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Identity card number:. Centre number :

964/2 TRIAL 2012

BIOLOGY

PAPER2

STRUCTURE AND ESSAY

Two and a half hours

JABATAN PELAJARAN NEGERI KELANTAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

Instructions to candidates:

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU

ARE TOLD TO DO SO.

Answerall questions in Section A. Write your answers in the

spaces provided.

Answer any four questions in Section B. Write your answers on

the answer sheets provided. Begin each answer on a fresh sheet

of paper and arrange your answers in numerical order. Tie your

answer sheets to this question paper.

Answers should be illustrated by large and clearly labeled

diagrams wherever suitable.

Answers may be written in either English or Bahasa Malaysia.

For examiners use

Section A

1

2

3

4

Section B

Total

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* This question paper is CONFIDENTIAL until the examination is over. CONFIDENTIAL*

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Section A [40 marks]

Answerall questions in this section

1. Figure below is a diagrammatic model of a section of cell surface membrane.

(a) (i) State the name usually given to this model. [1 mark]

(ii) Give two reasons why it is so called. [2

marks]

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(b) Name the structures labeled A, B and C. [3 marks]

A :

B :

C :

(c) What is the function of the channel labeled in the diagram? [1 mark]

(d) Many substances are said to be actively transported across membranes. Explain what

is meant by active transport. [2 marks]

(e) Suggest one possible function of the carbohydrate chains attached to the surface of

the membrane. [1 mark]

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...

.

(c) Calculate the actual length of the mitochondrion , if the magnification is

x70000. Show your working and give your answer in suitable units. [3

marks]

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3 The graph below shows the change in membrane potential during the passage of a nerve

impulse.

(a) (i) What is meant by the resting membrane potential of an axon? [1 mark]

(ii) What is the value of the resting potential? [1 mark]

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(b) Which labelled part of the graph corresponds to:

(i) depolarization of the axon membrane?

[1 mark]

(ii) repolarisation of the axon membrane? [1 mark]

(c) Describe briefly what happens to the axon membrane:

(i) At P [2

marks]

........

(ii) At R [2

marks]

...........

.....

(d) Explain why the membrane is more negative at S. [2 marks]

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..

4. The pedigree below shows the inheritance of red-green colour blindness .Red-green

colour blindness is a sex-linked recessive condition. The gene for colour blindness is carried on

the X-chromosome.

(a) Why is X linkage more common than Y linkage? [2 marks]

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(b) Assuming that the alleles C and c control the expression of the trait, state the

genotype of the following individuals labelled A, B, C, D and E. [5 marks]

A:.

B:.

C:.

D:..

E:..

(c) By means of a genetic diagram , what is the probability of a red green colur blindness

father and a mother carrying the allele for red green colour blindness having a red

green colour blindness daughter? [3 marks]

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Section B [60 marks]

Answer any four questions in this section

5 (a) Draw and label the general structure of deoxyribonucleic acid (DNA). [2 marks]

(b) DNA replicates during interphase. Describe this process. [13 marks]

6 Describe the process that occurs at the granum of chloroplast. [15 marks]

7 (a) Describe the process of ultrafiltration and selective reabsorption that occurs in the

kidney. [11 marks]

(b) Explain how the kidney produces urine that is more concentrated than blood.

[4

marks]

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8 Some organisms overcome the harsh conditions of the environment by hibernation,aestivation or diapauses. Explain the importance of each of the phenomenom for the survival in

animals. [15 marks]

9 (a) What is meant by vector in DNA cloning? State the characteristics of this vector.

[5 marks]

(b) Describe how DNA cloning is carried out by bacteria. [10 marks]

10 (a) Describe the genetic and environmental factors that cause variation in an organism.

[12 marks]

(b) Explain the importance of variation in a population. [3 marks]

JABATAN PELAJARAN NEGERI KELANTAN

PEPERIKSAAN KENDALIAN BERASASKAN SEKOLAH (TRIAL)

TINGKATAN 6 TAHUN 2012

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MARKING SCHEME

BIOLOGY (PAPER 2)

STRUCTURE AND ESSAY

Marking Scheme

1(a)(i) fluid mosaic model 1

[1 mark]

(ii) scattered mosaic arrangement of proteins that float in the phospholipid layer 1

The lipid and protein molecules can move laterally (and rotate on their axis) 1

[2marks]

(b) A : phopholipid bilayer 1

B : extrinsic/peripheral protein 1

C : intrinsic protein 1[3 marks]

(c) To allow transport of polar molecules and ions across the membrane byfacilitated diffusion

1

[1 mark]

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Magnification

= 79 x 103 m

70000

= 1.13 m

1

1

[3 marks]

Total :10

3 (a)(i)

Is the potential difference between the inside and outside of an axon membranewhen the axon is in a resting state

