stpm trial 2009 bio q&a (n9)

8/14/2019 Stpm Trial 2009 Bio Q&A (n9)

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CONFIDENTIAL*JABATAA :GERISEMBILAN JA WBILANJABATAA 96411 'GERISEMBILAN JA STPM 2009 ITBILANJABATAA :GERISEMBILAN JA WBILANJABATAf\.. ~ ' - ' ov . . . " . . . . . G E R I S E M B I L A N JAw .. .... ~ ' - ' 'V I .. " ..... ~ ~ ~ , , , ~ ~ . I I B I L A N JABA TANPELAJARANNEGERISEMBILAN JABA TANPELAJARANNEGERISEMBILANJABATANPELAJARANNEGERISEMBILAN JABATANPELAJARANNEGERISEMBILAN

.JABA TANPELAJARANNEGERISEMBILAN JABA TANPELAJARANNEGERISEMBILANJABATANPELAJARANNEGERISEMBILAN JABA TANPELAJARANNEGERISEMBILANJABATANPELAJARANf'v BIOLOGY -AJARANNEGERISEMBILANJABA TANPELAJARANN .AJARANNEGERISEMBILANJABATANPELAJARANf'v- __ ... __ ... _._ .. . _ . . .. .... __AJARANNEGERISEMBILANJABATANPEL A . I A R A N N r : : r . : ; F R I r : : M R I I AN .IARATANPFI A.IARANNEGERISEMBILANJABA TANPEL Paper 1 EGERISEMBILANJABATANPEL . EGERISEMBILANJABATANPEiJ-IJt-IrV..,VIVC\JC/\10C1VIDILJ1IV JrlDrll rlIVrCLrlJrI/\t-IIVlvEGERISEMBILANJABA T A N P E L A J A R A N N E G E R I . ~ . . . . . . ANPELAJARANNEGERISEMBILANJABATANPELAJARANNEGERf. (1 % hours) -ANPELAJARANNEGERISEMBILANJABATANPELAJARANNEGERI. . -ANPELAJARANNEGERISEMBILANJABATANPELAJARANNEGERISEMBILAN JABA TA NPELAJA RANNEGERISEMBILAN

JABATAN PELAJARAN NEGERl SEMBILANPERCUBAAN BERSAMASIJIL TINGGI PERSEKOLAHAN MALAYSIA2009

Instructions to candidates:DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SOThere are fifty questions in this paper. For each question, Jour suggested answers aregiven. Choose one correct answer and indicate it on the multi-choice answer sheetprovided.Read the instructions on the multiple-choice ans'rjler sheet very careJully.Answer alJ questions. Marks will not be deductedJor wrong answers.

This question paper consists of 16 printed pagesSTPM TRIAL 964/1 CONFIDENTIAL*

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CONF ID EN TI AL* 2I. "Vhi ch of th e follovving are th e prope rti es of water?

ABCD

Specific heatcapacityHi ghLowHi ghLow

Latent heat of Surfacevapo uri sa tion ten sionHi gh HighLow HighHigh LowHigh Low

2. Diagram belO\v shows a type of monosaccharide.

OHWhich of the foll ow in g pol ymers can be formed from the condensation of themol ec ul e shown in the diagram above?

GlycogenII Amylose111 AmylopectinIV StarchA II and IIIB I, II and IIIC II , III and IVD J, 11, III and IV

3. A mitochondrion in (I mammalian mu scle cell measures 1.2 f-lm. In the elec tronmicrograph, th e length of the organelle is 48 mm. What is the magnification of theelec tron micrograph?

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A 40000 XB 12000 XC 4400 XD 4000 X

*This question paper is CONFIDENTIAL until th e examination is over CONFIDENTIAL*

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C:ONF IDENTI AL* 3

4. Which of th e following structures of membran e bound organelles is correctlymatched with its function?

A

BCD

StructureAn ex tens ive network of tubes; each tube isbound by a single menlbrane. .A sta ck of elongated, curved sacs; each sac isbound by a single membrane.A spheri ca l sac bound by a single membrane.A sac bound by two membranes, the inner ishighl y folcled.

FunctionLipid synthesisPhotosynthesis

Protein synthesisPackaging of proteins

s. The graph below shows th e rate of reaction with and without an inhibitor.

Rate ofreaction

Concentration of substrateWhich of th e following is true regarding the above graph?

Curve J Curve 2

A Compet itive inhibi tor Non-competitive inhibitorCurve 3

Normal activity.

B Co mp etiti ve inhibitor Normal activity Non-competitive inhibitor

c Non-competitive inhibitor Competitive inhibitor Normal activity

D Normal activity Compet it ive inhibitor Non-competitive inhibitor

964/ 1*Thi s ques tion paper is CONFIDENTIAL until the examination is over CONF1DENTIAL*

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CONF IDEN ']'IAL* 4

6. Th e table below shows amin o acids and th eir ba se se qu ence on the mRNA codon.Amino ac id Base se qu ence on th e mRNA codon,Valine GUCGlyc ine GGUMethionine AUA

What is the ba se sequ ence on pnrt of a DNA strand , which wo uld code for the tripeptideva l ne-m eth ion ine-gl yc in e?

