Stpm Trial 2009 Bio Q&a Terengganu

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Name : …………………………………………………

NRIC : …………………………

JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR TRIAL JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR STPM 2009 JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARAN JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR NEGERI TERENGGANU JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR BIOLOGY JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR PAPER 1 JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR One hour and forty-five minutes JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR

964/1

Instructions to candidates :

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO. There are fifty questions in this paper. For each question, four suggested answers are given. Choose one correct answer and indicate it on the multiple-choice answer sheet provided. Read the instructions on the multiple-choice answer sheet very carefully. Answer all questions.Marks will not be deducted for wrong answers.The total score for this paper is the number of correctly answered questions.

This question paper consists of 16 printed pages.

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CONFIDENTIAL*

1

Which molecule is made up of or contains glucose molecules? A B C D

2

Secretion Protection Absorption Transportation

Which statement is true of transcription? A B C D

6

Lysosome Ribosome Microbody Mitochondrion

Which of the following is most important function of epithelium tissue? A B C D

5

The angle between hydrogen atoms is 104.3°. Oxygen is more electronegative than hydrogen. Hydrogen is covalently bonded to oxygen to form water. Polar compounds with partial charges tend to dissolve in water.

Which organelle, in an animal cell, is spherical in shape and bounded by a single membrane? A B C D

4

Fructose Cellulose Ribonucleic acid Deoxyribonucleic acid

Which statement best explain the polarity of water? A B C D

3

2

It begins with ATG and ends with TAG. The sense strand is used as a template. The DNA polymerase is used to synthesis DNA. It uses 70s ribosome in prokaryote and 80s ribosome in eukaryote.

Which statement is true of non-competitive inhibitor ? A B C D

Its mode of action is reversible. It binds directly to enzyme at the active site. Its binding to enzyme lowers the activation energy. Its inhibitory effect can be reduced by increasing the substrate concentration.

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Which plants that Rubisco binds with both carbon dioxide and oxygen ? A B C D

8

C3 plants C4 plants CAM plants C4 and CAM plants

The rate of photosynthesis of a fresh water plant is measured by different colours spectra. Which of the following is the correct order of colour that gives an increasing rate of photosynthesis? A B C D

9

3

Blue- Red-Orange-Yellow-Green Blue-Green-Yellow-Orange-Red Green-Yellow-Orange-Red-Blue Green -ed-Blue-Yellow-Orange

Which of the following molecules contains the most energy? A B C D

ATP Glucose Sucrose Starch

CO2

A

Ribulose biphosphate D

Glycerate 3-phosphate B Glyceraldehyde 3- phosphate

Ribulose phosphate C

Hexose

10

The diagram above shows the stages in the dark reaction. At which stage is the reduced NADP reoxidised?

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4

CH3 C=O

CH3 CO2

Pyruvate decarboxylase

Pyruvate 11

NAD+

C=O

C=O OH

NADH

H

CH3 H C OH

Alcohol dehydrogenase

Acetaldehyde

H Ethanol

The diagram above shows the conversion of pyruvate to ethanol during anaerobic respiration. Which of the following allows glycolysis to continue? A B C D

The regeneration of NAD+ The regeneration of NADH The release of carbon dioxide The addition of yeast to ethanol

Acetyl coenzyme A

ADP

GTP X

ATP 12

GDP

The diagram above shows the Krebs cycle. What is process X ? A B C D

13

Krebs cycle

Phosphorylation Oxidation of GTP Oxidative Phosphorylation Substrate level phosphorylation

Which is an example of saprophytic organism? A B C D

Mucor sp. Taenia sp. Rafflesia sp. Periplenata sp.

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5

Lactose

Starch

Lactase

Sucrose Amylase

Sucrase

Maltose Maltase

X 14

Y

Z

The diagram above shows the digestion of lactose, starch and sucrose. What are the substances X,Y and Z?

A B C D

X Glucose and glucose Glucose and fructose Glucose and galactose Glucose and galactose

Y Glucose and fructose Glucose and galactose Glucose and fructose Glucose and glucose

Z Glucose and galactose Glucose and glucose Glucose and glucose Glucose and fructose

Percentage of oxygen saturation (%) 100

pH 7.4 pH 7.2

50

0 50 15

100

Partial pressure of oxygen (mmHg)

The graph above shows the oxygen dissociation curves for two values of pH. Which statement about the curve is true? A B C D

The increase in pH is due to vigorous activities. The increase in pH causes the curve to shift to the right. The percentage of oxygen-saturated haemoglobin decreases when pH increases. The shifting of the curve to the right is due to an increase in concentration of blood carbon dioxide.

