Biology Paper 2 Marking Scheme

SOLAF 2 4551/2

SKEMA JAWAPAN SOLAF 2 K2

1 a (i) Mitosis

(ii) C, B, E, A, D

1

1

b P : chromosomes

E : metaphase

1

1

c P1 sister chromatid separates

P2 moves to opposite poles

1

1

d (i) Increase the number of cells in organism 1

d (ii) 1

1

fMitosis Meiosis

Occurs in Somatic cells Occurs in reproductive cell

To replace damage/dead cell. To produce gamates

No cross over during prophase Cross over occurs during

prophase

Produce 2 daughter cells Produce 4 daughter cells

Daughter cells are diploid (2n) Daughter cells are haploid

(n)

Genetically identical Not genetically identical

No genetic variability Genetic variability occurs

Any 2 pairs

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4551/1 © 2011 Hak Cipta Jabatan Pelajaran Perak SULIT

1

SOLAF 2 4551/2

2 a (i)

Note : no. of water molecule correct – 1

1

(ii) osmosis 1

(iii) Able to explain what is osmosis

P 1 movement of water molecule across the plasma membrane

P2 from low concentration area to the high concentration area // following concentration gradient

P3 through semi permeable membrane

1

1

1Max 2

b (i) Solution A : hypotonic

Solution B : hypertonic

1

1

(ii) in solution A : causes water to diffuse into the cells by osmosis

the cell expands and burst // undergoes hemolysis

in solution Bcauses water to diffuse out from the cells

cell shrink / undergoes crenation

1

1

1

1Max 3


2

A B

SOLAF 2 4551/2

c P1 concentrated salt solution is hypertonic to the cells

P2 water diffuses out from the cells by osmosis // fish is dehydrated

P3 not suitable for the growth of microbes

1

1

1

12

3 a (i) Able to name the hormone secreted by gland P

P : ADH // FSH // LH Q : Thyroxine

11

(ii) Able to state the condition caused by the growth of gland Q

Goiter 1(iii) Able to suggest how to overcome the problem in (a)(ii)

Taking enough iodine in our diet 1b Able to label adrenal gland with letter S correctly.

Answer

1

c Able to explain the role of gland R in regulating theperson blood glucose concentration from 0 minuteto 90 minutes

Sample Answer

P1 : From 0 to 60 minutes, the blood glucose level increases more than the normal level

P 2 : Islet cells in gland R is stimulated to secrete insulin

P3 : Insulin stimulates the conversion of excess glucose to glycogen (in the liver)

1

1

1


3

S

S

SOLAF 2 4551/2

P4 : This cause the glucose level to return to the normal level at the 90 th

minute1

Max 3

d (i) Able to state the person’s blood osmotic pressure based on the situation given

The blood osmotic pressure increases 1

(ii) Able to explain how gland Q involves in returning the osmotic pressure of the blood to normal levels.

P1 : The osmoreceptor detects the increase in the osmotic blood pressure

P2 : Gland Q is stimulated to release more ADH

P3: ADH is transported by blood to the kidneys

P4 : ADH increases the permeability of the wall of distal convoluted tubule and collecting ducts

P5 : More water is reabsorbed from the filtrate into the blood

1

1

1

1

1

Max 3

12

4 a Able to name substance R

R : ammonium compounds 1

b (i) Able to name process P

Nitrogen fixation

(ii) Able to explain the process P in leguminous plants

P1 nitrogen fixing bacteria lives in the root nodules of leguminous plants.

