Mark Scheme Kertas 2 Sarawak Zone A

4551/2 PAPER 2

SECTION A: MARK SCHEME No

Marking criteria

Mark

Total

QUESTION 1 1(a)

(b)

(c)

Able to name the cell in Figure 1.1 P: Guard cells Q: Smooth muscle / muscle cell R: Efferent neuron / nerve cell/ neuron

1 1 1

Able to state the function of P, Q and R Function of P: Regulate size of stoma.

1

Function of Q: Contraction of Q causes movement of substances.

1

Function of R: transmission of impulses (from CNS to effector)

1

3

1 1

2

1,1

2

Able to complete the cell organization M and N in the spaces provided in Figure 1.2. M: epithelial tissue N: small intestine / stomach

(d)

3

Able to name two types of tissues Sample answer: Epithelial tissue// connective tissue// muscular tissue// nerves tissue. Any two tissues Able to state the function of each tissue named in (d)(i) Sample answer: i.

Epithelial tissue Function: as a protective layer/ excretion/heat regulation

1

ii. Connective tissue Function: such as blood, supplies oxygen and nutrients. iii. Muscle tissue

1

Function: regulate body temperature 1

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iv. Nerve tissue Function: receive and send impulses from and to the central 1

nervous system.

2

( Any two types and functions) TOTAL MARKS

12

QUESTION 2 2.(a)(i)

Able to label the structure P and Q in the spaces provided in Figure 2.1. Sample Answer: P: Nucleus Q: Cytoplasm

(a)(ii)

1 1

2

Able to tick ( √ )the non organelles that are found in plant and animal cells in Table 2.1 Sample Answer: cell wall, plasma membrane and cytoplasm

3 3

(b)(i)

Able to state the characteristics of the molecule that can pass through channel protein of plasma membrane. Sample answer: Small / water a soluble molecule / discharged particle/ion / dissolved gaseous.

(ii)

1 1

Able to explain how molecules pass through channel protein Sample answer Small molecules / ions move through the protein pore in the channel protein by simple diffusion. The molecules are at higher concentration (outside) diffuse the cell cause the molecules to diffuse to a lower concentration into the cells through the pore/ down the concentration gradient.

1 2 1

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(c)

Able to explain the plant cell shown in Figure 2.2 and 2.3 Sample answer: Figure 2.2 F1: the sucrose solution is hypertonic to the plant cell sap

1

E1: water molecules diffuse out by osmosis E2 :the cell undergoes plasmolysis/ cytoplasm shrink.

1 2 1

F1 with E1 / E2 Figure 2.3 F2: the sucrose solution is hypotonic to the plant sap 1

E3: water molecules diffuse in by osmosis E4: the plants become turgid

1 1

F2 with E4 / E5

2

12

TOTAL MARKS

Question 3 3(a) (i)

Able to name the interaction P, Q and R in the spaces provided in diagram 3.1. P: Simbiosis Q: Mutualisme R: Prey-predator

(ii)

1 1 1

3

1

1

Example of saprophytism Mucor sp / mushroom / fungus / saprophytic bacteria on dead organism matter.

(iii) Able to explain the interaction of Q and R by using suitable example. Interaction of Q P1: Example lichen ( algae and fungus) P2: relationship between two species of organism in which both benefit. P3: The alga produces food for itself and also for the fungus and the fungus supplies carbon dioxide and nitrogenous

1 1 1

Maxs: 2

4551/2 PAPER 2 products for the algae to produce its food. Interaction of R P1: Example: Owl (predator) , rat/( prey) P2: control each other population. P3: When the number of predators increase, the number of preys will decrease P4: When the number of preys decrease, the number of predators will decrease.( predators will die of starvation) P5: the number of preys will start to increase.

1 1 1 1 1

Maks: 2

(b) Able to state the differences between the interaction of commensalisms and parasitism. Able to classify the organism in one set of interaction. Commensalism

Parasitism 1, 1

Relationship between two organisms in which one organism benefits while the other neither derives any benefits nor is harmed. ( c )(i)

Relationship between two organisms in which one organism benefits and the other is harmed. 2 1

Interpecific competition

(ii) P1:The population of Staphylococcus aureus decreases at a slower rate P2: but eventually species X will still out compete Staphylococcus aureus

TOTAL MARKS

1 1

3

13

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QUESTION 4 (a)(i)

Able to name hormone P and Q 1 1

Hormone P: Oestrogen Hormone Q: Progesterone (ii)

Able to state the function of hormones X and Y

(b)

Function of hormone X: State the function of hormone X and Y // stimulates growth and development of several primary follicles in the ovary to become a Graafian follicle.

