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SOLAF 2 4551/2
SKEMA JAWAPAN SOLAF 2 K2
1 a (i) Mitosis
(ii) C, B, E, A, D
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b P : chromosomes
E : metaphase
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c P1 sister chromatid separates
P2 moves to opposite poles
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d (i) Increase the number of cells in organism 1
d (ii) 1
1
fMitosis Meiosis
Occurs in Somatic cells Occurs in reproductive cell
To replace damage/dead cell. To produce gamates
No cross over during prophase Cross over occurs during
prophase
Produce 2 daughter cells Produce 4 daughter cells
Daughter cells are diploid (2n) Daughter cells are haploid
(n)
Genetically identical Not genetically identical
No genetic variability Genetic variability occurs
Any 2 pairs
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SOLAF 2 4551/2
2 a (i)
Note : no. of water molecule correct – 1
1
(ii) osmosis 1
(iii) Able to explain what is osmosis
P 1 movement of water molecule across the plasma membrane
P2 from low concentration area to the high concentration area // following concentration gradient
P3 through semi permeable membrane
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1
1Max 2
b (i) Solution A : hypotonic
Solution B : hypertonic
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(ii) in solution A : causes water to diffuse into the cells by osmosis
the cell expands and burst // undergoes hemolysis
in solution Bcauses water to diffuse out from the cells
cell shrink / undergoes crenation
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1Max 3
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A B
SOLAF 2 4551/2
c P1 concentrated salt solution is hypertonic to the cells
P2 water diffuses out from the cells by osmosis // fish is dehydrated
P3 not suitable for the growth of microbes
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1
1
12
3 a (i) Able to name the hormone secreted by gland P
P : ADH // FSH // LH Q : Thyroxine
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(ii) Able to state the condition caused by the growth of gland Q
Goiter 1(iii) Able to suggest how to overcome the problem in (a)(ii)
Taking enough iodine in our diet 1b Able to label adrenal gland with letter S correctly.
Answer
1
c Able to explain the role of gland R in regulating theperson blood glucose concentration from 0 minuteto 90 minutes
Sample Answer
P1 : From 0 to 60 minutes, the blood glucose level increases more than the normal level
P 2 : Islet cells in gland R is stimulated to secrete insulin
P3 : Insulin stimulates the conversion of excess glucose to glycogen (in the liver)
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4551/1 © 2011 Hak Cipta Jabatan Pelajaran Perak SULIT
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S
S
SOLAF 2 4551/2
P4 : This cause the glucose level to return to the normal level at the 90 th
minute1
Max 3
d (i) Able to state the person’s blood osmotic pressure based on the situation given
The blood osmotic pressure increases 1
(ii) Able to explain how gland Q involves in returning the osmotic pressure of the blood to normal levels.
P1 : The osmoreceptor detects the increase in the osmotic blood pressure
P2 : Gland Q is stimulated to release more ADH
P3: ADH is transported by blood to the kidneys
P4 : ADH increases the permeability of the wall of distal convoluted tubule and collecting ducts
P5 : More water is reabsorbed from the filtrate into the blood
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1
1
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Max 3
12
4 a Able to name substance R
R : ammonium compounds 1
b (i) Able to name process P
Nitrogen fixation
(ii) Able to explain the process P in leguminous plants
P1 nitrogen fixing bacteria lives in the root nodules of leguminous plants.
