mark scheme bio 2 trial perak 2007 biology spm paper

Mark scheme Bio 2 trial perak 2007 sulit

BIOLOGY SPM PAPER 2 - ANSWER SCHEME Section A

No. Of Question

Answer Mark Remark

1(a) P: cell wallQ: vacuoleR: mitochondriaS: nucleus

1111

1(b)(i) Cellulose 11(b)(ii) Permeable 1

1(c) (i) Synthesis energy(ii) glucose + O2 ATP // energy + CO2

12

*Product – 1 m*Subtrats – 1m

1(d)(i) M: cell of a root hair N: red blood cell

11

1(d)(ii) M: small // tiny // large number to increase TSA /V for efficient absorption of water and minerals.

N: no nucleus // biconcave disc (N have large surface area) so absorption of O2 more efficient.

1

1

TOTAL 13

No. Of Question

Answer Mark Remark

2(a)

1

1

1

2(b)(i) Turgid / deplasmolysed 12(b)(ii) Flaccid / plasmolysed 1

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epidermis

Strip AStrip B

Strip C


2(c) In (b)(i) distilled water is hypotonic to the plant sap. Water molecules moves into the plant cell thus cause the cell to becomes turgid.

In (b)(ii) sucrose solution is more concentrated compare to the plant sap. Water molecules will move out from the plant cell to the solution outside causes the cell to become plasmolysed

F-1E-1

F-1E-1

2(c)ii Deplasmolysis. Concentration of water molecules outside is greater compare to the plant sap, thus water molecules move into the plant cells causes it to change back to normal shape // turgid.

11

2(d) The red blood cells undergo crenation 1TOTAL 12

No. Of Question

Answer Mark Remark

3(a)(i) P : replication of DNA // S phaseQ : cytokinesis

11

3(a)(ii) S phase - to ensure the daughter cells receive the same number of chromosomes as the parent cell. - to maintain diploid condition in the new generation.

2E – 2 mark

3(a)(iii) division of cytoplasm 13(b)(i) Anaphase 13(b)(ii) 4 13(c)(i)

1

1

3(c)(ii) 1. To maintain number of chromosomes in the daughter cells equal to parent cell.

2. To maintain diploid number of chromosomes in daughter cells.

1

1

3(d) Tissue culture // cloning // stem cutting 1TOTAL 12

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No. Of Question

Answer Mark Remark

4(a)(i) Place for gaseous exchange // store the oxygen gas before gaseous exchange occur

1

4(a)(ii) Have very large total surface area // have moist surface all the time // have very thin wall ( only one single layer of cell)

1

4(b)(i) - have C-shaped cartilage rings // cartilage rings- keep the trachea open permanently- avoid the trachea from collapse when the outside

pressure is higher than inside pressure during inhalation

- oxygen can continuously flow through trachea to the alveoli / lung

111

1

F – 1 markE – 1 mark

Max – 2

4(b)(ii) L-1A-1

A –Arrow L- Label

Arrows must be drawn inside the trachea, bronchus & bronchiole

4(b)(iii) - external intercostal muscles contract/ internal intercostal muscles relax caused the rib cage moves outwards and upwards

- diaphragm muscles contract, the diaphragm lower and flatten

- the volume of thoracic cavity increase but the pressure decrease (lower the atmospheric pressure)

- air forced into the lung // alveolus

1

1

1

1

Max – 3

4(c) - gill is the respiratory organ for fish but lung is for human

- Gill have filament and lamella to increase the surface area, but lung have alveoli to increase the surface area

- Gill touch / surrounded by water all the time, but lung never touch/ surrounded by water

- Gill received oxygen directly from water, but lung received oxygen from atmosphere via trachea, bronchus and bronchioles

1

1

1

1

Max – 3

TOTAL 12

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Oxygen goes in Carbon dioxide goes out


No. Of Question

Answer Mark Remark

5(a)(i) P – GlomerulusQ – Bowman capsule

11

Max – 2

5(a)(ii) - have very large surface area- have many podocyte (that enable the

ultrafiltration occur tremendously)

11

Max - 1

5(b)(i) Ultrafiltration // filtration 15(b)(ii) - Blood in Glomerulus have very high

hydrostatic pressure (Because the diameter of aferent arteriole is larger then diameter of eferent arteriole)

- All the component of blood plasma diffuse into Bowman capsule except the bigger molecule (such as protein)

- This fluid known as glomerular filtrate

1

1

15(c) - If blood osmotic pressure lower than normal

level, it can be detected by osmoreceptor cell in hypothalamus

- Hypothalamus will produce less ADH ( and send to kidney)

- Convoluted tubule and collecting duct become less permeable to water

- Less water will be reabsorb by the tubule thus- Osmotic pressure of blood increase back to

normal pressure.

