modul ka f5 2011 combine (t)

7/28/2019 Modul Ka f5 2011 Combine (t)

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Topic 10 Transport

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No Mark SchemeSubMark

TotalMark

1(a) Able to describe the role of leucocytes in body defence.

P1 - The process is called phagocytosis

P2 - Phagocytes move towards pathogens using its pseudopodium

(Approaching)

P3 - Phagocytessurround the pathogens ( Engulfing)

P4 - Pathogen is hydrolysed by lysosome (Digestion) and reabsorbed

P5 - Destroyed pathogen is removed from the phagocyte

Any 4 4

1(b) Able to explain how individual could achieve immunity level.

P1 - The graph shows Artificial Active Immunity

P2 - The person has been injected with a vaccine

P3 - The vaccine contain killed or weakened antigens/

bacteria/viruses

P4 - antigens/bacteria stimulates lymphocyte / WBC to produce

antibodies

P5 - 1st dose usually induces a slow production of antibody

(and shorter lasting)

P6 - Booster dose (2nd and 3rd ) are needed to stimulates more

antibody to achieve immunity level ( and larger lasting response).

P7 - any invasion of the pathogenic microorganisms, the body

is able to destroy them immediately

P8 - Eg of vaccination: BCG / Hepatitis / Polio / HPV (cervix cancer)

Any 6 6

2(a) Able to differentiate the composition between fluid R and fluid S

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1

1

1

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2(b) Able to explain how fluids is formed

P1 - (When the blood flows from arteries into capillaries) there is Any 7 7

Fluid R (blood plasma) Fluid S (lymph)

1. Has less lymphocyte1. Has a larger numbers of

lymphocyte

Explain : Lymphocyte is produced by the lymph node.

2. Contain erythrocyte &

plasma protein / eg:

fibrinogen

2. No erythrocyte, no plasma

protein / eg: fibrinogen

Explain: RBC & plasma protein are too big molecule to pass

through

3. Has high content of

oxygen3. Has lower contents of oxygen

Explain : oxygen has been used up by the cell


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Chapter 11 Locomotion and Support

No Mark SchemeSubmark

TotalMark

1(a) Able to explain how earth worm support itself.

P1 Has hydrostatic skeleton

P2 Body wall consist of outer circular muscle

P3 (and) inner longitudinal muscle.

P4 Body cavity is filled with a fluid which is held in compartments.

P5 The muscles act antagonistically.

Any 4 4

1(b) Able to explain the mechanism of locomotion in earthworm.

P1 (When earthworm is crawling over a surface), the chaetae inposterior end of the body pushed into the ground to anchor it.

P2 The muscle in the anterior end of the body contracts, while thelongitudinal muscle relaxes.

P3 (Hence) the anterior end of the body elongates

P4 The hydrostatic pressure builds up in the body

P5 The body fluid is pushed backward.

P6 The chaetae in the posterior end of the body are withdrawnwhile the chaetae in the anterior end of the body are push intothe ground.

P7 The longitudinal muscle in the anterior end of the body contract,while the circular muscle relax.

P8 causes the anterior end of the body become short and thick.

P9 The body fluid flows into the anterior end of the body

P10 causing the posterior end of the body pulled forward.

P11 The earthworm moves on the ground by alternately lengtheningand shortening its body, assisted by chaetae.

Any 6 6

1(c) Able to explain the role of muscles, tendons and ligaments

in the movement of forearms.

P1 Forearm has two sets of muscles; biceps and triceps Any 10 10

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P2 acts antagonistically

P3 muscles connected to bone by tendons.

P4 Bones are held together by ligaments.

P5 When the biceps contracts, the triceps relaxes.

P6 Biceps becomes shorter (and thicker), triceps becomes longer

(and thinner).

P7 This exerts a pulling force which transmitted to the radius

through the tendons.

P8 The radius is pulled upward and the fore arm is bent.

P9 When the triceps contracts, the biceps relaxes.

P10 The triceps becomes shorter and thicker while the biceps

becomes longer and thinner.

P11 This exerts a pulling force on ulna through tendons.

P12 The ulna and radius pulled downward, causing the forearm to

straighten.

2(a) Able to explain the adaptation of the plant which enable it to

float on water surface.

P1 Stem and enlarged petiole with many air sacs

E1 Provide buoyancy

P2 Many fibrous roots can trap air

E2 allow the plant to float

P3 Stem and roots have aerenchyma tissues

E3 makes the plants light and enable plants to float.

E dependent on P

Any 4

4

2(b) Able to explain the modification found in the woody plant

tissues which giving support to the plant.

