Skema jawapan biologi percubaan spm 2008 1 perlis/skema...Skema jawapan biologi percubaan spm 2008 1...

Skema jawapan biologi percubaan spm 2008 1

Structured question: No. 1 Item No. Suggested Answer: Mark

1(a) P: Cell wall

Q: Plasma membrane R: Vacuole S: Nucleus

Award 1 mark for two correct answers: maximum:

2

(b) (i)

(ii)

(iii)

correct label of chloroplast (T) Palisade mesophyll Chloroplast contains chlorophyll for absorption of light energy for photosynthesis

1

1

2

(c)(i)

(ii)

-Smooth diagram -correct labels

Root hair cell does not have chloroplast while the cell in Diagram 1 has chloroplast Root hair cell has a protruding structure but the cell in Diagram 1 does not have a protruding structure

1

1

1

1

(d) Water is absorbed by the root hair through osmosis Cell sap of the root hair has a higher solute concentration than the soil water.

TOTAL

1

1

12

Peperiksaan Percubaan Biologi SPM 2008 Panitia Biologi Negeri Perlis


No. 2 Item No. Suggested Answer: Mark

2 (a)(i)

(a)(ii)

(a)(iii)

X : Aerobic respiration Y : Anaerobic respiration in plants Aerobic respiration is the oxidation of glucose to release energy in the presence of oxygen Anaerobic respiration is the process whereby glucose is broken down to release energy in the absence of ( without using ) oxygen.

1 1

1

1

(b)

Plants – Alcohol (ethanol), carbon dioxide.

Animals – Lactic acid, energy

1 1

(c)

Glucose is broken down completely in aerobic respiration.

In anaerobic respiration glucose is not broken down completely but partially to give lactic acid or ethanol and carbon dioxide.

Some of the energy is still stored in the lactic acid molecule and ethanol molecule.

[Any two correct answers, award 2 marks ]

1

1

1

(d)

Glucose zymase ethanol + carbon dioxide + energy (210 kJ)

1

(e)

Muscle is in the state of oxygen deficiency Blood cannot supply oxygen fast enough to meet the demand of ATP// muscles are in the state of oxygen deficiency

Muscles obtained extra energy from anaerobic respiration

During anaerobic respiration/ O2 debt, lactic is released and cause fatigue.

When this occurs the oxygen debt is said to have been paid.

[Any three correct answers, award 3 marks ]

1

1

1

1

1

3 max



No. 3 Item No. Suggested answer Mark

3 (a) The capture and recapture technique. 1

(b)(i)

(ii)

Mark the specimens using a non-toxic permanent ink marker.

The mark must not be lost and must not inhibit normal body activities.

The mark does not prevent the rat from randomly mixing with the other

unmarked rats.

1

1

1

(c)

PS =(no. of individuals in 1st sample x no. of individuals in 2nd sample)

(no. of marked individuals recaptured)

Population size (PS) = (100 x 140) / 40

= 350 rats

1

1 (d)

To give sufficient time for the random dispersal and mixing among the

rats in the population.

1 (e) Changes in the size of population after three months can be caused

by:

•increase in number of the rats due to increase in birth rate.

•decrease in number of the rats due to death of old rats, diseases or

eaten by predators.

•migration (immigration or emigration) of the rats.

1

1

1 2 max

(f) •The nitrate fertilizer in the river water is absorbed by the algal cells.

•Eutrophication occur

•The algae grow and reproduce rapidly that they completely cover the

water.

•They block out the light for plants growing beneath them, which

causes death.

•Decomposing bacteria acting on the dead plants and algae compete

for the oxygen in the water.

•As a result, fish and other organisms in the river die due to the lack of

oxygen.

[Any three correct answers, award 3 marks ]

1

1

1

1

1

1

3 max



No. 4

Peperiksaan Percubaan Biologi SPM 2008

Item No. Suggested answers Marks

4(a)

Correct label of axes and

unit

Points transfer (all

correct)

Smooth curve

1

1

1

(b)(i)

(ii)

The blood glucose level rises sharply between 0700 and 0800 hours.

The blood glucose level decreases gradually between 0800 and 1200

hours.

1

1

(c) • The blood glucose rises sharply because the carbohydrates in the

meal are digested.