1

[1mark]

(ii) - 70mV 1

[1mark]

(b)(i)

P 1

[1mark]

(ii) R 1

[1mark]

(c) (i) More Na+ ions channels open //influx/rapid diffusion of Na+ ions into the axon 1

Membrane is polarized/ potential inside the membrane is positive 1

[2 marks]

(ii) K +ions channels open//K+ ions diffuse out of the axon 1

Membrane is repolarised/potential inside the membrane becomes negativeagain

1

[2 marks]

(d) Sodium channels are closed and potassium channels remain opened 1

K+ ions continue to flow out 1

[2 marks]

Total:10

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4 (a) The X chromosome is much larger than Y chromosome 1

Carries genes for non-sexual characteristics as well as those determining sex 1

[2 marks]

(b) A : XCY 1

B : XCXC 1

C : XCXC 1

D : XcXc 1

E : XcY 1

[5 marks]

(c) Female Male

P XCXc X XcY

(normal but carrier) red green colour blindness

Gamet XC Xc Xc Y

progeny XCXc XCY XcXc XcY

female carrier male normal female RGB maleRGB

probability of having a red green colour blindness daughter = 1

4

1

1

1

[3 marks]

Total: 10

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5 (a) Structure of DNA

Label

Diagram

1

1

[2 marks]

(b) the method of DNA replication is proposed by Watson and Crick 1

is known as semi-conservative replication. 1

the hydrogen bonds between the bases break, the DNA double helix unwind 1

forming a replication fork 1

process is catalysed by enzyme helicase 1

each strand acts as a template for the formation of a new strand 1

DNA polymerase catalyses the synthesis of new daughter strand on the template

strand continuously

1

in the 5 to 3 direction 1

free DNA nucleotides join up to the exposed bases by specific base-pairing 1

adenine pairs with thymine and cytosine with guanine 1

the leading strand is synthesized continuously in a direction towards the

replication fork

1

the lagging strand is synthesized as a series of segments called Okazaki fragments 1

in a direction away from the replication fork 1

DNA ligase join the sugar-phosphate backbones of Okazaki fragments to form a

single strand

1

specific base pairing has ensured that two identical copies of the original DNA

have been formed

1

each new DNA molecule contains one old and one new strand 1

16 max 13

6 process that occurs at the granum of chloroplast is light reaction 1

reactions requiring light

i.

1

Light reaction takes place in two pathways i.e cyclic photophosphorylation and 1

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non-cyclic photophosphorylation

Non-cyclic photophosphorylation

PSII/P680 absorbed light energy and the electrons become excited

1

electrons are accepted by the primary electron acceptors 1

electrons are then transferred to PS1/P700 1

through electrons transport system 1

in chemiosmosis process 1

PS1/P700 absorbed light energy and the electrons are excited 1

electrons are transferred to the primary electrons acceptors 1

which is then transferred to NADP+ to produce NADPH 1

water molecules split up/photolysis of water take place 1

releasing electrons and H+ 1

to replace electrons in PSII/P680 1

Cyclic photophosphorylation

PSI/P700 absorbed light energy which excite the electrons to a high energy

level

1

electrons are accepted by the primary electron acceptors1

through electrons transport system 1

electrons are recycled back to PS1/P700 1

ATP is produced from ADP 1

19 max 15

7 (a) Ultrafiltration

blood in the glomerulus undergoes ultra-filtration

1

in the glomerulus hydrostatic pressure is higher compared to the osmotic pressure 1

caused by the diameter of efferent arteriole which is smaller than the diameter of

afferent arteriole

1

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the numerous tiny pores or fenestrations (0.1m in diameter) between the

endothelium cells of capillary walls permits many smaller molecules to pass

through

1

bigger molecules such as plasma protein and red blood cells are retained

in the blood

1

glomerular filtrate has a chemical composition similar to that of the blood plasma /

contains glucose, amino acids, vitamins, some hormones, urea, uric acid

,creatinine, ions and water but without white and red blood cells, platelets and

plasma protein molecules such as albumins and globulins

1

6 Max 4

Reabsorption

all glucose and amino acids are actively reabsorbed from the filtrate into the

proximal convoluted tubule cells

1

some sodium ions are actively reabsorbed here 1

a large volume of water are reabsorbed by osmosis from the tubule to the blood 1

(the descending limb is permeable to water)

water is drawn out by osmosis to the interstitial fluid in the medulla 1

the loss of water concentrates Na+ and Cl- in the descending limb 1

(the ascending limb is impermeable to water)

sodium and chloride ions diffuse out from the thin segment 1

the thick ascending limb actively transport sodium and chloride ions from the

filtrate into the medulla

1

this raises the osmotic concentration of the tissue fluid in the medulla 1

as a result water is reabsorbed from the collecting duct and descending limb by

osmosis

1

the loop of Henle functions as a countercurrent multiplier. 1

active reabsorption of sodium ,chloride and calcium ions ions occur in the distal

convoluted tubule

1

11 max 7

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(b) Production of urine which is more concentrated than blood is

