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Suggested answers to Self test

Chapter 2

Section A (1 mark each)

1.D 2.C 3.A 4.C 5.A

Q3 Only cells that contain chloroplasts can carry out photosynthesis.

Section B

6. A bacterial cell does not have a true nucleus,but a palisade mesophyll cell does. / The DNA of a

bacterial cell lies freely in the cytoplasm, but the DNA of a palisade mesophyll cell is enclosed in

the nucleus. (1)

The cell wall of a bacterial cell does not contain cellulose, but the cell wall of a palisade mesophyll

cell does. (1)

A bacterial cell does not have endoplasmic reticulum / organelles bounded by a double

membrane, but a palisade mesophyll does. (1)

Bacterial cells are prokaryotic cells while palisade mesophyll cells are plant cells.

7.

(a)

(i) X: cytoplasm (1)

Y: mitochondrion (1)

(ii) Cytoplasm holds the organelles / is a site for many chemical reactions. (1)

Mitochondrion is the main site where the energy-releasing stage of respiration takes

place. (1)

(b) Chloroplast (1)

palisade mesophyll cells / guard cells (or other cells that contain chloroplast) (1)

(c) The cell is a eukaryotic cell. (1)

It has a true nucleus. / Its nucleus is surrounded by a nuclear membrane. (1)

It contains mitochondria / chloroplasts. (1)

8.

(a)

(i) The light microscope makes use of light to form an image, (1)

while the electron microscope makes use of electron beams. (1)

(ii) Advantages:

An electron microscope can produce images with a higher magnification and resolution

than a light microscope. (1)

More details of the specimen can be seen. (1)

Disadvantages:

An electron microscope is very expensive / can only be used to study dead specimens /

produces black and white images, (any 2) @(1) x2

(b) The images produced by the transmission electron microscope are two-dimensional, (1)

while the images produced by the scanning electron microscope are three-dimensional. (1)


Chapter 3 Section A (1 mark each)

1.A 2.A 3.B 4.C 5.A

Q3 The temperature of the solutions and the surface area of the differentially permeable

membrane affect the rate of osmosis only.

Section B

6.

(a) The Davson-Danielli model suggests that protein molecules are aligned on each side of the

phospholipid bilayer. (1)

The fluid mosaic model suggests that protein molecules are interspersed among the

phospholipid bilayer. (1)

(b) This is because the Davson-Danielli model cannot explain some properties of the cell

membrane. (1)

(c) Scientific knowledge is tentative and subject to change. / Scientists build on the work of other

scientists. (1)

7.

(a) large (1)

(b) phagocytosis (1)

(c) energy (1)

(d) pit (1)

(e) pseudopodia (1)

(f) Enzymes (1)

8.

(a) To remove any sucrose or glucose solution on the surface of the tubing, (1)

so as to avoid contamination of the water in the beaker, (1)

which would affect the result of the experiment. (1)

(b) The water potential of sucrose solution is lower than that of distilled water. (1)

The dialysis tubing is differentially permeable, (1)

so water entered the dialysis tubing by osmosis. (1)

(c) Glucose molecules are small enough to diffuse out of the dialysis tubing, (1)

while sucrose molecules are too large to pass through the dialysis tubing. (1)

The concentration gradient in set-up Y is lower than in set-up X, (1)

so less water enters set-up Y by osmosis. (1)



1.B 2.C 3.B 4.A 5.D

Q2 • 37 °C is the temperature of the human body. This is not a universal optimum

temperature for all enzymes. Many enzymes in other organisms have an optimum

temperature that is quite different from that of the human body.

• Enzymes catalyze both catabolic and anabolic reactions.

Section B

6.

