TRIAL PERAK 2009| No | Explanation | Marks | Remarks |
| 2(a)(i) | S : Ribosomes T : Golgi apparatus / Golgi body | 1 1 | Total = 2 |
| (ii) | Function of T : Processes, modifies and packaging the enzyme | 1 | Total = 1 |
| (b)(i) | DNA / In the forms of genes. | 1 | Total = 1 |
| (ii) | P1 – DNA replicates/ opens/ unwinds for the synthesis of RNA/ messenger-RNA P2 – (Genetic information) in DNA is copied to RNA P3 – RNA moves out of the nucleus and attaches itself on the ribosomes. | 1 1 1 | Any 2 P Total = 2 |
| (c)(i) | F : Suddent and random change in nucleotide sequence of DNA/ genetic composition of a cell / particular gene // Permanent changes in DNA structure E1 : can occur in cells E2 : can be inherited causing abnormal development in offspring. | 1 1 1 | F and any 1E Total = 2 |
| (ii) | Down’s Syndrome / Klinefelter’s Syndrome / Turner’s Syndrome / Trisomy X | 1 | Total = 1 |
| 2(d)(i) | Sickle cell anaemia | 1 | Total = 1 |
| (ii) | F : Red blood cells contains less haemoglobin E1 : Less oxygen will combined with haemoglobin E2 : Less oxygen to be transported to the body cell / can be used in cellular respiration | 1 1 1 | F and any 1E Total = 2 |
| | TOTAL | 12 MARKS | |
| No | Explanation | Marks | Remarks |
| 5(a)(i) | Pepsin / Rennin | 1 | Total = 1 |
| (ii) | Pepsin -  Protein Polypeptides / peptone OR Rennin -  Caseinogen(Soluble) Casein(insoluble) | 2 2 | (a)(i) and (a)(ii) are dependent Total = 2 |
| (b)(i) | Amino acids. | 1 | Total = 1 |
| (ii) | P1 – Build news cells / growth P2 – Repair or renew damaged cell P3 – Producing enzyme / antibody / hormone | 1 1 1 | Any 1 P Total = 1 |
| (c)(i) | Rice / any source of carbohydrate | 1 | Total = 1 |
| (ii) | Amylase / pancreatic amylse | 1 | Total = 1 |
| (d) | Aids in peristalsis // avoid constipation | 1 | Total = 1 |
| 5(e) | F : Q / pancrease secretes insulin E1 : Insulin stimulate the conversion of glucose into glycogen / the rate of respiration is increased E2 : Reduce the blood glucose level into normal | 1 1 1 | Total = 3 |
| | TOTAL | 11 MARKS | |
TRIAL TERENGGANU 2010| No | Explanation | Marks | Remarks | |||
| 3(a) |
 *Correct label & name of cell | 1 | Total = 1 | |||
| (b) | P1 – Palisade mesophyll (cells) contain a lot of chloroplasts to absorb maximum sunlight. P2 – Palisade mesophyll (cells) are closely arranged to absorb maximum sunlight. P2 – Spongy mesophyll (cells) are loosely arranged for gaseous/oxygen and carbon dioxide exchange P3 – Many stomata located at lower epidermis to allow the exchange of gaseous. **INTERNAL FEATURES answers only accepted, because question : “ Based on the diagram” | 1 1 1 1 | Any 2 P Total = 2 | |||
| (c)(i) | Grana / Granum | 1 | Total = 1 | |||
| (ii) | P1 – Chlorophyll absorbed light energy P2 – Energy is used to split water molecules into hydrogen ion and hydroxyl ion. P3 – Electron is received by hydrogen ion to form a hydrogen atom. P4 – Hydroxyl ion loses electron to form hydroxyl group P5 – Produce water and oxygen | 1 1 1 1 1 | Any 3 P Total = 3 | |||
| (iii) | D1 – Light reaction occur in grana BUT dark reaction occur in stroma. D2 – Light reaction require light energy BUT dark reaction does not require light energy Accept any suitable answer | 1 1 | Total = 2 | |||
| (d) | P1 – Factors affecting the rate of photosynthesis can be controlled at high level all the time P2 – Artificial lighting can be used when natural light intensity is low P3 – Concentration of carbon dioxide can be increased by burning paraffin P4 – Temperature can be controlled by using temperature controller | 1 1 1 1 | Any 3 P Total = 3 | |||
| | TOTAL | 12 | ||||
NO 5 - EARLOBE
a) (i) ee
(ii) - The attached earlobe characteristic is controlled by the recessive allele.
- An individual who has attached earlobes must be a recessive homozygote to
show/exhibit the attached earlobe characteristic, that is ee.
- If the genotype is Ee or EE, free earlobe will be expressed.
b) (i) Ee
(ii) - Because the genotype of V’s husband(U) is ee, which means he has attached
earlobe.
- and from the pedigree, two of V and U’s children have the genotype Ee and
another one has the genotype ee.
- If the genotype of V is EE, none of their children will have the genotype ee.
**actually this question should need 3 marks
c)
 SpermOvum | e | e |
| E | Ee | Ee |
| e | ee | ee |
d) (i) EE
(ii) - The genotype of T is ee because she has attached earlobe.
- If she wants all of her children to have free earlobe, her future husband MUST have a
genotype that is homozygote dominant; EE.
- lf the genotype of her husband is Ee, there is a possibility that her child might have
attached earlobe.
- If the genotype of her husband is ee, then all her children will have attached earlobe.
NO 5 – SEX OF OFFSPRING
(a) i) 50%
ii) - according to the type of sperm involved in fertilisation.
- if the sperm which carries the Y chromosome fertilises the ovum, the baby will be a boy.
- if the sperm which carries the X chromosome fertilises the ovum, the baby will be a girl.
(b) P = 44 + XHXH
Q = 44 + XHY
R = 44 + XHXh
S = 44 + XhY
(c) i) Offspring S
ii) - Haemophilia is a genetic disease caused by sex-linked recessive allele on X chromosome.
- A female has two X chromosomes, so she will only display / express haemophilia when
both of the alleles linked to the X chromosomes are recessive alleles.
- A male has only one X chromosome and one Y chromosome. The Y chromosome does
not carry haemophiliac gene. Hence, he will be a haemophiliac even if one recessive
allele linked to his chromosome.
(d) i) Alleged father N
ii) - A few bands in the DNA fingerprinting of the child are the same as those in the DNA
fingerprint of the alleged father N while…
- the other bands are similar to those of the mother.
- Hence, the alleged father N is the biological father of the child.
OBJECTIVE QUESTION:
1. B
2. C
3. C
4. D *clue=phenotype ratio
5. A
G I R L s……………. Pray to Allah, may success be yours. ALL THE BEST! …me : Teacher NH
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