Stpm Trial 2009 Bio Q&a (n9)

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BIOLOGY

JABATAN PELAJARAN NEGERl SEMBILAN PERCUBAAN BERSAMA SIJIL TINGGI PERSEKOLAHAN MALAYSIA

2009

Instructions to candidates: DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO There are fifty questions in this paper. For each question, Jour suggested answers are given. Choose one correct answer and indicate it on the multi-choice answer sheet provided. 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 pages

STPM TRIAL 964/1

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2.

2

"Vhi ch of th e follovving are th e properti es of water?

Specific heat capacity

Latent heat of vapo uri sation

Surface ten sion

A

Hi gh

Hi gh

High

B

Low

Low

High

C

Hi gh

High

Low

D

Low

High

Low

Diagram belO\v shows a type of monosaccharide.

OH

Which of the foll owin g pol ymers can be formed from the condensation of the mol ec ul e shown in the diagram above? Glycogen II Amylose 111 Amylopectin IV Starch A B C D 3.

II and III I, II and III II , III and IV J, 11, III and IV

A mitochondrion in (I mammalian musc le cell measures 1.2 f-lm. In the electron micrograph, th e length of the organelle is 48 mm. What is the magnification of the electron micrograph ? 40000 X B 12000 X C 4400 X D 4000 X A

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4.

Which of the following stru ctures of membran e bound organelles is correctly match ed with its function? Structure A

B C D

s.

Function

An ex tens ive network of tubes; each tube is bound by a single menlbrane . . A stack of elongated, curved sacs; each sac is bound by a single membrane. A spheri ca l sac bound by a single membrane. A sac bound by two membranes, the inner is hi ghl y folcled.

Lipid synthesis Photosynthesis Protein synthesis Packaging of proteins

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

Rate of reaction

Concentration of substrate Which of th e following is true regarding the above graph? Curve J A

Competiti ve inhibi tor

Curve 3

Curve 2

Non-competitive inhibitor Normal activity

. B

Competiti 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

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6. The table below shows amin o acid s and th eir base sequ ence on the mRNA codon. Amino ac id Valine Glycin e Methionin e

Base sequ ence on th e mRNA codon, GUC GGU AUA

What is the base sequ ence on pnrt of a DNA strand, which wo uld code for the tripeptid e va line-m eth ion ine-gl yc in e? C CAG TAT CCA D TAT CAG CCA

A GUC AUA GGU B GTC ATA GGA

7. Di agram below shows th e electron pathwny in cyc lic and non-cyc lic photophosphorylation during the li ght reacti on stage in ph otosynth es is.

ferrodoxin

i ;.,.

II ~

quinone

energy

leve l

\ 8\1

1

2e --

II

~

y

e- Z \

X

. 2e -

"\/"-"

~DP' -+NADPH /

2H I + 2e --

TI

mOlepcule

O 2 released

(PS 1\

",---;J

~~0 +-------------~ Which of the followin g statements are tru e for electron pathway in the above diagram?

II III IV

Th e electrons in PSI and PS II nre exc itecl to hi gher energy leve ls NADP+ is red uced in non-cyc li c ph otophosphorylation ATP is produ ced in steps Y and Z Molecule P is a water molec ul e

A I and II onl y B II and IV only

C I, II and III only D I, II , III and IV

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8. The grap h belo'w shows th e relati onship between the rate of photosynthesis with environmental fa ctor X.

rate of photosynthesis (arbitrary unit)

t

Factor X Which of the foll ow ing is fa ctor X? I II III IV

Light intensity Oxygen concentrati on Temperature Carbon dioxid e co ncentrat ion

A B C D

I and IV onl y II and III onl y " and IV onl y I, III and IV on ly

9. Which products are formed during anaerob ic respiration?

A B C D

Mu sc les Pyruvate, NAO+, ATP Lactate, NAO+, ATP Lactate, NAD-", ATP, CO 2 Ethanol , NADH , ADP, CO 2

Yeast cells Ethanol, NAD+, ATP Ethanol, NAO+, ATP, CO 2 Ethanol, NAO+, ATP Lactate, NADH, ADP, CO 2

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10. Diagram below shows a reaction in Krebs cycle.