OR

- If blood osmotic pressure higher than normal level, it can be detected by osmoreceptor cell in hypothalamus

- Hypothalamus will produce more ADH ( and send to kidney)

- Convoluted tubule and collecting duct become more permeable to water

- More water will be reabsorb by thus tubule- Osmotic pressure of blood become normal

again

1

1

1

11

1

1

1

11

Max-4

TOTAL 11

SECTION B

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No. Of Question

Answer Mark Remark

6(a) Day 0 – 7Follicle– very small– start to develop when receive FSH from pituitary– the wall of follicle will produce estrogenFSH - pituitary start to release FSH, FSH will go to the ovary- FSH stimulate development of follicle

Endometrium- stimulate by estrogen; undergo thickening / repairing

Day 8 – 14Follicle

- become larger, develop to form follicle Graaf FSH / LH/ Estrogen/progesterone

- FSH decrease, LH at maximum level , estrogen at maximum level

- LH stimulate ovulation / completion of meiosis I, estrogen stimulate the thickening of endometrium

- Progesterone level very low

Endometrium- endometrium become very thick (ready to

implantation (of embryo))-

Day 15 - 21Follicle

- Follicle undergoes ovulation/ released oocyte II- The remaining follicle tissue / corpus luteum

secreted small amount of estrogen but large amount of progesterone

FSH/ LH/ Estrogen/ Progesterone- Progesterone stimulated the thickening of

endometrium , halted the secretion of FSH/LH- Development of new follicle and ovulation

stop.

Endometrium- more thicker and highly vascular- ready for implantation of embryo

Day 22 – 28

1

1

1

1

1

1

1

1

Max 3 marks for each stage

- 9 marks

At least the answer shows

the relationship between 3 parameter i.e follicle,

hormone and endometrium

- 1 mark

Total : Max – 10

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Corpus luteum- if no fertilisation, corpus luteum become

disintegrate

FSH/ LH/ Estrogen / Progesterone- FSH, LH and estrogen at minimum level;

progesterone level also drop

Endometrium- endometrium become breakdown &

disintegrate- blood and tissue are shed / lining of uterus

discharge through vagina as menstrual flow.

1

1

1

1

6(b) Similarities - height of man / length of instar increases by

time- both show horizontal line / constant growth

during adult

Difference- Form of graph – Sigmoid form for human and

like series of steps in insect - Age of organism – the height measured yearly,

but in insect used day for measuring the length- Caused of different – human have endoskeleton

but insect have exoskeleton- Stages involve – in human, the curve has three

different phases, but there are five steps in insect // nymphal stages

- Vertical and horizontal line : curve for human did not shows different line (only the curve from continuous points), but there are five different horizontal and vertical lines each

- Zero growth – no point to show zero growth in human, but there are 5 time of zero growth (at horizontal line)

- Sudden growth : no sudden growth for human, but there are sudden growth in insect (at vertical line)

- Ecdysis : no ecdysis in human but ecdysis occurred in insect

- Mitosis : the cells in human undergo mitosis all the time, but in insect, mitosis only occurred at certain time (during ecdysis)

- Absorption of air : in human, there are no absorption of air, but in insect, during ecdysis

1

1

1

1

1

1

1

1

1

1

1

1

Max – 10 marks

*2 marks for similarities, 8

marks for differences

TOTAL 20

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No. Of Question

Answer Mark Remark

7(a) Transportation of respiratory gas.- oxygen enter the alveoli during inhalation- gaseous exchange occurred at alveoli; oxygen

from alveoli diffused to the blood capillaries while carbon dioxide diffused from blood capillaries to the alveoli

- diffusion of oxygen from alveoli to blood capillaries and carbon dioxide from blood capillaries to alveoli caused by different of partial pressure of both gases

- partial pressure of oxygen in alveoli is higher than partial pressure of oxygen in blood capillaries; while partial pressure of carbon dioxide in blood capillaries is higher than partial pressure of carbon dioxide in alveolus

- oxygen diffused in cytoplasm of red blood cell - oxygen combined with haemoglobin to form

oxyhaemoglobin- Oxygen in the form of oxyhaemoglobin then

send to all part of the body. - Heart pumped the oxygenated blood to all body

cells- Oxygen diffused from blood capillaries to the

cell because partial pressure of oxygen in the blood capillaries higher than in the cell

- Carbon dioxide diffused from cells to the blood capillaries because the partial pressure of carbon dioxide in cell is higher than in blood capillaries

- Carbon dioxide send to the lung as ion carbonate, carbonic acid or carbaminohaemoglobin in the blood plasma

- Deoxygenated blood going back to the heart by the vena cava and to the lung by pulmonary artery.

11

1

1

11

11

1

11

11

1

Any 5 from all points

Total ; max-10 marks

7(b) - There is a concentration gradient between soil and epidermal cell // soil water is hypotonic to the epidermal cell

- So water enter the cell of root hairs / epidermal cells by osmosis

- this will caused the osmotic pressure of the epidermal cell decrease / hypotonic compare to the adjacent cells

- water diffuse/ enter the adjacent cells by osmosis

- So water continuously move inward - The movement of water across epidermal cells

and cortex is through the cell wall, cytoplasm and vacuole by osmosis (until endodermal cell).