P2 i.e Rhizobium species

P3 convert nitrogen in atmosphere into nitrates/ ammonium compound

1

1

1

1


4

SOLAF 2 4551/2

c Able to explain the effect when there is no denitrifying bacteria

P1 Process Q / denitrification did not take place

P2 no nitrate will be broken down into nitrogen and oxygen

P3 causes imbalance of gases in atmosphere // atmospheric nitrogen decreses

1

1

1

d Able to explain the effect of using excessive fertilizers in agriculture

P1 run off of excess fertilizers will increase the nutrient in aquatic ecosystem

P2 causing excessive growth of algae // algae bloom

P3 algae covers the water surface and block sunlight

P4 aquatic plant could not carry out photosynthesis and die

P5 algae die, decomposing bacteria uses oxygen for its activity

P6 reduce the oxygen content in water // increase in BOD

1

1

1

1

1

1

Max 4

12

5 a (i) Able to name the type of twins

P : Siamese twins

Q : identical twin

(ii) Able to explain the formation of twin Q

P1 An ovum is fertilized an a sperm to produce zygote

P2 the zygote splits into two halves by mitosis

P3 each halves develop into foetus

P4 the two foetus sharing the same placenta

1

1

2

1

1

1

1Max 3


5

SOLAF 2 4551/2

b P1 Both twin in P and Q have similar genetic content

P2 the twins in P and Q have the same physical characteristics

P3 twin in P and Q have similar sex

P4 twins in Q are completely separated and grow as different individual twin in P are not completely separated, but joined at certain parts of the

bodies// sharing organs

1

1

1

1

Max 3

c Able to explain the differences between R and S

P1 R is formed from fertilization of an ovum by a sperm S is form from two different sperm that fertilize two different ovum

P2 in R, one zygote is formed and split into two by mitosis In S, two different zygote are formed

P3 R is identical twin S is fraternal twin

P4 Twin in R share the similar genetic content // have identical characteristics

Twin in S have different genetic content // have different characteristics

P5 Twin R have same sex Twin S may have same sex / different sex

1

1

1

1

1

Max 4

12

6 a(i)

F1P1

F2P2

F3P3

Able to describe the adaptive characteristics in leaves tissue to carry out photosynthesis

Have a layer of epidermispenetration of sunlight for photosynthesis

pallisade mesophyll cells are arranged upright and packedcarry out photosynthesis at maximum rate

spongy mesophyll are loosely arranged create air spaces for gases exchange

11

11

11


6

SOLAF 2 4551/2

4

a(ii) P1P2P3P4

P5P6P7

P8P9

P10

P11P12P13P14

At 12.00 night the concentration of CO2 is highno uptake of CO2 for photosynthesisthe light intensity is lowstomata closed

at 6.00am the concentration of CO2 started to decreaseincrease in light intensity cause stomata starts to openCO2 is absorbed by the leaves for photosynthesis

at 12.00 noon, the concentration of CO2 is the lowestleaf receives maximum sunlight to carry out photosynthesis // rate of photosynthesis is maximumstomata open widely

at 6.00pm, the light intensity started to decreaseCO2 concentration starts to increaseless intake of CO2 for photosynthesis small opening of stomata

1111

111

11

1

1111Max 10

bS1S2

D1

D2

D3

D4

D5

Similarity both digest celluloseboth have microbes/ bacteria/ protozoa in alimentary canal to produce cellulase

Animal P Animal Q

The stomach consists of four chambers

The stomach consists of one chamber

Size of caecum is small Size of caecum is big and long

bacteria and protozoa are found in rumen and reticulum

bacteria and protozoa are found in caecum

cellulose is digested in rumen and reticulum

cellulose is digested in caecum

food passes the alimentary canal once

food passes the alimentary canal twice

11

1/0

1/0

1/0

1/0

1/0

2s+4dMax 6

20


7

SOLAF 2 4551/2

7 a(i) F1P1

F2P2

F3P3

F4P4

Small and large number of alveoliincrease the surface area for gases exchange

Covered with large network of capillariesease the gases exchange and transport of gases to body tissues

moist surfacegases can easily dissolved and diffuse into/ out blood

alveolus has thin wall ie one cell thickenhance the gases exchange

11

11

11

11

3F+3P6

a(ii) P1P2

P3

P4P5

P6

oxygenated blood flows through blood capillarieshave high partial pressure of oxygen compared to body tissuesoxygen diffuses into the body tissues down the concentration gradientthe cells carry out cellular respiration to produce energythis will increase the partial pressure of carbon dioxide in body tissues compared to blood.carbon dioxide diffuses out from body tissues into blood capillary to be sent to lungs