P2 i.e Rhizobium species
P3 convert nitrogen in atmosphere into nitrates/ ammonium compound
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1
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SOLAF 2 4551/2
c Able to explain the effect when there is no denitrifying bacteria
P1 Process Q / denitrification did not take place
P2 no nitrate will be broken down into nitrogen and oxygen
P3 causes imbalance of gases in atmosphere // atmospheric nitrogen decreses
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1
1
d Able to explain the effect of using excessive fertilizers in agriculture
P1 run off of excess fertilizers will increase the nutrient in aquatic ecosystem
P2 causing excessive growth of algae // algae bloom
P3 algae covers the water surface and block sunlight
P4 aquatic plant could not carry out photosynthesis and die
P5 algae die, decomposing bacteria uses oxygen for its activity
P6 reduce the oxygen content in water // increase in BOD
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1
1
1
1
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Max 4
12
5 a (i) Able to name the type of twins
P : Siamese twins
Q : identical twin
(ii) Able to explain the formation of twin Q
P1 An ovum is fertilized an a sperm to produce zygote
P2 the zygote splits into two halves by mitosis
P3 each halves develop into foetus
P4 the two foetus sharing the same placenta
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2
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1Max 3
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SOLAF 2 4551/2
b P1 Both twin in P and Q have similar genetic content
P2 the twins in P and Q have the same physical characteristics
P3 twin in P and Q have similar sex
P4 twins in Q are completely separated and grow as different individual twin in P are not completely separated, but joined at certain parts of the
bodies// sharing organs
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1
1
1
Max 3
c Able to explain the differences between R and S
P1 R is formed from fertilization of an ovum by a sperm S is form from two different sperm that fertilize two different ovum
P2 in R, one zygote is formed and split into two by mitosis In S, two different zygote are formed
P3 R is identical twin S is fraternal twin
P4 Twin in R share the similar genetic content // have identical characteristics
Twin in S have different genetic content // have different characteristics
P5 Twin R have same sex Twin S may have same sex / different sex
1
1
1
1
1
Max 4
12
6 a(i)
F1P1
F2P2
F3P3
Able to describe the adaptive characteristics in leaves tissue to carry out photosynthesis
Have a layer of epidermispenetration of sunlight for photosynthesis
pallisade mesophyll cells are arranged upright and packedcarry out photosynthesis at maximum rate
spongy mesophyll are loosely arranged create air spaces for gases exchange
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SOLAF 2 4551/2
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a(ii) P1P2P3P4
P5P6P7
P8P9
P10
P11P12P13P14
At 12.00 night the concentration of CO2 is highno uptake of CO2 for photosynthesisthe light intensity is lowstomata closed
at 6.00am the concentration of CO2 started to decreaseincrease in light intensity cause stomata starts to openCO2 is absorbed by the leaves for photosynthesis
at 12.00 noon, the concentration of CO2 is the lowestleaf receives maximum sunlight to carry out photosynthesis // rate of photosynthesis is maximumstomata open widely
at 6.00pm, the light intensity started to decreaseCO2 concentration starts to increaseless intake of CO2 for photosynthesis small opening of stomata
1111
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1111Max 10
bS1S2
D1
D2
D3
D4
D5
Similarity both digest celluloseboth have microbes/ bacteria/ protozoa in alimentary canal to produce cellulase
Animal P Animal Q
The stomach consists of four chambers
The stomach consists of one chamber
Size of caecum is small Size of caecum is big and long
bacteria and protozoa are found in rumen and reticulum
bacteria and protozoa are found in caecum
cellulose is digested in rumen and reticulum
cellulose is digested in caecum
food passes the alimentary canal once
food passes the alimentary canal twice
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1/0
1/0
1/0
1/0
1/0
2s+4dMax 6
20
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SOLAF 2 4551/2
7 a(i) F1P1
F2P2
F3P3
F4P4
Small and large number of alveoliincrease the surface area for gases exchange
Covered with large network of capillariesease the gases exchange and transport of gases to body tissues
moist surfacegases can easily dissolved and diffuse into/ out blood
alveolus has thin wall ie one cell thickenhance the gases exchange
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3F+3P6
a(ii) P1P2
P3
P4P5
P6
oxygenated blood flows through blood capillarieshave high partial pressure of oxygen compared to body tissuesoxygen diffuses into the body tissues down the concentration gradientthe cells carry out cellular respiration to produce energythis will increase the partial pressure of carbon dioxide in body tissues compared to blood.carbon dioxide diffuses out from body tissues into blood capillary to be sent to lungs
