stpm trial 2009 bio q&a terengganu

8/14/2019 STPM Trial 2009 Bio Q&A Terengganu

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Name : NRIC :

JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR



JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJARJABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR











JABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJARJABATAN PELAJARANTERENGGANUJABATANPELAJARAN TERENGGANUJABATANPELAJAR









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.

Answerall 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.

TRIAL

STPM 2009

BIOLOGYPAPER 1

One hour and forty-five minutes

964/1

JABATAN PELAJARAN

NEGERI TERENGGANU

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

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1 Which molecule is made up of or contains glucose molecules?

A FructoseB CelluloseC Ribonucleic acid

D Deoxyribonucleic acid

2 Which statement best explain the polarity of water?

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

3 Which organelle, in an animal cell, is spherical in shape and bounded by a single

membrane?

A LysosomeB RibosomeC MicrobodyD Mitochondrion

4 Which of the following is most important function of epithelium tissue?

A SecretionB Protection

C AbsorptionD Transportation

5 Which statement is true of transcription?

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

6 Which statement is true of non-competitive inhibitor ?

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

concentration.

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11 The diagram above shows the conversion of pyruvate to ethanol during anaerobicrespiration. Which of the following allows glycolysis to continue?

A The regeneration of NAD+B The regeneration of NADH

C The release of carbon dioxideD The addition of yeast to ethanol

12 The diagram above shows the Krebs cycle. What is process X ?

A PhosphorylationB Oxidation of GTPC Oxidative PhosphorylationD Substrate level phosphorylation

13 Which is an example of saprophytic organism?

A Mucor sp.B Taenia sp.C Rafflesia sp.D Periplenata sp.

CH3

C=O

C=O

OH

CH3

C=O

H

CH3

H C OH

HPyruvate decarboxylase Alcohol dehydrogenase

CO2 NAD+NADH

Pyruvate EthanolAcetaldehyde

Acetyl coenzyme A

GDP

GTP

X

ADP

ATP

Krebs

cycle

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

X Y Z

A Glucose and glucose Glucose and fructose Glucose and galactoseB Glucose and fructose Glucose and galactose Glucose and glucoseC Glucose and galactose Glucose and fructose Glucose and glucoseD Glucose and galactose Glucose and glucose Glucose and fructose

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

A The increase in pH is due to vigorous activities.B The increase in pH causes the curve to shift to the right.C The percentage of oxygen-saturated haemoglobin decreases when pH

increases.D The shifting of the curve to the right is due to an increase in concentration of

blood carbon dioxide.

Lactase

Lactose Starch

Amylase

Maltose

Maltase

Sucrose

Sucrase

X ZY

100

0

50

10050

pH 7.2

pH 7.4

Percentage of oxygen

saturation (%)

Partial pressure of oxygen (mmHg)

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16 The diagram above shows the diffusion of carbon dioxide from respiring cells intothe blood involving steps P,Q,R and S. Which step requires carbonic anhydraseto proceed to the next?

A PB QC RD S

17 What is the volume of air that can be forced out following the deepest possibleinspiration?

A tidal volumeB vital capacityC residue volumeD expiratory reserve volume

18 Which response occurs when a person loses a lot of blood?

A A decerase in renin secretionB An increase in the secretion of sodium ionsC An increase in the production of angiotensinD A decrease in the production of aldosterone

CO2 CO2 + H2O H2CO3

O2 + Hb HbO2-

HHb H+

+ H2CO3-

Red blood cell

Tissue cells

Capillary wall

P

Q

R

S

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19 Which condition causes the closing of a stoma?

A The influx of potassium ions into the guard cellsB The increase in the concentration of glucose in the guard cellsC The decrease in the concentration of carbon dioxide in the guard cells

D The increase in the concentration of absicisic acid when plants are exposedto stress

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

Lactate Pyruvate glucose glycogen

What is the pathway known?

A Cori cycle

B Krebs cycleC Calvin cycleD Ornithine cycle

21 The diagram above shows the control of sodium ions level in the blood plasma.

What is hormone X?