• High glucose level in blood stimulates the secretion of insulin.

• Insulin promotes the conversion of glucose to glycogen in the liver

(and the uptake of glucose for cellular respiration).

Thus the blood glucose level decreases gradually between 0800 and 1300

hours.

1

1

1

1

3 max

(d) • The blood glucose level drops during the period between 1800 to 1900

hours because the last meal was at 1600 hour.

• Between 1900 hours to 2200 hours the blood glucose level maintains

at about 4.6 mmoldm-3.

• This occurs because of the secretion of glucagon from the pancreas.

• Glucagon promotes the conversion of glycogen to glucose.

• As a result the blood glucose level rises again at 2200 hours.

1

1

1

1

1

3 max

(e)

• Blood glucose level needs to be maintained within the normal range to

provide a constant environment for the optimal functioning of cells.

1

Panitia Biologi Negeri Perlis


No. 5 Item No. Suggested answers Marks

5

(a)(i)

1

(a)(ii) Coloured and smooth seed Biji berwarna dan licin

1

(b)(i)

(ii)

Crossing over/ Pindah silang

Prophase I meiosis/ Profasa I Meiosis

1

1

(c)

1

1




(d) Second Probability/

Kemungkinan kedua. Cross over occurs that produced more gametes that carry different genetics characters. Berlaku proses pindah silang yang akan menghasilkan lebih banyak gamet yang membawa ciri genetik yang berbeza. The fusion of these gametes will produce variety of individuals. Percantuman gamet-gamet ini akan menghasilkan individu yang bervariasi.

1

1

1 2 max

1

(e)

To produce haploid gamete. Untuk menghasilkan gamet yang haploid. The fusion male and female haploid gametes will produce diploid zygote which has the same number of chromosomes with the parent’s. Therefore, the number of chromosomes in the same species of organism remains the same in each generation.

Percantuman antara gamet jantan yang haploid dan gamet betina yang

haploid akan menghasilkan zigot yang diploid, yang sama bilangan kromosom induk. Oleh itu bilangan kromosom dalam spesis organisma yang sama dapat dikekalkan dalam setiap generasi.

1

(f)

OR any other possible answers

1

1

1



No. 6 Item No. Suggested Answer Mark

6(a)(i)

Able to list the general characteristic of enzymes Sample answers: P1 – Enzymes are proteins which are synthesized by living Organisms P2 – Enzymes bind to their substrates and convert them into product in the enzymatic reaction P3 – Enzymes have specific site called active sites to bind to specific substrates // enzymes are highly specific in their reaction. P4 – Enzymes speed up the rate of biochemical reaction but remain unchanged // are not destroyed at the end of the reaction. P5 – Enzymes are needed in small quantities because they are not used up P6 - Most enzyme- catalysed reaction are reversible

1

1

1

1

1

4 max 6(a)(ii)

Able to discuss the uses of enzymes in industrial processes and our daily life, using suitable examples Sample answers: Type of industry/ Application (A)

Enzyme used (E) Uses (U)

Rennin solidifies milk Lipase Ripening of the cheese

1. Food processing industry (a) dairy product

Lactase Hydrolysed lactose to glucose in the making of ice cream

Amylase Convert starch flour into sugar in making bread

(b) bakery products

Protease Convert protein in the making of biscuits

Amylase Convert malt into glucose for the fermentation of yeast

(c) Alcoholic drinks(beer/wine industry) Zymase Convert sugar into ethanol

during fermentation of yeast (d) Fish products Protease Removes the skin of fish (e) Meat products

Protease Tenderises meat

(f) cereal grain products

Cellulose Break down cellulose and removes seed coats from cereal grain

1 1

1

1

1

1

1

1



(g) Seaweed products

Cellulose Digest cell wall and extract agar from seaweed

Amylase Change starch into sugar in making syrups

(h) Starch products

Glucose isomerase

Convert glucose into fructose/production of high fructose syrup

2 Leather products

Trypsin/protease Removal of hair from animal hides

3. Medical products

Pancreatic trypsin

Treats inflammation

4. Biological washing powder/detergent

Protease and amylase

Dissolve protein and starch stains in clothes

1

1

1

1

1

6 max


6(b) Able to explain how extra cellular enzyme is produced by emphasizing

on the role of P,Q, R and S.