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6

P CO2

CO2 + H2O

H2CO3 Q

S

H+ + H2CO3-

HHb

Tissue cells

HbO2-

O2 + Hb R

Capillary wall

Red blood cell

16

The diagram above shows the diffusion of carbon dioxide from respiring cells into the blood involving steps P,Q,R and S. Which step requires carbonic anhydrase to proceed to the next? A B C D

17

What is the volume of air that can be forced out following the deepest possible inspiration? A B C D

18

P Q R S

tidal volume vital capacity residue volume expiratory reserve volume

Which response occurs when a person loses a lot of blood? A B C D

A decerase in renin secretion An increase in the secretion of sodium ions An increase in the production of angiotensin A decrease in the production of aldosterone

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19

Which condition causes the closing of a stoma? A B C D

20

7

The influx of potassium ions into the guard cells The increase in the concentration of glucose in the guard cells The decrease in the concentration of carbon dioxide in the guard cells The increase in the concentration of absicisic acid when plants are exposed to stress

The biochemical pathway which converts lactate into glucose and later into glycogen in the liver is as follows. Lactate

Pyruvate

glucose

glycogen

What is the pathway known? A B C D

Cori cycle Krebs cycle Calvin cycle Ornithine cycle

Adrenal cortex Hormone X Reabsorption of Na+ ions increase

Inhibition ( negetive feedback)

Concentration of Na+ ions in blood plasma decreases

21

The diagram above shows the control of sodium ions level in the blood plasma. What is hormone X? A B C D

Adrenaline Aldosterone Angiotensin Antidiuretic

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22

Which of the following correctly explains the distribution of ions on either side of the membrane of an axon in its resting state? A B C D

23

nodes of Ranvier presynaptic membrane postsynaptic membrane membrane of the synaptic vesicles

Which statement is not true of auxin? A B C D

25

A high concentration of organic anions outside and a low concentration of K+ ions inside. A high concentration of organic anions inside and a low concentration of Na+ ions outside. A high concentration of K+ ions and organic anions outside and a high concentration of Na+ ions inside. A high concentration of Na+ outside and a high concentration of K+ ions and organic anions inside.

Where are the receptor sites for neurotransmitters situated? A B C D

24

8

It stimulates the division of cell in a stem. It stimulates the elongation of coleoptile. It promotes the formation of lateral shoot. It inhibits the elongation of root at the high concentration.

Oestrogen and progesterone are used in contraceptive pills. What is the effect of this hormones on mestrual cycle? A B C D

Its maintain the endometrium of the uterus. Its stimulate the release of luteinising hormone. Its inhibit the production of gonadotropic hormones. Its stimulate the release of folicle stimulating hormone.

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9

The diagram below shows the findings of an experiment on the effect of light on the flowering of a plant.

Treatment

Result

Dark P

Flowering Light Red Light

Q

Non flowering Light Red Light

R

Flowering Light Far –red light 24 hours

Which of the following are true of the plant?

27

l ll lll lV

The plant is a long day plant. The plant is a short day plant. Far-red light cancels off the action of red light. Red light can replace the requirement of dark period.

A B C D

l and lll l and lV ll and lll ll and lV

What do W,X and Y represent in the following simplified flow chart of the humoral response?

Antigen Stimulate W

X

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Plasma cell

Y

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A B C D

28

29

X Memory B cell Antibody Memory T cell Memory T cell

Y Antibody Memory B cell Helper T cell Cytotoxic T cell

Which of the following are true of B cell? l ll lll lV

It forms immunity through the humoral response. It forms immunity through the cell mediated mechanism. It is produced and it achieves maturity in the bone marrow It is produced in the bone marrow and it achieves maturity in the thymus gland

A B C D

l and ll l and lll ll and lV lll and lV

Which of the following is true of an oviparous animal? A B C D

30

10

An individual hatches from the egg outside the female parent’s body An individual hatches from the egg in the uterus of the female parent An individual is born before maturity and continues to develop in the sac of the female parent. An individual develop in the uterus of the female parent and the embryo obtains the nutrient from the placenta.

The hormone which plays an important role in seed germination is A B C D

ethene auxin cytokinin gibberellin

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11

Which of the following is the absolute growth curve of maize plant?

Gain as percentage of previous mass

A

Daily gain

Age/weeks

B

Gain as percentage of previous mass

C 32

Age/weeks

Daily gain

Age/weeks

D

Age/weeks

The following are events that occur during seed germination. l ll lll lV V

Synthesis and secretion of enzymes Activation of the aleurone layer Flow of sugars to the embryo Release of gibberellin Hydrolysis of starch

Which of the following is the correct sequence of events during seed germination? A B C D

ll lV lV V

lV l ll lll

V V l lV

lll ll V ll

l ll lll l

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In a species of flowering plant, CR CR genotype produces red flowers, Cw Cwgenotype produces white flowers and CR Cwgenotype produces pink flowers. What is the percentage of the progeny that have pink flowers if a cross is made between CRCw and CRCw ? A B C D

34

36

XO XXY XYY XXX

Which of the following mutations in humans is /are trisomic? l ll lll

Down Syndrome Turner Syndrome Thalassemia major

A B C D

l only ll only l and lll ll and lll

Which of the following is not true of mutation? A B C D

37

0% 25 % 50 % 75 %

Which of the following is the genotype of klinefelter syndrome? A B C D

35

12

A chromosomal mutation of the deletion type involves the deletion of a base pair from gene. Genetic desease called cri-du-chat syndrome is caused by a deletion in chromosome 5 The deletion of two bases causes frame-shift mutation during triplet coding in transcription. Allopolyploidy is a chromosomal mutation which involves chromosome doubling caused by different genomes.