P1 Xylem tissues

E2 Xylem vessels and tracheids are strengthened with lignin

P2 Parenchyma tissues

E3 Store starch and sugar and water.

E4 Turgid cells give support to the plant.

Any 6 6

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P2 Collenchyma tissues

E5 Walls thickened with cellulose and pectin

P3 Schlerenchyma tissues

E5 Wall thickened with lignin to provide support2(c) Able to compare and contrast between the two vertebrae.

Similarities:

S1 Both have centrum

E1 Gives support and able to withstand compression force

S2 Both have neural canal

E2 to contain spinal nerve

S3 Both have neural spine

E3 For muscle attachment

S4 Both have transverse process

E4 For muscle attachment

S5 Both have neural arch

E5 Form neural canal which protect the spinal cord

E6 Both have zygapofisis

E7 To articulate with another vertebra

Differences:

Vertebra cervical Vertebra lumbar

D1: Flat (small) centrum Large and thick centrum

E1: Give more supportD2: Short neural spine Long neural spine

E2: Attachment more muscles

D3: Broad transverse

prosess

Well develop transverse process

E3: For attachment more muscles

D4: Has to vertebrarterial

canals

No vertebrarterial canal

E4: Enable blood supply to headD5: Bigger neural canal Small neural canal

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E5: Contain bigger spinal cord/ brain trunk

E dependent on S/D

ANY 5 Correct S & E similarities : 5 marks

ANY 5 Correct D & E differences: 5 marks3(a) Able to describe the mechanism of locomotion of a bird flying in

the air.

P1 Bird fly by flapping their wings / gliding

P2 The wings of bird is in the shape of aerofoil

During flying:

P3 (To fly) the pectoralis major contract

P4 The pectoralis minor relax

P5 The pectoralis muscles are antagonistic muscles

P6 The wings moving downward and backward

P7 The air resistance produced as a result of moving wing downward

P8 provide an upthrust on the wings

P9 The thrust is transmitted from wings to the coracoids until sternum

P10 (As result) the whole body is lifted up

P11 (then) the pectoralis minor contract

P12 The wings are pulled up

P13 The air resistance is very low

P14 The wings are ready to move downward.

During gliding:

P15 The wings spread (to act as aerofoil)

P16 The air move faster on the upper of the wings compared to the

lower of the wings

P17 The air pressure is lower in the upper surface than below the

wings

P18 Upward thrust produced enable the birds to glide.

Any 10 10

3(b) Able to discuss how someone could has a healthy

musculoskeletal system.

Any 10 10

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P1: Having a well-balanced diet

E1: contain sufficient calcium and phosphorus

E2: Contain sufficient vitamin D

E3: To build strong bones / prevent osteoporosis

P2: Having a good posture

E4: While standing, our body should be erect straight, so that the

weight of our body is supported by both our feet.

E5: While sitting, the thorax is vertical/the thigh is comfortable/ almost

all muscle relaxed

E6: While walking, our body should be upright and straight

E7: While lying down, use a mattress that is firm so that the body is

evenly supported

E8: Bend both knees when lifting heavy object from the floor.

P3: Using proper attire for daily activities

E9: Wearing tight could restrict the movement

E10: Wearing high-heeled shoes could injure the back bone.

P4: Taking appropriate precautions during vigorous activities

E15: Consistent and moderate exercise can increase the bone mass

and prevent osteoporosis

E16: Very vigorous activity could results in pain/strain/ dislocation/

fractures.

P5: practicing correct and safe exercise techniques

E17: Warming up before exercise can raise the temperature of our

muscle to enabling them to make more efficient use of energy/

preventing injuries.4(a) Able to explain how coordination of myotomes and fins of a fish

contributes to swimming activities in the aquatic habitat.

P1: Myotomes are muscle block

P2: arranged in segments on both sides of the body / vertebral column.

Any 10 10

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P3: The muscles acts antagonistically / contraction of myotomes on

one side of vertebral column and relaxation of the myotomes on

the other side.

P4: the contraction of myotomes on the right side of the body will bend

the tail to the right // the contraction of myotomes on the left side of

the body will bend the tail to the left.

P5: Alternate contraction of the right and left myotome block causes

the body to bend side to side.

P6: This produces the forward thrust which propel the fish forward

P7: The paired fins and unpaired fin used to maintain the balanced of

body during swimming.

P8: The pectoral fins used to steering and brake.

P9: The pelvic fin are used to prevent diving and rolling movements

P10: Dorsal and ventral fins used to stay on course without yawing.

P11: Tail/caudal fin used to propel the fish.