1

P1 – Q is nucleus , contain DNA which store genetic information for enzyme synthesis in chromosome.

1 P2 – The genetic information is transferred to ribosome by RNA P3 – R is mitochondrion, carry out cellular respiration and produces

energy which is needed in enzyme synthesis. P4 – Protein that are synthesized at ribosome are transported in the space within the rough endoplasmic reticulum P4 – Protein depart from the rough endoplasmic reticulum wrapped in vesicles that bud off from the membrane of rough endoplasmic reticulum P5 – The transport vesicles then fuse with the membrane of the golgi apparatus/ P . P6 – The proteins are further modified during their transport in the golgi apparatus/ P. P7 – For example carbohydrates are added to proteins to make

glycoprotein P8 – Secretory vesicles, S containing these modified proteins bud off from the golgi apparatus and travel to the plasma membrane P9 – These vesicles fuse with the plasma membrane before releasing the proteins as enzyme outside the cell Grant marks: Student mention the role of P,Q ,R and S before or after the explanation.

1

1

1

1

1

1

1

1

1



No. 7

Item No. Suggested answer

Mark

7(a)(i) • The type of animal is ruminant • Examples – cow and buffalo/ goats (any two correct answers)

1

1


(a)(ii) The differences between ruminant digestive system and human digestive system are: • A ruminant digestive system has four-chamber stomach namely

rumen, reticulum, omasum and abomasum whereas a human digestive system has only one stomach.

• A ruminant is aided by bacteria and protozoa (that produce cellulose enzyme) to digest cellulose (whereas) human digestive system cannot digest cellulose/ is not aided in digestion of cellulose, (thus cellulose is egested without digested).

• Cellulose is digested in ruminant but cellulose is not digested in human.

• A ruminant can vomit out the content of its rumen to regurgitate and rechew food (rumination) but in human the food is chewed only once/ passes through the digestive tract only once.

1

1

1

1 2 max

7(b)(i) Functions of the liver/ T

• The liver secretes bile which is stored in the gall bladder and released

into the duodenum when required. • The liver regulates the blood sugar level by converting excess glucose

to glycogen and storing it.

1

1 • The liver removes excess amino acids by breaking them down first into

ammonia and then into urea which is excreted; and this process is known as deamination.

1 1 • The liver stores and metabolises fats.

• The liver synthesises fibrinogen and prothrombin which are blood-clotting substances and heparin which is an anticoagulant.

• The liver detoxifies many harmful chemicals.

1 1 • The liver breaks down worn out red blood cells and removes

circulating hormones from the blood when they are no longer required and breaks them down.

1 • The liver synthesises vitamin A and stores it together with vitamins D,

K and B 12 1 8 max



F1- T secretes bile that will emulsify lipids. 1 7(b)(ii) P1- bile is alkaline salt that will neutralise acidic chyme from stomach //

optimise the pH for enzyme action in duodenum. 1

P2- bile emulsify lipids by breaking them down into tiny droplet // provide

greater surface area. 1

F2- Pancreas secretes pancreatic juice which contains pancreatic amylase,

trypsin and lipase. 1

P1- Pancreatic amylase completes the digestion of starch to maltose 1 starch + water maltose


P2- Trypsin digests polypeptide into shorter chain of peptide polypeptide + water peptides P3- Lipase completes the digestion of lipids into fatty acid and glycerol lipid droplets + water glycerol + fatty acids

1 1 1 1 1

8 max

pancreatic amylase

trypsin

lipase



No. 8 Item No. Suggested answer Mark

8(a)(i) Salt 1

(a)(ii) • a large intake of salty food causes the osmotic pressure of the

blood to increase. 1 1 • the person becomes thirsty. • sensory cells in the hypothalamus / osmoreceptor cells detect

changes in the osmotic pressure of the blood. 1 1 • this triggers the pituitary gland to secrete ADH. • ADH carried by the blood to the distal convoluted tubule and

the collecting ducts 1 • ADH increases the permeability of the walls of these ducts to

water. 1 • more water is reabsorbed, causing the volume of urine to

decrease. 1 • no aldosterone is secreted from the adrenal glands because

of the high salt content in the blood. 1 1 • less salt is reabsorbed in the kidneys.