A GUC AUA GGUB GTC ATA GGA C CAG TAT CCAD TAT CAG CCA

7. Di ag ram below shows th e electron pathwny in cyc lic and non-cyclic photophosphorylation duringthe li ght reacti on stage in ph otosynth es is.

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energyleve l

ferrodoxinquinone II ;.,.1\ 8\1 y . X ~ D - + N A D P H 2e- /e- Z2e -- \ 2H + 2e --T OlepculeI "\I /"-" O2 released(PS 1",---;J~ ~ + - - - - - - - - - - - - - ~

Which of the followin g statements are tru e for elec tron pathway in the above diagram?Th e electrons in PSI and PS II nre exc itecl to hi gher energy leve ls

II NADP+ is reduced in non-cyc li c ph otophosphorylationIII ATP is produced in steps Y and ZIV Molecule P is a water molec ul eA I and II onl yB II and IV only

C I, II and III onlyD I, II , III and IV

*Thi s question paper is CONFIDENTIAL until the examination is over CONFIDENTIAL*

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CONF ID ENTIAL';' 5

8. The grap h belo'w shows th e relati onship between th e rate of photosynthes is with environment alfa ctor X.

rate of photosynthesis(arbitrary unit)t

Which of th e fo llow ing is factor X?I Light intensityII Oxygen concentrationIII TemperatureIV Ca rbon dioxide co ncentrat ionA I and IV onl yB II and III onl yC " and IV onl yD I, III and IV on ly

Factor X

9. Which products are formed during ana erob ic respiration?Mu sc les Yeast cells

A Pyruvate, NAO+, ATP Ethanol, NAD+, ATPB Lactate, NAO+ , ATP Ethanol, NAO+, ATP, CO2C Lactate, NAD-", ATP, CO2 Ethanol, NAO+, ATPD Et hanol , NADH, ADP, CO 2 Lactate, NADH, ADP, CO2

96411*T hi s question paper is CONFIDENTIAL until th e examination is over CONFlDENTIAL*

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CONFIDENTIAL * 610. Diagram below shows a reaction in Krebs cycle.

What is process X?A Chemiosmosis

Krebscycle

GTP

A D ~ G D P ATP

C PhotophosphorylationB Oxidative Phosphorylation D Substrate level phosphorylation

11. Which of the following organism is/are not saprophytes?Afucor

II RhizopZlsIII Ta eniaIV AmoebaA I onlyB II and III only C

III and IV onlyD I, III and IV only

12. Which of the following sequences brings about Bohr's effect?

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I I-t causes oxyhaemoglobin to release its oxygenII CO2 enters red blood cellIII Carbonic anhydrase catalyses the formation of 1-1 2C03IV Respiration of tissues gives out CO2A I, III, IV and IIB II , Il l , I and IV C IV , III , II and Io IV, II , III and I

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

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C O N F 1 E N A 713. Whi ch of the fo llow ing expbins why myoglobin IS foun ci III large amounts II I th e mu sc les of

bird s?A Myoglob in ca n ') Ll ppl )' th c encrgy needed for nyingB Myoglob in can supp ly oxyge n to th e Illll scul ar ti ssues when birds fly in areas w ith low

parti al pressure of oxygen.e Wh en compared to haemoglob in , myoglobin has higher affinity towards oxygen ane! can bett erretain oxygen in ac ti ve mu sc les .

D Myoglobin has a sma ll er Illolecul ar size to h3emcg lob in .14. Which proce ss does not occur durin g th e opening of stomata?

A Su ga r is converted into starch.B Starch is converted into Illa li c ac id .e Water enters int o guard cells by osmosis.D Potass iulll ions diffuse into guard ce ll from adj ace nt ce ll s.

15 . Whi ch of the follow ing in creases the rate of im pul se relea sed from s inoatrial node?

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I Impulse from the sy mpatheti c nerveIl Impul se from the paras ymp ath eti c nerveIII Thyrox ine horm oneIV Adrena line horm oneA I and IVB II and IV

e I, I II andl VD II , lIl andl V

*This question paper is CONFIDENTI AL until the ex amination is over CONFlDENTIAL*

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CONF ID ENTIAL* 816. Diagram belo\-\' sho'vvs a i'vlullch model.

x

Pure water

Semipermeable membraneI f A is fill ed with concentrated sucrose so lution, while B is fill ed \vith dilute sucrose so lution,which of tile foll owing statements are true?1 A is eq uivalen t to leaves in translocat ion .II Hi gh turgor pressure is created in A.III X is analogoLls to phloem in plants.IV Y is analogous to xy lem vessels in plants.

A 1, II and IIIB 1, II and IVC I, III and IVD I, II, 11I and IV

17. What is tile function of ad renaline in th e negative feedback mechanism ifbody temperature ofendotherms decreases?A Stimulate secret ion of sweatB Increase convers ion of glycoge n into g lucose in li verC Trigge rs a shiverin g processD Contract erector mu sc les

18. Liver is an organ in vo lved in homeostasis bec ause it

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I produces bile sa ltII performs gluconeogenesisII I performs detoxi fi cati onIV produces a hi gh amount of hea tA III and IVB I, II I and IVC II , III and IVD I, II , II I and IV

*This question paper is CONFIDENTIAL until th e examination is over CONFIDENTIAL*

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CONFIDENTIAL* 919. Which of the foll owing are th e characte risti cs of an impul se?