(a) Correct title: The effect of temperature on enzyme activity (1)

Choice of axes (1)

With labels and units (1)

Correct plotting and joining of line (1)

Temperature (°C) 5 10 15 20 25 30 35 40 45 50 60

Rate of production (mg dm-3 hr-1) 14 19 24 31 40 51 68 93 98 89 33

(b) At 5 °C to 40 °C, the enzyme activity increases. (1)

At 40 °C to 45 °C, the increase in the enzyme activity slows down. (1)

At 45 °C to 50 °C, the enzyme activity gradually decreases. (1)

At 50 °C to 60 °C, the enzyme activity decreases sharply. (1)

45 °C is the optimum temperature for the enzyme. (1)


(c)

(i) As the temperature rises, enzyme and substrate molecules have more kinetic energy. (1)

The molecules move around more rapidly and collide with each other more frequently.

(1)

This increases the chance of forming enzyme-substrate complexes. (1)

Thus the enzyme activity increases.

(ii) High temperatures cause conformational change in the active site / change in the shape of

the active site of the enzyme. (1)

The substrate is then unable to fit into the active site of the enzyme to form the

enzyme-substrate complex. (1)

Therefore, the enzyme activity decreases.

(d) pH (1)

As unsuitable pH may also cause conformational change in the active site of the enzyme. (1)

7.

(a) Lipase (1)

(b) Lipase catalyses the breakdown of lipid in an oil stain (1)

into soluble products. (1)

The soluble products can be removed by water. (1)



1.C 2.D 3.C 4.D 5.B 6.A

Q3 Beef is rich in iron and broccoli is rich in dietary fibre.

Q4 The protease in solution (3), which is a protein, gives a positive result

Section B

7.

(a) nitrogen (1)

(b) carboxyl (1)

(c) dipeptide (1)

(d) polypeptide (1)

(e) non-essential (1)

(f) essential (1)

8.

(a) (273 kJ/100 ml) x 250 ml x 4 (1)

= 2730 kJ (1)

(b) Transfer a drop of milk to a piece of filter paper. (1)

A translucent spot will remain after drying. (1)

Immerse the paper in an organic solvent and the spot will disappear. (1)

It is necessary to immerse the paper in an organic solvent, e.g. ether or acetone, to check

whether the formation of the translucent spot is caused by fats (lipids) or other substances.

9.

(a)

(i) There was a continuous increase in the total energy intake and the intake of proteins and

lipids. (1)

The intake of carbohydrates remained steady / about the same. (1)

(ii) Carbohydrates:

(330 x 17.1)/11300 x 100% = 50% (1)

Proteins:

(99 x 18.2)/11300 x 100% = 16% (1)

Lipids:

(99 x 38.9)/11300 x 100% = 34% (1)

(b)

(i) The average lipid intake is higher than the recommended intake. (1)

(ii) High lipid intake may lead to overweight / obesity. (1)

High lipid intake may increase the risk of high blood pressure / heart disease. (1)



1.C 2.C 3.B 4.A 5.C 6.D 7.C

Q5 There is a drop in pH value over time. If W is the answer, the pH value of the reaction

mixture will be very low right at the beginning. But the question gives no information about

the pH value at the beginning of the experiment. What we can notice is that acids must be

produced in the reaction to lower the pH. So fatty acids produced in the digestion of milk fat

are the cause of the drop in pH value.

Section B

8.

(a) X: capillary (1)

Y: lacteal (1)

(b) X: monosaccharides / amino acids / minerals / water-soluble vitamins (1)

Y: fatty acids / glycerol (1)

(c) The presence of microvilli increases the surface area for absorption. /

They have thin epithelium which is one-cell thick to allow rapid absorption due to the short

distance for diffusion of food molecules into the blood. /

The presence of the lacteal / capillaries allows absorbed food molecules to be carried away

rapidly. (any 2) @(1) x2 = (2)

9.

(a)

Correct connection (arrows) of the three blood vessels and correct positions of labels (1)x3

(b) Excess glucose is converted to glycogen and stored in the liver. (1)

(c)

(i) Glycogen stored in the liver is converted to glucose when needed. (1)

The glucose leaves the liver through the hepatic vein. (1)

(ii) Excess amino acids are deaminated in the liver. (1)

The amino groups are converted to urea, (1)

which leaves the liver through the hepatic vein. (1)

(iii) They are digested food products which are absorbed into the blood in the small intestine.