Krebs cycle GTP

AD~GDP ATP What is process X? A Chemiosmosis B Oxidative Phosphorylation

C Photophosphorylation D Substrate level phosphorylation

11. Which of the following organism is/are not saprophytes? Afucor II RhizopZls III Taenia IV Amoeba

A I only B II and III only

C III and IV only D I, III and IV only

12. Which of the following sequences brings about Bohr's effect? I II III IV

I-t causes oxyhaemoglobin to release its oxygen CO 2 enters red blood cell Carbonic anhydrase catalyses the formation of 1-1 2 C0 3 Respiration of tissues gives out CO 2

A I, III, IV and II B II , Ill , I and IV

C IV, III , II and I o IV, II , III and I

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13. Whi ch of the fo ll ow ing exp bins why myog lobin

IS

foun ci

III

large amounts

III

th e mu sc les of

bird s? Myog lob in ca n ') Ll ppl )' thc encrgy needed for nyin g Myog lob in can supp ly oxyge n to th e Illll scul ar ti ss ues when birds fl y in areas w ith low parti al pressure of oxyge n. e When compared to hae mog lob in , myoglobin has hi gher affinit y toward s oxyge n ane! can better retain oxyge n in acti ve mu sc les . D Myog lobin has a sma ll er Ill olecul ar size cO ~ ~l p!"e d to h3emcglob in . A

B

14. Which process does not occur durin g th e openin g of stomata? A Su ga r is co nverted into starch. B Starch is converted into Ill ali c acid .

e Water enters int o guard cell s by osmosis. D Potass iulll ions diffuse into guard ce ll from adj ace nt cell s. 15 . Whi ch of the foll owin g in creases the rate of im pul se released from sin oatri al node?

I Il III IV

Impul se from the sympatheti c nerve Impul se from the paras ympath eti c nerve Thyrox in e horm one Adrenalin e hormone

A I and IV B II and IV

e

I, III andl V D II , lIl andl V

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16. Diagram belo\-\' sho'vvs a i'vlull ch model.

x Pure water

~.

Semipermeable membrane

If A is fill ed with conce ntrated sucrose so lution, while B is fill ed \vith dilute sucrose so lution , which of tile foll ow ing statements are true? 1 A is eq uivalen t to leaves in tran slocat ion .

II Hi gh turgor press ure 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 III B 1, II and IV

C I, III and IV D I, II, 11I and IV

17. What is tile function of ad renalin e in th e negative feedback mechanism ifbody temperature of

endotherms decreases? A B C D

Stimulate secret ion of sweat Increase co nvers ion of glycoge n into glu cose in li ver Triggers a shiverin g process Contract erector mu sc les

18. Liver is an organ in vo lved in homeostasis because it

I II III IV

produces bile sa lt performs gluconeogenesis performs deto xi fi cati on produ ces a hi gh amount of heat

A B C D

III and IV I, III and IV II , III and IV I, II, III and IV

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19. Which of the foll owin g are th e charac teri sti cs of an impul se? I II III IV

It foll ows the a ll or nothin g law Speed of impul se transmi ss ion is faster in mye linated axo n Speed of impul se tran smi ss ion is fas ter in unm ye linated axon 'with a small er diam eter In th e abso lute re fractory peri od, a new impul se ca nnot be generated even if a stron g stimulus is rece ived

A L II and !II B I, IlandlV C I, III and IV D I, II , III and IV

20. Curare effects neurolllLl scul ar juncti on by A binding to receptors on postsynap tic membran e B preventin g exocytosis of acetylcholine into the synaptic cleft C preventin g action of cholin esterase D preventing di ffu sion of calcium ions into presynaptic membrane 21. The follovvings are the events th at occur in th e act ion of a steroid based hormone .

I II III IV

Hormone-receptor co mplex diffuse into the nucleus Steroid based hormone diffu se into the ce ll through the cell membrane Hormone bind s to the receptor in th e cytoplasm Specific gene is activated in DNA

The correct sequence for the events above is A I, II , Ill, IV B I, Ill , II , IV CII , III , I, [V )) II , [, [II , IV 22 . Ifplant X is a short da y pl ant with a critical da y length of8 will pl ant X not flow er'?

1/2

hours, in which of th ese conditions

Exposed to red li ght for 9 hours II Kept in darkness for 8 hours III Exposed to far red Iight for 12 hours IV Kept in darkn ess for I S hours before ex posed to red li ght for I hour A B C D

I and II III and IV I, II and IV II , III and IV

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23. Which of the followings are invol ved in ce ll-m ed iated response of the il11rnun e system? I involves B lymphoc ytes II mature T lymphocytes req uire macrophages to prese nt th e anti ge n for them to bind III in vo lves the th ymu s gland IV mat ure lymph ocytes can directly bind with ant igens. A

I and IV

B II and 11I C I and 11I D II and IV

24. HIV lies dorm ant in th e body because A its reverse transcriptase is engulfed by T4 cell

B its particle rest in the cytoplasm ofT4 cell C the replication of its RNA occurs later D its DNA is incorporated into the DNA ofT4 ce lI

25. The sporangium of Marchantia where spores are pl:oduced is a

A B C

diploid structure haploid structure dioecious structure D ov iparous stru cture 26. Which orthe foll ow in g terms app ly to birds?

ov iduct II vIvIparous III internal fertilisation IV hermaphrod ites A II and III

B I and III C III and IV D II and IV 27. In an amniotic egg of a bird, the rese rvo ir fo r waste and th e part th at provides for gas diffu sion is the A chorion B yo lk C amnIon o allc1lltois

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28.