I It foll ows th e a ll or nothing lawII Speed of impulse transmi ssion is faster in mye linated axonII I Speed of impul se transmi ssion is fas ter in unm ye lin ated axon 'with a sma ller diameterIV In th e absolute re fractory period, a new impulse cannot be generated even if a strongstimulus is rece ivedA L II and !I IB I, IlandlVC I, III and IVD I, II , III and IV

20. Curare effects neurolllLl scul ar junction byA binding to receptors on postsynap tic membraneB preventing exocytosis of acetylcholine into the synaptic cleftC preventing action of cholinesteraseD preventing di ffu sion of calcium ions into presynaptic membrane

21. The follovvings are the event s th at occur in th e act ion of a steroid based hormone.I Hormone-recept or complex diffuse into the nucleusII Steroid based hormone diffuse into the ce ll through the cell membraneIII Hormone binds to the receptor in th e cytoplasmIV Specific gene is activated in DNAThe correct sequence for the event s abo ve isA I, II , Ill, IVB I, I ll , II , IVC II , III , I, [V)) II , [, [II , IV

22 . Ifplant X is a short day pl ant with a critical da y length of8 1/2 hours, in which of th ese conditionswill pl ant X not flower '?

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Exposed to red light for 9 hoursII Kept in darkness for 8 hoursIII Exposed to far red I ght for 12 hoursIV Kept in darkness for IS hours before exposed to red li ght for I hourA I and IIB III and IVC I, II and IVD II , III and IV

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

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CONFIDENTIAL* 1023. Which of the followings are invol ved in cell-m ed iated response of the il11rnun e system?

I inv olves B lymphocytesII mature T lymphocytes req uire ma crophages to prese nt th e anti ge n for th em to bindIII in vo lves th e th ymu s g landIV mat ure lymphocytes can directly bind with antigens.A I and IVB II and 11IC I and 11ID II and IV

24. HIV lies dormant in th e body becauseA its reverse transcriptase is engulfed by T4 cellB its parti cle rest in the cytoplasm ofT4 cellC the replication of its RNA occurs laterD its DNA is incorporated into the DNA ofT4 ce lI

25. The sporangium of Marchantia where spores are pl:oduced is aA diploid structureB haploid structureC dioecious structureD ov iparous structure

26. Which orthe fo ll ow in g terms app ly to birds?

oviductII vIvIparousIII internal fertilisationIV hermaphrod itesA II and II IB I and IIIC III and IVD II and IV

27. In an amniotic egg of a bird, the reservo ir fo r waste and th e part th at provides for gasdiffusion is the

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A chorionB yo lkC amnIono allc1lltois

*Thi s question paper is CONFIDENTIAL until the exan1in ation is over CONFIDENTIAL'"

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CONfIDENTIAU' II28.Th e formation 01blastul a frolll n zygote in vo lves C\ success iOI1 of" rnp id ce lldivision s. Thi s speciall ype orcell di vision creates Clill uiti ce llul ar emb ryo

Th e above statement refers to whi ch stage of embryo llic development?A BlastomereB CleavageC GastrulationD Organogenesi s

29. The hormone that plays an imp ortant role in seed germ inati on and ea rly seeelinggrowth isA auxll1B gibberelinC cytok ininD absc isic acid

30. The tab le below shows four patterns of growth curv es and their exampl es.Growth pattern Examples

(a) Isometric grov"th I (i) 1lulllan organs(b) Allometric growth (i i) Coral ree fs(c) Intermittent growth (iii ) fi sh(eI) unlimited growth (iv) Grasshopp er

Which of the fo llow ing is correct ly p ~ l i r e d-(a) (b) (c) (d)

A (i) (i i ) (iv) (ii)B (i ) (i ) (i ') (i)C (i ii) (i) (i i) (iv)D (iii) (i) (iv) (i i)

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*Thi s question paper is CONfIDENTIAL un til the exam ination is over CONFIDENTIAL*

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CONF ID ENT IA L* 123 1. Which of the fol low in g statement s me tru e ilb out neur osec retion secreted by th e neuro sec retoryce ll in in sects?

I It is stored in corpus carel iacull1II It inhibi ts th e effect of juveni le hormoneII I It is also known as prothoracictrop hi c hormoneIV It stimulates th e secretion of ecdysone from prothorac ic hormoneA I, II and IIIB I, III and IV

C II , III and IVD I,II , lll and lV32. In pea plants, red flovvers (R) are dominant to white fl owers (r) and ta ll (T) to shOl1 (t). The tabl e

shows th e ga metes and poss ibl e offspr in gs of a dihybr id cross . The numbers 1- 16 represent th egenotypes of each ind ividua J.

RT Rt rT I' tRT I 2 3 4Rt 5 6 7 8rT 9 iO I I 12rt 13 14 15 16

If plants 4 and 13 are crossed , what proportion of th eir offspr in g \;vi ll show at least one recessivetrait?