(1)

(d) Bile (1)

Vitamin A (1)

(or other correct answers)



1.C 2.A 3.A 4.B 5.C 6.C

Q3 Although exhaled air contains a higher proportion of carbon dioxide when compared with

that in inhaled air, it contains only 4% carbon dioxide. The amount of nitrogen in the exhaled

air is about 78%.

Section B

7.

(a) hairs (1)

(b) capillaries (1)

(c) trachea (1)

(d) mucus (1)

(e) cilia (1)

Distinguish between hairs and cilia. Hairs are structures on the skin that can be seen with the

naked eye. Cilia are found on the surface of the cells and can only be seen under a

microscope.

8.

(a) Gas X dissolves in the water film lining the air sac (1)

and then diffuses down the concentration gradient across the walls of the air sac and the

capillary. (1)

(b) The water film lining the air sac allows gas X to dissolve. /

The walls of the air sac and the capillary are thin / one-cell thick to reduce the diffusion

distance of gas X. /

The capillary is close to the air sac to reduce the diffusion distance of gas X. /

The biconcave disc shape of cell Y increases the surface area to volume ratio. This allows

gas X to diffuse into the cell more efficiently. /

The flow of blood removes gas X to maintain a steep concentration gradient of gas X

between the air sac and the blood for more efficient diffusion.

Any 3 @(1)+(1) x3 = (6)

9.

(a) The rib cage will move upwards and outwards. (1)

The diaphragm will become flattened. (1)

(b) The movement of the rib cage and the diaphragm causes the volume of the thoracic cavity

to increase. (1)

The volume of the lungs increases and the air pressure in the lungs decreases. (1)

The air pressure becomes lower than the atmospheric pressure. (1)

Air rushes into the lungs. (1)



1.B 2.A 3.B 4.D

Q2 Chamber X has a thicker wall. It is the left ventricle.

Section B

5. @(1) x5 = (5)

Letter Name of blood vessel

D Renal vein

B (Posterior) vena cava

C Hepatic vein

I Hepatic portal vein

F Pulmonary vein

6.

(a)

(i) Proteins are found in plasma but not in tissue fluid. (1)

This is because proteins are too large. (1)

They cannot pass through the capillary wall. (1)

(ii) Sodium ions are very small. (1)

They are being forced out of the capillary wall into the tissue fluid. (1)

(b) Tissue fluid provides a constant environment for body cells. /

Tissue fluid serves as an important link for the exchange of materials between capillaries and

body cells. (1)

7.

(a) The blood pressure is high (1)

due to the pumping action of the heart. (1)

The blood pressure changes periodically (1)

due to the contraction and relaxation of the heart. (1)

(b) The diameter of the capillaries is small. (1)

The blood has to overcome the great resistance of the blood vessel walls as it travels over a

long distance. (1)

(c) The blood stops flowing (1)

due to very low blood pressure. (1)

The skeletal muscles lying next to the veins squeeze the veins when they contract. This

provides a force for blood flow. (1)

The presence of valves prevents the backflow of blood in the veins. (1)



1.A 2.A 3.C 4.C 5.B 6.D

Q2 When there is no light, photosynthesis stops and only respiration occurs.

Q4 When light intensity is at 3 arbitrary unit, all carbon dioxide produced in respiration is

used by the plant in photosynthesis.

Section B

7.

(a) broad (1)

(b) water (1)

(c) spongy mesophyll (1)

(d) Stomata (1)

(e) guard cells (1)

8.

(a) X: palisade mesophyll (1)

Y: vascular bundle (1)

Z: spongy mesophyll (1)

(b) Leaves are broad and flat, (1)

which provides a large surface area for gas exchange. (1)

There are stomata on the epidermis. (1)

This allows gases to pass into and out of the leaves freely. (1)

9.

