Post on 07-Apr-2018
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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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6. The process of transcription/ RNA
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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6. The process of transcription/ RNA
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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6. The process of transcription/ RNA
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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6. The process of transcription/ RNA
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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7a) the role of liver in the metabolism
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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7a) the role of liver in the metabolism
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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7a) the role of liver in the metabolism
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