A 1/16 B 611 .6 C 7/ 16 o 911633. The colour of onions is controlled by tw o pairs of all eles Ss and Rr, whi ch segregate

independently. The a ll ele S is dOlninant anci 1l1u st be present to a ll ow development of pigment inth e sk in . In its absence, the on ion is wh ite. A ll ele R is dDmin ant ond gives a red co lour, therecessive r g ives a ye llow co lou i-. Y\ hat wi ll be th e ratio of phenotypes in th e offspring of a crossbetween plants of genotypes SSR,R and ss r!"!

I white 3 red J wh teB

1 redI red I ye ll ow

CD I whit e : 2 red I ye llow

34 . Which of the following statement's are no t tru e about mutation s?

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mutat ions are spont aneo us changes in th e amount or structure of DNA Orchanges in th e sequence of Ilucleo tide ba ses of a gene.

II the changes that occurred are non-randomIII all mutations lead to production of non -functi ona l pr ote in sIV germ-line mutations can be -nh eritedA I and IIIB I and IV C II and II Io II and IV

*Thi s qu estion paper is CONF ID ENTIAL until the exa min ation is over CONFIDENTIAL'"

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CONF IDENTIAL* 1335. Chromoso mal deletion occurs

A when th e deleted DNA segment becomes rea ttac hed in an in ve rted po sition.B wh en the chromClsome brea ks and a segment of it is los tC wh en a segmen t of a chromoso me co ntainin g a se qu ence of lltlcleoticles is repeated.D when the chromosome segment bec omes deleted an d rejoins at a different pos iti on on th e

same chromo so me or another non-h omol ogous chromosome.36. Th e Hardy-Weinberg eq uati on does not apply if

I th ere is migrationII there is natural se lection.II I mutations occur.IV th ere is non-random matin g.V there is a large population .A I, II , III and IVB I, II , III and VC I, Ill , IV and VD II , III , IV and V

37. The maintenance of the allele for sickl e-ce ll anaemia in hum an popul ations in malar ia-end emicregion in Africa is an exa mple ofA ge netic driftB ge ne flowC founder effectD heterozygote advan tage

38. What is th e function of th e induce r of th e lac operon?A Bind to the promoter and prevents th e represso r from bindij!g to th e operator.B Bind to the op erator and prevents the repressor from binding to th e promoter.C Binds to the repressor and prevents it from bi nd ing to th e' PI:Oll1oter.D Binds to the represso r and prevents it from binding to th e operato r.

39. The statemen ts below re fer to different stages in th e production of human insulin by genet iceng inee ring techniques. Wh at is the co rrect sequence productio-il?

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I DNA cut "vith res triction enzymes.II mRNA extracted frolll pancrea tic ce ll sIII Plasmid DN A and human DNA joined using li gase enzy mes.IV DNA copy made us ing reverse transcriptaseV Recombinant plasmid inc orporated into bacterial ce ll.A I, Il l , IV , II , VB II , IV , II I , V, I

C II , IV , I, Ill , VD II , I, IV , Ill , V

*Thi s qu es tion paper is CONFIDENTI AL until th e examin at ion is over CONFIDENTIAL*

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CONFIDENTIAL* 1440. Which of the following are characteristics of pl asmid ?

Found in the bacteriaII Integral par! of bacterial chromosomeIII Confers special characteristics to th e organi smIV Self-replicatingV Small, linear molecule

A I, ll , III and I VB I, III, IV and VC I, III and IVD III , IV, and V4 J. The following are uses of recombinant DNA technology except

A screening for carriers of genetic diseasesB identification of badly-decomposed victimsC treatment of diabetic patients with synthetic insulino prevention of inheritance of genetic disorder by offsprin g

42. The table bdO\v shows the taxonomic groups and taxa for the hou se ny.

I11IIIJV

Taxonomic groupKingdomPhylumClassOrder

Taxon(a) Arthropoda(b) Insecta(c) AnimaIia(d) Diptera

Which of the following combinations is correct ?

ABCD

IXc)(c)(e)(c)

11(a)(b)(d)(a)

III(b)(a)(a)(d)

43. Which of the following is likely to be radially symmetrical?A A chordate C An arthropod13 A cnidarian D An annelid

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

IV(d)(d)(b)(b)

CONFIDENTIAL*

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CONF ID ENT IAL* IS44. Whi ch of th e fo ll owing statements are tru e about Marchanlia?

I Mono ecio usII Asexual reprod uction th ro ugh ge mm a cup sIII Ga metop hyte is domin antIV ga metop hyte is depend ent on sporoph yte

A I and IIB I and IIIC II and IIID 11 and IV45. Which of th e follow ing ca nn ot cause th e ge netic va ri ation?

I Mutati onII Ca moufl age1II RecombinationIV DominancyA II and IIIB III and IVC I and IIID II and IV

46. Which one of th ese is the definiti on fo r th e.:biogeochemi ca l cycle?A Circul ation of chemi ca l elements th ro ugh th e biotic co mp onen t of an ecosystem.B Circul ati on of chemical elements th rou gil th e ab iotic componen t of an ecosystem.C Circulation of chemi ca l element s through th e bio tic and ab iotic co mpo nent of an ecosystem.D Circul ation of organi c molecules through th e bio tic components of an ecosyste m.

47. Th e main reservo ir of ph osphate is

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A inorga ni c ph osph ate ions in soil.B inorgani c phosph ate in rocksC inorga ni c phosphate in orga ni sms.D Orga ni c phosphate in orga ni sms.