Th e formation 0 1- blastul a frolll n zygo te in vo lves C\ success iOI1 of" rnp id ce ll divi sion 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 lli c deve lop ment? A B C D

Blastomere Cleavage Gastrul ation Organogenesi s

29. The hormone that plays an imp ortant role in seed germ inati on and ea rl y seeelin g growth is A auxll1

B gibberelin C cytok inin D absc isic acid 30. The tab le below shows four patterns of gro wth curves and their exampl es. Examples

Growth pattern

(a) (b) (c) (eI)

I (i) 1-lulll an orga ns

Isometric grov·"th All ometric growth Intermittent growth unli mited growth

Which of the fo ll ow in g is correct ly

(i i) Coral ree fs (iii ) fi sh (iv) Grasshopp er p~lired ? -

A

B C

D

(a)

(b)

(i) (ii) (i ii) (iii)

(i ii) (ii i) (i) (i)

(c) (iv) (i\') (i i) (iv)

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(d)

(ii) (i) (iv) (i i)

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Which of the fol low in g state ments me tru e ilbout neurosec reti on secreted by th e neurosecretory ce ll in in sects? I II III IV

It is stored in corpus carel iac ull1 It inhibi ts th e effect of juveni le hormon e It is also known as prothoraci ctrop hi c hormon e It st imul ates the secretion of ecdysone from prothoracic hormone

A I, II and III B I, III and IV

C II , III and IV D I,II , lll and lV

32. 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 gametes and poss ibl e offspr in gs of a dih ybrid cross . The numbers 1- 16 represent the genotypes of each ind ividu aJ.

RT Rt rT rt

RT

Rt

rT

I't

I 5

2

3

4

6

9 13

iO

7 II

14

15

8 12 16

If plants 4 and 13 are crossed , what propo rtion of th eir offsprin g \;vi ll show at least one recessive trait? A

1/16

B

611 .6

C

7/ 16

o

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33. The colour of onion s is controlled by two pairs of all eles Ss and Rr, whi ch segregate independently. The all ele S is dOlnin ant anci 1l1ust be present to all ow developm ent of pigment in the sk in . In its absence, the on ion is wh ite. All ele R is dDmin ant ond gives a red co lour, the recessive r gives a ye ll ow co lou i-. Y\ hat wi ll be th e rat io of phe notypes in th e offsprin g o f a cross between plants of genotypes SSR,R and ss r!"! A

B

1 red I red

I w hite I ye ll ow

C D

3 red J wh ite I white : 2 red I yell ow

34. Which of the following statement's are not tru e abo ut mutation s?

II III IV A

B

mutations are spontaneo us changes in th e amount or stru ctu re of DNA Or changes in the sequence of Ilucl eoti de bases of a gene. the changes that occurred are non-random all mutations lead to prod uction of non -fun cti ona l protein s ge rm-line mutatio ns can be -inherited I and III I and IV

C

o

II and III II and I V

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35. Chromoso mal deleti on occurs when th e deleted DNA segment beco mes rea ttac hed in an in ve rt ed position. B when the chromClsome brea ks and a segment of it is lost C when a segment of a chromoso me co ntainin g a sequ ence of lltl cleoticles is repeated. D when the chromosome segment becomes deleted an d rejoins at a different pos iti on on th e same chromoso me or another non-h omol ogo us chromosome. A

36. Th e Hardy- Weinberg eq uati on does not app ly if I th ere is migration II there is natural selection. III mutation s occur.

IV th ere is non-random matin g. V there is a large population . A I, II , III and IV

B I, II , III and V C I, Ill , IV and V D II , III , IV and V

37. The maintenance of the allele for sickl e-ce ll anae mia in hum an p opul ati ons in malaria-end emic region in Africa is an exampl e of A genetic drift

B gene flow C founder effect D heterozygote advan tage

38. What is th e function of th e indu cer of th e lac operon?

A Bind to the promoter and prevent s th e represso r from bindij! g to th e operator. B Bind to the operator and prevents the represso r 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 bindin g to th e operato r. 39. The statemen ts below refer to different stages in th e production of hum an in sulin by genet ic engin eering techniqu es. Wh at is the correct sequence productio-il? I DNA cut "vith restriction enzymes. II mRNA extracted frolll pancreati c cell s

III Plasmid DN A and human DNA joined usin g li gase enzymes. IV DNA copy made usin g reverse transcriptase V Recombinant plasmid incorporated into bacteri al ce ll. A I, Ill , IV, II , V B II, IV , III , V, I

C II , IV , I, Ill , V D II , I, IV , Ill , V

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40. Which of the following are characteristics of pl asmid ?

