SPM EXA]VIINATION PAPER 2OO3

Paper IID 2A 3B 4A

Paper 2Section Af (a) (i) Mitosis

(ii) 1. Process that plays a role in plant andorganisms growth

2. Process that plays a role in cellreplacement in an organism

II, IIL ITelophase MetaphaseAnaphase ProphaseInterphase- Chromatin will undergo replication

process

- Enzyme and protein which needed in celldivision is syrthesised

- Cytoplasm and the number of organellesuch as mitochondria will increase

(d) - Cellfromcowwhichisrequiredischosen- Then, the nucleus is removed from the cell- The ovum from another cow is taken and also

its nucleus is removed

- The nucleus from the cow is then injected to thedenucleated ovum and ovum is left to undergomitosis

* The cells that are produce are transferred tothe surrogate mother to develop normally untilbirth

Digestion of starch, peptone, peptides,process of hydrolysis of fat to fatty acidsand glycerol is inhibited.Glucose level in blood cannot be controlledat normal range because excess glucosecannot be converted to glycogen.The patient has to reduce intake of foodthat rich in starch because of the absenceof pancreatic arnylase and starch cannotbe digested in the small intestine.The patient need to reduce intake ofsugarbecause when the pancreas is removed,the hormone insulin cannot be producedanyrnore.

(i) Whorl: Composite{ii) Genetic factor(iii) Anangement and recombination of genes

during crossover will produce gametes thatare different genetically.

(i ) Discontinuous variation(ii) - Ability to roll the tongue

- BloodgroupThumbprint is discontinuous variation and thephotograph is continuous variation for individualthat has same face i.e for the identical twin.

(d) - The body mass is continuous variation whereasthumbprint is discontinuous variation

- Body mass is influenced by genetic factorand environment whereas thumbprint is onlyinfluenced by genetic factor.

(i) Photosynthesis(ii) Light energy received 300 000 000 - 10070

60 000 - 100 x 60 000%rrnerg)'relarneo 300 0o0 000

= 0.003Vo(iii) Because some ofthe light energy is reflected

by the outer surface ofthe source.(i) Total energy from the tissues of herbivore

that cannot be used is= (5000-30)kJ= 4970kJ

(ii) Some of the energy is lost during theprocesses of respiration, metabolic andgrowth of the herbivore.

The organism P can oxidise ammonia to nitrate, byusing the plant nitrate to make protein. Because ofthat, if organism P is destroyed by the use ofpesticide, the nitrogen-flrxation process of plantswill be unhindered and producers (plant) cannotmake its own protein.(i) Nematodes(ii) Parasites obtain its nutrient from its host.

Hence, when the parasite in destroyed:producer, herbivore and carnivore that arehosts can get suffrcient nutrient for growth.

5C 6D7C8C9Cl0ArrA12c

13D 14A l5C 16D t7 B 18A19B 20 A 2L B 22 D 23C 24 B25 C 26 C 27 B 28 A 29 C 30.D31D 32 B 338 34 B 35D 36A37 B 38D 39A 40 D 4L D 12 D43 B 44 D 45 D 46 C 47 A 4a A49 D 50A

(a)

(b)

(c)

(b) (i)(ii)

(c) (i)(ii )

(ii) -

(a)

(b)2 (a) (i)

(b)(ii) P

Epitlrelialcells

(c) (i) Absorptionofdigestedfood (d)(ii) Surface area of T has structure that are

folded, increase surface area for absorption ofdigested food.

(d) (i) - S is pancreas, removal causes enzymelipase, amylase, t4rpsin, hormone glucagonand insulin cannot be produced.

(c)Bloodcapill

242

(c)Trophic level 3

Trophic levcl 2

Trophic level 1

5 (a) (i)(ii)

(b) (i)

(ii)

(c) (i)

Parent

genotype

Gamete

Oflspringgenotype

Offspring phenohype

Carnivore 30 kJ

5x103kJ

6x10{kJ

46

- The male karyotype will have a pair ofhomologous chromosome that havedifferent sizes, i.e. chromosome pair XY.

- Thc female karyotype will consist of allchromosome pairs with the samc sizeand homologous.

Gamete Q: 22 + XGamete R: 22 + YGamcte J: 22 + XProcess P is meiosis, i.e. the homologouschromosome pairs is separated to differentdaughter cells.

Father

- A concentration gradient between thepotassium permanganate(Vll) at thcbase of the beaker with the distilledwater at the top.

- The diffusion of potassiumpermanganate(Vll) molecule will occurfrom the region of high concentration tolow concentration, which is in accordanceto the concentration gradient to achieveequilibrium of concentration.