*T hi s qu es tion paper is CON FID ENTIAL until the examinati on is over CON FID ENTIA L*

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CONFIDENTI AL* 1648. Th e success of organisms in the ecosystem can be shown by the

1 population di stributionIl population si zeIII prey-predator relationshipIV position of trophi c level in th e ecosystem.A I and IIB I and IIIe l a nd IVD II and III

49. A quadrate with a measurement of SO cm x SO cm is L1 sed to ascerta in the dens ity ofa type of herbaceous plant in a farmin g ar ea. The bar chart below shO\vs th eresults of the experiment.10987

Number 6ofplant 5

432

o 2 3 4 5 6 7 8 9 10Quadrat number

From the bar above, th e dens ity of th e herb8ceous plants per squ are metre or farm isA 16B 20C 80D 160

SO. Which of the following statement is not true about cmr ying cap ac ity?

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A Total number of organi sms that can be support ed by enviro nm ental resources .B Carrying capacity is limited by limited resources.C Carrying capacity of an ecosystem is constant.D Affected by environmental conditions.

END OF QUESTION PAPER

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

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CONFIDENTIAL*Narne: ...................................


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CONFIDENTIAL 2Section A[40 marks]

Answer all questions in this section.I. In some plants, Dowering is induced by a critical photoperiod. Cocklebur (Xanthium

strumarium) is a short-day plant. The photoperiodic response is controlled by a specificlight-sensitive pigment.(a) State what is meant by the terms:(i) photoperiod

(ii) photoperiod ism

[2 marks](b)(i) Name the pigments responsible for photoperiodism and their interconvertible

fomls.

[2 marks](ii) With an aid of a diagram, explain the process involved II I the interconversion

betvveen the two fonns of the pigment.

[3 marks](c)(i) State what is meant by a short-day plant.

[1 mark]

CONFTDENTIAL

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CONF l lmNTIAL 3(ii ) Explain huw fl owering is controlled in short-day plant.

[2 marks]

2. Graph A shows the changes in dry mass (g) of the tuber, leaves and stems of a potatoplant. Graph B shO\vs four growth patterns of various parts of the human body plotted inpercentage or size against time (in year).

"'":Ju'"(l )N'U'jatV(cdC


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CONFIDENTIAL 4(ii) E.\p la in th e dec line of th e dry mass of the tuber.

[2 marks]

(b) State the age at which the human reproductive organs are growing most rapidly.

[1 mark](c) Sta te the type of growth pattern shown in the human's organ growth curve.

Ex plain.

[2 marks](d)(i) Describe the growth of the human thymus gland.

[2 marks]( ii ) Ex pl a in the importance of the human thymus gland?

[2 marks]

CONFIDENTIAL

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CONFIDENTIAL 53. An experiment to determine the distribution of plants in a study area wa s done by lls lnga sampling method. Each quadrat size is 1m x 1m. The number of organisms from eachplant species in each quadrat is shown in the table below.

r- .. --- . . ~ ...._--- - - - - - -Quadrat 1 2 3 4 5 6 7 -_. 9 10 Total -Mimosasp_ , - 5 - 10 5 2 3 9 - - 34f----

15 18 6 12 14mperata sp_ - - 9 15 15 104(a) Name the sampling method used in this experiment.

[ I markl

(b) By using the data given in the preceding table, calculate the:(i) frequency ofMimosa sp_

[ I mark](ii) relative frequency ofMimosa sp.

[3 marks](iii) density of Imperata sp.

[I mark1(iv) relative density of Imperata sp_

[3 marks]

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CONFIDENTIAL 6(c) Name another sa mpling me th od that ca n be used to plot the distribution of

pla nts in a s tudy a rea.

[1 mark](d) I f a sample or so il was takc n from this study ficld , name a method to obtain

the orga ni sms present in the so il.

[I mark]

4. The karyoty pe (M ) below was obtained from a pcrson suffering from a certaingenetic diseases.

.. "' ..... IIIIU UUUHI I2 4 6 7 8 9

! ! ~ 9 U n U I I H ~ ~ a a 10 I I 12 D 14 15 16 17 18

~ ~ o o ~ ~ 8 ~ 19 20 2 I 22 X X X

(a) How many c hromosomes arc round in the somatic cell of this person.?

[I mark](b) Name the gene tic disorder cau se by the chromosomal abnormality as shown in

the Karyotype M.

[I mark]

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CONFIDENTIAL 7(c) Describe the chromosomal events which may cause this genetic disorder.

[4 marks]

(d) State two features which will be shown by a person with the karyotype M asshown above.

1 ........ ..... ...... .... .....................................................................

2...... .... .. . .. . . . . . . . . . . . .. .... ........ .. ........ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

[2 marks]

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CONFIDENTIALSec ti on B[60 marks]

Answer OilY four questions /rom this section.5. (a) Stale struct ural dil"fcrcnces between RNA and DNA. [3 marks]

(b) Disc uss huw lhe ) f " ( 1 1 a t i o n from DN!\' is used to fonn a correct sequence ofam in a acids in th e polype ptide. [12 marks]6. (a) Describe the properties orhaemoglobin that make them efficient in transporting

oxygen.(b) Ex pl a in how gaseous exc han ge takes place in the alveolar surface.

