[edu.joshuatly.com] pahang stpm trial 2011 biology paper 2 (w ans) (1)

PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

NEGERI PAHANG DARUL MAKMUR

2011

BIOLOGY

PAPER 2

Two and a half hours

Instructions to candidates DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO Answer all questions in section A. Write your answers

in the spaces provided.

Answer any four questions in section B. Write you

answers on 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

labelled diagrams wherever suitable.

For examiner’s use

Section A

1

2

3

4

Section B

5

6

7

8

9

10

Total

This question paper consists of 12 printed pages .

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

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This question paper is CONFIDENTIAL until the examination is over.

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Section A [40 marks]

Answer all the questions in this section.

1 During research into the mechanism of DNA replication, bacteria were grown for many generations in a medium containing only “heavy” isotop of nitrogen, 15N. This resulted in all the DNA molecules containing only 15N.This is illustrated in Fig 1.1.

Fig 1.1

These bacteria were then grown in a medium containing only “light” nitrogen, 14N. After the time taken for the DNA to replicate once, the DNA was analysed.The results are shown in Fig 1.2

Fig 1.2

(a) Explain how these data support the semi-conservative hypothesis of DNA

replication. …….………………………………………………………………………………..

………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

[ 3 marks ]


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The bacteria continued to grow in the “light”nitrogen, 14N, medium until the DNA had replicate once more. The DNA molecules were analysed. The result are shown in Fig 1.3.

Fig 1.3


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Fig 1.4 shows simple diagrams of DNA molecules,including the nitrogen content of each.

Fig 1.4

(b) With reference to Fig 1.4,select the letter or letters which best represent the

bacterial DNA in Fig 1.1, Fig 1.2 and Fig 1.3

Fig 1.1………………………………………………………………………………. Fig 1.2 ……………………………………………………………………………… Fig 1.3……………………………………………………………………………….

[ 3 marks ]


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The bacteria continued to grow in the “light”nitrogen , 14N, medium until the DNA had replicated once more. The DNA molecules were analysed (c) Complete the bar chart below to indicate the expected results of the composition of

these DNA molecules

[ 3 marks ]

(d) Name the enzyme mainly responsible for the synthesis of a new strand of DNA. …………………………………………………………………………………………

[ 1 mark ]


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2 Figure 2.1 shows the possible pathways taken by water across the root of a plant.

Fig 2.1

(a) Name cell Q.

…………………………………………………………………………………..

[ 1 mark ]

(b) Using only the information in the diagram,explain how cell Q is adapted for its

function. …………………………………………………………………………………..

[ 1 mark ]

(c) Name the process by which water enters cell Q from the soil.

…………………………………………………………………………………..

[ 1 mark ]

(d) How could nitrate ions enter the root if the concentration of nitrate ions outside the plant is less than the concentration inside cell Q?

…………………………………………………………………………………..

[ 1 mark]

(e) Pathway 1 is known as the vascular pathway,as the water passes into and through the cell vacuoles.Name pathway 2 and pathway 3

Pathway 2……………………………………………………………………….

Pathway 3……………………………………………………………………….

[ 2 marks]


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(f) What makes up the Casparian strip in cell type S and what is its function in water transport ?

……………………………………………………………………………………….

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

[2 marks]

(g) Describe two ways in which xylem vessels are adapted for their function of water transport.

………………………………………………………………………………………

………………………………………………………………………………………

……………………………………………………………………………………….

………………………………………………………………………………………

[2 marks ]


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3 In the banded snail (Cepaea nemoralis) the colour of the shell is either yellow (appearing green with the live snail inside) or pink. In a sample of 300 snails from a population in an area of sand dune near Magilligan strand, 192 had yellow shells. The colour of the shell is genetically determined and pink is the expression of the dominant allele.

(a) Calculate the proportion of the population that have yellow shells. ………………………………………………………………………………………….

[1 mark ] (b) Assuming the population to be in Hardy–Weinberg equilibrium for shell colour,

calculate the relative frequencies of the yellow and pink alleles, and the number in the sample of 300 that might be expected to be heterozygous. (Show your working in the space below.)

Relative frequency of the yellow allele …………………………. Relative frequency of the pink allele …………………………… Number of heterozygous snails in the sample …………………..

[3 marks ]

(c)These snails are predated by song thrushes. The thrush, having captured a snail, breaks open the shell by hammering it against a stone, so that examination of the shells found at these “anvil sites” reveals the frequencies of the forms predated. Of 35 shells found at such sites, 28 were yellow.