4(b) Able to justify that unbalanced diet, an unhealthy lifestyle and the

process of aging may cause diseases such as osteoporosis and

arthritis and explain how such diseases can be avoided

P1: (unbalanced diet) such as diet less in calcium /

P2: less in phosphorus could lead

P3: less in vitamin D

P4: unhealthy lifestyle such as consume liquor

P5: Process of ageing such as life after menopause

P6: (could) cause osteoporosis / bone becomes porous/ soft and brittle

P7: (could) cause arthritis / inflammation of the joints.

Any 10 10

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Way to overcome osteoporosis:

P8: Optimize calcium intake to increase the bone mass

P9: Optimize vitamin D intake to enhance calcium absorption

P10: Exercise regularly to help strengthened the muscle and bone

P11: undergo hormone replacement therapy during menopause to

prevent osteoporosis

Way to overcome gouts:

P12: Reduction of offal and protein in diet

P13: Taking medication to lessen the joint inflammation and to reduce

the level of uric acid in the body.

P14: Less/stop consumption of liquor.

P15: Massage on the surrounding muscles using heat therapy.

Chapter 12 Coordination And Support


TotalMark

1(a) Able to explain the transmission of a nerve impulse from

neurone P to neurone R across Q

P1 Q is a synapse/ synaptic cleft.

P2 The transmission of information across a synapse involves

the conversion of electrical signal into chemical signal in the

form of neurotransmitter.

P3 Neurotransmitter is produced in vesicles in the axon

terminal called synaptic knob.

P4 Synaptic knob contains abundant mitochondrion to generate

Max 8 8

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energy for the nerve transmission.

P5 When an impulse arrived at the synaptic knob, the

vesicles release the neurotransmitters into the synapse.

P6 The neurotransmitters molecules diffuse across the synapse

P7 to the dendrite of another neurons.

P8 Reaching R, impulse is converted back into electrical

signal.

P9 The transmission of impulse in one way direction

P10 since the vesicle containing neurotransmitter is only found in

pre-synaptic membrane.1(b) Able to explain the causes and the effects of the Alzheimers

and Parkinsons diseases on victims.

Alzheimers Parkinsons

Caused by

- the shrinkage of brain

tissues and lack of

neurotransmitter.

- usually affects the elderly

Caused by

- the reduced level of

neurotransmitter in the brain

caused tremors and weakness

of the muscles

- the hardening of the

cerebral arteries

Effect:

- Loss of intelligent

- Loss of memory

- Poor concentration

Effect:

- The muscle cannot function

smoothly and become stiff and

jerky in their action

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1 + 1

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1(c) Able to draw the reflect arch and describe the pathway involved

in the transmission of nerve impulses which result in the reflect

action.

D -1

L -1

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P1 - The knee jerk action involves two types of neurons named

afferent and efferent neurons.

P2 - As the hammer strike, the force stretches the quadriceps

muscle and stimulates the stretch receptors in the muscles

triggering a nerve impulses

P3 - Afferent neurons transmit the information to the efferent

neuron in the spinal cord

P4 - The efferent neurons transmit the information to the

quadriceps muscle as an effector and the muscle contracts

thus swing the leg forward

P5 - If the patient is able to swing the leg forward, it indicates that

the patients nerve system is still functioning

P6 - If there is no response, it shows that the patients nervous

system fails to function properly

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2(a) Able to explain why the pituitary gland is known as the

master gland

P1 - because it secretes several hormones that control other

endocrine glands

P2 - for example, TSH is secreted to stimulate thyroid gland to

release thyroxine

//accept any correct hormones and their function

2 2

2(b) Able to explain the involvement of both the nervous system

and the endocrine system in this situation

P1 - The situation is called fight or flight situation

P2 - Nerve impulses from the eyes (receptors) travel to the brain

P3 - The information is interpreted and the brain sends nerve

impulses to the adrenal glands

P4 - The adrenal glands are stimulated to release adrenaline

P5 - The hormone increases the heartbeat rate, blood pressure

and blood flow to the muscle

P6 - The breathe become faster and deeper

P7 - metabolic activity and glucose level in blood increase

P8 - The skeletal muscles become more energized and enable

a person to fight off an attacker or flee immediately

8 8

3(a) Able to explain how the body of a healthy person restores the

blood sugar level to normal if the level drops too low

P1 - The islet cells in the pancreas are stimulated to release

glucagon

P2 - Glucagon stimulates the liver to break down glycogen to

glucose

P3 - This restores the blood sugar level to normal

P4 - Glucagon also promotes lipid breakdown

P5 - This releases fatty acids that can be metabolized to

generate energy

P6 - This restores the blood sugar level to the normal range

6 6

3(b)(i) Able to complete the diagram and state the role of the nephron

D 1

L 1

4

Bowmanscapsule

Proximalconvoluted

tubule

Loop ofHenle

Distalconvolutedtubule

Collectingduct


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Chapter 13 Reproduction and Growth


TotalMark

1(a) Able to describe the processes which occurred from P to Q

P1 Ovulation releases a secondary oocyte , which enters the

oviduct.P2 The secondary oocyte starts meiosis II which progresses until

metaphase II.