(a) In pot X, the average height of seedlings increases from day 0 to day 25. (1)

In pot Y, the average height of seedlings increases for the first 5 days. (1)

and remains unchanged from day 5 to day 25. (1)

(b) The seedlings may be short. (1)

They may have yellow leaves. (1)

(c) Nitrogen is needed for the synthesis of proteins. (1)

Amino acids that make up proteins contain nitrogen. (1)



1.A 2.C 3.B 4.C 5.D

Q2 Organic nutrients are made using radioactive carbon dioxide in leaf Y. Radioactive organic

nutrients are translocated to the other parts of the plant in the sieve tubes of the phloem.

However, as phloem is removed in the ringed portion, the radioactive organic nutrients

cannot reach leaf Z.

Section B

6.

Xylem vessels Sieve tubes

X �

X X

� �

X �

� X

@(1) x5 = (5)

7. Strip X consists mostly of wood. (1)

The cell walls of the cells in wood are thick and lignified. (1)

They provide strong mechanical strength to resist bending. (1)

Strip Y consists of thin-walled cells of the pith. (1)

The cells lose water by osmosis and become flaccid. (1)

The flaccid cells cannot provide support to resist bending. (1)

8.

(a)

(i) Root hair (1)

(ii) The presence of root hairs provides a large surface area for the absorption of water and

minerals. (1)

Long and fine root hairs can easily grow between soil particles / are in close contact with

soil particles. This helps absorb water and minerals around them. (1)

(b) The epidermis is not covered with cuticle. (1)

This allows water to pass through the epidermis into the root easily. (1)

OR

The root is highly branched. (1)

This provides a large surface area for the absorption of water. (1)

(c) Water is drawn up the xylem vessels by the transpiration pull from the roots to the leaves. (1)

Water in the cortex cells near the xylem vessels enters the xylem vessels. A water potential

gradient is set up across the cortex. (1)

Water moves inwards from cell to cell by osmosis, or along the cell walls across the cortex.

(1)

This lowers the water potentials of the root hair cells. Water in the soil enters the root hair

cells by osmosis. (1)



1.B 2.C 3.D 4. C 5.B 6.D

Q1 The relative time spent in a certain stage of the cell cycle is proportional to the number of

cells at that stage. Among the four stages of mitosis, the largest number of cells was at

prophase.

Section B

7.

Mitotic cell division Meiotic cell division

X �

� �

� X

X �

(1) x 4

8.

(a) (2) x 2

(b) It produces gametes with different genetic combinations (1)

and causes genetic variations among individuals of the same species. (1)

This increases the chance for the species to survive when environmental conditions change.

(1)

9.

(a) A cell cycle is the sequence of events that takes place in a body cell from one cell division to

the next. (1)

(b) X: interphase (1)

Y: mitosis (1)

Z: cytoplasmic division / cytokinesis (1)

(c)

(i) The amount of DNA in the cell doubles when DNA replicates at stage X. (1)

The amount of DNA in the cell is halved when chromatids separate and move into two

daughter cells. (1)

(ii) The two daughter cells formed (1)

are genetically identical to their parent cell. (1)



1.A 2.C 3.D 4.C 5.B

Q2 The wings catch the wind and carry the seeds away from the tree.

Section B

6.

(a) Bulb (1)

(b) Stem tuber (1)

(c) Rhizome (1)

(d) Corm (1)

7.

(a) The anthers hang outside the flower to release pollen grains into the air. (1)

The stigmas are feathery / hang outside the flower to catch pollen grains in the air. (1)

(b) The loss of pollen grains in the wind is great. (1)

Releasing a large number of pollen grains helps increase the chance that some pollen grains

will reach the stigmas / of successful pollination. (1)

(c) The offspring show greater genetic variations (1)

which may help the plant species to adapt / survive to changes in the environment. (1)

8.

(a) P: sepal (1)

It protects the inner parts of the flower when the flower is in the bud stage. (1)

(b)

(i) Q, S (1) x2

Meiotic cell division takes place in the pollen sacs inside the anther and the ovary to

form the male gametes and the female gametes respectively.