A AdrenalineB AldosteroneC AngiotensinD Antidiuretic

Hormone X

Concentration of Na+

ions in

blood plasma decreases

Adrenal cortex

Inhibition

( negetivefeedback)Reabsorption of Na+

ions

increase

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22 Which of the following correctly explains the distribution of ions on either side ofthe membrane of an axon in its resting state?

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

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

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

D A high concentration of Na+ outside and a high concentration of K+ ions andorganic anions inside.

23 Where are the receptor sites for neurotransmitters situated?

A nodes of RanvierB presynaptic membrane

C postsynaptic membraneD membrane of the synaptic vesicles

24 Which statement is not true of auxin?

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

25 Oestrogen and progesterone are used in contraceptive pills. What is the effect ofthis hormones on mestrual cycle?

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

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Antigen

Stimulate

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

Which of the following are true of the plant?

l The plant is a long day plant.ll The plant is a short day plant.lll Far-red light cancels off the action of red light.

lV Red light can replace the requirement of dark period.

A l and lllB l and lVC ll and lllD ll and lV

27 What do W,Xand Y represent in the following simplified flow chart of the humoralresponse?

Treatment Result

Dark

P Flowering

Red LightQ Non flowering

Red LightR Flowering

Far red light

24 hours

Li ht

Li ht

Li ht

W

X Plasma cell

Y

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W X YA B cell Memory B cell AntibodyB B cell Antibody Memory B cellC T cell Memory T cell Helper T cellD T cell Memory T cell Cytotoxic T cell

28 Which of the following are true of B cell?

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

gland

A l and llB l and lllC ll and lV

D lll and lV

29 Which of the following is true of an oviparous animal?

A An individual hatches from the egg outside the female parents bodyB An individual hatches from the egg in the uterus of the female parentC An individual is born before maturity and continues to develop in the sac of

the female parent.D An individual develop in the uterus of the female parent and the embryo

obtains the nutrient from the placenta.

30 The hormone which plays an important role in seed germination is

A etheneB auxinC cytokininD gibberellin

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

32 The following are events that occur during seed germination.

l Synthesis and secretion of enzymesll Activation of the aleurone layerlll Flow of sugars to the embryolV Release of gibberellinV Hydrolysis of starch

Which of the following is the correct sequence of events during seedgermination?

A ll lV V lll lB lV l V ll ll

C lV ll l V lllD V lll lV ll l

Gain as percentage of previous mass Daily gain

A Age/weeks B Age/weeks

Gain as percentage of previous mass Daily gain

C Age/weeks D Age/weeks

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33 In a species of flowering plant, CR CR genotype produces red flowers, CwCwgenotype produces white flowers and CR Cwgenotype produces pink flowers.What is the percentage of the progeny that have pink flowers if a cross is madebetween CRCw and CRCw ?

A 0 %

B 25 %C 50 %D 75 %

34 Which of the following is the genotype of klinefelter syndrome?

A XOB XXYC XYYD XXX

35 Which of the following mutations in humans is /are trisomic?

l Down Syndromell Turner Syndromelll Thalassemia major

A l onlyB ll onlyC l and lllD ll and lll

36 Which of the following is not true of mutation?

A A chromosomal mutation of the deletion type involves the deletion of a basepair from gene.

B Genetic desease called cri-du-chatsyndrome is caused by a deletion inchromosome 5

C The deletion of two bases causes frame-shift mutation during triplet codingin transcription.

D Allopolyploidy is a chromosomal mutation which involves chromosomedoubling caused by different genomes.

37 A study on 400 mice about their resistance towards a type of poison has beencarried 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. Calculatethe number of mice expected to have Rr genotype.

A 16B 72C 128D 256

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38 Which of the following is true of repressor protein?

A It is coded by lacY.B It binds with the promoter and inhibits transcription.C It binds with gene that codes for -galacosidase.

D It changes is conformation after binding with lactose

39 A mutation in the lactose operan occurs which causes the repressor protein notbeing able to bind with the operator region. Which of the following statements istrue of the mutation?