(i) Calculate the proportion of the predated snails that have yellow shells. …………………………………………………………………………………

[1 mark]


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(ii) Given this information about predation, can it be assumed that the population is in Hardy–Weinberg equilibrium with respect to the shell colour? Explain your answer.

……………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

[ 2 marks ]

(d) Other studies have suggested that predators tend to take proportionally more of the more common forms when faced with the choice of prey. Suggest how such predation would maintain both forms (for example yellow and pink snails) in a polymorphic population.

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

[ 3 marks]


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4 The following diagram is a simplified representation of a technique used in genetic engineering .

(a) Give the name of Enzyme A :…………………………………………………………………. Enzyme B : ………………………………………………………………….

[ 2 marks ]


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(b) Suggest why the same enxyme ( enzyme A) is used to cut both the vector DNA and the chromosomal DNA fragment . …………………………………………………………………………………………..... …………………………………………………………………………………………..... …………………………………………………………………………………………….

[ 2 marks ] (c) What is the function of enzyme B ? ………………………………………………………………………………………….....

[ 1 mark]

(d) What name is normally given to the molecule C ? …………………………………………………………………………………………….

[ 1 mark] (e) Describe two ways in which the chromosomal DNA fragment can be obtained . …………………………………………………………………………………………..... ………………………………………………………………………………………….....

………………………………………………………………………………………….....

………………………………………………………………………………………….....

[ 2 marks]

(f) What can be added to molecule C to aid in selection and identification of the cells that have successfully taken up the fragment? ………………………………………………………………………………………….....

[ 1 mark] (g) Suggest a possible medical use for this technology ………………………………………………………………………………………….....

[ 1 mark]


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Section B [60 marks]

Answer any four questions in this section

5 (a) Describe the structure of the cell membrane and state the functions of the component parts.

[9 marks] (b) Write an account outlining the similarities and differences in the way that

mitochondria and chloroplasts generate a proton gradient and synthesis ATP.

[6 marks ]

6 (a) Plants are a diverse group of eukaryotic organisms. Describe the different characteristics of the Bryophyta, Filicinophyta, Coniferophyta and Angiospermophyta.

[ 8 marks ] (b) With the aid of a diagram, explain fully the function of the loop of Henle and the collecting duct in osmoregulation by the kidney.

[ 7 marks]

7 Describe the sequence of events which occurs as an impulse is propagated along an axon and is subsequently transmitted across a synapse.

[15 marks]

8 Describe how the human body defends itself against infection with reference to both humoral and cell mediated responses.

[15 marks ]

9 (a) Describe the development of the mammalian embryo from the zygote stage to the formation of the extra embryonic membranes.State the functions of these membranes

[ 9 marks ]

(b) Describe the role of hormones in ecdysis and metamorphosis in insects.

[ 6 marks ]

10 (a) Describe the energy transfer from the sun to producers through the trophic levels.

[11 marks ]

(b) Describe and explain the shape of a bacterial population growth curve.

[4 marks]


1

PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA

NEGERI PAHANG DARUL MAKMUR BIOLOGY

2011

MARKING SCHEME PAPER TWO


2

Section A

Q1 (a) 1) 1 template/original/old, and 1 new ; 2) Complementary base pairing; 3) 2 isotopes in molecules; 4) half/1(original) strand with , “heavy” N/ 15N; 5) half/1(new) strand with ,”light” N/ 14N; 6) no molecules with only 1 isotope; * any 3 (b)A; C; B and C; (c) Bar drawn in 14N column and 14N/15N column; 75% for 14N ; 25% for 14N/15N;

(d) DNA polymerase

1 1 1 1 1 1 max=3 1 1 1 1 1 1 1 total =10


3

Q2 a) Root hair(cell) b) Large /increase surface area c) Osmosis d) Active transport/uptake (not: facilitated diffusion) e) 2= symplast(pathway) / symplastic 3 = apoplast (pathway)/ apoplastic f) Contains waterproof material (suberin)/makes cell wall impervious

to water; 1) Diverts water from apoplast route to symplast route/ 2) Offers plant some control over what enters pericycle/ 3) Forces water to pass through semi-permeable membrane; *any one (g) i) Hollow lumen with no living cytoplasm reduces resistance to

water movement/ ii) Lignified walls prevent collapse/ iii) Small lumen to enhance capillarity iv) Pits to allow lateral movement of water; *any two

1

1

1

1

1 + 1

1

1

1

1

total = 10

Q3 (a) (i) 192 ÷ 300 = 0.64 (64%);