(ii) S (1)

(c) P will wither and fall off. (1)

Q will wither and fall off with the filament. (1)

R will wither and fall off. (1)

S will become the seed. (1)

T will grow into the fruit wall. (1)



1.C 2.D 3.C 4.C 5.A

Q1 Her last menstrual cycle was from 2nd to 29th July.

Q4 The amniotic fluid is secreted from the amnion and it contains cells shed from the

foetus.

Section B

6.

(a)

(i) V (1)

(ii) R (1)

(b) Secretions from structures Q, T and U activate and nourish sperms, (1)

provide a medium for sperms to swim in (1)

and neutralize the acidity of the female's vagina. (1)

7.

Letter

C (1)

A (1)

E (1)

P (1)

8.

(a) The presence of numerous embryonic villi on the placenta

greatly increases the surface area for the exchange of materials. /

The walls of the embryo's capillaries and the embryonic villi are very thin.

This provides a short distance for the diffusion of materials. /

There are a lot of blood vessels which can carry materials away rapidly.

Thus, a steep concentration gradient of materials between the embryo's blood and the

maternal blood is maintained.

(any 2) @(1) + (1) x2 = (4)

(b) To prevent the higher blood pressure of the mother from damaging the blood vessels of the

embryo. (1)

To avoid problems brought by any blood group incompatibility between the embryo and the

mother. (1)

If the blood group of the mother is not compatible with that of the embryo, clumping

and breakdown of red blood cells may occur in the embryo. The embryo may die.

(c) Urea and carbon dioxide (1) x2

(d) The cells of the placenta have a high rate of respiration (1)

to release a large amount of energy (1)

for the synthesis of hormones. (1)



1.A 2.C 3.B 4.D 5.D

Q2 Normally, light is not essential for seed germination.

Section B

6.

(a) mitotic (1)

(b) apical (1)

(c) length (1)

(d) lateral (1)

(e) thickness / diameter (1)

7.

(a)

(i) 1.6 mm (1)

(ii) 5.2 mm (1)

(b) Cell Y is in the region of elongation of the root. (1)

Cell Y takes in water and elongates. As a result, the distance between cells X and Y increases

greatly after 10 hours. (1)

(c) This is because new cells are continuously formed by mitotic cell division in the region of

cell division. (1)

(d) 10 mm from the tip of the root (1)

Mature xylem is found in the region of differentiation. The region of differentiation is

behind the region of elongation.

8.

(a) (35 - 10) cm2 / (30 - 20) day

=2.5 cm2 / day (1)

(b) Place the leaf on a piece of graph paper. (1)

Trace the outline of the leaf on the graph paper using a pencil. (1)

Count the total number of squares within the outline. (1)

Multiply the area of one square by the total number of squares. (1)

(c) The measurement of the leafs surface area is not a suitable method for measuring the overall

growth rate of the tree. (1)

This is because the tree grows mainly in height and thickness rather than in surface area of its

leaves. (1)

(d) This is because the increase in the height / length of the tree only occurs at the shoot tip (1)

and root tip. (1)



1.C 2.D 3.A 4.C 5.A 6.C

Q1 There are no photoreceptors on the blind spot. For a person with normal vision, although sharp

images can be formed on the blind spot, no nerve impulses can be generated to the brain and

hence images formed on it cannot be seen.

Q2 If the amount of light entering the eyes is not enough, the images cannot be seen clearly. If the

visual centre does not function properly, the brain cannot interpret the images correctly.

Q4 When the person took his ride in the lift, the pressure on both sides of the eardrum was unequal. The

eardrum bulged and therefore he had an uncomfortable feeling in it.

Q6 Glucose is not a plant hormone that affects the growth of oat coleoptiles.

Section B

7. the lens is too thick the eyeball is too long

Ray diagrams to show corrected path of light rays from a distant object

Correct diagram to show:

concave lens (1)

lens of the eye is too thick / eyeball is too long (1)

parallel light rays before passing through the concave lens (1)

diverging light rays after passing through the concave lens but before entering the eye (1)

converging light rays after entering the lens (1)

(all light rays must be drawn with arrows to indicate directions)

image formed on the retina (1)

Appropriate title to the diagram (1)

Keep in mind that light rays from a distant object should be parallel to each other and the

object should not be shown.

