Question 7:
Diagram 7.1 shows the parts of blood circulatory and lymphatic system in human.

Diagram 7.1 

(a) Explain how fluid Q is formed. [4 marks]

(b) Describe similarities and differences between fluids P and R. [10 marks]

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(c) Diagram 7.2 shows a type of worm which is transmitted by mosquitoes to human. The mosquito is the vector of disease X.

Diagram 7.2 

Discuss how disease X occurs.
Suggest ways to prevent the disease. [6 marks]

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Answer:
(a)

• Fluid Q is the interstitial fluid / tissue fluid.
• It is formed from blood plasma.
• When blood flows through the capillaries, the higher blood pressure at the arterial end of the blood capillary network causes blood plasma to diffuse out into the intercellular spaces.
• This fluid that fills the intercellular spaces is termed the interstitial fluid.

(b)
Similarities:
1. Both blood plasma and lymph are pale coloured fluids.
2. Both the fluids are aqueous solutions containing various types of solute molecules.
3. Like glucose, amino acids, lipids, respiratory gases and antibodies.
4. Both the fluids contain agranulocytes / lymphocytes.


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(c)

• Disease X is usually called elephantiasis / filariasis.
• The disease is caused by filarial worms which become lodged in lymph nodes and block the flow of lymph fluid.
• The microscopic worms are spread by mosquitoes / Cules Mosquitoes.
• The mosquitoes suck the microscopic worms when they feed on blood of an infected person.
• The worms are then transmitted into blood of another person when the mosquito bites to suck blood.
• The microscopic worms migrate into the lymph vessels and live in the lymph nodes. Here the worms grow, reach adulthood and breed.
• The worms block the flow of lymph fluid, causing a higher hydrostatic pressure to build up within the lymph vessels.
• Interstitial fluid accumulates in the intercellular spaces and cause swelling of the legs/ arms/ breast/ testicles.
• The skin in the swollen region becomes dry and scaly and pored/ tears. The condition causes severe pain.
• Ways to prevent the disease:

 1. Avoid being bitten by mosquitoes
 - Sleeping with mosquito netting
 - Wearing clothing to cover the body when in outdoors
 - Applying / using mosquito repellents.

2. Destroy breeding grounds of mosquitoes like preventing accumulation of water in rubbish bins and containers.

3. Kill adult mosquitoes by fumigation, mosquito larvae by spraying insecticide or petroleum oil on water surfaces of stagnant water.

Question 8:
(a) The wife of a newly married couple is diagnosed with an inheritance disease. If this couple had children, the children might have genetic disorder. The doctor advised her to use some birth control methods.

Table 8.1 shows the comparison of some birth control methods.

Table 8.1

(i) Based on Table 8.1, choose the best two effective methods to prevent pregnancy.
Explain your reasons.

(ii) Beside the contraceptive methods in Table 8.1, the consumption of contraceptive pills is also being practised.
Discuss the advantages and disadvantages of this contraceptive method.
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(b) A flower is the reproductive organ for a plant. Fertilization process occurs in the ovule to form a fruit. The fruit is the ripened ovary of a flower.
Diagram 8.2 shows the process of a flower which turn into a fruit.
Diagram 8.2

Based on Diagram 8.2 explain how a fruit is formed from a flower.

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Answer:
(a)(i)

• The two effective methods to prevent pregnancy are the use of intrauterine device (IUD) and tubal ligation.
• IUD – A loop made of plastic or copper is inserted into the uterus by a doctor / trained medical nurse.

- The ovum is prevented from being fertilized by preventing the sperm from reaching the ovum.
- It is highly effective as the chance of pregnancy is only 1%.

 

• Tubal ligation

- The Fallopian tube is clamped / cut and sealed in the surgical procedure
- The ovum is prevented from entering the Fallopian tube, hence there will be no fertilization
- It is very effective method as there is 0% pregnancy

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(a)(ii)

Advantages

• An easy method to prevent pregnancy without the use of physical devices like IUD, uterine diaphragm, condom
• Menstrual flow is lighter
• Menstruation becomes regular
• Avoids menstrual cramps

Disadvantage

• May use vaginal bleeding / spotting
• Missed menstrual periods
• Dizziness, headache
• Breast tenderness
• Enlargement of breasts
• Body weight gain
• Mood swings

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(b)