1


4

(b) Relative frequency of the yellow allele (q) = 0.8 (80%) [consequential to answer provided in (a)(i)]; Relative frequency of the pink allele (p) = 0.2 (20%) [consequential to answer above]; Number of snails in the population heterozygous for shell colour (2pq)N = (2 X0.2 X0.8) X 300 = 96 (c) (i) 28 ÷ 35 = 0.8 (80%); [1] (ii) (1) No, as the thrush is selecting proportionally more yellow (less pink)/ only 0.64 of the population is yellow yet yellow shells make up 0.8 of the predated snails (0.36 pink though only 0.2 are predated); (2) Selection pressure alters the allele frequency/the yellow form will decrease (pink will increase); [2] (d) (1)The more common form would decrease (e.g. yellow would

decrease); (2) the less common form would increase (e.g. pink would increase);

(3) until it becomes the more common form, when the cycle would

continue (e.g. pink would become more common in which situation it would be more heavily predated);

1

1

1

1

1

1

1

1

1

total=10

Q4

(a) (i) restriction enzyme /(restriction) endonuclease (not: specific example)

1


5

(ii) DNA ligase

(b) (1)to give matching/the same(sticky) ends

(2) so that the vector/plasmid can join with the fragment;

(3) complimentary bases,(not codon)

*(Any two)

(c) catalyzes the covalent bonding of the DNA fragment

(d)Recombinant (DNA molecule)

(e) (1) Can be cut directly from the host DNA/ use of restriction enzyme

(2)RNA can be extracted/use of messenger RNA and treated with reverse transcriptase which turns RNA into DNA

(f) antibiotic resistance gene/sequences/(radioactive ) marker genes.

(g) insulin production/production of factor viii/interferon production /human growth hormone production

(not: liposomes/cystic fibrosis treatment)

1

1

1

1

max=2

1

1

1

1

1

1

Total 10

Q5(a) 1) Singer Nicholson / fluid mosaic model; 2) Bilayer of phosplipids 3) Hydrophobic / water hating tails face each other; 4) Hydrophilic / water loving heads face water / outwards; 5) Integral /transmembrane protein firmly inserted in the lipid bilayer

1 1 1 1 1


6

6) Peripheral protein are not embedded in the lipid at all 7) Membrane carbohydrates found on the external surface of plasma

membrane 8) Glycolipids(carbohydrate are covalent bonded to lipids) 9) Glycoprotein (carbohydrate are covalent bonded to protein) *Any five

10) Phospholipids / lipid bilayer;

Separate contents from outside / acts as barrier; /Phospholipid allows fat soluble substances through / selective;

11) Carrier protein; Used for active transport; /Specific substances transported;

12) Cholesterol affects fluidity;

13) Channel/carrier protein for facilitated diffusion;

14) Glycoprotein / glycolipid; For cell

recognition/signalling/hormonerecognition. *Any four

1 1 1 1 max=5 1 1 1 1 1 max=4

5(a) 9

5(b)

ATP Synthesis Similarities Both need high energy electrons Both need stalked particles/ATP synthetase

1

1


7

Both need proton pumps Both involve a series of carriers/pumps in electron transport chain Both need protons which are pumped into a cavity Both are involved in creating an electrochemical/ chemiosmotic/H-ion/H+/proton gradient

Both involve redox reactions

*Any three

1

1

1

1

1

max=3

Differences 1) Cristae/(folded) inner membrane carries stalked particles in mitochondria Or Intermembrane cavity in mitochondria Thylakoid membrane in chloroplasts carries stalked particles Or Thylakoid cavity in chloroplasts 2) Water formed during ATP synthesis in mitochondria Water broken down (in non-cyclic photophosphorylation) in chloroplasts 3)Protons derived from glucose breakdown Protons derived from water/photolysis produces H ions 4) High energy electrons are derived from chlorophyll in chloroplasts High energy electrons are derived from glucose/reduced NAD/FAD in mitochondria

1

1

1

1

max=4

5(b) 7

6(a)

Bryophyta have no roots / only have rhizoids; Bryophyta have simple leaves/stems / only a thallus; Bryophyta produce spores in capsule; Byrophyta are nonvascular; Bryophyte exhibit (pronounced) alternation of generations / a significant gametophyte generation;

1

1

1

1

1


8

*any two Filicinophyta have roots, stems and leaves; Filicinophyta (often) have divided/pinnate leaves; Filicinophyta produce spores in sporangia/spores on the undersides of leaves; Filicinophyta exhibit alternation of generations; Filicinophyta have primitive vascular tissue / no true xylem and phloem; *any two Coniferophyta have woody stems; Coniferophyta (often) have narrow leaves/needles/scales; Coniferophyta produce seeds in cones/unenclosed seeds; *any two Angiospermophyta have flowers; Angiospermophyta have ovules in ovaries; Angiospermophyta produce seeds (with hard coats) in fruits;