- Hence, at the end of the experiment, thepurple colour of potassiumpermanganate(\lll) can be seen throughoutthe water in the beaker because thepotassium permanganate molecules havemoved by simple diffusion to a region oflow concentration of potassiun-rpermanganate(VlI).

Milk- Pasteurisation is a method ofpreservation

of milk.- Fresh milk is heated to 63 "C for 30

minutes and then cooled in3tantly.- Or milk is heated to 72 "C for 15 scconds

and then cooled instantly.- The method of preservation will destroy

the microorganisms but will not changethe nutrient value and colour of milk.

Fish- The process ofdehydration is a method of

preserving fish.- The fish is dried with the use of fire,

smoke or is left in the hot sun.

- Food that is dried will have very lowcontent of water and also is covered withcarbon.

- The water content is which is low willcause the microorganisms which ispresent to be destroyed or change intospores which are not active.

(b) - When the plant cell is put into 1Vo of sucrosesolution, the solution is isotonic to the plantcell sap.

- Hence, there is no concentration gradientbetween the osmotic pressure ofthe cell sap ofthe plants with the environment.

- So the rate of water molecules moving into theplant cell is equal with the rate of watermoving out from the cell to the surrounding.

- Hence, there is no change in the structure orthe size ofthe vacuole observed.

- Then the cell is put into 307o of sucrose solution,that is a hypertonic solution compared to thecell sap ofthe plants.

- There is a osmotic concentration gradientbetween the cell sap of the plants with thesurroundings.

- The water molecules will move out of thevacuole in the plant cytoplasm of to thesurrounding to achieve an osmotic equilibrium.

- This will cause the volume of water in thevacuole in the cytoplasm to decrease, hence the

cell membrane will be detached from the cell

wall and the vacuole will contract.

(ii)

Girl:Normal

Girl:Normal

Boy:

Normal

Section BI (a) (i) The process shown in Figure 8 is simple

diffusion.At the beginning of the experiments thebase ofthe beaker has a high concentrationof potassium permanganate(Vll) whereasin the distilled water, the concentrationof potassium permanganate(\4l) is low.

243

Pyramid ofenergy

Mother -

@

Boy:

Colour-blind

I(ii) = -xlOOVo4

= 25%(e) - The female carrier of colour-blind gene has

genotype XBXb and the male colour-blind hasgenotype ofXY.

- A great possibility that during fertilisation ofgametes that carries the allele Xb is notfertilised because fertilisation that occurs israndom.

- Hence, genotype XBX" and XbY cannot beobtained.

The cell undergoes plasmolysis.When tlre cell is put back into 0.lVo of tt'esucrose solution, the solution is hypotonic tothe plant cell sap.There exists an osmotic concentrationgradient between the plant cell sap with thesurrounding solution.This situation will cause a lot of watermolecules from the surrounding move into theplant cell compared with water that moves outfrom the plant cell to the surrounding.The volume of the water in the cell increase,the vacuole enlarges, and the c5,toplasm andthe cell membrane will be pushed towards thecell wall.The cell becomes turgid.

Stage 1:

- The insulin concentration in the bloodof the individual Y is 0.02 arbitraryunit and at this instance, the glucoseconcentration is at normal concentration,that is 0.09 dmr.

Stage 2:

- At stage 2, i.e. 15 number afterindividual Y taken the glucose solution,the glucose concentration in his bloodwill start to increase beyond the normalconcentration, i.e. nearly achieving theconcentration of 0.014 g/dmr.

- The insulin concentration in the blood ofthe individual Y will increase comparedwith before taking the glucose solution,i.e. nearly achieving 0.06 arbitrary unit.

Stage 3:

- At stage 3, i.e. 45 minutes after taking theglucose solution, the glucose concentrationin the blood of the individual Y will comeback to normal concentration, i.e. thesame as before the taking of glucosesolution i.e- 0.09 g /dmr.

- The insulin concentration in the blood ofindividual Y back to normal just as in thebeginning of the experiment, i.e. 0.02arbitrary unit.

- A{ter 15 minutes of taking the glucosesolution, the concentration of glucose inthe blood of individual Y increase untilnearly 0.13 g/dmr.

- The high blood sugar level in individualsY will stimulate the Langerhans cells tosecrete insulin.

- The insulin is carried by the blood willreach the liver and increase the oxidationofglucose and change glucose to glycogen-

- At the same time, insulin will hinder theformation of glucose from glycogen andcarbohydrate sources.

- The effect is insulin will cause the bloodsugar level to decrease to normalconcentration.