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6
b P1
P2P3P4P5P6P7P8
the muscle respired by anaerobic respiration during 100 meters eventBecause oxygen supply cannot support the oxygen needs / not enough / no oxygenso the glucose is not oxidised completely and will form lactic acidsmuscle is said to experience oxygen deptwill inhale air and exhale air very fast// breathing rate is faster oxygen is send more to the cells.the lactic acid is oxidised to carbon dioxide, water and energy
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8 a P1P2
P3
P4
Pioneer species for disused pond is submerged plant activities of pioneer plants cause a change in habitat, make it suitable for another speciesThe dead pioneer plant will decay and deposits in the bed of pondincrease the nutrient content in pondpond becomes shallower
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SOLAF 2 4551/2
P5P6
P7
P8P9P10P11
P12
this condition is suitable for the growth of floating plants which replace the pioneer speciesfloating plants cover the surface of pond, preventing the light from penetrating the waterthe rate of photosynthesis decreasessubmerged plants die and add humus to the bedfloating plants is replaced by amphibian plantpond becomes drier and suitable for the growth of herbaceous plantssoil becomes drier and more fertile and more suitable for teresterial plants soil becomes drier and more fertile and more suitable for teresterial plants
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1
1111
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Max 10b P1
P2
P3
P4
P5P6P7P8
P9
P10P11
Preservation refers to the management of ecosystems and the environment to ensure a healthy and balanced natural environment // take care of the environmentConservation is the wise use of natural resources for environmental protection // very careful use / control use of natural resources (This includes) the protection, management and renewal of natural resources.Need to ensure that the flora and the fauna are not extinct / lost foreverFuture generation can learn about the natural ecosystem.To maintain the quality of the environmentdo not disturb the energy flow / food web / food chainto keep the plant / animal which have high medicinal value // source of food help to maintain the natural cycles ( eg. Water cycle, nitrogen cycle, carbon cycle)provides natural catchment areaspreserves natural ecotourism / recreational activitiesprovides natural catchment areaspreserves natural ecotourism / recreational activities
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1111
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11Max 10
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9 a P1
P2P3P4
P5
haemophilia is caused by gene mutation for synthesis of blood clotting factor has undergone mutationthe no clotting factor for the patient.It is caused by a recessive allele which is linked to X chromosomemore common in male because the trait will express itself even if there is only one recessive allele present on his X chromosome.for female, two recessive alleles have to be present on her X chromosomes in order for her to become haemophiliac.
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SOLAF 2 4551/2
P6
P7
P8
P9
P10
P11
P12 Ali has 25% probability to have haemophilia.
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1
1
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1Max 8
b F1F2
S1
D1
D2
D3
D4
D5
Blood group – discontinous variationheight – continous variation
similarityS1 - both creates varieties among organism
differences
Blood group Height
D1 - Graph shows discrete distribution
D1- Graph shows normal distribution
D2- Influenced by genetic factor D2- Influenced by genetic and environmental factor
D3 - Traits are controlled by a single gene
D3 -Traits are controlled by more than one gene
D4 - Characters cannot be measured and graded // qualitative
D4 - Characters can be measured and graded // quantitative
D5 – the difference between individual is distinct
D5 – the difference between individual not distinct
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1/0
1/0
1/0
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1/0
1S+3D
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XHY XHXh
XH Y XH Xh
XHYXHXhXHXH XhY
x
Normalfemale
Carrierfemale
Normalmale
Haemophiliac male
P6 - Parent
P7 - Meiosis
P8 - Gamete
P9 - Fertilisation
P10 - F1
generationP11 -
Phenotype
SOLAF 2 4551/2
c P1
P2P3
P4P5
P6P7
P8
Gene mutation takes place when there is a change in the sequence of nucleotide basesBase deletion removal of a base from a normal gene sequence
Or
Base insertioninsertion of an extra base into a normal sequence
Or
Base substitution replacement of one base with another.
Or
Gene mutation results in a defective protein being produced or no protein is produced at all
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