a woody perennial plant, and G raph N shows thegrowth curve of an insect.

(a)(i) With reference to G raph M and N , name the pattern and explain the form of the twotypes of growth curves.

Graph M

………………………………………………………………………………………...…

………………………………………………………………………………………..…[2 marks]

Graph MGraph N



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Graph N

……………………………………………………………………………………….….

……………………………………………………………………………………….….

[2 marks]

(ii) The growth curve of the insect uses length as a measure of growth. Give a reason whythis growth curve cannot be considered as a true measurement of the insect’s growth.

….………………………………………………………………………………………

.…………………………………………………………………………………………[1 mark]

(iii) If a growth curve is plotted for the same insect using dry mass as a measure of growth,what form of growth curve would be produced?

…………………………………………………………………………………..………[1 mark]

(b)(i) State the role played by the ecdysone and juvenile hormone in the metamorphosis ininsects.

Ecdysone: ………………………………………………………………………………….

…………………………………………………………………………………………….[1 mark]

Juvenile hormone :........……………………………………………………………….....

…..………………………………………………………………………………………… [1mark]

(ii) Explain why synthetic juvenile hormone can be used as insecticides.

……………………………………………………………………………………………..

………....…………………………………………………………………………………..

...…………………………………………………………………………………………..

…………………………………………………………………………...………………..[2 marks]



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3 The bacterium Eschericia coli produces β- galactosidase only in the presence of lactose.

β- galactosidase hydrolyses the lactose sugar to glucose and galactose. The diagram below represents a part of DNA which controls the synthesis of β- galactosidase.

(a) Draw an arrow from the box T to show the new position of the repressor molecule if the

lactose molecule is absent. [1 mark]

(b) Explain why there is no β- galactosidase synthesis when molecule T is in the new position.

…………………………………………………………………………………………

………………………………………………………………………………………… [2 marks]

(c) Explain how the presence of the lactose molecule induces the production ofβ- galactosidase.

…………………………………………………………………………………………..

………………………………………………………………………………………….

…………………………………………………………………………………………..

………………………………………………………………………………………… [3 marks]

(d) State the function of lac Y gene.

…………………………………………………………………………………………..

…………………………………………………………………………………………..[1 mark]



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(e) State the functions of the following structures in the lactose operon.

(i) Regulator gene

…………………………………………………………………………………………….[1 mark]

(ii) Promoter

……………………………………………………………………………………………..[1 mark]

(iii) Operator

.....………………………………………………………………………………………….[1 mark]

4 The diagram below shows three types of unicellelular organisms.

(a) Name the taxonomic kingdom of the organisms above.

……………………………………………………………………………………………[1 mark]

(b) State the scientific name and the phylum of each of the organisms given in the diagramabove.

Organism Name Phylum

A

B

C

[6 marks]



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(c) Organism B has two nuclei, namely macronucleus and micronucleus. State the functionof micronucleus.

…..………………………………………………………………………………………… [1 mark]

(d) State two different characteristics between organism A and C.

......…………………………………………………………………………………………

….………………………………………………………………………………………… [2 marks]



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TRIAL STPM 2010OBJECTIVE ANSWER SCHEME

1. D 11.B 21.C 31.D 41.D

2. D 12.D 22.B 32.A 42.D

3. C 13.A 23.D 33.C 43.D

4. C 14.A 24.D 34.C 44.C

5. C 15.C 25.C 35.C 45.B

6. D 16.D 26.C 36.B 46.A

7. B 17.D 27.B 37.A 47.D

8. B 18.D 28.C 38.B 48.C

9. C 19.D 29.D 39.A 49.D

10. A 20.D 30.D 40.B 50.D



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1

TRIAL STPM 2010 – ANSWER SCHEMESECTION A

NO ANSWER SUB-TOTAL

TOTAL

1(a)(i) Sarcomere 1 1(ii) A-Z line

B-ActinC-myosin

111 3

(b)(i) I band 12(ii) I band and H zone 1

(c)(i) Provides energy to break cross- bridges/rotatesmyosin head (reject muscle contraction)

Provides energy to pump actively calcium ions back into thesarcoplasmic reticulum

1

12

(ii) Calcium ions bind to troponin causing reorientation of thetropomyosin

Exposing myosin head binding site on the actin filament

1

1 2

TOTAL 102a(i) Graph M unlimited growth It is produced following an annual serial of smaller sigmoid

curvesGraph N intermittent growth Growth only takes place for a short period of time during

ecdysis before the new skin hardens

11

11

2

2

(ii) Because growth is represented by the increase in organicmaterial in the body.