A study on 400 mice about their resistance towards a type of poison has been carried out. The resistance characteristic is controlled by a the dominant allele R. 36% of the mice population is found to be resistant towards the poison. Calculate the number of mice expected to have Rr genotype. A B C D

16 72 128 256

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38

Which of the following is true of repressor protein? A B C D

39

41

It is coded by lac Y. It binds with the promoter and inhibits transcription. It binds with gene that codes for β-galacosidase. It changes is conformation after binding with lactose

A mutation in the lactose operan occurs which causes the repressor protein not being able to bind with the operator region. Which of the following statements is true of the mutation? A B

β -galacosidase enzyme is not produced at all. β -galacosidase enzyme is produced continuosly with or without lactose.

C

β -galacosidase enzyme is produced continuosly in the absence of lactose only. β -galacosidase enzyme is produced continuosly in the presence of lactose only

D

40

13

Which of the following are true of restriction enzymes? l ll lll lV

It restricts transcription. It is found in all eukaryotic cells. It acts on palindromic sequences. It is sensitive to changes in temperature and pH.

A B C D

l and ll l and lll ll and lV lll and lV

Which of the following are the products of the translation of the lactose operon? l ll lll lV

Permease Transacetylase β -galacosidase RNA polymerase

A B C D

l, ll, and lll l,ll and lV l,lll and lV ll, lll and lV

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Which of the following regarding taxa and their examples not correct? A B C D

43

Phylum: Chordata Class: Mammalia Order: Hominidae Species: sapiens

Which of the following statements is not true of an artificial classification system? A B C D

44

The system is based on phylogenetic relationship. The system can be used toconstruct dichotomous keys. Organisms are placed into groups for specific purposes. Organisms are placed into group according to their different characteristics which are arbitrarily chosen.

Based on the table below, match phyla of organisms to their characteristics.

l

Phylum Cnidaria

P

ll lll lV

Artropoda Mollusca Nematoda

Q R S

A B C D

45

14

l P Q R S

ll Q R S R

Characteristic Body divided into head,muscular foot and visceral mass Diploblastic body, polymorphism Segmented legs, chitinous exoskeleton Body covered with thin and elastic cuticle, pseudocoelom lll S P Q P

lV R S P Q

Which of the following energy flows in an ecosystem involves the transfer of the greatest amount of energy? A B C D

Plant Plant Herbivore Carnivore

Herbivore Decomposer Carnivore Decomposer

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15

Based on the table below, match the evidence of evolution theory to its example. Evidence of evolution theory l ll lll lV

A B C D

47

P Q R S

l

ll

lll

lV

P P R R

Q S P Q

S R Q P

R Q S S

Marsupial animal Homolog structure Fossil recordC Cytochrome

the number of individuals in a population the population size when the mortality rate is more than the natality rate the population size when the natality rate is more than the mortality rate the population size of species which can be supported by resources available in a habitat

A total of 50 squirrels were caught from a forest, marked and released. A few days later, 40 squirrels were caught from the same region and 4 of them had marks. Estimate the population size of the squirrels in that forest. A B C D

49

Paleontology Geographical distribution Comparative anotomy Comparative biochemistry

Carrying capacity of a population is A B C D

48

Example

160 200 500 2000

Which of the following are true of an ecosystem? l ll lll lV

Phytoplanktons are producers. The last consumer obtains the highest energy. Ecosystem is an open system with input and output of energies. Heterotrophs include herbivores, carnivores, decomposers and detritivores.

A B C D

l and ll lll and lV l, lll and lV ll, lll and lV

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50

16

The graphs below show the effects of three types of ecological selections.

P

Q

R Initial population Final population

Which of the following is correct regarding Biston betularia in industrial areas, human birth weight in developed countries and rabbit population in the Andes Mountains?