II III IV V

Found in the bacteria Integral par! of bacterial chromosome Confers special characteristics to th e organi sm Self-replicating Small, linear molecule

A B C D

I, ll, III and IV I, III, IV and V I, III and IV III , IV, and V

4 J. The following are uses of recombinant DNA technology ex cept A screening for carriers of genetic diseases B identification of badly-decomposed victims C treatment of diabetic patients with synthetic insulin o 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.

I 11 III JV

Taxonomic group Kingdom Phylum Class Order

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

Which of the following combination s is correct ? I

A B C

D

Xc) (c) (e) (c)

11

III

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

(b) (a) (a)

(d)

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

43. Which of the following is likely to be radially symmetrical? A 13

A chordate A cnidarian

C D

An arthropod An annelid

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44. Whi ch of th e fo ll ow ing statements are tru e about Marchanlia? I Monoecious II Asexua l reprod ucti on th ro ugh ge mm a cups III Gametophyte is domin ant IV gametophyte is depe ndent on spo roph yte

A B C D

I and II I and III II and III 11 and IV

45. Which of th e foll ow in g cann ot cause th e geneti c vari ation? Mutati on Camoufl age 1II Recombinati on IV Dominancy I II

A B C

D

II and III III and IV I and III II and IV

46. Which one of th ese is the definiti on fo r the.:biogeochemi ca l cyc le? A B C D

Circul ati on of chemi ca l elements th ro ugh th e bioti c co mponen t of an ecosystem. Circul ati on of chemical elements th rou gil th e ab ioti c componen t of an ecosystem. Circulati on of chemi ca l elements through th e biotic and ab iotic co mpo nent of an ecosystem. Circul ati on of organi c molecul es through the bioti c components of an ecosystem.

47. The main rese rvo ir of ph osphate is A B C D

inorgani c ph osph ate ions in soil. inorgani c phosph ate in rocks inorgani c phosphate in organi sms. Organi c phosphate in organi sms.

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48. The success of organi sms in the ecosystem can be shown by th e

1 population di stribution Il population si ze III prey-predator relati onship IV position of trophi c level in th e ecosystem. A I and II B I and III eland IV D II and III 49. A quadrate with a measurement of SO cm x SO cm is L1 sed to ascertain the density of a type of herbaceous plant in a farmin g area. Th e bar chart below shO\vs th e results of the experiment. 10 9

8 7

Number

6

of plant

5

4 3 2

o

2

3

4

5

6

7

8

9

10

Quadrat numbe r

From the bar above, th e density of th e herb8ceous plants per squ are metre or farm is A 16 B 20 C 80 D 160 SO. Which of the following statement is not true about cmryin g capac ity?

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 964/1

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STPM 2009

BIOLOGY

II

•••

• . . . . . . "T"

JABATAN PELAJARAN NEGERl SEMBILAN PERCUBAAN BERSAMA SIJIL TINGGI PERSEKOLAHAN MALAYSIA 2009 Instructions to candidates:

For examiner's use

1 DO NOT OPEN TIDS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO.

2

3 Answer all questions in Section A. Write your answers in the spaces provided

4 5

Answer any four questions in Section B. Write your answers on your own test pad Begin each answer on a fresh sheet ofpaper. Answers should be illustrated by large and clearly labeled diagrams wherever suitable.

6 7

8

Arrange your answers in numerical order and tie the answer sheets to this question paper.

9 10 Total

This question paper consists of 8 printed pages STPM TRIAL 964/2

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Section 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 specific light-sensitive pigment. (a) State what is meant by the terms: (i)

(ii)

photoperiod

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 betvveen the two fonns of the pigment.

III

the interconversion

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

[1 mark]

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C ONF llmNTIAL (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 potato plant. Graph B shO\·vs four growth patterns of various parts of the human body plotted in percentage or size against time (in year).

Graph A

,, ,, ,

12

~ 10

Leaves:'


E1 B c:-

o

6 4

2

o

4

3

2

8

7

6

5

Time (weeks) Graph B 200% I

180

"'" ::J u

160

.~

140

'"

\

I

\ Thymus gland

I

\

\ \

I

I

(l)

N

'U'j

a tV

(»

120

I

/

100%

cd

C


/

80 /

~

(l)

0...