Length of the exoskeleton is not a true reflection of growth forinsects

1

1 Max 1(iii) (Normal) sigmoid curve // S-shaped curve 1 1

(b)(i) Ecdysone stimulates ecdysis / moulting of exoskeleton // stimulates

development of adult characteristicsJuvenile hormone maintains the juvenile characteristics of an insect

1

1 2(ii) Synthetic juvenile hormone can block the production of adult

characteristics in insects. Thus insects don’t mature to breed.

1

1 2TOTAL 10

3(a) T molecule binds to the operator 1 1(b) The operator and part of the promoter are covered, preventing

RNA polymerase from binding to the promoter. The lactose operon system is switch off.//Transcription of

structured genes to produce enzyme does not occur.

1

12

(c) A small amount of lactose is converted to the allolactoseisomer. Allolactose bind to the repressor, changing itsstructure

1



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2

NO ANSWER SUB-TOTAL

TOTAL

Allolactose repressor complex detaches from the operator.RNA polymerase binds to the promoter.

Transcription of structured gene occurs, mRNA is produced,translated and enzymes like ß-galactosidase are synthesize

1

1 3

(d) The lac Y gene encodes the synthesis of permease enzyme (themembrane protein that transports lactose molecules into cell) 1 1

(e) (i) codes synthesis of repressor molecule(ii) binding site for RNA polymerase(iii) binding site for repressor molecule

111 3

TOTAL 104(a) Protoctista// Protista 1 1(b)

Organism Name Phylum

A Amoeba sp. Rhizopoda

BParamecium sp.

Ciliophora

C Euglena sp. Zoomastigina

(Scientific names must ended with sp.e.g Reject Amoeba )

1+1

1+1

1+1 6

(c) Function of micronucleus: controls the function of sexualreproduction 1 1

(d)Characteristic Organism A Organism C

Cell wall Absent Present

Flagella Absent PresentStructure used forlocomotion

Pseudopodium Flagellum

(accept any other suitable answers)

1/0

1/01/0

Max 2

TOTAL 10



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3

SECTION B

NO ANSWER SUB-TOTAL

TOTAL

5.(a)Carbon dioxide concentration

Required in light-independent reaction Increasing carbon dioxide concentration, increases the rate of photosynthesis

Rate of photosynthesis becomes constant after carbon dioxideconcentration exceeds reaches certain level

At higher concentration, the rate of photosynthesis decreasesas CO 2 absorption is reduced by stomatal closure(Any 3)

Temperature Photosynthesis process is controlled by enzymes The rate of photosynthesis increases with an increase in

temperature provided light intensity and concentration of CO 2are not limiting factors

Rate of photosynthesis doubles for every 10 ° C increase intemperature until optimum temperature

Above optimum temperature, rate decreases as the enzymesare denatured

(Any 3)Water availability Water is required in light dependent reaction//photolysis of

water where electrons/ H atom are given out to reduce NADP

Without water plants close their stomata and would prevententry of CO 2 into the plants for photosynthesis

(Any 2)

Oxygen concentration Relatively high O 2 level inhibits photosynthesis

O2 competes with CO 2 for active site of RuBisCo whichwill decrease the overall rate of photosynthesis

1

1

1

11

11

1

1

1

1

1

1

11

Max: 3

Max :3

Max :2

2

(b) Photosynthesis Aerobic respirationAn anabolic process whichresults in the synthesis ofcarbohydrate from simpleinorganic molecules

A catabolic process whichresults in the breakdown ofcarbohydrate molecules tosimple inorganic molecules

Energy is accumulated andstored in the form ofcarbohydrates

Energy is stored in the formof ATP for future use.