A B C D

Biston betularia in industrial areas P Q R R

Human birth weight in developed countries Q R P Q

Rabbit population in the Andes Mountains R P Q P

964/1 This question paper is CONFIDENTIAL until the examination is over. CONFIDENTIAL*

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CONFIDENTIAL* Name : ………………………………………………… NRIC : ………………………… JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN964 PELAJARAN JABATANPELAJAR / 2 ERENGGANUJABATANPELAJARAN TERENGGANU OTI 2 JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR STPM 2009 JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN NEGERI TERENGGANU TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR BIOLOGY JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR PAPER 2 JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR STRUCTURE AND ESSAY JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PENLAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

Two and a half hours

Instructions to candidates :

Answer all the questions in Section A in the spaces provided. Answer any four questions from section B. For this section, write your answers on the answer sheets provided. Begin each answer on a fresh sheet of paper. Answers should be illustrated by large, clearly labeled diagrams wherever suitable. Answers may be written in either Malay or English. Arrange your answer in numerical order and tie the answer sheets to this booklet.

For examiner’s use Section Marks Marks Obtained 1 10 2 10 A 3 10 4 10 40 5 15 6 15 B 7 15 8 15 9 15 10 15 60 TOTAL 100

This question paper consists of 10 printed pages STPM 964/2 *This question paper is CONFIDENTIAL until the examination is over.

[Turn over CONFIDENTIAL*

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2

SECTION A Answer all questions in this section 1. The diagram below shows a carbohydrate molecule formed from two sugar units CH2OH

CH2OH O

C

H

O

C

C O

(a) Name the carbohydrate

[1 mark]

…………………………………………………………………………………….. (b) Name the type of chemical bond between the two sugar units

[ 1 mark ]

…………………………………………………………………………………….. (c) State the chemical reaction involved.

[ 1 mark ]

………………………………………………………………………………………. (d) State one function of this type of carbohydrate in living organism.

[1mark ]

………………………………………………………………………………………… (e) Name the storage carbohydrate in humans.

[ 1 mark ]

……………………………………………………………………………………………

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3

(f) Explain why the substance mention in (e) is suitable for storage in humans. [ 3 marks] ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… (g) Another type of carbohydrate which is found in plant cell wall is cellulose. Explain why cellulose cannot be digested in humans.

[ 2 marks]

……………………………………………………………………………………………. …………………………………………………………………………………………….. …………………………………………………………………………………………….. 2.

(a) The diagram below shows protein synthesis in an eukaryote cell.

A C

Step 1

Step 2 Ribosome

Nucleus

B

B diffuses out from the nucleus tRNA

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4

Based on the diagram above: (a)(i)

Name Step 1 and Step 2.

[ 2 marks ]

Step 1 : ........................................................................ Step 2 : ........................................................................ (ii) Name the structures labeled A, B and C.

[ 2 marks ]

A : ......................................................................... B : ........................................................................ C : ........................................................................ (iii) Explain what happens in Step 1.

[ 2 marks ]

.................................................................................................................... ..................................................................................................................... .....................................................................................................................

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5

The diagram below shows two methods in which enzymes can be produced for industrial processes. Microorganism

Extracellular enzyme

Enzyme extracted

Enzyme is purified

Immobilised enzyme (i)

Give two advantages using immobilized enzymes.

[ 2marks ]

..…………………………………………………………………………………. …………………………………………………………………………………. …………………………………………………………………………………. (ii)

Name three ways in which an enzyme or a cell can be immobilized. [ 3marks] …………………………………………………………………………….. …………………………………………………………………………….. ……………………………………………………………………………..

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6

P

S

R

Q

3. (a) The diagram above shows the Krebs cycle occurs in a cell. (i) Name the substances labelled P,Q,R and S.

[ 2 marks ]

P…………………………………… Q………………………………….. R………………………………….. S………………………………….

(ii)

State the place where the Krebs cycle occurs in cell.

[1 mark ]

………………………………………………………………………………………….

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7

Explain the role of NAD+ in Kreb cycle.

[2 marks ]

……………………………………………………………………………………………. ……………………………………………………………………………………………. (b) The diagram shows the main stages in light-independent reaction in photosynthesis.

Ribulose biphosphate RuBP

Carbon dioxide

Reduced NADP NADP

Glycerate 3-Phosphate GP ATP

Triose Phosphate ADP

Ribulose phosphate

(i) Write in the boxes provided in the diagram the number of carbon atoms in each of the relevant substances. [ 1 mark ] (ii) What are the roles of ATP in the conversion of glycerate 3-phosphate to triose phosphate and ribulose phosphate to ribulose biphosphate? [ 2 marks ] . ............................................................................................................................... ................................................................................................................................. (iii) A plant was allowed to photosynthesis normally. The light was then switch off. There was a rise in the amount of glycerate 3-phosphate present in the chloroplast of this plant. Explain why. [ 2 marks ] ................................................................................................................................. ................................................................................................................................. ................................................................................................................................. 964/2 * This question paper is CONFIDENTIAL until the examination is over

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4. (a) Diagram below shows two growth curves of different species.

Height/cm

Height/cm

`

8

Time/years

A broad bean plant

Time/years

An insect

(i) Name the type of the growth curve represents by each species.