(b) (i) - For individual Y, the reading ofthe bloodglucose concentration at the beginning

of the investigation is 0.09 g/dmr. i.e. atthe norrnal range, on the other hand, forindividual X, the initial reading of bloodglucose level is 0.03 g/dmr. i.e. more thanthe normal concentration ofblood glucose.

- The normal concentration of glucose inthe blood is 0.09 g/dm-3.

- After the two individuals have taken theglucose solution, their blood glucoseconcentration increases, i.e. until0.03 g/dmr in the blood of individual Yand achieving 0.02 g/dmr in the blood ofindividual X.

- 45 minutes after taking the glucosesolution, the blood glucose concentrationin both of the individual starts todecrease, i.e. for individual Y the glucoseconcentration achieving 0.09 g/dmr again.

- However for individual X, the bloodglucose concentration did not return to thesame concentration as in the initial stagei.e. the reading of the blood glucose is lessthan 0.17 g/dmr, still more than theinitial blood glucose concentration.

- Individual Y shows negative feedback inthe control of blood glucose, whereasindividual X shows a positive feedback forthe control ofblood glucose.

(ii) - At the beginning of the investigation, theinsulin concentration of the blood ofindividual Y is only 0.01 arbitrary unitwhereas the insulin concentration inindividual X is 0.02 arbitrary unit.

- This shows that individual X has aproblem of producing insulin by theLangerhans cells in his pancreas.

- After taking the glucose solution, theinsulin reading of individual Y increasevery quickly achieving 0.04 arbitraryunits, but not for individual X.

- At stage 2, the blood insulin concentrationof individual continue to increase until iireaches the maximum level i.e. 0.06arbitrary units a{ter 30 minutes ofconsuming the glucose solution and afterthat return back to shou' the decrease ofthe concentration. On the other hand, forindividual X the glucose concentrationshows slightly increase after 45 minutes ofinvestigation and the maxinum readingis only nearby to 0.02 arbitrary units.

- At stage 3, the insulin concentration in theblood ofindividual Y starts to decrease tothe initial value at the beginning of theinvestigation, i.e.0.02 and achieve the levelof the initial investigation in 75 minutesafter the investigation has began, but isindividual X there is no change of insulinconcentration.

(c) {i) - Benedict's test to test the presence ofglucose in the urine ofindividual X and Y.

Procedure (Benedict's test)

- About 1cm of urine sample of individual Xand Y is put into different test tubes.

(a) (i)

(ii)

244

-

Section C3 (a) (i)

(b) -

- Add 1 ml of Benedict's solution into eachofthe test-tube.

- The mixtures are heated in a water bathfor 5 minutes.

- The change that happen to the contents ofthe test tube is observed and recorded.

- Ifthe result is the colour change from theblue solution to green, yellow, orange andfinally brick red, there are presence ofreducing sugar, i.e. glucose is present inthe urine sample.

- It is predicted that the Benedict's testfor individual X will produce brick-redprecipitate but for individual Y, thecolour of the mixture in the test tube willremain blue.

- If brick-red precipitate is obtained, thisshows that the individual has diabetesmellitus.

(ii) - Individuals who has to control the glucoseconcentration in his blood is individual X.

- Individual X has to control his diet, i.e.reducing food intake that contains glucose.

Yes, because it consist of all the differentgroups offood at the current proportion.

- The products ofdigestion includes glucose,amino acid, fatty acid and glycerol.

- Glucose and monosaccharides will be usedat the beginning of cellular respiration byour body cells. The formation of ATP aschemical energy will occur.

- 38 ATP will be produced during aerobicrespiration but only 2 ATP will beproduced in anaerobic respiration. Lacticacid will also be produced.

- Amino acid will combine with other aminoacid to change the protein after thenucleus gives translocation to form therequired protein.

- Fatty acid will combine with glucose toform fatty acid droplets which turns as thebasic unit in the nervous system andlymphatic system. Others are used forthe cell when respiration.

- Excess fats will be sent to the liver andthen to the adipose tissues to be storedas fat.

High possibility that the teenager will haveseveral diseases such as obesity, constipation,poisoning, high dosage of mineral salt, kidneyfailure and if continues will become cancer.Obesity is caused by excessive intake of food.Hence, high blood pressure, diabetes, heartdisease and arteriosclerosis can be causedindirectly.Constipation is caused by lack of fibre.Intestinal disease will occur ifit persists.Excessive intake of mineral salt will causepoisoning of high dosage of mineral salt. If themineral salt is excessively taken, cirrhosis ofthe liver will occur, high intake of calcium and

phosphorus will result in the formation ofkidney stones.