Oxygen is released Oxygen is used

1/0

1/0

1/0



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4

NO ANSWER SUB-TOTAL

TOTAL

Results in an increase in drymass

Results in a decrease in drymass

Occurs in chloroplast Occurs in the mitochondrionOccurs only in cells that havechlorophyll and only in the

presence of light

Occurs in all cellscontinuously throughout thelifespan of the cell

(Any 5)

1/0

1/0

1/0Max :5

TOTAL 156. Atrial Systole

both atria contract both atrioventricular valves open blood flows into ventricles pulmonary and aortic valves remain closed

Ventricular systole both ventricles contract increased ventricular pressure of blood causes atrioventricular

valves to close semilunar valves open blood flows to aorta and pulmonary artery

Diastole Ventricles relax Semilunar valves close

Blood flows into atria from vena cava and pulmonary veins.

1111

1111

111

Any 2

Any 2

3

(b) Structure : the heart is composed of striated muscle cells the cells are cross-connected by branching fibres and

intercalated discs separating individual muscle cells permitdiffusion of Na+ and K+

this allows rapid spread of impulses from cell to cell cardiac muscle is myogenic

(Any 3)

Heartbeat: initial stimulus originate from the sinoatrial node (SAN)

which is the pacemaker found on the right atrium wall nearthe entrance of the vena cava

membrane of SAN cells are permeable to Na+ Na+ enter cells and depolarizes cell membranes An excitatory wave is generated, and spread rapidly from the

SA node across the two atria making both atria contract Waves reach the AVN ( atrioventricular node) located

11

1

1

1

111

1

Max:3



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5

NO ANSWER SUB-TOTAL

TOTAL

between atria. Impulses from AVN are conducted by the bundle of His that

branches into Purkinje fibresThe waves conducted throughout the ventricular walls and theventricles contract (Any 5)

1

1Max :5

TOTAL 157.(a) i

any foreign/non-self substance in the body that trigger an immune response against it

1 1 2

ii. a substance produced by plasma cells as a response to the presence of antigen

1 1 2

(b) Response of T-lymphocytes is cell mediated response cytotoxic T-lymphocytes binds and causes lysis of the cell membranes of infected/malignant/non-

self cells Helper T-lymphocytes produce chemicals

to activate/promote growth and multiplication of otherlymphocytes response of B-lymphocytes is a humoral response B-lymphocytes response to antigen by dividing and

differentiating into plasma cells these plasma cells are synthesized in large quantities, produce antibodies specific to the antigens

that will destroy the antigens

1 11

1

11

11 8

(c) in the immune system, cells/lymphocytes have the ability todetect whether or not a foreign substances present in the bodyis compatible with it

if compatible it will be accepted as self if not it will treated as non-self and will trigger an immune response

(Any 3)

1

111

Max: 3

TOTAL 158. After double fertilisation, the triploid nucleus divides rapidly

to form the multicellular endosperm. The multicellular endosperm functions as a nutrient storage. The zygote divides through mitosis to form two cells: basalcell

and terminal cell. The basal cell multiplies and develops into the suspensor thatanchors the embryo to the endosperm.

The terminal cell develops into the embryo The embryo cells further develops into one or two cotyledon The embryo cells between the cotyledons differentiate to formthe shoot apical meristem

The embryo cells near the suspensor differentiate to form theroot apical meristem

1

11

11

11

1

1



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NO ANSWER SUB-TOTAL

TOTAL

The root apical meristem develops to become the radicle The shoot apical meristem develops to become the plumule

(Any 10)

1

1 Max:10

(b ) Monocotyledonous seed Dicotyledonous seed

Has only one cotyledon

Cotyledon reduced toshield-like structurecalled as scutellum

Aleurone layer is present. Coleoptiles protect the plumule, the coleorhiza

protects the radicle Testa and pericarp arefused together

Has two cotyledon

Cotyledons are fleshy,no scutellum

No aleurone layer

Coleoptiles andcoleorhiza are not

present Testa and pericarp arenot fused together

1/0

1/0

1/0

1/0

1/0 5

TOTAL 159.(a) Types of gene mutation are base/nucleotide substitution,

base/nucleotide deletion, base/nucleotide inversion and base/nucleotide insertion