*any two

max=2

1

1

1

1

1

max=2

1

1

1

max=2

1

1

1

max=2

6(b) 8

6(b)

1) Loop of henle passes from cortex down into medulla 2) Where high salt concentration is maintained in the bathing fluid/ low water potentia created 3) Increased concentration achieved by movement of salt/Na+/CI- From ascending limb into surrounding fluid. 4) This is carried out by active transport 5) The ascending limb is impermeable to water (at bottom)

1 1 1 1 1


9

6) Osmotic movement of water/out of permeable descending limb 7) Diagram of loop with arrows showing movement ion and water And relative concentration (ie numerical concentration changes) 8) Collecting duct passes from second tubule(in cortex) down through medulla (may be on diagram) 9) Where salt levels concentration detected by osmoregulation cells in the hypothalamus / hypothalamus monitor concentration in blood 10) Rise in osmotic pressure/ causes regulator cells to stimulate increased release of ADH (or vice versa) 11) Hormone rise increases permeability of walls of collecting duct so increasing water retention

1 1 1 1 1 1 1 max=7

Q7

Propagation along an axon: 1) at rest an axon membrane has a resting potential/polarised membrane 2) whereby it is negative on the inside 3) when stimulated the potential difference is reversed/it becomes positive on the inside/it becomes depolarized 4) an action potential is evoked 5) this causes a depolarisation of the neighbouring part of the axon membrane/ local circuits

1

1

1

1

1


10

6) the original part of the axon recovers its resting potential/repolarised 7) during this period of recovery that part of the axon cannot be stimulated/there is a refractory period 8) reference to absolute and relative refractory period 9) a transfer of action potentials represents an impulse/an impulse is a self-perpetuating action potential 10) impulses are either “fired” or not “fired”/an “all-or-nothing” law applies/ threshold level required for stimulation 11) impulses are speeded up if the axon has a large diameter (greater surface area for ion exchange) 12) Impulses are also speeded up if the axon is myelinated 13 ) Due to the association of Schwann cells 14) Since transmission “jumps” from node to node (of Ranvier)/transmission is salutatory Transmission across a synapse: 1) synaptic knob contains vesicles 2) when an action potential reaches a synaptic knob Ca2+ ions enter 3) vesicles are caused to move towards and fuse with the pre-synaptic membrane/resulting in exocytosis of a synaptic transmitter chemical 4) which in most peripheral nerves is acetylcholine (Ach) 5) the transmitter diffuses across the synaptic cleft 6) the synaptic cleft is only about 20 nm in width 7) at the post-synaptic membrane the transmitter attaches to receptors 8) causing depolarisation of the post-synaptic membrane

1

1

1

1

1

1

1

1

1

max=8

1

1

1

1

1

1


11

9) at excitatory post-synaptic potential occurs 10) summation of synaptic knobs (spatial/temporal) may be needed to promote an EPSP 11) If this is sufficient in magnitude an action potential is evoked 12) continued stimulation of the post-synaptic membrane is prevented 13) by the enzyme (e.g. cholinesterase) which causes breakdown of the transmitter substance/transmitter substance reabsorbed into the synaptic knob 14) Reabsorbed breakdown products (e.g. choline and ethanoic acid) synthesise transmitter substance using ATP [13]

1

1

1

1

1

1

1

1

max=7

8(a)

1)Antigen adhere to marcophage 2)Macrophage forms pseudopods that eventually engulf the antigen 3) Present fragment of antigen of membrane,acts as antigen presenter 4) Interleukin costimulated helper T cell 5) Humoral response involves production of proteins (globulins) called antibodies 6) They are specific to the antigen 7) Antibodies are Y shaped / formed from 4 polypeptide chains 8) The antibody antigen complex inactivates the antigen

1

1

1

1

1

1

1

1

1


12

9) An example of how / identifying for phagocytosis / agglutination / lysis / immobilisation / neutralisation 10) Humoral response involves (lymphoid tissue to form) B lymphocytes / B cells 11) Each B cell (or T cell) has receptors for its specific antigen 12) Antigen stimulates proliferation of plasma cells and memory cells 13) Memory cells remain ready to divide if same antigen is encountered again (i.e. increase number rather than output) 14) cell mediated response involves T lymphocytes / T cells 15) T cells develop / mature in thymus gland 16) There are 3 types of T cell/ref. To cytotoxic T cells /killer cells, helper cells and memory or suppressor cells 17) Function of killer cells, directly kill pathogen/ bacterium cell infected by virus 18) function of helper cells, stimulate phagocytosis / make T or B cells competent / assist cytotoxic T cells /killer cells 19) role for suppressor cells - suppress B cells / lymphocyte action after infection / switch off immune system when antigen not present