• as a result, the urine produced is concentrated and the volume is low until osmotic pressure of the blood returns to normal

1

9 max


8(b) F1-The drop in enviromental temperature is detected by

thermoreceptor // cold receptor in the skin P1-Nerve impulse is sent to the hypothalamus. P2- Nerve impulses are transmitted along the afferent neurone to the

hypothalamus. F2 – Hypothalamus acts as the temperature regulatory centre P3 - Thermoreceptoris in hypothalamus detect the drop of temperature

of the blood flowing past it P4- Nerve impulses are sent to the efector through efferent neurone. F3- The effectors respond by physical means (involving muscle) P5- Voluntary muscle activity is increased such as rubbing the hands

to keep warm. P6- Involuntary muscles contract and relax frequently leading to

shivering to produce heat F4- by the process of metabolisme ( which involves the endocrine

glands) P7- Adrenal glands stimulate the secretion of adrenaline which causes

an increase in the metabolisme rate P8- More heat is produced

1 1 1 1 1 1 1 1 1 1 1 1

10 max



No. 9

Item No. Suggested answer Mark

9(a) Variation may occur through crossing over during meiosis I , independent assortement of chromosomes, random fertilisation and mutation

Crossing over -occurs during prophase I of meiosis 1 -non-sister chromatids of homologous chromosomes break at the chiasma

1

-segment of chromatids exchange places 1 -segment of the maternal chromatids become attached to the paternal chromatids

1

-New combination of genes are produced 1 3 max

Independent assortment of chromosomers - At mataphase I the homologous pairs of chromosomes are arranged on the metaphase plate at random

1

- Each homologous pair of chromosoms is positioned to the other pairs

1

- At the first meiotic division, there is an independent assortment of maternal and paternal chromosomes into daughter cells

1

- results in variety of gametes. 1 3 max

Fertilisation -Gametes with diverse combination of homologous chromosomes fused together to form a zygote

1

-zygotes formed have greater variety of gene combination 1 Mutation -change of particular genes/chromosomes in gamets 1 - mutation in gamete can be inherited causing abnormal development

in the offspring 1

Jumlah 10 max

9(b) P1 – penyakit hemofilia di kawal oleh gen resesif yang terdapat pada

kromosom seks // Xh 1 . P2 – kehadiran satu gen hemofilia pada anak lelaki menyebabkan ia

menjadi pengidap // XH 1 Y. Peperiksaan Percubaan Biologi SPM 2008 Panitia Biologi Negeri Perlis


P3 – kehadiran satu gen hemofilia pada perempuan menjadikannya

pembawa // XH


Xh

END OF ANSWER SCHEME END OF ANSWER SCHEME

Xh. P3 – kehadiran satu gen hemofilia pada perempuan menjadikannya pembawa // XH

P4 – kehadiran sepasang gen hemofilia / 2 gen pada kedua-dua kromosom seks menjadikannya pengidap // Xh Xh . P4 – kehadiran sepasang gen hemofilia / 2 gen pada kedua-dua kromosom seks menjadikannya pengidap // X X h Y XH XH X

P5 P5

Xh 1 .

h Xh . 1

h Y XH XH

1 P6 XH Xh XH Xh XH Y XH Y

1 P7 Female female male male

1 Carrier carrier normal normal Perempuan perempuan lelaki lelaki Pembawa pembawa normal normal XH Y Xh Xh

P8

1

1 P9 XH Xh XH Xh Xh Y Xh Y Female female male male Carrier carrier haemophiliac haemophiliac Perempuan perempuan lelaki lelaki

1 Pembawa pembawa hemofilia hemofilia P10 – Couple A is advised to proceed with marriage as their children

will be free from haemophiliac. P10 – Pasangan A dinasihatkan boleh meneruskan perkahwinan

kerana anak yang dihasilkan adalah normal dan bebas dari penyakit hemofilia.

1 P11 – Couple B is advised not to proceed with marriage as their

children will suffer from haemophilia. P11 – pasangan B dinasihatkan tidak meneruskan perkahwinan

kerana anak lelaki mendapat hemofilia. 1 10 max


Skema jawapan biologi percubaan spm 2008 1 perlis/skema...Skema jawapan biologi percubaan spm 2008 1...

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