In base/nucleotide substitution, a nucleotide/base is replaced by another base/nucleotide in the DNA sequence of a gene

Causes missence mutation// this new base/nucleotide altersone genetic code to a different genetic code which may codefor different amino acid

In base/nucleotide inversion, two or more nucleotides/baseshave been reverse in the DNA sequence within the gene.

The altered genetic code results in a different amino acid inthe polypeptide chain and the formation of non- functional

protein In base /nucleotide insertion, and extra base/nucleotide is

inserted into the DNA sequence of a gene Causes the whole base sequence to be shifted to one place

backward

In base/nucleotide deletion, a base/nucleotide is deleted fromthe DNA sequence of a gene Causes the whole base sequence to be shifted to one place

backward Insertion and deletion are more harmful than substitution and

inversion Both base /nucleotide insertion and deletion are frameshift

mutation as every single triple code after the mutation point isaltered

1

1

1

1

1

1

1

1

1

1

1



33/34

7

NO ANSWER SUB-TOTAL

TOTAL

Because frameshift mutation leads to production of non-functional protein (Any 10)

1Max:10

(b) Down’s syndrome is an example of aneuploidy with 47

chromosomes instead of 46 chromosomes in an individual It is a result of non-disjunction during meiosis( I and II) Two chromosomes of no 21 failed to separate during

Anaphase I/II of meiosis The gametes produced contain 23 + X / 23 + Y When a sperm with 23 + Y fuses with normal ovum / When a

sperm with 23 + X fuses with normal ovum/ When n ovumwith 23 + X fuses with normal sperm, the zygote formedcontains 45 + XY/XX //trisomy//47 chromosomes

1

1

1

1

1 5

TOTAL 1510.(a)

Gene therapy

Genetic disorders which are caused by a single defective genecould be cured by inserting a functional gene into certain

body cells such as bone marrow cells. Adenosin deaminase (ADA) is important in the proper

functioning of the immune system. It is used by white bloodcells to produce antibodies.

The normal gene for ADA is cloned in a vector. The vector containing the ADA gene is introduced into a

retrovirus. A small amount of bone marrow cells are isolated from the patient and cultured in the lab.

The bone marrow cells are infected with the ADA genecontaining the retrovirus.

The retrovirus integrated the normal ADA gene into the bonemarrow cells.

Bone marrow cells with the normal ADA gene are injectedinto bone marrow of the patient.

These cells continue to divide thus producing sufficient ADAenzymes. (Any 7)

1

1

1

1

1

1

1

1

1 Max: 7

(b) DNA fingerprinting techniques

DNA fingerprinting is used for identifying criminals and toidentify dead victims beyond recognition and for solving

paternity cases. In every person, the sequence of bases on the intron is unique

and it usually contains a repeating sequence calledminisatellite/ VNTR (Variable Number Tandems Repeat)Minisatellite can be used to form a DNA fingerprint.

1

1



34/34

NO ANSWER SUB-TOTAL

TOTAL

DNA extracted from cells and treated with restriction enzyme to produce

RFLP(restriction fragment length polymorphism) so thattheir lengths are unchanged.

Gel electrophoresis is performed to separate the fragments. The southern blotting procedure is carried out to transfer the

fragments to a nitrocellulose filter. A radioactive probe with a base sequence complementary to

the mini satellite repeat sequence is used to hybridise withthe fragments

The remaining DNA fragments are exposed to an X-ray filmto produce a unique DNA fingerprint that consists of anumber of bands //(autoradiography)

DNA fingerprints from different specimens can be comparedto look for differences and similarities regarding the profileof bands (Any 8)

1

1

1

1

1

1

1

Max:8TOTAL 15

[edu.joshuatly.com] n9 stpm trial 2010 biology [w ans] [7d789a37] (1).pdf

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