[2 marks ]

A broad bean plant : …………………………………………… An insect : ………………………………………………………. (ii) State one characteristic of each growth curve.

[ 2 marks]

A broad bean plant : ……………………………………………………… An insect : …………………………………………………………………. (iii ) The growth curve of the insect uses length as a measure of growth. This growth curve cannot be considered as a true measurement of the insect growth. Explain why. [ 2marks] …………………………………………………………………………………… ………………………………………………………………………………….. …………………………………………………………………………………. (b )

In an experiment, seedlings (x) of 100 g of corn grains were grown in darkness for 10 days. After 10 days, the dry mass of the seedlings was analysed and then compared with a sample of 100 g of ungerminated corn grains (y).

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9

The table below shows the results of the experiment. Dry mass of ungerminated corn grains (y)

Dry mass of corn seedlings after 10 days (x)

Cellulose

1g

4g

Starch Other organic substances

65 g

8g

15 g

31 g

Total dry mass

81 g

43 g

(i) Explain why the total dry mass of seedling y decreased after 10 days. [ 2marks ] ………………………………………………………………………………………. ………………………………………………………………………………………. ………………………………………………………………………………………. (ii) Seedlings x contain more cellulose than corn grain y which did not germinate. Explain why. [ 1mark ] ……………………………………………………………………………………… ……………………………………………………………………………………… (iii) Name a source a carbon used for synthesising new cellulose molecules. [ 1mark ] ……………………………………………………………………………………….

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10 Section B [60 marks] Answer any four questions from this section.

5 (a) Protein ca n be di vided into t wo g roups, na mely f ibrous proteins and g lobular proteins, based on the shape of the molecule. Describe the differences between the two types of proteins. [7 marks] (b) i. Distinguish between diffusion and osmosis [2 marks] ii.Distinguish between phagocytosis and pinocytosis. [3 marks] iii. State three features of a substance, which influence its ability to pass through a cell membrane. [3 marks] 6 (a) Draw and labeled the structure of a cell membrane based on Singer’s model [2 marks] (b) Explain the roles of the structures of the cell membrane in the transportation of substances into the cell. [13 marks] 7. (a) With the aid of labelled diagram, explain i The structure of a stoma ii. The mechanism of stomata opening and closing.

[ 10 marks]

(b) Outline t he env ironmental f actors which i nfluence t he openi ng and cl osing o f a stoma. [5 marks] 8 (a) Explain briefly the experiment conducted by Meselson and Stahl to prove the DNA replication method. [8 marks] (b) Explain DNA replication. [7 marks] 9 (a) Describe how t he nephr on i n t he kidney r egulates the w ater co ntent of body fluids. [9 marks] (b) Explain how selective reabsorption occurs in the proximal convoluted tubule. [6 marks] 10 (a) A population of fruit flies consists of 250 individuals. 195 individuals have grey bodies. The allele for grey is dominant (K) and the allele for ebony body is recessive (k). Assuming the population is in a genetic equilibrium, calculate the frequency for the alleles and the ratios for the three genotypes in the population. [8 marks] (b) The ABO blood group is governed by a set of three multiple alleles, IA , IB and IO. I and I B are codominant, I O is resessive. A man of blood group B married a w oman of unknown A BO bl ood group. They had t hree ch ildren. O ne o f t he children had bl ood group A, one had blood group AB and one had blood group O. (i) State the genotypes of the parents and give an explanation for your answer. [5 marks] (ii) Draw a g enetic diagram to show the inheritance of ABO blood groups in this family. [2 marks] A

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ANSWER SCHEME BIOLOGY STPM PAPER 1 TRIAL/ OTI 2 2009 Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

ANSWER B B A B B A A C D B B C A D D A B C D A B D C C C

Q 26 27 28 29 30

ANSWER C A B A D

41 42 43 44 45 46 47 48 49 50

A C A B A C D C C C

31 32 33 34 35 36 37 38 39 40

B C C B A A C D B D

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Marking Scheme- Trial STPM 2009 Paper 2 1(a) (b) © (d) (e)

Dissacharide 1,4 glycosidic bond Condensation Act as energy source/ storage Glycogen

1 1 1 1 1

(f)

It is not dissolve in water/compact Does not increase the osmotic pressure of the cell Store more energy

1 1 1

(g)

-Humans do not have the cellulose enzyme - that can digest the 1,4 glycosidic bonds between the glucose monomer that made up glucose

1 1

TOTAL

2(a)(i)

(ii)

(b)

Step 1 : Transcription Step 2 : Translation

10M

1 1

A : DNA B : mRNA C : Polypeptide

1 correct = 0m 2 correct = 1m 3 correct = 2m Max =2 m

-The double helix DNA unzip and one of the strand act as the template. -(The template) is used to form (a single stranded) mRNA. -The free nucleotides are attached together based on the complementary base pairing principles ( between DNA and RNA ) - by the role