- Excessive continued intake of protein will resultin kidney failure. The kidney cannot excretethe urea that is formed; hence toxin willaccumulate in the blood. This will cause death.

- The use of preservative, seasoning and foodcolouring will cause cancer because excess ofthem is carcinogenic.

(a) (i) - The source of pollution that occur at thebeach is domestic discharge.Food remnants from houses and hotelsare discharged to rivers/ drain andsubsequently they flow to the sea, causingthe dense growth ofE. coli.bacteria.Waste disposal has to be treated firstbefore being discharged to the sea.Water containing food remnants has to befiltered; added bacteria decomposers offungi decomposers which are not harmful,left for several days to decompose foodremnants to mineral salts before beingdischarged to the sea.Enforcement of Environment QualityLaw 1979 that controls the disposal oftoxic waste.Sea water at beach X is polluted by thedense growth of microorganisms suchas E. coli. This is because of the disposalof non biodegradable waste to the seaby inhabitants.

Problem statement:What is the level of water pollution atbeach X?

Ilypothesis:The level of water pollution is dependent onthe value of BOD.

Apparatus and Materials:Reagent bottles with stoppers, methyleneblue solution. (1cm3), water sample (100 cms).

Procedure:1. 5 samples of water (100m3) are obtained

from 5 different station along the beach X.2. All the water samples is put into

different reagent bottles until full.With the use of syringe, 1 ml ofmethylene blue solution is added to thewater samples.The reagent bottles are closedimmediately and make sure that thebottles are not shaken.

5. Repeat steps 2 to 5 with distilled water.6. The bottles are observed every hour

until the blue colour is bleached or aslong as 6 hours. The time for thebleaching ofthe colour is recorded.

R.esultsThe water sample that is obtained near to ahousing estate and hotel is the fastest to be

bleached.

(ii) -(ii)

3.

4.

245

(b)

Conclusion:Time for bleaching of the blue colour isfaster if the oxygen content is less. Thismeans that the area is the most pollutedwith microoiganisms.The higher BOD value indicates that thewater is the most polluted.

The diagrams show human activity on theenvironment to enable the growth ofeconomy anddevelopment of the country.Among what is shown is industrialisation, loggingand organisation.The qualities of life of humans will increaseswith the above activities. Job opportunities willincrease, unemployment will decrease, theeconomy of the country will improve because offoreigrr investment, infrastructure will improveand transportation from one place to another willbe more advanced. Communication will be moresophisticated enabling global communicationwill improve. Hence, the life style of human willimprove. In spite of all the advancements, theecosystem will be affected. This is because ofmore areas are being developed to the aboveactivities. The decreases number of trees willcause the definite ofoxygen and air pollution willoccur because the concentration ofcarbon dioxidewill increase. The earth temperature will increaseand the gleenhouse effect will occur. Besides this,acid rain, soil erosion, floods and water reservoirand diseases will occur. Animals will lose theirhabitats. Flora and fauna species will faceextinction.

1. When white cloth is used, as many as6 black buttons are taken comparedwith other coloured buttons.When black cloth is used, as many as7 white buttons are taken comparedwith other coloured buttons.When white cloth is used, the blackbutton is clearly seen compared withthe other colours, hence more blackbuttons are taken.When black cloth is used, the whitebuttons are more clearly seen comparedwith other colours, hence more whitebuttons are taken.

2.

(ii) 1.

2.

Colour ofcloth areaColour ofbuttons takenSize ofcloth used.Colour of different clothused.

(c) (i) _

(rr) - that is being

- Taking the buttons which are on top ofthe cloth quickly.

- Size of the cloth is fixed at 50 cm x 50 cm.When the cloth colour and colour of the buttonsare different, hence the buttons are more clearlyseen and is taken quickly.The more different the colours of the tloth andthe colour of the buttons, the clearer the buttonson top of the cloth are the more differentcoloured buttons are to be taken.The black colour button will be taken the mostfollowed by the green buttons because thesebuttons are clearly seen on top ofthe yellow cloth.When the tree is more polluted, the colour of thebark of the tree becomes darker. Hence, thebutterflies with bright coloured will be morevisible to their predators. As time passes, thenumber of butterllies which are bright colouredwill decreases.The change in the physical structure so that it issame as the environment hence it is not toootrvious to the predators for survival.Effect: The prey will be safe from the predators.Variegated leafRed coloured leaf I H"bit"t of..terpillars andGreencoloured leaf J environmentofcaterpillars

Green-coloured caterpillarsGrey-coloured caterpillars

Posses colour closclylinked to theenvironment toprotecL it lrornits predators

Objective ofstudyTo investigate the relation between the conr:cntrtrtionofcarbon dioxide with the rate ofphotosyntlresis.

