biology 2009 stpm

8/4/2019 Biology 2009 Stpm

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BIOLOGY

STPM 2009

PAPER 2


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SECTION A

1a) X: Chloroplast

Y: Vacuole

b) P: Glucose

Q: PEP /phosphoenol pyruvateR: malate


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CAM plants vs C4 plants

Involves one cell type/mesophyll cells only// Krants

anatomy absent

Carbon fixation fromatmosphere occurs at night

Involves chloroplast andvacuole

Stomata/lenticles open atnight

Involves two cell types/mesophyll and bundle

sheath// Krantz anatomypresent

Carbon fixation fromatmosphere occurs duringthe day and night

Involves chloroplast only

Stomata open during theday and night


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Section A

1d) (i) reduce water loss/transpiration during

the day//adaptation to/ canwithstand arid condition//carbon fixation moreefficient

(ii) pineapple/ cactus/ orchid/ dragon fruit


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Section A

2a) A: glycolysis

B: Lactate/ Lactic acid fermentation

C: Alcohol/ alcoholic fermentation

D: Krebs/TCA/ Tricarboxylic acid/ Citric acid cycle

E: Electron transport chain/ETC/Electron transport

system// oxidative phosphorylationNote: spelling must be correct


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Section A

2b) A: in cytoplasm/cytosol

D: in mitochondrial matrix/ mitochondriamatrix

c) 22 ATP

d) NADH/reduced NAD+ cannot be oxidizedoxygen is the final electron/proton acceptor


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Section A

3a) (i) as solutes in plasma/by dissolving in

plasma// carbonic acid in the plasma(ii) by binding to protein portion of

haemoglobin// as carbaminohaemoglobin

(iii) as bicarbonate ions/HCO3- in the plasma


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Section A

b) the diffusion of chloride ions into the

erythrocyte (from the plasma)- to maintain electrical neutrality/ balance

electrochemical charge

c) (i) Enzyme A: carbonic anhydrase

(ii) CO2 + H20 H2C03 H+ + HC03

-


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Section A

3d) combine with CO2 to form

carbaminohemoglobin decreasing theacidity/ increasing the pH

- to combine with H+/proton forming HHb/haemoglobinic acid decrease the acidity/

increasing the pH


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Mutant anticodon vs Mutant mRNA

codons

5-ACG-3

5-AUU-3

5-AUC-3

5-UUG-3

5-CUG-3

5-AAG-3 5-AGG-3

5-AUA-3

5-CGU-3

5-AAU-3

5-GAU-3

5-CAA-3

5-CAG-3

5-CUU-3 5-CCU-3

5-UAU-3


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Section A

4b) 5-CAA- 3 and 5-CAG-3

- One amino acid can be coded by morethan one codon// several codons encode forone amino acid


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Section B

5(a)

Starch

The bond links -glucose monomers

Linked by -1,4 bondsfor linear structure

Linked by -1,6 bondsfor branched structure

Cellulose

The bond links -glucose monomers

Linked by -1,4 bondsfor linear structure

-1,6 bonds are absent


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5(a)(ii) why are the differences are

biological important

Starch:

Forming storage polysaccharide Insoluble in water//does not affect the

osmotic pressure water potential//osmotically inactive

Helical/branched structure//compact// spacesaving

The bond can be digested by amylase


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5b) The role of carbohydrate in

membrane

i) Conjugate with lipid/forming glycolipid

ii) Conjugate with protein/formingglycoprotein

iii) Glycolipid/glycoprotein for tissuerecognition// self recognition// cell to cell

recognitioniv) Glycolipid/glycoprotein for cell adhesion

v) Glycolipid/glycoprotein act as receptor ontarget organ


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6. The process of transcription/ RNA

synthesis

Diagram:

Initiation (promoter, RNA polymerase, DNA) Elongation (RNA exit, DNA open, polarity of

mRNA)

Termination (all structures detached)

Bonus: 1 mark for all correct/all 3 correct


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synthesis

Description of transcription:

i)

It is the process by which RNA issynthesised

ii) Using DNA as a template//only one of thestrands is transcribed

iii) Transcription proceeds/RNA molecules issynthesised in the 5 3 direction

iv) The process involves 3 stages: initiation,elongation and termination


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synthesis

Initiation:

v) RNA polymerase recognise and binds to thepromoter

vi) Near the transcription start site/beginning ofthe gene// DNA strands unwind


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synthesis

Elongation:

vii) RNA polymerase travels/moves along the DNA

templateviii) Catalysing the addition of polymerising RNA

nucleotides one at a time

ix) Complementary to the DNA template// the newly

synthesise RNA strand begins to separate from thetemplate

(after about the 10-12 nucleotide are synthesised)


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synthesis

Termination:

x) The RNA polymerase encounters/meets/reaches a terminator/stop sequence

xi) RNA transcript is released// RNApolymerase detaches from DNA// the two

DNA strands reform/rewind


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7a) the role of liver in the metabolism

of protein

The role of liver:

i) Protein is broken down into amino acidsii) The excess amino acid is brought to the

liver for removal/regulation

iii) Removal/regulation of excess amino acidis by deamination and transamination


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of protein

In deamination

(iv) The amino group is removed from theamino acid

(v) The amino group/ammonia then enters theornithine cycle

(vi) If then combine with carbon dioxide(produced by the Krebs cycle)