(c )(i)

-

Enzyme can be reuse/repeatedly The product is not contaminated with enzyme The enzyme can be used at a wider temperature and pH

(ii)

-

Trap in an carrier matrix such as resin Place in gel like silica Bind by covalent bond in matrix like cellulose

1 1 1 1 Any2=2m 1 1 1 Any2= 2m 1 1 1

TOTAL

10M

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3 (a)

(b) (c)

P :aceytil coA Q : α ketoglutarat R : malat S : 0xaloacetat The matrix of the mitochondria -

(d)

3-4= 2m 1-2= 1m 1

it function as coenzyme,to carry out oxidation-reduction reactions acts as hydrogen acceptor to remove hydrogen atom and electron from a substrate Then passed to the electron transport system for ATP production

5 , 3, 3

NADPH and ATP /products of light-dependent reaction needed to convert -glycerate 3-phosphate to triose phosphate TOTAL

(b)

©

(d) (i)

1 Any2=2m 1

Glycerate-3-phosphate - transfer energy Ribulose phosphate - supplies phosphate (and transfer energy)

4 (a)

1 1

1 1 1 1 10M

M : Sigmoid curve // Limited growth N : Intermittent growth M : The growth of the organism continues throughout life N : The growth patern shows periods of extremely rapid growth follow by periods where there is little or no growth // Discontinuos growth

1 1 1 1

Growth as represented by increase in organic materials such as proteins is continuous Growth curve which uses length as a parameter is therefore not a true reflection of growth

1

- Germination occurred in darkness ( not photosynthesis ) - Starch had been hydrolysed into sugar - (which was) used for cellular respiration / other metabolic Activities

(ii)

- Embryo needs energy

(iii)

- Cellulose was synthesized to make new cell wall (during the process of growth )

1

1 1 1 Any 2 = 2m

1 1

TOTAL

10M

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NO. 5(a)

5(b)i

NO. 5(b) ii.

SUGGESTED ANSWER Fibrous proteins Globular proteins -do not have a tertiary structure. The -have a tertiary structure. Quaternary secondary structure is the most structure may or may not be present important -polypeptide chains are cross-linked at -polypeptide chain is tightly folded to form a spherical shape interval (to form long fibers//sheet) -dissolve in water (to form colloidal -insoluble in water, due to the large solutions), due to the hydrophilic R groups number of hydrophobic R groups -amino acid sequence may vary slightly -amino acid sequence is highly specific (never varies) between two samples -the length of polypeptide is identical in two -the length of polypeptide may vary in samples two samples of the same fibrous protein -amino acid sequence rarely exhibit -amino acid sequence is remarkable regularities regular -perform metabolic functions -perform structural function. -e.g. enzymes/I -e.g. keratin/fibroin/collagen

MARK

total

8M max 7

diffusion -net movement of solute /solvent molecules down a concentration gradient -membrane may or may not be present. If present it is a fully permeable

1/0 1/0

1/0 1/0 1/0 1/0

1/0 -involves a partially permeable membrane (permeable to water but not solute molecules total

Phagocytosis

1/0 2M MARK

Pinocytosis -material taken into cell is in liquid form -not selective (substances dissolve in surrounding medium will be taken into cell

1/0 1/0

-liquid is taken into cell by invagination of membrane total

5(b) iii.

1/0

osmosis -net movement of water molecules down a water potential gradient

SUGGESTED ANSWER

-material taken into cell is in solid form -selective process (cell can discriminate between particles taken into the cell and those not taken into cell -particles are taken into cell by invagination of membrane or by pseudopodia

1/0

-molecular size of the substance -solubility of the substance in lipid -charge on the particle of the substance total

1/0 3M 1 1 1 3M

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6 (a)

A

G

C

F D–1 L–1

B

D E (b)

-the biological cell membrane acts as barrier and are selectively permeable -the membrane consists of a fluid bilayer of phospholipids and various protein molecules embedded in it. Some of this protein molecules act as ion channels, carrier protein or pumps -the phospholipids bilayer has a hydrophobic middle region made up of hydrophobic fatty acids tails -the phospholipids bilayer is permeable to very small uncharged molecules like oxygen, and carbon dioxide, Steroid based hormone, fatty acids and alcohol(simple diffusion) - simple diffusion of water molecules across the semipermeable cell membrane is called osmosis. -some integral membrane protein form hydrophilic ion channels. This allows diffusion of various charged ions e.g. K+, Na+, Ca+, and HCO3-, down their concentration gradient. -some of this ion protein channels can open or close and are called gated channels e.g. voltage-gated channels and ligandgated channels -other large sized hydrophilic molecules such as glucose are transported across the cell membrane through facilitated diffusion using a protein carrier molecules - in facilitated diffusion, the binding of substances to the specific protein carrier causes the carrier to changes its shape and the substance is released into the cell -transport protein on the cell membrane can also transport substance across the cell membrane against the concentration gradient through active transport. -in active transport, the shape of protein carrier changes using energy (ATP) -exocytosis and endocytosis are active transport processes that move material in bulk across the cell membrane -excocytosis involve the transportation of substances out of the cell in bulk through the fussion of vesicle membrane with the cell membrane