2

1 1

Function of hormone Y: State the function of hormone X and Y// Causes ovulation to occur // promotes development of the corpus luteum // stimulates corpus luteum to produce progesterone and oestrogen// stimulate completion of meiosis I by changing primary oocytes into a secondary oocyte. Able to explain the effect humanal imbalance that causes the disruption of ovulation process. Sample answer: P1: If P secretion is excessive, secretion of Y is stimulated. P2 : Inhibits X, no follicle development / ovulation is Hastened Answer 2 P1: Insufficient of P secretion inhibits secretion of Y. P2: no ovulation ( c ) (i)

Corpus luteum

(c ) (ii)

Able to state the relationship between the structure T and the

2

1 1

2

1

1

level of hormone Q from16th to 28th day P1: On the 16th day, T is formed, hormone Q is secreted. Continuous development of T increases hormone level Q. P2: On the 24th day, T degenerates, the level of hormone

1 1 1

3

Q declines. (d)

Able to state the importance of menstrual cycle P1: Produce female gamete (ovum for fertization) P2: Thickening endometrium – prepare for implantation of the embryo TOTAL MARKS

1 1

2 12

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Question 5 (a) (i)

Able to name the types of variation shown in Diagram 5.1 and Diagram 5.2. Diagram 5.1: Continuous variation Diagram 5.2: Discontinuous variation

(ii)

(b) (i)

Continuous Variation Phenotype influenced by genetic and environmental factors

Discontinuous Variation Phenotype generally determined by genetic factors only

1

Frequency distribution is a bellshaped curve (normal distribution)

Frequency distribution is a bar chart, pie chart or histogram

1

(ii)

2

Able to state the factor that cause the variation in diagram 5.3. 1

Able to state how the factor in b(i) cause the variation. P1: Independent assortment of chromosomes during meiosis // metaphase 1. P2: Crossing over during meiosis I // prophase II. P3: Random fusion of gametes ( during fertilization)

c(i)

2

Able to state two differences between the variation in a(i)

Sample answer: genetic factor (ii)

1 1

Grey coloured Biston betularia

1 1 1 2 1

1

Able to state the reason for c (i) P1: The grey coloured Biston betularia is well camouflaged against lichen- covered tree trunks in unpolluted environment.

1

P2: giving them protection from predators.// avoid being eaten by predator.

1

2

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(iii)

Sample answers P1: Dark melllanic Biston betularia increase. P2: and well camouflaged in polluted environment P3: giving them protection from predators

1 1 1

TOTAL MARKS

2 11

Question 6 (a)(i)

a.iii.

6.b.i.

Able to explain three structural adaptations of the ileum for effective absorption of food F1 – length of intestine is long (6 m) E1 – increase time for food absorption increase SA for absorption

1 1

F2 – inner surface is (highly) folded E2 – to increase the SA for absorption

1 1

F3 – numerous villi (on inner surface of ileum) E3 – to increase the SA for absorption

1 1

F4 – numerous microvilli (on the surface of epithelial cell) E4 – to increase the SA for absorption

1 1

F5 – numerous blood capillaries and lacteal E5 – to transport absorbed nutrients (away)//to maintain concentration gradient (from diffusion of nutrients)

1 1

Any three F and corresponding E Able to describe the digestion of milk in stomach P1 – (Gastric glands in the wall of )stomach secrete gastric juice P2 – Gastric juice (contain mucus, HCl), pepsin and rennin P3 – Rennin coagulates milk by converting soluble milk protein caseinogen to the insoluble casein P4 – Casein is then hydrolyses (digested) by pepsin to peptones P5 - HCL optimises pH in the stomach for action of enzyme.

6

1 1 1 1 1

Able to explain the effects of the removal of organ X on enzyme and hormones and how these affect the digestion and the level of glucose in the body F1 – no insulin secretion E1 – excess blood glucose cannot be converted to glycogen E2 – Blood sugar level increases

1 1 1

Max 4

4551/2 PAPER 2 F2 – no glucagons secretions E 3 – stored glycogen cannot be converted to glucose E4 – low blood glucose level cannot be increased to normal level

1 1 1

F3 – no trypsin secretion. E5 – protein digestion not completed

1 1

F4 – no amylase secretion E6 – Starch digestion not completed F5 – no lipase secretion E7 – lipid not digested. No lipid digestion in body E8 – body do not have enough amino acids and glucose E9 – no fatty acids and glycerol.

1 1 1 1 1 1

All the 5F and any corresponding E 10

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7a.