(vii) And converted to urea


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of protein

In transamination

(viii) An amino group from one amino acid istransferred to other organic acid

(ix) To form other amino acid

(x) Which will be used for synhesis of plasmaprotein/albumin/globulin/prothrombin


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of protein

The non-nitrogenous

(xi) The non-nitrogenous part of amino acid/organic acid/keto acids/the remaining carbonchain enters glycolysis/Krebs cycle

(xii) Converted to carbohydrates/energy

production

(xiii) Some will be converted to fat


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7(b)The Cori cycle and its function in

the metabolism of carbohydrates

(i) The function of Cori cycle is to convert

lactate to glucose(ii) Lactate/lactic acid produed in the muscle is

brought to the liver

(iii) In the liver, lactate/lactic acid is converted

to pyruvate

(iv) Pyruvate is converted to glucose viagluconeogenesis


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7(b)The Cori cycle and its function in

the metabolism of carbohydrates

(v) Glucose is returned to the muscle

(vi) Glucose is converted to pyruvate throughglycolysis

(vii) Pyruvate is converted to lactate underanaerobic condition


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8(a) interaction of phytochromes in

flowering

(i) Phytochromes present in the (young) leaf/ leaves

(ii) Stimulated by light

(iii) Exist in two forms Pr/R and Pfr/FR(iv) Pfr/FR is the active form of pytochrome

(v) Day light/red light converts Pr/R to Pfr/FR

(vi) the process is rapid (dependent on point (v))(vii) In darkness/in shade/at sunset far red light

converts Pfr/FR to Pr/R


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8(a) interaction of phytochromes in

flowering

(viii) The process is slow (dependent on point (vii))

(ix) High Pfr/FR/low Pr (ratio) promotes flowering in long

day/short night plants// inhibits flowering in shortday/long night plants

(x) High Pr/R//low Pfr/FR (ratio) promotes flowering inshort day/long night plants// inhibits flowering inlong day/short night plants

(xi) Pfr/FR converts precursors to florigen

(xii) Florigen is translocated to flowering part/organ/buds


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8(b) the involvement of hormones in an

apical dominance

(i) The presence of a growing apical budinhibits the growth

(ii) Controlled by the interaction of severalhormones

(iii) Auxin produced by the shoot

(iv) Auxin inhibits the growth of lateral bud//promotes apical growth


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8(b) the involvement of hormones in an

apical dominance

(v) Gibberelins enhance the action ofauxin//gibberellins and auxin worksynergistically

(vi) Cytokinin is produced by the root

(vii) cytokinin promotes the growth of lateral

bud/inhibit apical growth

(viii) Gibberellins is produced by the apicalportion of stem and root


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Assumptions that maintain the Hardy-

Weinberg equilibrium

(i) large population*

sampling error and other random effectsnegligible//no effect of genetic drift(dependent on*)

(ii) No mutation**

no creation/formation of new allele(dependent on**)


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Assumptions that maintain the Hardy-

Weinberg equilibrium

(iii) No migration***

no gene flow (dependent on***)

(iv) No natural selection****

all genotype/phenotype have equal changeof survival (dependent on****)

(v) Random mating*****equal reproductive potential/chance to mate(dependent on*****)


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9(b) Phenylketonuria- genetic disorder

(i) p represents the dominant allele frequency// qrepresents the recessive allele frequency

(ii) P2 + 2pq + q2 = 1// p+q = 1

(iii) q2 = 1/15000 = 0.000067

(iv) q= 0.000067= 0.0082/0.00816

(v)

p= 1-0.0082=0.9918/0.992(vi) Carriers=2pq=2X0.9918X0.0082 = 0.0163/ 0.0162

(vii) 15000 X 0.0163 = 244/245/242


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10(a) Procedure of quadrat sampling

(i) Procedure of quadrat sampling

1. Quadrat of chosen size2. Laid down randomly//systematically placed

(along the transect)

3. Count all individuals in the quadrat

4. Extrapolate the average count/parametersof the whole study area


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10(a)(ii) requirement for reliable

estimates

1. The population of each quadrat examinedmust be determined accurately

2. The size/area and shape of quadrat mustbe consistent

3. The quadrat counted must be represented

of the whole area


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10(b)(i) procedure of capture-recapture

technique

1. Applicable for mobile animals

2. Animal captured, recorded/counted/ calculate,marked and released (at time 1)

3. Animal will be captured again and recorded/counted/calculate for marked and unmarkedanimals (at time 2)

4. Estimate total population size = marked animals in1st sample x total animal caught in 2nd samplemarked animals in 2nd sample


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10(b)(ii) assumptions

1. Random mixing of animals within the population

2. Closed population//no migration//no emigration and/or

immigration//no changed in population size3. Sufficient time must elapse between capture and recapture

4. Marking does not hinder movement/changebehavior/endangered of the marked animals/ permanent

5. No losses/mortality

6. No natality during sampling period7. Marked and unmarked animals are captured randomly

biology 2009 stpm

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