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

C -galacosidase enzyme is produced continuosly in the absence of lactoseonly.

D -galacosidase enzyme is produced continuosly in the presence of lactoseonly

40 Which of the following are true of restriction enzymes?

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

A l and ll

B l and lllC ll and lVD lll and lV

41 Which of the following are the products of the translation of the lactose operon?

l Permeasell Transacetylaselll -galacosidaselV RNA polymerase

A l, ll, and lllB l,ll and lVC l,lll and lVD ll, lll and lV

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42 Which of the following regarding taxa and their examples not correct?

A Phylum: ChordataB Class: MammaliaC Order: HominidaeD Species: sapiens

43 Which of the following statements is not true of an artificial classification system?

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

which are arbitrarily chosen.

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

Phylum Characteristicl Cnidaria P Body divided into head,muscular foot and visceral

massll Artropoda Q Diploblastic body, polymorphismlll Mollusca R Segmented legs, chitinous exoskeletonlV Nematoda S Body covered with thin and elastic cuticle,

pseudocoelom

l ll lll lVA P Q S RB Q R P S

C R S Q PD S R P Q

45 Which of the following energy flows in an ecosystem involves the transfer of thegreatest amount of energy?

A Plant HerbivoreB Plant DecomposerC Herbivore CarnivoreD Carnivore Decomposer

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

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

Biston betulariainindustrial areas

Human birth weight indeveloped countries

Rabbit population inthe Andes Mountains

A P Q RB Q R PC R P QD R Q P

PQ

R

Initial

population

Final

population

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Name : NRIC :

JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR















JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJARJABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR








JABATAN PENLAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

Instructions to candidates :

This question paper consists of 10 printed pages

Answerallthe 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 examiners use

Section Marks Marks

Obtained

A

1 10

2 10

3 10

4 10

40

B

5 15

6 15

7 15

8 15

9 15

10 15

60

TOTAL 100

Two and a half hours

OTI 2

STPM 2009

JABATAN PELAJARAN

NEGERI TERENGGANU

964 / 2

BIOLOGY

PAPER 2

STRUCTURE AND ESSAY

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964/2* This question paper is CONFIDENTIAL until the examination is over [ Turn over ]

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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

(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 ]

H OO

C

CH2OHCH2OH

O

CC

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(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.

NucleusA

B

Ribosome

C

tRNA

Step 1

Step 2 Bdiffuses outfrom the nucleus

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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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(b) The diagram below shows two methods in which enzymes can be produced

for industrial processes.

( i ) Give two advantages using immobilized enzymes. [ 2marks ]

...

.

.

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

..

..

..

Microorganism

Extracellularenzyme

Enzyme extracted

Enzyme is purified

Immobilisedenzyme

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(iii) Explain the role of NAD+ in Kreb cycle. [2 marks ]

.

.

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

(i) Write in the boxes provided in the diagram the number of carbonatoms in each of the relevant substances. [ 1 mark ]

(ii) What are the roles of ATP in the conversion of glycerate 3-phosphateto triose phosphate and ribulose phosphate to ribulose biphosphate?[ 2 marks ]

.

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

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

(iii) A plant was allowed to photosynthesis normally. The light was thenswitch off. There was a rise in the amount of glycerate 3-phosphate present inthe chloroplast of this plant. Explain why. [ 2 marks ]

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

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

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

Ribulosebiphosphate

RuBP

Ribulosephosphate

TriosePhosphate

Carbondioxide

Glycerate3-Phosphate

GP

ReducedNADPNADP

ATP ADP

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`

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

(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 growthcurve 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 darknessfor 10 days. After 10 days, the dry mass of the seedlings was analysed andthen compared with a sample of 100 g of ungerminated corn grains (y).

Time/years Time/years

Height/cm

Height/cm

A broad bean plant An insect

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The table below shows the results of the experiment.

Dry mass of ungerminatedcorn grains (y)

Dry mass of cornseedlings after 10 days

(x)

Cellulose 1 g 4 g

Starch 65 g 8 gOther organicsubstances

15 g 31 g

Total dry mass 81 g 43 g

(i) Explain why the total dry mass of seedling ydecreased after 10 days.