Variables(i) Manipulatedvariable:

concentration ofcarbon dioxide that is used.(ii) Respondingvariable:

number of bubbles released per minutcs(iii) Controlledvariable:

distance between the light source, t,he pl.,rrrt an,iwater-bath.

Hypothesis:The increase in carbon dioxide conccritrrrlirrt s;llincrease the rate ofphotosynthesis until the maxitlumlevel.

Apparatus:Boiling tutres, 20 ml beakers, measuring c.r.lindtr I t) ;i:i.light bulb (60 W), retort stand and clamp, tripod stanri,spatula, thermometer, stop watch.

Materials:Sodium hydrogen carbonate, aquatic plarit (.Ei r,&u )

(d)

(e)

(fl

G)

(h)

(i)

PaperSI (a) (i)

(b)

246

Colouringcloth

Numberof buttonsaccording to colour

Totaltaken

Idtr

q).t#

Boo

&

oq)lr

White 6 0 I I 1 10

Black 0 0 I I 0 10

Coloured 1 4 4 1 I l1

Technique used:Locating of the number of air bubbles which arereleased in l minute

Procedure:(i) A plant section is cut approximately 6 cm.(ii) The plant is put into a boiling tube filled with

sodium hydrogen carbonate ZVo (source of carbondioxide).

(iii) The cut-section of the plant is arranged by facingupwards.

(iv) A part of the boiling tube is immersed in thewater-bath at warm temperature and thetemperature is maintained.

(v) A light bulb of 60 watt is placed at a distance of10 cm from the surface ofthe beaker.

(vi) The bulb is lighted up and the stopwatch started.(vii) When the air bubble is released through the end of

the cut section ofthe plant is constant, the numberofair bubbles produced in 1 minute is recorded.

(viii)Steps (ii) - (iv) is repeated for concentration ofsodium hydrogen carbonate of concentration 37o,4Vo,1Vo andGVo.

(ix) Then, draw a graph of number of this bubblesreleased versus concentration of sodium hydrogencarbonate.

Presentation of results:

Predicted graph:

3.0 4.0 5.0 6.0

Concentration of sodium hydrogen carlmnate (7o)

Conclusion:Concentration of carbon dioxide affects the rate ofphotosynthesis, i.e. when the concentration of carbondioxide is increased; the rate of photosynthesis alsoincreases until a maximum value-

SPM EXAMINATION PAPER 2OO4Paper 1

lc 2c 37B 88 913D 14A t519D 20c 2r25 A 26 D 273lD 32 B 3337B38B3943 A 44 A 4549 A 50A

D4D5D6Dc 10A 11c 12cc 16B nD 18Bc 22 D 23B- 24 AB 2A D 29 B 30BA34D35C36CB 40 D 4tB 42 Cc 46 A 47 C 48C

Paper 2Section A1 (a) (i) P - enzyme

Q - substrate(ii) 1. Actspesifically

2. Can be reused

R - fattyacidS - glycerol

(b)

(b) (i) Lockandkeyhypotheis/theory(ii) As a catalyst which increase the rate of

digestion of fats to fatty acid and glycerolduring reaction. The enzyme will combinewith substrate at the active site to form acomplex enzyme-substrate, then the productswill leave the active site of the enzyme.

(c) Making of skin digestion of fat.(d) Rate ofreaction decreases. This is because a lot of

enzyme lipase has been destroyed at hightemperature.

(e) Because the enzyme is most effectively at atemperature of less than 37 oC. The low temperaturemakes the enzyme inactive.

J : hydrophilicheadM : proteinpore(i) Molecules which are very minute such as

small molecules and ions(ii) Small molecules and ions move across the

pore between the protein pore by simplediffusion. The molecules are at the higherconcentration outside the cell cause themolecules to be pushed into the cells throughthe pore.

Specifrc rehabitation(i) The glucose with high concentration is acts

as a hypertonic solution causing the waterin the food to move by exosmosis through theplasma membrane. This causes dehydrationoccurs to the food. Hence, the condition is notsuitable for the growth of microorganisms.