1

1 1

1 1

1

1

1

1

1 1 1

1

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-in endocytosis the bulk substances is transported into the cell through the invagination of the cell membrane -pinocytosis occurs when the cell membrane invaginates to actively transport a small amount of fluid into the cell -in receptor- mediated endocytosis, ligand (cholesterol molecules) bind to specific receptors in coated pits on cell membrane. -all these structures and its related process enable the cell membrane to function as semipermeable membrane as well as enable the cell membrane to regulate the movement of substances in and out of the cell

1 1 1

1 max13

7(a) (i)

H2O

H2O K+

K+

H+

H2O

H+ H2O

Diagram Label (a) (ii)

- 1 marks - 1 marks

The mechanism of stomatal opening (during day) -potassium ion (K+) are pumped from subsidiary cells into the guard cell, H+ are pumped out of the subsidiary cells to maintain the electro neutrality -the increase of ion K+ and sugar(from photosynthesis) concentration makes the water potential of the guard cells more negative (lower), therefore -the water from subsidiary cells moves into the guard cell -the resultant increase in hydrostatic pressure causes the guard cells to become turgid -the uptake of water causes increased bowing of the guard cell (owing to the greater expansion of the outer walls than the inner wall ) and the stoma open the mechanism of stomatal closing (during night) -K+ ion are actively transported out from the guard cells into the subsidiary cells, H+ ions are transported into the guard cells -photosynthesis does not occur and the carbon dioxide concentration increases and the pH of the guard cell fall -sugar is converted into insoluble starch, therefore the water potential of the guard cell increases

2m

Max =8m

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(b)

8(a)

--light/blue light stimulate guard cells to accumulate potassium and become turgid, stoma open; or by driving photosynthesis in guard cells chloroplast, making ATP available for active transport of H+ -temperature – increased temperature stoma opens -air movement -dehydration(water stress) –in case of water deficiency, guard cells lose turgor and stoma closes. Mesoplhyll cells produce hormone abscisic acid which signals the guard cells to close. -concentration of carbon dioxide – depletion of CO2 within the air spaces of the lesf causes the stoma to opens -moisture/humidity

Max = 5m

Three hypotheses were suggested to explain how DNA replication occurs: • Semiconservative replication • Conservative replication • Dispersive model The procedure of the Meselson and Stahl experiment are as follows: - Escherichia coli were cultured for many generations in medium containing heavy nitrogen isotope 15N in order to label all DNA in E. coli with the heavy (15N ) nitrogen isotope. - Bacteria with 15N -DNA were then transferred to medium containing normal nitrogen isotope 14N. - Samples were removed at fixed intervals corresponding to the generation time of E.coli at a specific temperature - DNA from different generations were extracted and centrifuged in a solution containing caesium chloride ( CsCl) to separate denser DNA containing 15N from the ordinary DNA containing 14N - The position of DNA with 15N and DNA with 14N was measured in ultraviolet The results of the Meselson and Stahl experiment are as follows: - Generation 0 : All the DNA molecules contain 15N on both strands of the double helix,forming a dark band near the base of the centrifuge tube. - Generation 1 Generation 1: All the DNA were hybrids containing 15N in one strand and 14N in another strand,forming a band between the heavy and light DNA band - Generation 2: Half of the DNA were hybrids and another half were light DNA with 14N - Generation 3 : Third generation onwards,DNA with 14N increases but the number of hybrid DNA remain unchanged The result of the first generation eliminated the conservative hypotheses because this hypothesis does not explain the presence of hybrid DNA.

Max 8m

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15

15

N

N

Gen 0 Heavy DNA 15

14

N

14

N

15

N

N

Gen 1

Hybrid DNA

15

N

14

N

14

N

14

N

14

N

14

N

14

N

15

N Light DNA

Gen 2

Hybrid DNA

Hybrid DNA

•

• b

• • • • • • • • • •

Light DNA

Hybrid DNA

The result of the second generation eliminated the dispersive hypothesis because this hypothesis does not explain the presence of light DNA in the second generation Meselson and Stahl proved that DNA replicates semiconservatif DNA helicase enxyme unwinds the parental double helix by breaking the weak hydrogen bonds between the complementary base pairs Exposed base sequence on DNA strand acts as a template to enable the assembly of new complimentary DNA strand DNA polymerase elongates the DNA strand by adding new deoxyrbonucleotides one by one through complementary base pairing Adenine (A) base is paired with thymine (T) base, while guanine (G) paired with cytosine (C) base The leading strand is formed continuously from 5’ to 3’ Replication on the other complementary strand occurs discontinuously The short strands of DNA formed are called Okazaki fragments Okazaki fragments are joined by ligase enzyme to form lagging strand When replication is complete, two molecules of DNA are produced,each with one parental strand and one new complementary strand DNA replication occurs semiconservatively