Remember to add arrow signs to the light rays.

Drawings should be clear and in proportion.

8.

(a) B: ear bones

(b) C: semicircular canals

(c) G: eardrum

(d) A: auditory canal

(e) E: cochlea

@ (1) x5


9.

(a)

Correct diagram to show:

the root elongates (1)

the root bends towards the side where the agar block is put (i.e. to the right) (1)

The bending of the root is a result of uneven growth of the two sides of the root. The

length of the root should be longer than that before the experiment

(b) Auxins produced at the shoot tip diffused to the agar block and then to the root. (1)

High concentration of auxins inhibited the growth of the root on the side where the agar block

was put. (1)

Therefore, the root bent towards the side where the agar block was put. (1)

(c) This enables the plant to grow deep into the soil. (1)

to get better anchorage. (1)



1.A 2.D 3.B 4.C 5.A 6.C

Q3 Movement of the diaphragm during breathing is under the control of medulla

oblongata and cerebrum. The medulla oblongata is responsible for involuntary

breathing and the cerebrum for voluntary breathing.

Q6 In voluntary actions, nerve impulses are initiated by the brain directly without involving

any stimuli.

Section B

7. D (1)

C (1)

A (1)

8.

(a) Synaptic cleft (1)

(b) Under stimulation by a nerve impulse, (1)

vesicles containing serotonin fuse with the membrane of the presynaptic neurone. (1)

Serotonin is released into gap X. (1)

It diffuses across gap X down a concentration gradient. (1)

Serotonin is then taken up by the serotonin receptors on the postsynaptic neurone. (1)

It stimulates the generation of a nerve impulse in the postsynaptic neurone. (1)

(c) To generate energy for the synthesis of serotonin / formation of vesicles / movement of

vesicles / release of serotonin from the vesicle

(any 2) @(1) x2 = (2)

(d) Antidepressants work by increasing serotonin synthesis (1)

or inhibiting serotonin transporters on the presynaptic membrane. (1)

Thus more serotonin remains in gap X and binds with the receptors on the postsynaptic

neurone. (1)

A low level of serotonin is caused by impaired synthesis or the reabsorption of

serotonin before it reaches the postsynaptic neurone.

9.

Diagram to show:

correct position of sensory neurone (1)

correct label and direction of transmission for sensory neurone (1)

correct position of motor neurone (1)

correct label and direction of transmission for motor neurone (1)

The knee jerk reflex does not involve any interneurone.


Chapter 17

Section A (1 mark each)

1.C 2.D 3.B 4.C 5.A

Q1 The cartilage does not contain blood vessels. The living cells in cartilage obtain

nutrients from the synovial fluid.

Q4 When the man lifts up the dumb-bell, muscle X contracts and muscle Y relaxes to allow

the forearm to move upwards.

Q5 Muscles 1 and 3 are extensors while muscles 2 and 4 are flexors. When the frog jumps,

muscles 1 and 3 contract to straighten the leg.

Section B

7.

(a)

(i) Ball-and-socket joint (1)

(ii) Shoulder joint between the upper arm bone and the pectoral girdle / hip joint between the

upper leg bone and the pelvic girdle (or other correct answers) (1)

(b) Lubricating fluid is absent in the model, while synovial fluid is present in the human body. (1)

Elastic string is absent in the model, while ligament is present in the human body. (1)

8.

(a) Appendicular skeleton (1)

(b) Structure X attaches the muscles of the forearm to the finger bones. (1)

It is inelastic. (1)

It transmits the pulling force generated by muscle contraction of the forearm to the finger

bones when the person is typing. (1)

This allows the muscles to control the movement of the fingers.

Do not confuse tendon with ligament. Structure X is a tendon. Tendons are inelastic

fibres that hold a bone and a muscle together. Ligaments are elastic fibres that hold

bones together at joints.

9.