• During pollination, mature pollens land on stigma.
• Each pollen contains a generative nucleus and a tube nucleus.
• The stigma secretes sugary fluid which induces the germination of the pollen.
• Tube nucleus directs the formation of pollen tube.
• The tube nucleus and generative nucleus move into the pollen tube.
• The pollen tube grows through the style and reaches the micropyle of the ovule.
• The tube nucleus breaks down just before the pollen tube reaches the ovule.
• The generative nucleus divides once by mitosis to form two male nuclei.
• The pollen tube enters the ovule through the micropyle and its tio dissolves to release the two male nuclei.
• One of the male nuclei fuses with the two haploid polar nuclei in the embryo sac to form a triploid cell called the endosperm.
• The other male nucleus fuses with the egg cell nucleus to form the diploid zygote.
• This phenomenon is termed double fertilization which induces the following changes and developments within the flower, resulting in the formation of fruit and seed.

- The ovary of the flower develops into the fruit.
- The ovules form the seeds.
- The zygote in the seed develops into the embryo and endosperm develops into the nutritive endosperm tissue, surrounding the embryo.
- The sepals and style dry up and form scar tissues of the fruit.
- The petals dry and wither-off.
- The flower stalk becomes the fruit stalk.

Question 6:
(a) Diagram 6.1 shows a gaseous exchange in the alveolus.

Diagram 6.1

Explain how oxygen is transported to the body tissues for cellular respiration. [4 marks]

(b)

Based on the statement above, suggest and explain how to keep our lungs healthy. [6 marks]
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(c) Diagram 6.2.1 and 6.2.2 show the respiratory organ from two different organisms.


Explain similarities and differences in structures of the respiratory organ between the two organisms. [10 marks]

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Answer:
(a)
- Oxygen gas in the alveoli is at higher partial pressure / higher concentration than in blood.

- Oxygen molecules diffuse across epithelium of alveoli and blood capillaries into blood.

- Oxygen binds with haemoglobin in the red blood cells / erythrocytes to form oxyhaemoglobin.

- The oxygenated blood is then transported from the lungs to cells of body tissues.

- In the tissues, oxyhaemoglobin dissociates into oxygen and haemoglobin.

- Oxygen diffuses down a concentration gradient across the epithelia of blood capillaries into the cells of body tissues.

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(b)



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(c)


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Question 3:
Diagram 3 shows a human digestive system.

Diagram 3 

(a)(i) Name gland X. [1 mark]
(ii) Bread contains starch.
Explain the role of gland X in the digestion of bread in P. [3 marks]
(iii) Explain why the process of digestion of bread does not occur in Q. [3 marks]

(b)(i) Name organ R. [1 mark]
(ii) A person has a gallstone as in Diagram 3. Explain the effect of this condition to the digestion of lipid in S. [2 marks]

(c) Explain one difference in the digestion of protein that occurs in Q and S. [2 marks]

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Answer:
(a)(i)
Salivary gland

(a)(ii)
- Gland X produces saliva
- Saliva contains the salivary amylase enzyme.
- Salivary amylase hydrolyses / digests starch into maltose.

(a)(iii)
- Q (stomach) produces gastric juice.
- Gastric juice contains hydrochloric acid.
- Hydrochloric acid denatures / inhibits the action of salivary amylase. (Hence the digestion of starch in bread is stopped)

(b)(i)
Gall bladder

(b)(ii)
- Gall stone in the bile duct prevents the flow of bile from liver to duodenum.
- Bile emulsifies fat hence with the absence / lower amount of bile, the digestion of lipid is slowed down.

(c)
- In Q (stomach), protein is digested by pepsin enzyme in an acidic medium to form peptides.
- In S (duodenum), peptides are digested by trypsin enzyme in an alkaline medium to form dipeptides.

Question 4:
Diagram 4.1 shows the gland J and adrenal glands which secrete hormones that involve in regulating the content of water and salt in blood.

Diagram 4.1

(a)(i) Name the hormones L and M. [2 marks]
(ii)The water content in blood is lower than the normal range.
Explain how gland J regulates the water balance in the blood. [3 marks]

(b)

Explain the action of adrenal glands to overcome the situation above in regulating the body temperature. [3 marks]
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(c) Diagram 4.2 shows the process of haemodialysis.

Diagram 4.2

Explain what will happen if the semi-permeable tubing is straightened. [2 marks]

(d) The information below shows the symptoms of a patient with kidney failure.