"

/ /

60

/

40 20 O~-r~~~-r~~~~~~~-r~~

o

2

4

6

8

10

12

14

16

18

20

Time (years)

(a)(i)

Explain the shape of the curve for the tuber of the potato plant.

[2 mark]

CONFIDENTIAL

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CONFIDENTIAL (ii)

4

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)

Desc ribe the growth of the human thymus gland.

[2 marks] (ii )

Ex pl ain the importance of the human thymus gland?

[2 marks]

CONFIDENTIAL

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5

CONFIDENTIAL

3.

An experiment to determine the distribution of plants in a study area was done by llslng a sampling method. Each quadrat size is 1m x 1m. The number of orga nisms from each plant species in each quadrat is shown in the table below. r -.. -

Quadrat Mimosasp_ , f---Imperata sp_ (a)

-

4 10

5 5

18

-

6

1

2

3

-

5 15

-

6 2 12

7

3

14

---

-_.£ 9 9

.- .~-

...._---

------

9

10

15

-

Tota l 34

15

104

-

Name the sampling method used in th is experiment.

[ I markl

(b)

By using the data given in the preceding table, calculate the: (i)

frequency of Mimosa sp_

[ I mark] (ii)

relative frequency of Mimosa sp.

[3 mark s] (iii) density of Imperata sp.

[I mark 1 (iv) relative density of Imperata sp_

[3 marks]

CONFIDENTIAL

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6

CONFIDENTIAL (c)

Name a noth e r sa mplin g meth od that ca n be used to plot the di stributi o n of pla nts in a stud y a rea.

[1 mark] (d)

or

If a sample so il was takc n from thi s study ficld , name a method to obtain the o rga ni sms present in the soil.

[I mark]

4.

The karyoty pe (M)

be low

was o btained from a pcrson suffering from

a certain

genetic di seases.

. "' ..·IIIIU UUUH I

I

2

4

6

7

8

9

!!~9UnU IIH~~aa 10

II

12

D

14

15

16

17

18

~~oo~~8~§ 1; "; U" 19

(a)

20

2I

22

X

X

X

How ma ny c hromosomes arc round in the so matic cell of this perso n.?

[I mark] (b)

Name the ge ne ti c disord er cau se by the chromosoma l abnormality as shown in the Karyo type M.

[I mark]

CONFIDENTIAL

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CONFIDENTIAL (c)

7

Describe the chromosoma l events which may cause this genetic disorder.

[4 marks]

(d)

State two features which will be shown by a person with the karyotype M as shown above. 1......... ..... ...... .... ..................................................................... .

2 ...... .... .. . .. ........... .. .... ........ .. ........ . . . . ................... . ................... . [2 marks]

CONFIDENTIAL

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CONFIDENTIAL Sec ti o n B [60 marks]

Answer OilY four questions /rom this section. 5.

(a)

Stale stru ct ural dil"fcrcnces between RNA and DNA.

[3 marks]

(b)

Disc uss huw lhe inl~)f"(11ation from DN!\' is used to fonn a correct sequence of

[12 marks]

am in a acids in the pol y pe ptide. 6.

(a)

Describe the properties

or haemoglobin that make them efficient in transporting

oxyge n.

[6 marks]

(b)

Ex pl a in how gaseous exc han ge takes place in the alveolar surface.

[6 marks]

(c)

Explain br iefly th e effect of carbon mo nox ide on the etTiciency of human [3 marks]

h<1emog lo bin in the tran sporta tion of oxygen.

7.

8.

[10 marks]

(a)

Describe how the he al1 beat is initiated and how it is regulated.

(b)

Describe the causes of atherosc lerosi s and state its effects.

(a)

With refere nce to a labe lled diagram , describe the structure of a sarcomere.

[5 marks]

Exp lain what happen s to the myofibrils during contraction and relaxation of

[10 marks]

mu sc le.

9.

(b)

Expl a in th e effect o f" cu rare o n the contraction of skeletal muscle.

(a)

Describe how Human Immun ode ficiency Virus (HI V) causes Acquired ImrnLln ode l~ c i e nc y

(b)

Syndrome (A ID S) .

Desc rihe the dcve lupme nt of th e ce ll-mediated response

[8 marks] In

human immunity

[7 marks]

sys tem .

10.

[5 marks]

With the aid of graphs and suita hl e examples, explain briefly: (a )

Stab ilisi ng se lection

(b)

Directional selectio n

(c)

Disruptive se lectio n

[15 marks]

END OF QUESTION PAPER CONFIDENTIAL

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ANSWER TRIAL EXAM BIOLOGY PAPER 1 2009

NO 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 A D A A D C D A B D C D C A C D B C B B C C B D A

QUESTION 18 The answer given by teacher is II and IV.