P3 The nuclei of a sperm cell (n) and the ovum (n) fuse and form

a diploid zygote (2n). // A sperm fertilize the ovum to form a

zygote.

Max 4 4

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P4 Zygote begins to divide repeatedly by mitosis as it travels

along the fallopian tube towards the uterus.

P5 Morula is form followed by blastula.

P6 Implantation occur / The blastocyst attaches itself to the

endometrium.1(b) Able to explain the statement

P1 Cigarette contain nicotine / DDT / lead particles.

P2 The wall of maternal blood capillaries and the wall of foetal

blood capillaries are semi-permeable.

P3 Nicotine, drugs and alcohol are small in size.

P4 Nicotine, drugs and alcohol can diffuse from maternal blood

capillaries to foetal blood capillariesP5 through the placenta

P6 The substances carried by umbilical vein to the foetus.

P7 Nicotine, drugs or alcohol can affect the development of

foetus

P8 (example) cause disable / miscarriage . birth defect/ illness in

the resulting baby.

Max 6 6

2(a) Able to explain under what type of condition Method A can be

used to help Mr. and Mrs. Ali.

P1 Method A is use if the fallopian tubes of Mrs. Ali are blocked.

P2 sperm cannot reach the ovum, fertilization fail to occur.

P3 fertilization has to be done outside the body.

P4 developed zygote/embryo then retransfer and implant in the

uterus of Mrs. Ali.

P5 the embryo then undergo normal development in the uterus of

Mrs. Ali as normal pregnancy.

5 5

2(b) Able to discuss the role of Madam X in Method B and the issue

that may arise.

P1 Method B is used if the uterus of Mrs. Ali fail to carry the

implanted embryo because of damaged or abnormal uterus.

P2 Madam X is the woman who is willing / hired to carry the

Max 5 5

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implanted embryo to full term.

P3 Madam X is known as surrogated mother.

P4 Genetically the baby belongs to Mr. and Mrs. Ali.

P5 Who is the real biological mother of the baby, Mrs. Ali or

Madam X?

P6 There are cases that the surrogated mother refuse to return

the baby to the couple after giving birth.

3 Able to explain how the formation of the embryo sac and pollen

grain process occurs.

The formation of the embryo sac

F1 - The ovule develops from the ovarian tissue. It has a diploidembryo sac mother cell(2n)

F2 - Embryo sac mother cell undergoes meiosis to form a row of

four haploid cells called megaspores

F3 - Three of the four megaspores degenerated, leaving one in the

ovule

F4 - (The megaspore continues to grow and enlarges, filling up

most of the ovule). The nucleus of the megaspore then

undergoes mitosis three times to form eight haploid nuclei.

F5 - Three of the eight nuclei (migrate to one end of the cell) to form

antipodal cells, another two nuclei to form polar nuclei and

one of the three nuclei develops into an egg cell/female

gamete/ovum and flanked two synergid cells

The formation of pollen grain

F1 - Pollen grain are formed in the anther, an anther has four pollen

sacs.

F2 - Each pollen sac contains hundreds of cells called pollen mother

cells (2n)

F3 - Each pollen mother cell undergoes meiosis to produce four

haploid microspores(n)

F4 - The nucleus of each microspores then divided by mitosis to form

10 10

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a tube nucleus and generative nucleus.

F5 - The microspores develop into pollen grains4 Able to describe how pollination leads to the formation of fruit

and seed in a flowering plant (angiosperms).

F1 - Pollen grains have been released from the anther to the stigma

for pollination by insects or wind

F2 - The sugar solution (sucrose) secreted by the stigma stimulates

the pollen grain to germinate and form a pollen tube

F3 - The pollen tube grows down the style towards the ovule

F4 - The generative nucleus divides by mitosis to form two male

gamete nuclei

F5 - The male gamete nuclei move down the pollen tube led by thetube nucleus

F6 - When the pollen tube reaches the ovary, it penetrates the

ovule through the micropyle

F7 - The tube nucleus degenerates, leaving a clear passage for the

male nuclei to enter the embryo sac

F8 - Double fertilization occurs in the ovule. One male nucleus

fuses with the egg nucleus to form a diploid zygote(2n)

F9 - The other male nucleus fuses with the two polar nuclei to form

a triploid nucleus(3n

F10 - (After fertilization), the triploid nucleus divides rapidly by

mitosis to forms an endosperm and zygote divides by mitosis

develops into suspensor and embryo.