(ii)

Outer cell

o

oo

n.,it7Z

'=o

!E.ootrd2.,

(c)(d)

2.01.0 fiRNINQNNRORRUUUIUTJ\JUUUDU

-

Air moves outlnside the cell

Concentrationofsodiumhydrogen

Carbonate (7o)

Number of air bubbles releasedin l minute

1stl minute

2ndI rninute

3rdI minute Average

2.O

3.0

4.O

5.0

6.0

247

(a) (i) R: Pulmonary circulationOxygenated blood will be pumped intothe heart through the pulmonary vein.Deoxygenated blood will be pumped outfrom the heart to the lungs through thepulmonary artery.Oxygenated systemic circulation fromthe heart Blood is pumped throughoutthe body through the aorta andoxygenated blood to the heart throughvena cava.

\r-I

I

\

(c) An electronic gadget will give the frrst heartbeatto act as sino-atrial node and then spread to allthe atrium muscle and then to the atrioventricalnode until the whole heartbeats.

(d) (i) TherateofbloodflowinXishigherthaninybecause X pumps blood from the'heart to thewhole body whereas y returns blood to theheart

(ii) Vigorous exercise will cause defrciency ofoxygen. Hence, the heart will beat faster toincrease the blood flow to carry more oxygenand supply to the cells.

(a) Temperature, water, soilDecomposers, producers, consumers

(b) (i) Producer: kiambangplant(ii) Consumer: bird, earthworm(iii) Decomposer: bacteria

(c) Duckweed *Butterfly ----- Sparrow ...- Owl(producer) (primary (secondary (tertiary

consumer) consumer) consumer)(d) (i) Getridofthebirdsfromthehabitat

(ii) When the owl is get rid off, the population ofthe birds will increase because the predator(owl) is decreases.

(e) Trophic level 210 x 150000

800 = 15 000 J

Trophic level 310

100 x 15000 = 1500J= r.5kJ

(a) FollicleStimulatingHormone(FSH)Role: Sent from the pituitary gland to stimulatethe formation offollicles in the ovary

(b) Hormone P is oestrogen whereas hormone y isluteinising hormone (LH). Hormone p cause therejuvenation of uterus wall, whereas hormone y

will cause ovulation. Ifhormone p is not balancedwith Y, hence the ovulation process will be delavedor increased.(i,@

day 26(ii) Structure T is corpus luteum that secrete

progesterone, Q and oestrogen. When reachesd,ay 26, the structure will degenerate.Hormone P and Q will decrease.

(d) The pituitary gland will secrete FSH, LH, oxytocinand prolactin. If the menstrual persists, henceFSH and LH will play their part. If fertilisationoccurs, and during pregrrancy, FSH and LH willnot be secreted. Progesterone will be high. Themenstrual cycle will not occur. Oxytocin andprolactin hormone will be produced when the babyis born.

Section BI (a) Process P: aerobic respiration

Process Q: anaerobic respirationProcess P uses oxygen. Respiration produces a lotof ATP. Process e does not use oxygen.Respiration produces less ATp because of lacticacid formation.

(b) The athlete wears a tracksuit so that there is nosudden decrease in temperature in his muscles. Asudden decrease in temperature will cause theaccumulation of lactic acid and muscle pain/cramp. The athlete draws a deep breath severaltimes to oxidise lactic acid that has beenaccumulated and pays back the oxygen detrtduring vigorous activity. He walks freely so thatthe muscles takes time to oxidise 1/6 of lactic aci,-lto carbon dioxide, water and energy and 5/6 oflactic acid is converted back to glucose.

(c) Difference of respiratory system in human andfish:

(ii)

(b)

248

lfuman Fish

Organ Lungs Gills

Habitat On land In the water

Function Lungs do notneed a medium tofunction

Gills need flow ofwater through thegills for gaseousexchange

Adaptation Human lungshave a lot ofalveolus andinterweavedwithblood capillaries.Gaseousexchange occursby diffusionthrough alveolus.

Water is drank inand flows throughthe gills forabsorption ofoxygento blood capillarieswhich are cross-linked withfilaments of gills.Then, water flowsout through theoperculum and

absorbs carbondioxide f'rom theblood capillaries tothe water to berelease.

Characteristics Total surfaccarea is bigwith presenceof 700 millionalveolus.Surface ismoist.Thin wall.A lot ofbloodcapillariesnetrvork.

A lot ofgillsfilament withthin plates(gills lamella) toincrease surfacearea for gaseousexchangeeffectively.A lot ofbloodcapillariesnetwork.

Section C3 (a) Female ( p )

HHMale(6/)

hh

(a) (i) When the finger is poked with a needle, thereceptors at the tip of the frnger will receivestimulation and sends impulse to nervoussystem through the afferent nerve. Theimpulsc is forced to move in one directionthrough the synapses and then through theefferent ncrve to the brain for rcaction.

(ii) Figure 7(a) is reflex action whereas Figure7(b) is involuntary action.