Diagram = 2m

Total =10m 1 1

1

1 1 1 1 1 1 1 Any 5 = 5m

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9 (a)

(b)

- The kidney controls the blood osmotic concentration equilibrium through its control on the amount of water expelled from the body as urine. - When blood osmotic concentration increases, for example when there are inadequate water in the diet or excessive sweating occurs or excess salt is ingested, - the cells of the osmoreceptors in the hypothalamus are stimulated to secrete the antidiuretic hormone (ADH). - ADH increases the permeabilities of distal convoluting tubules and collecting ducts resulting in an increase in the reabsorption of water. - This increase the water content in the blood and body fluids and the osmotic concentration of blood and body fluid decrease. - A small volume of concentrated urine is produced. - When there is a high intake of water, the osmotic concentration of blood decrease. - The posterior pituitary gland secretes less ADH. - The collecting ducts and distal convoluted tubule remain impermeable to water. - Less water is reabsorbed as the filtrate passes in the distal convoluted tubule and collecting ducts. - Excess water is expelled through the kidney - and a large volume of dilute urine is produced.

- Almost 80% of the glomerular filtrate is reabsorbed at proximal convoluted tubules. - Al the glucose, amino acids, vitamins and hormones are reabsorbed actively into the peritubular capillaries. - Almost 70% of the sodium and chloride ions in the filtrate are reabsorbed actively into the peritubular capillaries. - This reduces the solute potential in the tubular filtrate. - Hence, 70 – 80% of the water is reabsorbed through osmosis. - About 50% of the urea in the filtrate diffuses into the peritubular capillaries. - This urea is then transported to all over the body. - The remaining urea in the tubule is excreted in the urine.

- Small protein molecules that have passed into the tubule during ultrafiltration are digested into amino acids that can diffuse into the peritubular capillaries.

1 1

1 1 1 1 1 1 1 1 1 1 max: 9

1 1 1 1 1 1 1 1

1 max: 6

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10(a) Carriers for the disorder: For the first cousin:

0.02 /2/100 2(0.02) = 0.04 =1/25 probability that one of the two common grandparents is a carrier

if one is the carrier, the offspring of a first-cousin marriage have a 1/16 probability of being homozygous for the disorder. The overall probability is : =1/25(1/16) = 1/400 For second-cousin offspring : p = (1/25)(1/64) = 1/1600 For the population at large, p = 1/10000 = 0.0001

1 1

1

1 1 1

1

1 Max 4

The number of Drosophilla with ebony body = 250-195 = 55 The frequency for homozygous recessive genotype (ebony body) q2 = 55/250 = 0.22 q = 0.22 = 0.469 p

= 1–q = 1 – 0.469 = 0.531

Frequency for KK (grey body) = p2 = 0.531 x 0.531 = 0.282 Frequency for Kk = 2 pq = 2 (0.282)(0.469) = 0.498 Genotype ratio = 0.282 (KK) : 0.489 (Kk) : 0.220 (kk)

1 1 1 1 1 1 1

1 1 1 1 1 Max 10

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10(b)(i)

Possible genotypes of blood groups of the childrens Blood group Possible genotypes A IA IA or IAIO AB IA IB O IO IO

1

Allele A causes production of antigen A on red blood cell Allele B causes production of antigen B on red blood cell Allele O causes no production of antigens on red blood cell Alleles A and B are codominant and allele O is recessive to both

1

As the first child is group A, its only possible genotype is IAIA or IA IO It must therefore have inherited one IA allele from one parent and IO allele from the other parent (IA IO)

1

As the second child is group AB, its only possible genotype is IAIB. It must therefore have inherited one IA allele from one parent and the other IB allele from the other parent

1

As the third is group O its only possible genotype is IOIO. It must therefore have inherited one IO allele from each parent. The mother, if IAIO, could donate such an allele and so, the father possible genotype is IBIO

1

So, the genotypes of the parents are: father blood group B Phenotype Parents Genotype IBIO

mother blood group A I AI O

1 Max: 5m

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10(b)(ii)

A genetic diagram to show the inheritance of ABO blood groups in the family. father mother blood group B blood group A Phenotype Parents Genotype IBIO IAIO Meiosis Gametes Offspring:

IO

1

Meiosis

IA

IB

IO Male gametes

IA Female Gametes

25 % blood group A (IAIO) 25% blood group B (IBIO)

IO

IA

I AI B

IAI O

1

IO

IBIB

IOIO

1 Max: 2m

25% blood group AB (IAIB) 25% blood group O (IOIO)