F11 - The ovule develops into a seed while the ovary enlarges and

develops into a fruit

Max 10 10

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Chapter 14 Inheritance

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TotalMark

1 Able to explain the possibilities of the blood group and the genotypes of

the offspring when the fathers blood group is A and the mothers blood

group is B.

P1 The ABO blood group in humans is controlled by three alleles IA, , IB and Io.

P2 Alleles IA and IB are codominant but allele Io is recessive.

P3 There are four possibilities : Blood group AB, A, B, O

Father Mother

(a) Parental genotypes: IA IA X IB IB

Gametes IA IB

Genotype F1 IA IB

Phenotype F1 All offspring have blood group AB

Father Mother

(b) Parental genotypes: IA IA X IB IO

Gametes IA IB IO

Genotypes F1 IA IB IA IO

Phenotypes F1 AB A

Phenotipic ratio 1 : 1

@

Phenotype F1 50% of offspring have blood group AB

and 50% have blood group A

Father Mother

(c) Parental genotypes: IA IO X IB IB

Gametes IA IO IB

Genotype F1 IA IB IB IO

Phenotypes F1 AB B

Phenotypic ratio 1 : 1

@

Phenotype F1 50% of offspring have blood group AB

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Fertilisation

meiosis

meiosis

meiosis

meiosis

Fertilisation

Fertilisation

Fertilisation


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Chapter 15 Variation


TotalMark

1 Able to

(i) State the example of continuous variation and discontinuous

variation

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

and discontinuous variation

Example of continuous variation: Height or weight

Example of discontinuous variation: ABO blood group

Similarity:

- both create varieties in the population of species

- both type of variations are caused by environment factor or

genetic factors or both

Differences

Continuous variation Discontinuous variationP1 Graf distribution shows a

normal distribution

P1 Graf distribution shows

a discrete distribution

P2 The characters are quantitative

/ can be measured and graded

(from one extreme to the other)

P2 The characters are

qualitative / cannot be

measured and graded

1

1

1

1

1

1

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(from one extreme to the

other)

P3 Exhibits a spectrum of

phenotypes with intermediate

values between the highestand the shortest in the

population.

P3 Exhibits a few distinctive

phenotypes with no

intermediate values inbetween them.

P4 Influenced by environmental

factors

P4 Is not Influenced by

environmental factors/ Is

caused by genetic

factors and also by the

mutation of genes and

chromosomes.

P5 Two or more genes control the

same character

P5 A single genes

determines the

differences in the traits of

the character

P6 The phenotype is usually

controlled by many pair of

alleles

P6 The phenotype is

controlled by a pair of

alleles

1

1

1

1

2 Able to state the importance of variation

P1 Variation provides new genetic material for the survival of the

fittest,

P2 e.g. the mutated genes of the black peppered moth

P3 Variation prepares a species to survive when there are changes in

the external environment, like after a volcanic eruption, or in global

warming;

P4 e.g. the black peppered moths survive well in a soot-covered

environment.

P5 A population with a varied genotypes or genotypes is useful in

spreading the particular species over a wider range of habitats;

P6 e.g. the house sparrow.

P7 Produced phenotype/physical differences among individuals.

Max 6 6

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P8 E.g. No human are alike even though they are identical

3 Able to discuss genetic and environment factor affecting

variation

Genetic factors

F1 crossing over during prophase 1/meiosis 1

P1 occur between chromatid from a pair of homologous chromosomes

P2 the exchange of parts between chromatid results in new genetic

combination.

P3 produced a large number of gametes with different genetic

composition.

F2 independent assortment

P4 homologous part of chromosome are arranged randomly on

metaphase plate/during metaphase 1

P5 during anaphase 1,each homologous pair of chromosomes

separate.

P6 resulting in an independent assortment of maternal and paternal

chromosomes into daughter cells

F3 Random fertilization

P7 sperms and ovum with a variety of combinations of

chromosomes/ genetically different are randomly fertilized.

P8 Thus, variation exists between individuals from the same

species// zygote produces will have a variety of diploid

combination.

F4 Mutation

P9 mutation causes permanent change in the genetic

composition/genotype of an organism

Environmental factor

Max10 10

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F5 (can cause variation among individuals at same species) by

interacting with genetic factor.

P10 examples of factor at least 2 type of food/ exercise/ skill/

experience/ education/ sunlight/ climatic

Any 9 from genetic factor

And any 1 from environment factors

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modul ka f5 2011 combine (t)

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