1

1

1

1

1

1

1

1

1

1

1

1

1

1

max=15

Q9(b)

1) Prothoracicotrophic hormone (PTTH) is secreted by the neurosecretory cells ;

2) Stored in the corpora cardiaca/corpus cardiacum

3) Released and transported to corpora allata / corpus allatum and prothoracic glands

4) PTTH signals the prothoracic gland to produce hormone ecdysone

5) secretion of ecdysone is episodic,each release will stimulates a moult.

6) when concentration of juvenile hormone(JH) is relatively high,each moult only leads to a larger larva stages

7)when concentration of JH is very low/below threshold level,causes the larva to moult to become pupa

1

1

1

1

1

1

1


13

8) when production of JH ceases and concentration of ecdysone is high,the pupa metamorphoses to become an adult insect.

1

max=6

Q9(a)

1) the cells of the trophoblast secrete enzymes that enable the blastocyts to implant in th endomentrial lining of the uterus.

2) the trophoblast cells proliferate and forms the trophoblastic villi which grow into the endometrium

3)the trophoblast forms the chorion.

4) Function of chorion : form placenta

4)the inner cell mass of the blastocyst forms the three germ layers

5) the germ layers form three other extra-embroynic membranes called amnion,yolk sac and allantois.

6) the amnion enclose the entire embryo and the space between the embryo and amnion become filled with amniotic fluid secreted by the cell of the amnion

7)function of amnion:

i) protects embryo form desiccation

ii)enable embryo to move freely in the early stages of development

iii)absorbs shock

*any one

8)function of yolk sac; form part of the gut(digestive tube) ; early red blood cell formation

9)function of allantois : form umbilical cord.

1

1

1

1

1

1

1

1

1

1


14

10a) Transfer of energy from the sun to producers: Any five points 1) less than 1% of the total amount of incoming solar radiation is used in photosynthesis/primary production 2) solar radiation refl ected or absorbed (and radiated back into space) by ozone layer/dust particles/clouds 3) Most energy fails to strike green plants/absorbed by soil 4) of the solar radiation striking a leaf much is used in evaporating water from the leaf 5)sunlight is absorbed by the chlorophyll (during which it is converted to chemical energy) of green plants 6) some light striking the plant is not in the photosynthetically active range/chlorophyll absorbs mostly red and blue wavelengths 7) the total amount of energy captured in newly synthesised

1 1 1 1 1 1 1


15

carbohydrate/during photosynthesis represents the gross primary production (GPP) 8) a proportion of the GPP will be respired by the plant and will be lost as respiratory heat [allow in section below if not presented here] 9) this represents assimilated material available to the next trophic level/the difference between GPP and respiration is the net primary production (NPP) [allow in section below if not presented here] Transfer of energy through the trophic levels: Any five points 1) much of the producer material will die and enter the decomposer trophic level 2) In fact, generally over 50% of the plant material enters the decomposer food chain rather than the consumer food chain 3) some of the plant material eaten by primary consumers (herbivores) will be egested as faeces and will (along with urine) be available to decomposers 4) some of the plant material eaten and digested by primary consumers (herbivores) will be assimilated 5) some of the energy assimilated will be respired and so lost as heat (in all trophic levels) 6) some energy may be lost in urine 7) generally the biomass/productivity of one trophic level is less than that of the previous trophic level/in fact, only about 10% of the energy consumed by one trophic level is available for consumption by the next 8) this results in a limit in the number of trophic levels in an ecosystem to four or five/by the fi fth trophic level there is not enough energy to support another trophic level 10(b) (The following may be labelled on a graph but must include an explanation to gain a mark.)

I. Lag phase – period of preparation for growth / intense metabolic activity, notably enzyme synthesis / numbers are low therefore rate of multiplication is low.

II. Log / exponential phase – no limiting factors / abundant

resources / rapid reproduction

1 1 1 1 1 1 1 1 1 1 total =11 1 1 1


16

III. Decelerating phase- population growth begin to slow down

IV. Stationary phase – carrying capacity reached / rate cell division

= rate cell death / population has reached maximum size.

V. Competition for limited resources.

VI. due to a depletion of resources / accumulation of toxins. *Any 4

1 1 1 1 max : 4

total 10(b)=4


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[edu.joshuatly.com] pahang stpm trial 2011 biology paper 2 (w ans) (1)

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