11

1

11

1

6

b P1

P2P3P4P5P6P7P8

the muscle respired by anaerobic respiration during 100 meters eventBecause oxygen supply cannot support the oxygen needs / not enough / no oxygenso the glucose is not oxidised completely and will form lactic acidsmuscle is said to experience oxygen deptwill inhale air and exhale air very fast// breathing rate is faster oxygen is send more to the cells.the lactic acid is oxidised to carbon dioxide, water and energy

1

11111118

20

8 a P1P2

P3

P4

Pioneer species for disused pond is submerged plant activities of pioneer plants cause a change in habitat, make it suitable for another speciesThe dead pioneer plant will decay and deposits in the bed of pondincrease the nutrient content in pondpond becomes shallower

11

1

1


8

SOLAF 2 4551/2

P5P6

P7

P8P9P10P11

P12

this condition is suitable for the growth of floating plants which replace the pioneer speciesfloating plants cover the surface of pond, preventing the light from penetrating the waterthe rate of photosynthesis decreasessubmerged plants die and add humus to the bedfloating plants is replaced by amphibian plantpond becomes drier and suitable for the growth of herbaceous plantssoil becomes drier and more fertile and more suitable for teresterial plants soil becomes drier and more fertile and more suitable for teresterial plants

11

1

1111

1

Max 10b P1

P2

P3

P4

P5P6P7P8

P9

P10P11

Preservation refers to the management of ecosystems and the environment to ensure a healthy and balanced natural environment // take care of the environmentConservation is the wise use of natural resources for environmental protection // very careful use / control use of natural resources (This includes) the protection, management and renewal of natural resources.Need to ensure that the flora and the fauna are not extinct / lost foreverFuture generation can learn about the natural ecosystem.To maintain the quality of the environmentdo not disturb the energy flow / food web / food chainto keep the plant / animal which have high medicinal value // source of food help to maintain the natural cycles ( eg. Water cycle, nitrogen cycle, carbon cycle)provides natural catchment areaspreserves natural ecotourism / recreational activitiesprovides natural catchment areaspreserves natural ecotourism / recreational activities

1

1

1

1

1111

1

11Max 10

20

9 a P1

P2P3P4

P5

haemophilia is caused by gene mutation for synthesis of blood clotting factor has undergone mutationthe no clotting factor for the patient.It is caused by a recessive allele which is linked to X chromosomemore common in male because the trait will express itself even if there is only one recessive allele present on his X chromosome.for female, two recessive alleles have to be present on her X chromosomes in order for her to become haemophiliac.

1

111

1


9

SOLAF 2 4551/2

P6

P7

P8

P9

P10

P11

P12 Ali has 25% probability to have haemophilia.

1

1

1

1

1

1

1Max 8

b F1F2

S1

D1

D2

D3

D4

D5

Blood group – discontinous variationheight – continous variation

similarityS1 - both creates varieties among organism

differences

Blood group Height

D1 - Graph shows discrete distribution

D1- Graph shows normal distribution

D2- Influenced by genetic factor D2- Influenced by genetic and environmental factor

D3 - Traits are controlled by a single gene

D3 -Traits are controlled by more than one gene

D4 - Characters cannot be measured and graded // qualitative

D4 - Characters can be measured and graded // quantitative

D5 – the difference between individual is distinct

D5 – the difference between individual not distinct

112

1

1/0

1/0

1/0

1/0

1/0

1S+3D


10

XHY XHXh

XH Y XH Xh

XHYXHXhXHXH XhY

x

Normalfemale

Carrierfemale

Normalmale

Haemophiliac male

P6 - Parent

P7 - Meiosis

P8 - Gamete

P9 - Fertilisation

P10 - F1

generationP11 -

Phenotype

SOLAF 2 4551/2

c P1

P2P3

P4P5

P6P7

P8

Gene mutation takes place when there is a change in the sequence of nucleotide basesBase deletion removal of a base from a normal gene sequence

Or

Base insertioninsertion of an extra base into a normal sequence

Or

Base substitution replacement of one base with another.

Or

Gene mutation results in a defective protein being produced or no protein is produced at all

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1

8

20


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Biology Paper 2 Marking Scheme

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