• Swelling of leg
• Breathing difficulties due to extra fluid in lungs
• Blood in urine

Discuss the consequences of the kidney failure to the patient. [2 marks]

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Answer:
(a)(i)
L : Antidiuretic hormone
M : Adrenaline

(a)(ii)

• Gland J (pituitary gland) secretes excessive antidiuretic hormone (ADH).
• ADH increases the permeability of the distal convoluted tubules and collecting ducts to water.
• More water is reabsorbed from the glomerular filtrate into blood and the osmotic pressure of blood is returned to normal level.

(b)

• The adrenal glands secrete excess adrenaline.
• Adrenaline increases the rate of conversion of glycogen into glucose.
• Adrenaline also increases the heart rate, hence more blood sugar / glucose is provided to the muscles.
• Metabolic rate increases and more heat is produced, hence the body temperature raises.

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(c)

• The rate of blood dialysis will be lowered.
• Total surface area of the tubing decreases causing less blood is treated / purified per unit time.

(d)

• The swelling of legs is due to the accumulation of interstitial fluid in tissues of the legs as the fluid is not returned in blood stream.
• The accumulation of interstitial fluid in the lungs causes the rate of diffusion of respiratory gases across the alveoli to be reduced, resulting in breathing difficulties.

Since the nitrogenous waste in blood are not removed through urine, glomeruli and kidney tubules are damaged resulting in bleeding.

Question 1:
Diagram 1.1 shows the structure of a plant cell.

Diagram 1.1 

(a) Name the structures C and D. [2 marks]

(b)(i) State the component which forms structure C. [1 mark]
(b)(ii) Draw and label the basic unit structure of the main component which is named in 1(b)(i). [2 marks]

(c) The cell in Diagram 1.1 is immersed in distilled water for an hour.
(i) Explain what will happen to the structure B. [3 marks]
(ii) Explain why the cell does not burst. [2 marks]
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(d) Diagram 1.2 shows two examples of tissues in plants. 

Diagram 1.2 

Based on Diagram 1.2, state a characteristic of each tissue.
(i) Meristematic tissue: 
(ii) Collenchyma tissue: [2 marks]

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Answer:
(a)
C : Chromosome / Chromatin
D : Rough endoplasmic reticulum

(b)(i)
Deoxyribonucleic acid (DNA)

(b)(ii)


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(c)(i)
Structure B (vacuole) will swell or increase in size.
- The vacuole contains cell sap which is hypertonic (to distilled water).
- Water molecules diffuse into the cell by osmosis and fill the vacuole causing it to swell.

(c)(ii)
- Plant cells are surrounded by a strong and rigid cell wall.
- Cell wall is able to withstand the strong turgor / hydrostatic pressure that develops within the cell.

(d)(i)
- Small-sized cells
- Dense cytoplasm / lack of vacuoles
- Large nucleus / nucleus at various stage of division / mitosis
- Undifferentiated cell
(Any one of the characteristics)

(d)(ii)
- Cell wall is not uniformly thickened
- Corners of cell have thicker cell wall
(Any one of the characteristics)

Question 2:
Diagram 2.1 shows the foods which contain carbohydrates.

Diagram 2.1 

(a)(i) Based on diagram 2.1, name the type of carbohydrates in rice and honey. [2 marks]
(a)(ii) Explain why diabetic patients are advised not to consume excessive amount of rice in their daily diet. [3 marks]

(b) Diagram 2.2 shows the formation of molecule R in milk.

Diagram 2.2

Based on the diagram 2.2, describe:
(i) the formation of molecule R. [2 marks]
(ii) the breakdown of molecule R. [2 marks]

(c) When sucrose solution is heated with Benedict’s solution, the blue solution remains unchanged.
Explain why Benedict’s test gives negative result on sucrose. [3 marks]

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Answer:
(a)(i)
Rice : Starch
Honey : Fructose / glucose / sucrose

(a)(ii)

• When rice is completely digested, it produces glucose.
• Diabetic patients lack / produce lower amount of insulin hormone.
• They are unable to convert excessive glucose into glycogen, hence the higher concentration of glucose in blood.

(b)(i)

• Molecule R is disaccharide.
• It is formed when two molecules of monosaccharide are linked together by condensation reaction / with the loss of water molecule.

(b)(ii)

• Molecule R is hydrolysed / digested by the addition of a water molecule by an enzyme.
• To form its component monosaccharides / P and Q.

(c)

• Sucrose is non-reducing sugar.
• Sucrose unable to reduce the blue copper (II) ions / Cu²⁺ (aq) in Benedict’s solution.
• Red copper (I) ions / Cu⁺ (aq) are not produced / formed.