[ 2marks ]

.

.

.

(ii) Seedlings xcontain more cellulose than corn grain ywhich did notgerminate. Explain why. [ 1mark ]

(iii) Name a source a carbon used for synthesising new cellulose molecules.[ 1mark ]

.

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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 lobularproteins, based on the shape of the molecule. Describe the differences between the twotypes 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 Singers model[2 marks]

(b) Explain the roles of the structures of the cell membrane in the transportation ofsubstances into the cell. [13 marks]

7. (a) With the aid of labelled diagram, explain [ 10 marks]i The structure of a stoma

ii. The mechanism of stomata opening and closing.

(b) Outline the environmental f actors which influence the opening and cl osing of astoma. [5 marks]

8 (a) Explain briefly the experiment conducted by Meselson and Stahl to prove the DNAreplication 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 bodyfluids. [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 greybodies. 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 thealleles 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.IA and IB are codominant, IO is resessive. A man of blood group B married a woman ofunknown A BO bl ood group. They had t hree ch ildren. One o f t he children had bl oodgroup 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 genetic diagram to show the inheritance of ABO blood groups in this

family. [2 marks]

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ANSWER SCHEME BIOLOGY STPM

PAPER 1 TRIAL/ OTI 2 2009

Q ANSWER Q ANSWER

1 B 26 C

2 B 27 A3 A 28 B

4 B 29 A

5 B 30 D

6 A 31 B

7 A 32 C

8 C 33 C

9 D 34 B

10 B 35 A

11 B 36 A

12 C 37 C

13 A 38 D14 D 39 B

15 D 40 D

16 A 41 A

17 B 42 C

18 C 43 A

19 D 44 B

20 A 45 A

21 B 46 C

22 D 47 D

23 C 48 C

24 C 49 C

25 C 50 C

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Marking Scheme- Trial STPM 2009Paper 2

1(a)(b)(d)(e)

(f)

(g)

Dissacharide1,4 glycosidic bondCondensationAct as energy source/ storageGlycogen

It is not dissolve in water/compactDoes not increase the osmotic pressure of the cellStore more energy

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

that made up glucose

TOTAL

11111

111

11

10M

2(a)(i)

(ii)

(b)

(c )(i)

(ii)

Step 1 : TranscriptionStep 2 : Translation

A : DNAB : mRNAC : Polypeptide

-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 complementarybase pairing principles ( between DNA and RNA )- by the role

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

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

TOTAL

11

1 correct= 0m

2 correct= 1m

3 correct= 2m

Max =2 m

111

1Any2=2m

111

Any2= 2m

111

10M

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3 (a)

(b)

( c )

(d)

P :aceytil coAQ : ketoglutaratR : malatS : 0xaloacetatThe matrix of the mitochondria

- 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

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

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

TOTAL

3-4= 2m1-2= 1m

1

11

1Any2=2m

1

11

1

1

10M

4 (a)

(b)

(d) (i)

(ii)

(iii)

M : Sigmoid curve // Limited growthN : Intermittent growthM : The growth of the organism continues throughout lifeN : The growth patern shows periods of extremely rapid growth follow byperiods where there is little or no growth // Discontinuos growth

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

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

Activities

- Embryo needs energy

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

TOTAL

1111

1

1

111

Any 2 =

2m

1

1

10M

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NO. SUGGESTED ANSWER MARK5(a) Fibrous proteins Globular proteins

-do not have a tertiary structure. Thesecondary structure is the mostimportant

-polypeptide chains are cross-linked atinterval (to form long fibers//sheet)-insoluble in water, due to the largenumber of hydrophobic R groups-amino acid sequence may vary slightly

-the length of polypeptide may vary intwo samples of the same fibrousprotein-amino acid sequence is remarkableregular-perform structural function.-e.g. keratin/fibroin/collagen