[6 marks][6 marks]

(c) Explain br iefly th e effect of carbon monox ide on the etTiciency of humanh


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ANSWER TRIAL EXAM BIOLOGY PAPER 1 2009

QUESTION 18

The answer given by teacher is II and IV.

NO ANSWER NO ANSWER

1 A 26 B

2 D 27 D3 A 28 B

4 A 29 B

5 D 30 D

6 C 31 B

7 D 32 C

8 A 33 A

9 B 34 C

10 D 35 B

11 C 36 A

12 D 37 D

13 C 38 D

14 A 39 C

15 C 40 C

16 D 41 D

17 B 42 A

18 C 43 B

19 B 44 C

20 B 45 D

21 C 46 C

22 C 47 B

23 B 48 A24 D 49 A

25 A 50 C

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PAPER 2 TRIAL STPM 2009 - SECTION B ANS\VER SCHEMENO ANSWERS SUB- TOTAL

TOTALSea) Differences between RNA and DNA

RNA (Ribonucleic acid) DNA (Deoxyribon uclcic l c i d ) Consist of a single Con sist of two po lynucl eotidepolynucleotide strand. strand s wh ich co iI around eachother to form a double he li x. 1RN A molecule is shorter. DNA molec ul e is mu ch longe r. 1Conta ins the pentos e sugar, Contains th e pentose su ga r,ribose. de ox yribose. 1Th e n o e n o u bases are The nitrogenoLl s base aread enine, uracil , cy tos ine and adenin e, th yrn i!le, cytos ine and 1gu anlll e' guanin e.

I i

I Present in th e nu cleus and Mai nly present in thecy topl as m. chromosomes in the nu cleus. 1Sma ll amount is present in th emi tochondri a and chloropl as ts.

There are three main type of Only one type of DNA.RN A: 1me ssenger RNA (mRNA),transfer RNA .(tRNA),ribosomal RNA (rRNA),

Any 3 Max 3(b) Prote in sy nth es is invo lves transc ripti on of DNA, amino ac id

acti vati on and transl ation . 1 1

Transcription: The part of DN A (cistro n) which codes fo r the specifi c 1polypeptides unwinds as the hydrogen bonds between the basesare broken.

The RN A polymerase attac hes to the promoter site. 1

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As the RNA polymerase moves along the template strand in the5' 3' directiori, the free nucleotides in the nucleoplasm pairwith the complementary bases in the exposed DNA templatestrand.

1

The nucleotides are linked together by phosphodiester bonds. The 1process require energy from ATP .

The mRNA passes through the nuclear pore into the cytoplasm and 1binds to ribosome.

The exposed regions of DNA are closed by hydrogen bondsbetween the complementary bases

Amino acid activationAny 4

One end of the tRNA molecule present in the cytoplasm has threebases called anticodon which are complementary to the mRNAcodon triplet.

The other end (free 3' end) has triplet ba ses CCA for theattachment ofa specific amino acid.

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1

1

Amino acids in the cytoplasm are attached to specific transfer 1RNA molecules, using energy from ATP to form specific aminoacid - tRNA complexes.

This is known as amino acid activation and is catalysed by enzymeaminoacyl-tRNA synthetase

The transfer RNA molecules then bring specific amino acids to theribosome

Any 3Translation The ribosome contains binding sites for mRNA and tRNA

molecules.

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1

1

Messenger RNA enters into the cytoplasm. The 5' end of mRNA 1binds to the small ribosome subunit.

A tRNA-amino acid complex with anticodon UAC and carrying 1

I Max 4

Max3

2

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6(a)

the amino acid methionine binds to the codon AUG (start codon)on mRNA.

The large ribosome subunit then binds to the small subunit to form 1Ifunctional ribosome. I

A second tRNA- acid complex with complementary anticodon 1enters into site A.

A peptide bond is formed between the two amlllo acids. Thecondensation reaction IS catalysed by the enzyme peptidyltransferase.

tRNA at site P is released from the ribosome into the cytoplasmand ribosome moves one codon along mRNA.

tRNA in the A site ribosome moves to P site. The translation process is repeated to form polypeptide chain until

the ribosome reaches the termination codon . (UAA, UAG andUGA)

Any 4TOTAL

has four haem groups to bind with four molecules of oxygenforming oxyhaemoglobin

very little oxygen is released when oxygen is transportedthrough the arteries

can maintain a high (80%) saturation of oxygen because thechange in partial pressure of oxygen in the arteries is little

releases the oxygen to the tissues for respiration where there is asharp drop of saturation of oxygen

adapted not to deprive the tissues of sudden loss of oxygenbecause further drop in partial pressure would cause a slowrelease of oxygen

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1

11

1

11

1

1

releases more oxygen when there is a higher concentration of H+ 1caused by higher carbon dioxide concentration to supply more

Max 415

3

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oxygen to the respiring tissues foetal haemoglobin has a higher affinity for oxygen which enable

it to get oxygen from the mother's bloodAny 6

(b) air is drawn into the alveolus by the expansion of walls of thelungs because of the negative pressure in the thoracic cavity

changes take place in the blood due to the low concentration ofcarbon dioxide in the air