NO 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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

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PAPER 2 TRIAL STPM 2009 - SECTION B ANS\VER SCH EM E NO

ANSWERS

SUBTOTAL

TOTAL

Differences between RNA and DNA

Sea)

RNA (Ribonucleic acid)

Consi st of a single pol ynucl eotide strand.

RN A molecul e is shorter.

DNA (Deoxyribon uclcic

~lcid)

Con sist of two po lynucl eotide strands wh ich co i I aro und each oth er to form a double he li x.

1

DNA molec ul e is mu ch longe r. 1

I

Contains the pentose sugar, ribose.

Contain s th e pentose su ga r, deox yribose.

1

The n i~rog enou s bases are ad enine, uracil , cytosine and guanlll e'

The nitrogenoLl s base are adenin e, th yrn i!l e, cytos ine and guanin e.

1

Present in th e nucleus and

Mai nly present in the chromoso mes in the nu cleus. Small amount is present in th e mi tochondri a and ch loropl asts.

i

I cytopl as m.

There are three main type of RN A: messenger RNA (mRNA), transfer RNA .(tRNA) , rib oso mal RNA (rRNA),

1

Onl y one type of DNA. 1

Max 3

Any 3 (b)

•

Protein sy nth es is in vo lves transcripti on of DNA, amino ac id acti vati on and transl ation .

1

Transcription:

•

The part of DN A (ci stro n) whi ch codes fo r th e specifi c

1

polypeptid es un winds as the hydroge n bonds between th e bases are broken.

•

The RN A polymerase attac hes to th e promoter site.

1

1

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• As the RNA polymerase moves along the template strand in the 5'

~

3' directiori, the free nucleotides in the nucleoplasm pair

1

with the complementary bases in the exposed DNA template strand. • The nucleotides are linked together by phosphodiester bonds. The

1

process require energy from ATP . • The mRNA passes through the nuclear pore into the cytoplasm and

1

binds to ribosome. • The exposed regions of DNA are closed by hydrogen bonds

1

between the complementary bases

I

Any 4

Max 4

Amino acid activation • One end of the tRNA molecule present in the cytoplasm has three

bases called anticodon which are complementary to the mRNA

1

codon triplet. • The other end (free 3' end) has triplet bases CCA for th e

1

attachment ofa specific amino acid. • Amino acids in the cytoplasm are attached to specific transfer

1

RNA molecules, using energy from ATP to form specific amino acid - tRNA complexes. • This is known as amino acid activation and is catalysed by enzyme aminoacyl-tRNA synthetase • The transfer RNA molecules then bring specific amino acids to the

1

1

ribosome

Max3

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

1

molecules. • Messenger RNA enters into the cytoplasm. The 5' end of mRNA

1

binds to the small ribosome subunit. • A tRNA-amino acid complex with anticodon UAC and carrying

1

2

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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

1

I

functional ribosome.

I

• A second tRNA- acid complex with complementary anticodon

1

enters into site A. • A peptide bond is formed between the two amlllo acids. The condensation reaction

IS

catalysed by the enzyme peptidyl-

1

transferase. • tRNA at site P is released from the ribosome into the cytoplasm

1

and ribosome moves one codon along mRNA.

• tRNA in the A site ribosome moves to P site.

1

• The translation process is repeated to form polypeptide chain until

1

the ribosome reaches the termination codon . (UAA, UAG and UGA) Any 4 TOTAL 6(a) • has four haem groups to bind with four molecules of oxygen

Max 4 15

1

forming oxyhaemog lobin • very little oxygen is released when oxygen is transported through the arteries • can maintain a high (80%) saturation of oxygen because the

1 1

change in partial pressure of oxygen in the arteries is little • releases the oxygen to the tissues for respiration where there is a

1

sharp drop of saturation of oxygen • adapted not to deprive the tissu es of sudden loss of oxygen

1

because further drop in partial pressure would cause a slow release of oxygen • releases more oxygen when there is a higher concentration of H+

1

caused by higher carbon dioxide concentration to supply more

3

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

1

it to get oxygen from the mother's blood

Max6

Any 6

(b)

1

• air is drawn into the alveolus by the expansion of walls of the lungs because of the negative pressure in the thoracic cavity

1

• changes take place in the blood due to the low concentration of carbon dioxide in the air

• HCO)- ions enter the red blood cells , changed into H2CO J and 1

dissociate into water and carbon dioxide gas which diffuses into alveolar spaces • at the same time oxygen from the air di ssolves in the moisture