(a) The axon of some motor neurones branches off to form a number of motor end plates at the

endings. (1)

The motor end plates forms synapses with the muscle fibres of skeletal muscle. (1)

These synapses are called neuromuscular junctions. (1)

(b) When a nerve impulse reaches the motor end plate, it stimulates the the motor end plate to

release neurotransmitter. (1)

The neurotransmitter diffuses across the neuromuscular junction and binds with the

neurotransmitter receptors on the muscle fibre. (1)

The neurotransmitter stimulates the membrane of the muscle fibre to generate electrical

impulse. (1)

The electrical impulse spreads along the muscle fibre and brings about muscle contraction.

(1)

(c) The neurotransmitter receptors in a myasthenic patient are blocked or damaged. (1)

Neurotransmitter released from the motor end plate cannot bind with the receptors. (1)

No electrical impulses can be generated in the muscle fibre. (1)


The muscle fibre cannot contract as normal. (1)

(d) Drugs can be taken to remove the substances that block or damage the neurotransmitter

receptors. (1)



1.B 2.B 3.D 4.D 5.C 6.A

Q2 The pancreas detects and responds to changes in blood glucose level directly. The

conversion between glucose and glycogen takes place in the liver.

Q4 Sweating helps regulate body temperature which is a homeostatic response. Secretion

of saliva and dilation of pupil are reflex actions.

Q5 Insulin and glucagon are secreted by the pancreas all the time, but their amounts vary

with the blood glucose level.

Section B

7.

(a) In digestion, liver produces bile for emulsification of lipids into small droplets. (1)

In homeostasis, liver acts as a reserve of glycogen for the body in the regulation of blood

glucose level. (1)

(b) From the study of liver in digestion, Bernard recognized the role of liver in regulating blood

glucose level. (1) This led him to further investigate homeostasis. (1) /

Other scientists based on Bernard's work and further developed his idea on the maintenance

of the stability of the internal environment. (1)

8.

(a) Organ X: pancreas (1)

Messenger Y: insulin (1)

Organ Z: liver (1)

(b) A decrease in the blood glucose level initiates a response (1)

which causes an increase in the blood glucose level. (1)

In a negative feedback mechanism, the response acts against the change.

9.

(a) Vigorous exercise requires a large amount of energy for muscle contraction. (1)

Thus a large amount of glucose in blood was used up to release energy (1)

and the blood glucose level dropped rapidly. (1)

(b) (100 - 32) minutes (1)

=68 minutes (1)

Time when Mary finished the exercise: the 32nd minute Time when Mary's blood

glucose level returned to normal: the 100th minute

(c) The pancreas detected an increase in blood glucose level. (1)

More insulin was produced by the pancreas. (1)

Insulin stimulated liver cells to convert the excess glucose to glycogen, which was then stored

in liver and muscles. (1)

It also stimulated body cells to take in more glucose from the blood. (1)

As a result, the blood glucose level fell and returned to normal. (1)



1.C 2.A 3.B 4.B 5.C

Q1 The four trees belong to the kingdom of Plantae.

Q3 Owls have feathers (not hair) covering their bodies.

Q5 Jellyfish and fish belong to the kingdom of Animalia.

Section B

6. Organism P — Archaea (1)

Organism Q — Eukarya (1)

Organism R — Bacteria (1)

7.

Insect Order Key sequence for identincation

X R la → 2b → 3a

Y Q la → 2a

Z S la → 2b → 3b

@(1) x 6

8.

(a) Kingdom X — Plantae (1)

Group Y — Flowering plants (1)

Group Z — Dicots (1)

(b) They all have chlorophyll. (1)

Their cells have a cell wall composed of cellulose. (1)

(c) Mosses do not have vascular tissues. (1)

Mosses do not have true roots. (1)

(d) Similarity:

Conifers and group Y produce seeds. (1)

Difference:

Conifers produce naked seeds while the seeds of group Y are protected inside fruits. (1)

(e)

(i) The cells of mushrooms have a cell wall. (1)

(ii) Mushrooms do not have chlorophyll. (1)



1.D 2.A 3.C 4.B 5.A

Q2 Decomposers convert organic matter in fallen leaves to inorganic nutrients. A high

degree of activity of decomposers results in a low amount of organic matter in soil.