Able to explain how osmoregulation takes place in his body E1 running causes an increase in body temperature E2 sweating occurs to reduce body temperature E3 lost of water from the body increases the blood osmotic pressure E4 increase osmotic pressure detected by osmoreceptors E5 in hypothalamus of the brain E6 pituitary gland is stimulated to secrete Anti diuretic hormone(ADH) E7 ADH is transported by blood to the kidney E8 ADH increased the permeability of the distal convoluted tubule and collecting tubule to water E9 reabsorption of water increased in both tubules E10 increase in water reduces the blood OP E11 medulla oblongata is also stimulated to produce a feeling of thirst

1 1 1 1 1 1 1 Max 9 1

Synthesis – able to write six consecutive Es 7b.

1 1 1

10

Able to state the differences between the endocrine system and nervous system Nervous system

Aspect

External stimulus Sensory organs Electrical and chemicals Neurones and synapse Rapid/ fast

Stimulus Receptors Impulse // nature of impulse Medium of transmission Speed of transmission of impulse Target organs

Specific locations//organs Quick and short

Response

Endocrine system Internal stimulus Sensory cells Chemical // hormones Blood

2/0 2/0 2/0 2/0

Slow 2/0 Various organs Long lasting

2/0 2/0 Max 10

10

4551/2 PAPER 2

8(a)(i)

Able to describe how the phenomenon in Figure 8 occurs. P1: P2: P3: P4: P5: P6: P7: P8: P9: P10.

8(a)(ii)

The phenomenon is the greenhouse effect Carbon dioxide ,chlorofluorocarbons (CFCs), methane, nitrous oxide and water vapour make up the greenhouse gases. Greenhouse gases produced by burning of fossil also produced belching from cattle and anaerobic respiration in paddy field. effect of deforestation Sunlight enter the Earth’s atmosphere . Most of the radiation is absorbed by the Earth. Some radiation is radiated back into space. Accumulated /increased concentration of greenhouse gases trapped the radiated heat. Warm the atmosphere.

1 1 1 1 1 1 1 1

Max: 8

1 1 1 1

4

Able to suggest ways to minimize the effects of this phenomenon. P1: Reduce burning of fossil fuels. P2: Use alternative energy sources. P3: Slow down deforestation for farming or rice growing. P4: Replant trees which have been cut down.

8(b)

1 1

Able to describe the effect and consequences of deforestation to the ecosystem. F1: No roots system E1: Vegetation can rapidly succumb to soil erotion. E2: Sendimentation of the rivers. F2; No catchment area E3: landslide on steep hills E4: causes flash floods during rainy seasons. F3: Habitat destruction E5: Extinction of flora and fauna E6: Loss of biodiversity

1 1 1 1 1 1 1 1 1

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F4: E7: E8: E9:

9.

Increase of carbon dioxide in atmosphere Greenhouse effect Global warming Climatic change F1, F2,F3 and F4 - 4 marks E1 – E9 - Any 4 Es 4 marks

1 1 1 1 8

Able to illustrate the statement using a monohybrid cross between pure-bred tall pea plant and a pure-bred short pea plant.

(a)

Dominant allele for tall is represented by T Recessive allele for short is represented by t

Parents

:

pure bred Tall pea plant

Genotypes

:

TT

Gametes

:

All T

:

Gametes

:

pure bred short pea plant X

F1 phenotypes:

Genotypes

X

[1 mark]

tt

[1 mark]

All t

[1 mark]

All tall pea plant Tt Tt

T

F2 phenotypes:

X

Tt

t

TT Tall Pea plant

[1mark]

T Tt

Tall Pea plant

Tt

t tt

[1 mark] [1mark]

Tall Short pea plant pea plant 3:1

[1mark] [1 mark]

Total:

[8 marks]

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9(b) Able to explain the inheritance of colour-blindness F1:

A man has XY chromosomes and a woman has XX chromosomes

[ 1 mark]

F2 : Gene for colour blindness is recessive.

[ 1 mark]

F3:

Linked to the X chromosomes .

[ 1 mark]

F4:

The chromosomes has no sex-linked trait.

[ 1 mark]

F5:

Colour blind male ( X n y )

E1:

Xn -

from, mother

E2:

Y -

from father

F6:

Normal male ( X N y )

E3:

X N-

from mother

E4:

Y -

from father

F7:

Carrier female ( X N X n )

E5:

X N-

from mother/ father

E6:

X

n-

from father/ mother

F8:

Colour blind female (X n X n )

E7:

Xn-

E8:

n

X -

[ 2 mark]

[ 2 mark]

[ 2 mark]

from father from mother

[ 2 mark] TOTAL: [ 12 marks ]