1

lining the inner surface of the alveolus • then into the red blood cellsthrough the alveolar and capillary

1

wall • in the red blood cells, oxygen molecules bind with the haem

1

group of the haemoglobin to form oxyhaemoglobin • there is aiways a concentration gradient between the gas in the

1

blood and alveolar spaces by the rhythmic expansion and contraction of the alveolar walls (c)

• carbon monoxide

IS

Max 6

Any 6

considered to be a dangerous respiratory

1

pOison • it combines with haemoglobin more read iIy than oxygen to form 1

cnrboxyhaemoglobin • resulting in the inabil ity oC haemoglobin to take up oxygen

1

• decreasing the oxygen supply to the respIrIng tissues which

1

eventuall y stops the cellular respiration

Max 3

Any 3

TOTAL

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

1

mLiscles themselves. • The regular heart beat depends on the two nodes present in the 1

heart. The first node is the sino-atrial node (SAN) which is embedded in the wall of the right atrium close to the point where the anterior vena cava enters the heart. The second node is the atrio-ventricu lar node (A VN) which is embedded between the , right atrium and ri ght ventricle. • SAN functions as a pacemaker. There is a potential difference

1

across the membranes of the cells of SAN . As sodium ions enter the cells, they depolarise the SAN and produce a wave of excitation . • The wave of excitation that originates in the SAN sp reads across

1

both atria. It causes both atria to contract simultaneoLisly. There is a delay of about 0.1 s in the conduction of excitation from SAN to the A VN, which means that the atrial systole is completed before the ventricular systole begins. • The A VN conducts the wave ofexcitation to the bundle of His and

1

its finer branches known as the Purkyne/Purkinje tissue which then conduct to the apex and throughout the ventricular walls. The ventricles contract simultaneously from the apex upwards, squeezing blood out of the ventricles towards the aorta ancl pulmonary artery. In this way, atrial systole can occur rhythmically.

• •

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

/

repeated

Max 5

Any 5

5

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

1

pacemaker, its rate is regulated by the autonomic nervous system which is divided into the sympathetic nervous system and parasympathetic system. The sympathetic nerves, part of the sympathetic nervous system, have their origin in the cardiac acclerator centre of the medulla. • Stimulation of these nerves causes a release of noradrenaline

1

which results in an increase in the heart rate. The vagus nerves, part of the parasympathetic nervous system, originate in the cardiac inhibitory centre of the medulla, • Stimulation of the vagus nerves causes release of acetylcholine in

1

the SAN, A VN and the bundle of His. This reduces the heart rate • At times of stress, adrenaline is secreted by the medulla of the

1

adrenal glands. Adrenaline increases the heart rate. • An increase in the partial pressure of carbon dioxide (a drop in the

1

pH of blood) or the decrease of blood pressure increases the heart

5

rate. Artherosclerosis • The thickening and hardening of the arterial wall caused by the

1

deposition of lipids, e.g. cholesterol, triglycerides, fibres and calcium deposits beneath the inner walls of arteries known as the endothelium. • The deposits formed a plague or atheroma. Continuous deposition

1

causes an increase in the size of the plague . .It protrudes into the lumen, narrowing the lumen of the arteries and reduce the blood flow. • If the plague breaks through the smooth endothelium, its rough

1

surface causes a blood clot, called thrombus. If the thrombus increase in size, it narrows or blocks the lumen and prevents the

6

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

1

carotid arteries, iliac and coronary arteries. Any 3

Max3

Effects of artherosclerosis • Artherosclerosis causes thrombosis and embolism.

1

• Reduced blood flow to the heart can damage the heart tissues; it causes chest pain called angina pectoris or heart attack known as

1

myocardial infarction. • Reduced blood supply to the brain causes stroke. Narrowing of arteries causes hypertension or high blood pressure

1

Any 2

Max2 15

TOTAL 8(a) Dark band (A) ,~_ _~A

Light band (I) \~

--.-_ _~_ _[r-::::::~==:::::::;:==~M~m~embrane -Zmembrane

- -.....----.,...----t-- ThIck filament (myosin) -4----4-_ _. : - -....--~_ _-+-=_;....T..;...:.h,;;.;.ln.;...;.filamenl (actin)

\'--_-----~v_~---------I

One sarcomere

•

Diagram

1

Labels:

2

• •

Yz Yz Yz Yz Yz Yz Yz

• • • • •

H zone dark band/fA band I band//light band Actinl/thin filament Myosin//thick filament Z membrane//line M membrane/lline

Any 4

Max 3

7

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•

The sarcomere is the basic unit of the myofibril between two Z 1

lines.