HCO)- ions enter the red blood cells , changed into H2CO J anddissociate into water and carbon dioxide gas which diffuses intoalveolar spaces

at the same time oxygen from the air di ssolves in the moisturelining the inner surface of the alveolus

then into the red blood cellsthrough the alveolar and capillarywall

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1

1

1

1

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in the red blood cells, oxygen molecules bind with the haem 1group of the haemoglobin to form oxyhaemoglobin

there is aiways a concentration gradient between the gas in the 1blood and alveolar spaces by the rhythmic expansion andcontraction of the alveolar walls Any 6

(c) carbon monoxide IS considered to be a dangerous respiratory 1pOison

it combines with haemoglobin more read i y than oxygen to formcnrboxyhaemoglobin

resulting in the inabil ity oC haemoglobin to take up oxygen decreasing the oxygen supply to the respIrIng tissues which

eventuall y stops the cellular respiration

TOTAL

111

Any 3

Max6

Max 6

Max 3

15

4

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7(a) Initiation of heart beat The heart walls consists of cardiac muscles which are myogenic,

that is, their rhythmic contractions arise from within the heart 1mLiscles themselves.

The regular heart beat depends on the two nodes present in theheart. The first node is the sino-atrial node (SAN) which is 1embedded in the wall of the right atrium close to the point wherethe anterior vena cava enters the heart. The second node is theatrio-ventricular node (A VN) which is embedded between the,right atrium and ri ght ventricle.

SAN functions as a pacemaker. There is a potential difference 1across the membranes of the cells of SAN. As sodium ions enterthe cells, they depolarise the SAN and produce a wave ofexcitation.

The wave of excitation that originates in the SAN sp reads across 1both atria. It causes both atria to contract simultaneoLisly. There isa delay of about 0.1 s in the conduction of excitation from SAN tothe A VN, which means that the atrial systole is completed beforethe ventricular systole begins.

The AVN conducts the wave ofexcitation to the bundle of His and 1its finer branches known as the Purkyne/Purkinje tissue which thenconduct to the apex and throughout the ventricular walls. Theventricles contract simultaneously from the apex upwards,squeezing blood out of the ventricles towards the aorta anclpulmonary artery. In this way, atrial systole can occurrhythmically.

Sodium ions are pumped out of the cells and this repolarises the SAN. The atria are in a state of diastole. The whole process is then 1

/ repeatedAny 5 Max 5

5

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Regulation of heart beat Even though the rhythmic beating of the heart is initiated by the

pacemaker, its rate is regulated by the autonomic nervous systemwhich is divided into the sympathetic nervous system andparasympathetic system. The sympathetic nerves, part of thesympathetic nervous system, have their origin in the cardiacacclerator centre of the medulla.

Stimulation of these nerves causes a release of noradrenalinewhich results in an increase in the heart rate. The vagus nerves,part of the parasympathetic nervous system, originate in thecardiac inhibitory centre of the medulla,

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1

Stimulation of the vagus nerves causes release of acetylcholine in 1the SAN, AVN and the bundle of His. This reduces the heart rate

At times of stress, adrenaline is secreted by the medulla of the 1adrenal glands. Adrenaline increases the heart rate.

An increase in the partial pressure of carbon dioxide (a drop in thepH of blood) or the decrease of blood pressure increases the heartrate.

Artherosclerosis The thickening and hardening of the arterial wall caused by the

deposition of lipids, e.g. cholesterol, triglycerides, fibres andcalcium deposits beneath the inner walls of arteries known as theendothelium.

The deposits formed a plague or atheroma. Continuous depositioncauses an increase in the size of the plague. .It protrudes into thelumen, narrowing the lumen of the arteries and reduce the bloodflow.

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1

If the plague breaks through the smooth endothelium, its rough 1surface causes a blood clot, called thrombus. If the thrombusincrease in size, it narrows or blocks the lumen and prevents the

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6

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8(a)

blood flow (thrombosis occurs). Artherosclerosis often occurs in the arteries such as the aorta,

carotid arteries, iliac and coronary arteries.

Effects of artherosclerosis Artherosclerosis causes thrombosis and embolism. Reduced blood flow to the heart can damage the heart tissues; it

causes chest pain called angina pectoris or heart attack known asmyocardial infarction.

1Any 3

11

Reduced blood supply to the brain causes stroke. Narrowing of 1arteries causes hypertension or high blood pressure Any 2

TOTALDark band (A) Light band (I)__

--.-__ __ [ r - : : : : : : ~ = = : : : : : : : ; : = = ~ M ~ m ~ e m b r a n e -Zmembrane--.....- - - - . , . . . - - - - t - - ThIck filament (myosin)

-4- - - --4 -__ :--. . . . - - ~ __ +-=_;....T..;...:.h,;;.;.ln.;...;.filamenl (actin)

\ ' - - _ - - - - - ~ v _ ~ - - - - - - - - - I One sarcomere

Diagram 1Labels: 2 H zone Yz dark band/fA band YzYz

YzYzYzYzAny 4

I band//light band Actinl/thin filament Myosin//thick filament Z membrane//lin e M membrane/lline

Max3

Max2

15

Max 3

7

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The sarcomere is the basic unit of the myofibril between two Zlines. 1

The myofibril is made up of thick filament/myosin and thinfi lament/actin. II The thick and thin filaments overlap to form the darker band.