Reflex arc Involuntary action

Cannot be controlledby brain.

Controlled by medullaoblongata, part ofthebrain.

Reacts towardsexternal stirnulation.

R-eacts towardsinternal stimulation.

Reaction andmovement is very quickand not long lasting

Reaction andmovement is slowand last longer

When the condition is cold, the smooth muscle inthe arteriole of the skin will contract. Thediameter of the arteriole will decreasc and causeless blood to flow to the skin. This cause loss ofheat.

The erector muscle of the hair foliclecontracts causing the hair follicle to stand. Formammals with thick hair, this will increase the airtrapped which acts as heat insulator andpreventing heat loss.

The muscular skeleton contracts and relaxescausing one to shiver. This increases the bodytemperature.

The sweat gland is not stimulated. Hence,sweating does not occur. The adrenaline hormonesis secreted to increase the rate of glycogenconversion to glucose, hence the metabolic rateincreases.

The stimulation of surrounding temperatureand at low temperature is received bythermoreceptors in the skin and at the certainpart in the brain is sent by afferent nerve tocontrol part, hypothalamus.

(b)

(b)

(a)

Based on the differences above, can be

concluded that the child inherit a lot ofcharacteristics from the father who is dominant.

Parents

fBh-kJr I| *tit",t I

In the parents, a pair of alleles code forcolours at the same locus on the pair ofhomologous chromosomes. From the pair of allele,only one allele exists in the gametes due tomeiosis. This occurs in Mendel's First Law.

db JbGamete

Ifthc first generation is cross-breed betweenthemselvcs, the rat in Fl will obtain a pair ofalleles for black colour at the same locus atthe pair of homologous chromosomes afLerfertilisation.

From the pairs of alleles, only one alleleexists in the gamete.

The rat from F2 will obtain a pair of allelesfor coloured body at the same locus on the pair ofhomologous chromosome.Chemical K, a mutagen will cause mutation on thegenes or the chromosomes of the rat naturallyduring replication of the chromosome andsegregation of the chromosome to the gamete cellsin the reproductive organs. Mutation will bepassed down to the zygote. Mutation of thechromosome cause changes in the structure of thechromosome which includes:(i) loss ofseveral genes(ii) additionofgenes(iii) transferofgenes

Changes in the number of chromosomes isalso a part of mutation of chromosome, i.e. loss ofchromosome or addition of chromosomes. Genesmutation causes changes at the base ofthe DNAmolecule and this affects the production ofprotein.

Degeneration will exist in the rats. Therewill be sterile rats, their offsprings will defected.The rat population will decrease until a stateof equilibrium where they are able to adaptthemselves and live as mutants.

Characteristics

Father Mother chitd

Curly hairBlack hairHas dimplerThick lipsOval face

Straight hairBrown hairNo dimplerThin lipsRound face

Curly hairand blackHas dimplerThick lipsRound face

The combination of characteristics of thefather and mother can be inherited by the child by(i) cross-over at meiosis 1(ii) choosing of dorninant characteristics through

Mendel's First Law(b) Haemophilia is a sex-linked disease on

chmmosome X only. The characteristics is carriedby the recessive gene.

Haemophilia can occur in men because menonly need one recessive gene on chromosome X tocause the haemophilic characteristic.

A woman who has family pedigree ofhaemophiiia does not necessaiily has it becauseshe can still exist in heterozygous conditionwhich means that the female is normal but is acarrier for the next generation. Two recessivegenes are needed to show the haemophiliccharacteristic. The inheritance of haemophilia canbe avoided by making sure that every marriage ofthe grandchildren do not involve caniers orhaemophilic individuals, and eventually aftergenerations by elimination, the recessive geneswill disappear.

The more volume of juice that is needed todecolourise DCPIP, the less is the vitamin content.The volume will increase because oxygen fromthe atmosphere will reoxidise DCPIP which isreduced by an ascorbic acid.Vitamin C is easily oxidise by the oxygen.

(e)

(f)

(e)

(h)(i)

PaperBI (a) (i)

Figure Changes ofcolour ofDCPIP solution

1(b) ( Blue becomes colourless

I (b) (ii) Blue becomes colourless

1 (b) (iii) Blue becomes colourless

Papayajuice needs larger volume because thevitamin C content is very low. More juice isneeded to change the colour of methyleneblue to colourless.2ml4ml8ml

Rcagent Materials to be tested

DCPIPsolution 0.17o

Apple juicePineapplejuiceAscorbic acid 1.07o

Objective ofstudy:To investigate and estimate the population ofbirds.