-have a tertiary structure. Quaternarystructure may or may not be present

-polypeptide chain is tightly folded to form aspherical shape-dissolve in water (to form colloidalsolutions), due to the hydrophilic R groups-amino acid sequence is highly specific(never varies) between two samples-the length of polypeptide is identical in twosamples

-amino acid sequence rarely exhibitregularities-perform metabolic functions-e.g. enzymes/I

1/0

1/0

1/0

1/0

1/0

1/01/01/0

total 8Mmax 7

5(b)i diffusion osmosis-net movement of solute /solventmolecules down a concentrationgradient-membrane may or may not bepresent. If present it is a fullypermeable

-net movement of water molecules down awater potential gradient

-involves a partially permeable membrane(permeable to water but not solutemolecules

1/0

1/0total 2M

NO. SUGGESTED ANSWER MARK5(b)ii.

Phagocytosis Pinocytosis

-material taken into cell is in solid form-selective process (cell candiscriminate between particles takeninto the cell and those not taken intocell-particles are taken into cell byinvagination of membrane or bypseudopodia

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

-liquid is taken into cell by invagination ofmembrane

1/01/0

1/0

total 3M5(b)iii.

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

111

total 3M

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6 (a)

D 1

L 1

(b) -the biological cell membrane acts as barrier and are

selectively permeable-the membrane consists of a fluid bilayer of phospholipids andvarious protein molecules embedded in it. Some of this proteinmolecules act as ion channels, carrier protein or pumps-the phospholipids bilayer has a hydrophobic middle regionmade up of hydrophobic fatty acids tails-the phospholipids bilayer is permeable to very smalluncharged molecules like oxygen, and carbon dioxide, Steroidbased hormone, fatty acids and alcohol(simple diffusion)- simple diffusion of water molecules across thesemipermeable 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 arecalled gated channels e.g. voltage-gated channels and ligand-gated channels-other large sized hydrophilic molecules such as glucose aretransported across the cell membrane through facilitateddiffusion using a protein carrier molecules- in facilitated diffusion, the binding of substances to thespecific protein carrier causes the carrier to changes its shapeand the substance is released into the cell-transport protein on the cell membrane can also transport

substance across the cell membrane against the concentrationgradient through active transport.-in active transport, the shape of protein carrier changes usingenergy (ATP)-exocytosis and endocytosis are active transport processesthat move material in bulk across the cell membrane-excocytosis involve the transportation of substances out of thecell in bulk through the fussion of vesicle membrane with thecell membrane

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-in endocytosis the bulk substances is transported into the cellthrough the invagination of the cell membrane-pinocytosis occurs when the cell membrane invaginates toactively transport a small amount of fluid into the cell-in receptor- mediated endocytosis, ligand (cholesterolmolecules) bind to specific receptors in coated pits on cell

membrane.-all these structures and its related process enable the cellmembrane to function as semipermeable membrane as well asenable the cell membrane to regulate the movement ofsubstances in and out of the cell

1

1

1

1max13

7(a) (i)

Diagram - 1 marksLabel - 1 marks

2m

(a) (ii) The mechanism of stomatal opening (during day)-potassium ion (K+) are pumped from subsidiary cells into the guardcell,H+ are pumped out of the subsidiary cells to maintain the electroneutrality-the increase of ion K+ and sugar(from photosynthesis) concentrationmakes 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 cellsto 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 thestoma openthe mechanism of stomatal closing (during night)-K+ ion are actively transported out from the guard cells into thesubsidiary cells, H+ ions are transported into the guard cells-photosynthesis does not occur and the carbon dioxide concentrationincreases and the pH of the guard cell fall-sugar is converted into insoluble starch, therefore the water potentialof the guard cell increases

Max =8m

K+

H+

H2O

H2O

H+

K+

H2O

H2O

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(b) --light/blue light stimulate guard cells to accumulate potassium andbecome turgid, stoma open; or by driving photosynthesis in guard cellschloroplast, 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 hormoneabscisic acid which signals the guard cells to close.-concentration of carbon dioxide depletion of CO2 within the airspaces of the lesf causes the stoma to opens-moisture/humidity Max = 5m

8(a) Three hypotheses were suggested to explain how DNA replication occurs:

Semiconservative replication

Conservative replication

Dispersive modelThe procedure of the Meselson and Stahl experiment are as follows:

- Escherichia coliwere cultured for many generations in medium containingheavy 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 normalnitrogen isotope 14N.