(b) due to uptake of water by osmosis from the glomerular filtrate into the blood

cappilaries

1

occurs as a result of increased permeabilities of the walls of convoluted distal

tubule and collecting ducts to water

1

caused by the presence of ADH/anti-diuretic hormone 1

diffusion of urea out of the collecting ducts into the interstitial region of the

medulla

1

increases the osmotic concentration resulting in the removal of water. 1

5 max 4

Total 15

8 Hibernation

is a period of relatively low metabolic activity 1

associated with periods of low temperature/in winter 1

enables hibernating species of amphibian, reptiles, birds and

mammals

1

to survive at times when food is in short supply/to save its stored

food

1

by reducing their energy demands to a low level 1

involves a fall in body temperature to that of environment 1

and a reduction in heart rate, ventilation rate, metabolic rate,

growth and development

1

Stored fat in the mammals tissues are used to provide energy 1

for endotherm, hibernation is a way to maintain body temperature during cold

season

1

9 max 5

Aestivation

as response to drought season/hot/dry

1

eg: lungfish/amphibian/snail/worm/insect 1

can live in dry swamp/drought/summer by digging/move in mud 1

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undergo a period of metabolic inactivity 1

secrete a layer of slime which open on its lips to breathe atmospheric air 1

which enables them to survive conditions of extreme dehydration 1

during aestivation period, the stored fat and part of muscle tissue/protein is used 1

7 max5

Diapause

Is a form of arrested development shown by insects 1

Can occur at any stage in the life cycle : egg, larva, pupa oradult.

1

Main factors inducing diapause: daylength(photoperiod), food

availability,temperature and moisture.

1

Example: the Silkworm Bombyx mori has an embryonicdiapause (egg stage), enter diapause when expose to periods oflong daylength.

1

diapause occur to overcome unfavourable condition 1

organised its life cycle with environmental condition/food/light period/ organised

season/chances/mating period

1

6 max 5

Total :15

9 (a) Vectors are:

- DNA molecules that is used to carry foreign DNA into a host cell. 1

- Examples : plasmids and the bacteriophage . 1

[2 marks]

Characteristics

- must be able to replicate its own DNA. 1

- must at least has one specific restriction sites that can be recognized by

restriction enzymes

1

- must carry at least a selectable marker gene; eg ampR (antibiotic resistance) 1

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- can be easily extracted from its host/easier to insert into bacteria 1

- very small in size 1

5 max 3

Total: 5

(b) - target DNA from the donor cells are isolated /extracted 1

- target DNA cleaved/cut into fragments 1

- by specific restriction / endonuclease enzymes 1

- DNA fragments require vectors/plasmids 1

- plasmids are cut/cleaved by the same restriction enzyme 1

- DNA fragments are inserted into the plasmid 1

- forming recombinant DNA 1

- using DNA ligase 1

- recombinant DNA are inserted into bacterial host cells 1

- by transformation /transduction 1

- amplification of recombinant plasmids/ copies of recombinant plasmids are

produced by replication

1

- screening of recombinant DNA 1

- the recombinant DNA /required plasmids are extracted from the bacterial

cell

1

13 max 10

Total :15

10 (a) Genetic factors

Sexual reproduction 1

sexual reproduction involves the processes of meiosis during

gamete formation

1

meiosis allows genes to be reshuffled 1

occurs during prophase I and metaphase I 1

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during prophase I, crossing over occurs resulting in the exchange

of genetic materials between two homologous chromosomes

during metaphase I, segregation of chromosomes occurs

1

1

combination of these two processes produces a wide variety of

different gametes.

1

fertilization between the varied male and female gamete is

random

1

zygote produced contains numerous combination of genotypes 1

Mutation 1

Gene and chromosomal mutation 1

Produces a change in the genotypes 1

which causes the appearance of a new characteristics 1

Hybridisation 1

A cross between individual belonging to two different varieties 1

produces individuals with different traits from the parents 1

generating more variation 1

17 max 9

Environmental factors

Contributes to variable gene expression among individuals 1

can only cause changes in the phenotype 1

The characteristics are not inherited 1

Variation due to environmental sources is reversible 1

4 max3

Toatal: 12

(b) Importance of variation:

Causes some individuals in a population to be better adapted forsurvival than others

1

Enables a population to inhabit a larger range of habitat andniche

1

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A population with little variability has reduced capacity to

adapt to environmental changes

1

which may lead to the extinction of the population or the

whole species

1

is considered as the raw material for the process of natural

selection.

1

5 max 3

Total:15

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trial bio k2 2012

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