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.