- Samples were removed at fixed intervals corresponding to the generationtime of E.coli at a specific temperature

- DNA from different generations were extracted and centrifuged in a solutioncontaining caesium chloride ( CsCl) to separate denser DNA containing 15Nfrom 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 andlight DNA band

- Generation 2: Half of the DNA were hybrids and another half were light DNAwith 14N

- Generation 3 : Third generation onwards,DNA with 14N increases but thenumber 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. Max8m

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The result of the second generation eliminated thedispersive hypothesis because this hypothesis does notexplain the presence of light DNA in the secondgeneration

Meselson and Stahl proved that DNA replicatessemiconservatif

Diagram = 2m

Total =10m

b DNA helicase enxyme unwinds the parental double helix bybreaking the weak hydrogen bonds between thecomplementary base pairs

Exposed base sequence on DNA strand acts as a templateto enable the assembly of new complimentary DNA strand

DNA polymerase elongates the DNA strand by adding newdeoxyrbonucleotides one by one through complementarybase pairing

Adenine (A) base is paired with thymine (T) base, whileguanine (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 areproduced,each with one parental strand and one newcomplementary strand

DNA replication occurs semiconservatively

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111

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1Any 5 = 5m

Light DNA

Light DNA

Hybrid DNA

Gen 0

Gen 2

Gen 1

15N 15N

15N 14N14N 15N

15N 14N14N14N 14N14N14

N 15N

Heavy DNA

Hybrid DNA

Hybrid

DNA

Hybrid

DNA

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9 (a)

(b)

- The kidney controls the blood osmotic concentration equilibrium throughits control on the amount of water expelled from the body as urine.

- When blood osmotic concentration increases, for example when thereare inadequate water in the diet or excessive sweating occurs or excesssalt is ingested,

- the cells of the osmoreceptors in the hypothalamus are stimulated tosecrete the antidiuretic hormone (ADH).

- ADH increases the permeabilities of distal convoluting tubules andcollecting ducts resulting in an increase in the reabsorption of water.

- This increase the water content in the blood and body fluids and theosmotic 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 proximalconvoluted tubules.

- Al the glucose, amino acids, vitamins and hormones are reabsorbedactively into the peritubular capillaries.

- Almost 70% of the sodium and chloride ions in the filtrate are reabsorbedactively 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 duringultrafiltration are digested into amino acids that can diffuse into theperitubular capillaries.

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max: 9

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10(a) Carriers for the disorder: 0.02 /2/100For the first cousin: 2(0.02) = 0.04

=1/25 probability that one ofthe two commongrandparents is a carrier

if one is the carrier, the offspring of a first-cousin marriage have a1/16 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

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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 = 22.0

= 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

1111

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Max10

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10(b)(i) Possible genotypes of blood groups of the childrensBlood group Possible genotypes

A IA IA or IAIOAB IA IBO IO IO

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

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 andIO allele from the other parent (IA IO)

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

As the third is group O its only possible genotype is IOIO. It musttherefore have inherited one IO allele from each parent. Themother, if IAIO, could donate such an allele and so, the fatherpossible genotype is IBIO

So, the genotypes of the parents are:father mother

Phenotypeblood group B blood group A

Parents

Genotype IBIO IAIO

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10(b)(ii) A genetic diagram to show the inheritance of ABO blood groups in

the family.father mother

Phenotypeblood group B blood group A

ParentsGenotype IBIO IAIO

Meiosis Meiosis

GametesOffspring:

Male gametes

FemaleIAIB IAIO

GametesIBIB IOIO

25 % blood group A (IAIO) 25% blood group AB (IAIB)25% blood group B (IBIO) 25% blood group O (IOIO)

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IO

IB I

A IO

IOI

A

IO

IA

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stpm trial 2009 bio q&a terengganu

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