Hypothesis:The population size ofthe birds can be determined by:

Number of bird at lst capture x Number ofbird at 2nd capture

Number of marked birds at 2nd capture

Variables:(i) Manipulated: Birds which are capturrd and

markedResponding: Birds are markedControlled: Place of experiment, species of

birds, days of experiment

Apparatus and materials:Light ring, fruit-eating birds that has been determinedand special net that can cover the area ofsampling.

Technique used:Capture, mark, release and recapture

Procedure:1. A special net is put up at a sample area that has a

lot of birds to be sampled.2. A total 120 birds are captured and a small ring are

placed at their feet. Then they are released.3. After 2 days, 100 birds are recaptured randomly.4. The numberofbirds that tagged are countc.l.5. All the information are recorded.

Tabulation ofdata:Number of birds at lst capture, x = 120

the less Numberofbirdsat2ndcapture,y=100Number of birds tagged/with ring= g

Estimation of the number of birds in the sample area

250

(ii )

(b)

(c)

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

(iii)

,tii) Limejuicc = ixO.lVo

II xO.19"2

= 0.OS7cI

Papayajuice = t x0.lEa

1: xo.l704

= 0.14

= O.O2SVo(d) The more volume of fruit juice needed,

percentage of vitamin C is in the juice.

Variable Method to handlethe variable

1. Juiceneeded Use different juice i.e. limejuice and papayajuice

2. Bleaching ofDCPIP colour

BIue colour of DCPIPbecomes colourless

3. Concentrationofascorbic acidand its volume todecolouriseDCPIP

Concentration = 0.1%Volume = 2 ml

Material Volume (ml) todecolourise DCPIP

Ascorbic acidO.IVo

2ml

Lime juice 4ml

Papayajuice 8ml

_xyz

120 x 100z

120 000z

Conclusion:Number of birds can be determined by capture, mark,release and recapture technique

120 x 100z

SPM EXAMINATION PAPER 2OO5

CloningProblem: To identift quality genes that is beingclone resistant to diseases and changes in theenvironment.

Glass tube: tracheaBelljar: thoracic cage(i) Tochange thevolume ofbelljar(ii) - Balloon expands shows air goes into lungs

during inhalation- Balloon flatten shows air goes out from

lungs during exhalation

(e)

(a)

(b)

(c) (i)

Paper I1D 2B 3D 47A 88 98 10

13A 144 158 16r9A 20 D 2t c 2225C26A27C2a31C32A33D3437 A 38A 39A 4043C 44 C 45A 464SC50D

Paper 2Section AI (a) (i) P : mutualism

Q : commensalism(ii) Nitrogen-fixing bacteria gets its habitat for

survival and this bacteria fix atmosphericnitrogen to produce nitrate as organicfertilizer for the plants.

(b) (i) Organism X: epiphyteOrganism Y: saprophyte

(ii) Ability to make its own food withoutobtaining any minerals from the trees.

(c) X may be still survives if the tree height are tallenough to obtain enough sunlight.

(d) The hlpha of the mushroom/fungi produces anenzyme that can hydrolyse for the digestion ofcellulose on sawdust to produce glucose which isconsequently absorbed as nutrient for the plants,

2 (a) (i) K : chromatidL : spindle frbre

(ii) CellP : metaphaseCellQ: metaphasel

(b) 23=8(c) (i)

(ii) Testes /Ovary/Anther/Ovule(iii) 6Radioactive rays will kill cancer cells. When all thecancer cells are destroyed, they cannot reproduceby mitosis.

Balloon expands

Thin rubber sheet

(ii) When diaphragm flattens, it becomes level.Moves upwards creating a large thoracicvolume. Higher internal pressure thanatmospheric pressure force air to go intolungs.

(i) Because only CO, gas can be absorbed bypotassium hydroxide, leaves only O, which isinsoluble in potassium hydroxide but can beabsorbed by potassium pyrogallol.

(ii) Because exhalation produces CO, gas.Oxygen and glucose is used for cellularrespiration, produces COr.

P : GolgiapparatusR : vesicle/lysosome(i) ToproduceATP(ii) Q replicates section of DNA in chromosome

as m-RNA. Transports through nuclearmembrane to ribosome in rough endoplasmicreticulum for transcription and synthesis ofprotein enzymes.

(i) Protease enzyme used for digestion of hardprotein and to soft it. Soft protein is easy tocook.

(ii) Selulase enzJrme are used to digest cell wall ofseaweeds and let out the agar.

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C 5A 6CD TlD 12AB t7 D 18CB 23 D 24 Bc 29 A 308D35C36DB 4t C 42 BA 47 A 4A C

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