•

The myofibril is made up of thick filament/myosin and thin I

fi lament/actin.

•

The thick and thin filaments overlap to form the darker band.

•

During muscle contraction the position of thick filament remains 1

unchanged.

•

Thin filament slides past one another.

• •

Myofibril becomes shorter and shorter.

1 I

During muscle relaxation the position of thick filament remains unchanged.

• •

I

1

The thin filament slides out to the original position.

1

The muscle reverts to its original position.

1

Max 7

Any 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 1

prevented.

•

No impulse is transmitted to the ske letal muscles

1

•

causing paralysis.

1

•

A high concentration of curare causes death as the breathing process stops

1 Any 5

Max 5

TOTAL 9(a)

•

HIV enters the body via bodily fluids or blood.

•

Inside the body ,it binds with T helper cell s with corresponding

•

15 1

receptors.

1

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

1

8

•

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 DNA polymerase.

•

papercollection

1

Viral DNA enters nucleus and incorporates into DNA of host cell as provIrus.

•

Provirus replicates each time the host DNA replicates.

•

After about 6 years of dormancy, the provirus is transcribed into

I

1 1

mRNA.

•

Host cell synthesizes viral capsid, reverse transcriptase enzyme

1

and viral RNA.

•

New viruses formed in the host cell exit the cell by budding.

•

Virus kills T helper cells and destroys immune system, causing AIDS.

(b)

•

T cells formed in the bone marrow circulate in the blood

1

1 Any 8

Max 8

I

circulatory system until it reaches thymus glands. •

•

In thymus glands,T cells are differentiated to form T helper

1

cells(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 to

1

produce clones of effector Tc cells and memory cells. •

The effector Tc cells bind with antigen-MHC complex on 1

infected cells. •

•

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

1

The infected cells go through autolysis while the effector

1

cytotoxic T cell attacks other infected cells. •

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

1

Max 7

Any 7

TOTAL 10(a)

15

• Sketch of curves a) Stabilising Selection

Selecl,en again I bot"

~

, I

I I

/

,,

/ I

"

exl:el11e~

~

-

~.:~;~

\

\

:,,~A

;";1~~~

Popula tion

aller selecllon

:\

\'.

Or iginal

'< \"'" '" ,"~)OP\Jlallorl

The sketch must -

shows the normal distribution curve and the stabilising selection curve.ldirection of curve.

-

labell/selection pressure

2

2/0

10

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b) Directional Selection

_~_._.""_ Population

I

alter s,ele lJ tion

I

I I

I

"

The sketch must - shows the normal distribution curve and the stabi lisin g selection curve.ldirection of curve. - label//selection pressure

2/0

2

2/0

2

c) Disruptive Selection

S~,,'on"gr'~s~

' ', 'ne,"

, t ,,

, ,

\ '

, ;

,

, ,,

I I

The sketch must - shows the normal distribution curve and the stabil ising selection curve.ldirection of curve. - labell/selection pressure Sta bi/ising selection

• the bell-shaped curve

1

• selection favors the intermediate trait value over the extreme

1

values. • response to a stable environment// occurs when the environment

1

doesn't change

11

papercollection

• the mode stays the same

1

• the population graph gets narrower and taller as selection against

1

mutation takes place • experience a decrease in the amount of additive genetic variation for the trait under selection.

1

Max3

Any 3

Directional Selection

•

occurs whenever the environment changes in a particular way!!

1

selective pressure for species to change in response to the environmental change •

directional selection may favor one of the phenotypes at one of

1

the extremes of the normal distribution!! selection favors one extreme trait value over the other extreme!! selection favors the extreme trait values over the intermediate trait values'!! The average phenotype is selected against more than one other phenotype is selected for. •

the population's trait distribution shifts toward the other

1

extreme!! the mean of the population graph shifts!! One phenotype can gradually replace another.!/ results in a change in the mean value of the trait under selection. •

results in a population with new trait!resistant individuals begin

1

to occur and become the dominant type within the population.!/ the variance increases as the population is divided into two distinct groups.

Max3

Any 3

Disruptive Selection

•

occurs where an environment change may produce selection

1

pressures that favour two extremes of a characteristic!! selection pressures act against individuals In the middle of the trait distribution. •

the environment may favor two or more variant phenotypes at

1

12

papercollection

the expense of the mean.

•

a bimodal/two-peaked curve II the two extremes of the curve

1

create their own smaller curves.

•

results in two distinct populations/morphsllthese two forms may

1

become so distinct that they become new populations

•

plays an important role in speciation.

1 Any 3

Max3 15

TOTAL

13