During muscle contraction the po sition of thick filament remains 1unchanged. Thin filament slides past one another. 1 Myofibril becomes shorter and shorter. I During muscle relaxation the position of thick filament remains

unchanged. 1 The thin filament slides out to the original position. 1 The muscle reverts to its original position. 1

Any 7 Max 7(b) Curare binds to the receptor on the postsynaptic membrane of I

the neuromuscular junction where acethylcholine is supposed to ,bind.

Depolarization on the postsynaptic membrane is prevented, and 1 impulse transmission across the neuromuscular junction is also 1prevented. No impulse is transmitted to the ske letal muscles 1 causing paralysis. 1 A high concentration of curare causes death as the breathing 1process stops Any 5 Max 5

TOTAL 159(a) HIV enters the body via bodily fluids or blood. 1 Inside the body,it binds with T helper cell s with corresponding

receptors. 1 Lipoprotein membrane of HIV fuses with that of the T cell. 1

8

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Viral particle enters via endocytosis. I Capsid is removed I Content of viral RNA and reverse transcriptase enzyme released

into cytoplasrn of host cell. I Reverse transcriptase converts viral RNA into single strand DNA. 1 Single strand DNA convened into double strand DNA by DNApolymerase. 1 Viral DNA enters nucleus and incorporates into DNA of host cell

as provIrus. IProvirus replicates each time the host DNA replicates. 1After about 6 years of dormancy, the provirus is transcribed into 1mRNA.Host cell synthesizes viral capsid, reverse transcriptase enzyme 1 and viral RNA.

New viruses formed in the host cell exit the cell by budding. 1 Virus kills T helper cells and destroys immune system, causing 1AIDS. Any 8 Max 8

(b) T cells formed in the bone marrow circulate in the blood Icirculatory system until it reaches thymus glands.

In thymus glands,T cells are differentiated to form T helper 1cells(Th) and T cytotoxic cells (Tc).

Each has unique T cell receptor CTCR) on its surface. 1 Mature Th cell then circulates in the blood circulatory system I

until it meets an antigen presenting cell (APC ). The Th cell binds to APC provided the antigen-MHC complex on 1

APC is complimentary with the TCR on the Th cell. Intcrleu kin I(cytokine) is secreted from APC. 1

Interieukin 1 then stimulates the Th cell to secrete Interleukin 2. 1 Interleukin 2 stimulates division ofTh cell to produce clones of

9

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effector Th cells and memory cells and division of Tc cell toproduce clones of effector Tc cells and memory cells.

The effector Tc cells bind with antigen-MHC complex oninfected cells.

Perforin is released by effector Tc cells which will then perforateinfected cells to stimulate autolysis.

The infected cells go through autolysis while the effectorcytotoxic T cell attacks other infected cells.

Memory T cells respond for a second invasion of the samepathogen by actively dividing to form effector T cells.

10(a) Sketch of curvesa) Stabilising Selection

,

Selecl,en again I bot" e x l : e l 1 1 e ~

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I/

II

, -" \ \ Popula tion

" ; 1 ~ ~ aller se lecllon:\ \ ' . Or igina l'


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b) Directional Selection

I _ ~ _ . _ . " " _ Population

I"I

II

The sketch must

alter s,ele lJ tion

- shows the normal distribution curve and the stabi lisingselection curve.ldirection of curve.- label//selection pressurec) Disruptive Selection

,,,;

S ~ , , " g r ' ~ ''', 'ne,", t ,,

II,

, ,, \ '

The sketch must- shows the normal distribution curve and the stabil isingselection curve.ldirection of curve.- labell/selection pressureSta bi/ising selection the bell-shaped curve selection favors the intermediate trait value over the extreme

values.

2/0

2/0

11

response to a stable environment// occurs when the environment 1doesn't change

2

2

11

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the mode stays the same the population graph gets narrower and taller as selection against

mutation takes place experience a decrease in the amount of additive genetic variationfor the trait under selection.Directional Selection

1

1

1Any 3

occurs whenever the environment changes in a particular way!! 1selective pressure for species to change in response to theenvironmental change

directional selection may favor one of the phenotypes at one ofthe extremes of the normal distribution!! selection favors oneextreme trait value over the other extreme!! selection favors theextreme trait values over the intermediate trait values'!! Theaverage phenotype is selected against more than one otherphenotype is selected for.

1

the population's trait distribution shifts toward the other 1extreme!! the mean of the population graph shifts!! Onephenotype can gradually replace another.!/ results in a change inthe mean value of the trait under selection.

results in a population with new trait!resistant individuals begin 1to occur and become the dominant type within the population.!/the variance increases as the population is divided into twodistinct groups. Any 3

Disruptive Selection occurs where an environment change may produce selection 1

pressures that favour two extremes of a characteristic!! selectionpressures act against individuals In the middle of the traitdistribution.

the environment may favor two or more variant phenotypes at 1

Max3

Max3

12

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the expense of the mean.a bimodal/two-peaked curve II the two extremes of the curve 1create their own smaller curves.

results in two distinct populations/morphsllthese two forms may 1become so distinct that they become new populations

plays an important role in speciation. 1Any 3 Max3

TOTAL 15

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stpm trial 2009 bio q&a (n9)

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