1

1

1

1

11

Max – 10 marks

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- Endodermal cells have Casparian strip which is impermeable to water.

- So water moves to the xylem through the cytoplasm and vacuole of endodermal cells.

- The concentration gradient of water across the cortex and endodermis create a pushing force, water then move to xylem

- Water in xylem move upwards to the stem and leaves

- The movement of water is help by the root pressure and the transpiration pull

- Cohesive and adhesive properties of water molecules also helped the upwards movement of water along the xylem vessel

- The force form in the xylem vessel known as capillary action (caused by characteristic of xylem – xylem have long, narrow and hollow tubes).

- The capillary action enable the upwards movement of water along the xylem

- During transpiration, water loss to the atmosphere and thus create the transpiration pull.

1

1

1

1

1

1

1

1

1

TOTAL 12

SECTION C

No. Of Question

Answer Mark Remark

8(a) - This situation involved dihybrid inheritance- Adam is homozygous recessive for both hair (hh) and

eye (bb) traits.- His wife is homozygous dominant for hair trait (HH)

and heterozygous for the eye trait (Bb). (Assume H is the gene that controls the black hair and B is the gene that controls the black eye).

- Adam and his wife undergo meiosis I and II - All the gametes (sperms and ovum) are haploid- Adam will produce only one type of gamete i.e

brown hair and brown eye (hhbb), - His wife will produce two type of gametes i.e black

hair and black eye (HHBb)- The gametes (haploid) from Adam and his wife will

fertilise to produce zygote (diploid)- This zygote will receive dominant gene for hair trait

from his/ her mother and recessive gene for eye trait

11

1

111

1

1

1

Max – 10 marks

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from either his/ her mother- So Adam sons/ daughters will have black hair and

brown eye. - Their son / daughter will have heterozygous gene for

hair trait and homozygous recessive for eye traitOR

Parents Adam WifeGenotype (hhbb) (HHBb)

meiosis

hb HB Hb

Fertili-sation

PhenotypeSon/

daughter

HbBb Hhbb

GenotypeSon /

daughter

Black hair, Black hair Black eye. Brown eye

1

1

11

1

1

1

1

1

**From schematic diagram,

only 7 marks can be given

8(b) - Abdul have recessive gene for haemophilia at X chromosome (Xh)

- Genotype for Abdul is XhY- Aminah is heterozygous, she has both recessive gene

and dominant gene (XH Xh)- Abdul undergo meiosis, he will produce gamete Xh

and Y - Aminah also undergo meiosis, she will produce

gametes Xh and XH

- Fertilisation between gametes will produced diploid zygote

- The zygote will receive either XH Xh or Xh Xh or XHY or XhY

- The new generation will be one normal daughter but carrier, one haemophilic daughter , one normal son or one haemophilic son

- The phenotype probability is as follow : ¼ hemophilic girl : ¼ carrier girl : ¼ normal son : ¼ hemophilic son- In this situation, Abdul not suppose to marry Aminah

because they will have 50% hemophilic children

1

11

1

1

1

1

1

1

1

Max – 10 marks

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9(i)

(ii)

- That’s mean the next generation also will suffer this problem continously.

OR

Parent Abdul AminahGenotype XhY XH Xh

Meiosis

Gametes Xh Y XH Xh

Fertili-sation

F1 XH Xh Xh Xh XHY XhY

Genotype Girl Girl Carrier Hemophilia

Boy Boy Normal hemophilia

Proba-bility

¼ ¼ ¼ ¼

(a) Dynamic equilibrium is a state in which the population of an organism is fluctuating along the average value.

- The examples : owl and snake // snake and rat

- Owl is predator and snake is prey

- The predator eats the prey and the prey eaten by the predator.

1

11

1

1

1

1

1

1

2

2

1

1

1

From schematic

diagrammatic - only 8 Marks

Example correct=1 mark

Explanation max = 7

OR

Explanation 5 marks +Graph 2

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- The prey population increase due to foods availability

- The increase of prey population causes the increase of predator population.

- There are food (prey) // more food (prey) in the habitat

- When the predators hunting the preys as a food, the number of prey will be decreasing.

- When there is less number of preys, the population of predator decreases.

- Not enough food (prey) // less food (prey)

- The decrease in population of predator causes the prey to survive and breed, thus population of prey increases back.

- The population cycle repeated and thus the population is said to be in equilibrium state.

1

1

1

1

1

1

1

1

marks ( ie-pattern of the graph correct =1m-Parameter of both axis correct = 1m)

Total 10 marks

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mark scheme bio 2 trial perak 2007 biology spm paper

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