Q3 Producers capture energy from the sun and convert it to chemical energy stored in

organic compounds. Decomposers break down organic compounds into inorganic nutrients,

some of which enter food chains again.

Section B

6.

(a) [(1.8 x 104) / (3 x 10

6)] x 100% (1)

=0.6% (1)

(b) Some wavelengths of light are not absorbed by the chlorophyll of plants. /

Some light is reflected by the surface of plants. (1)

(c) While 1800 kJ m-2

yr-1

of energy is transferred from plants to caterpillars, only 100 kJ m-2

yr-1

of energy is transferred from caterpillars to insect-eating birds. (1)

Energy is lost from the caterpillars through uneaten body materials, egested materials,

excretory products and respiration. (1)

Less energy is available for converting to biomass of insect-eating birds than that for

caterpillars. (1)

Thus, the biomass of insect-eating birds is much smaller than that of caterpillars. (1)

7.

(a) Predator — snake

Prey — lizard

(all correct) (1)

(b) When the population of lizards increases, more food is available for snakes, so the population

of snakes increases rapidly. (1)

As snakes eat up lizards quickly, the population of lizards decreases. (1)

When food supply becomes limited, the population of snakes decreases. This allows the

population of lizards to recover, and the cycle repeats. (1)

8.

(a)

Symmetrical diagram (1)

Bars in correct position and proportion (1)

Correct title and labels (1)

(b) [(2250 - 240) / 2250] x 100% (1)

=89.3% (1)

(c) Accuracy is low because the underground parts of the plants and the animals in the soil are


not taken into account. (1)

Seasonal variations in the distribution of organisms are not taken into account. (1)

(d) Advantage:

No need to kill all the organisms and dry them. (1)

Disadvantage:

The water content in the organisms varies from time to time and this would affect the results.

(1)

Biomass is normally measured in terms of grams of dry mass per square metre.



1.A 2.B 3.D 4.B 5.C

Q2 Plants in the pond absorb carbon dioxide in the pond water and carry out

photosynthesis. Carbon dioxide is acidic Removal of carbon dioxide by the plants

results in an increase in the pH value of the pond water.

Q4 Maize converts some of the carbohydrates produced in photosynthesis to lipids and

stores them in its grains. Glycogen is the storage form of carbohydrates in animals.

Q5 20 arbitrary units of light intensity is the lowest light intensity required to produce the

greatest amount of crop.

Section B

6.

(a) To ensure that any starch detected in the iodine test is produced during the experiment. (1)

(b)

(i) To remove the chlorophyll in the leaf (1)

for observation of colour change when adding the iodine solution. (1)

(ii) Alcohol is flammable. (1)

7.

(a) From 120 to 210 seconds, the concentration of 3-C compound decreases from 8 arbitrary

units to 2 arbitrary units. (1)

From 120 to 150 seconds, the concentration of 5-C compound increases from 5 arbitrary units

to 7 arbitrary units. From 150 to 210 seconds, the concentration decreases to 3 arbitrary units.

(1)

(b) The 5-C compound acts as a carbon dioxide acceptor to ‘fix’ carbon dioxide into an organic

compound which starts the reactions in the Calvin cycle. (1)

As the concentration of carbon dioxide decreased from 1% to 0.03%, less 5-C compound was

used to ‘fix’ carbon dioxide into the cycle. (1)

At the same time, the 5-C compound was regenerated from triose phosphate molecules. (1)

Therefore, the concentration of 5-C compound increased.

8.

(a) Make sure the liquid levels in sides A and B are at the same level.

Make sure the bell jar is airtight.

(or other correct answers) @(1) x2

(b) The liquid level in side A will fall. (1)

Under bright light, the rate of photosynthesis is higher than that of respiration. (1)

There is a net production of oxygen by the green plant. (1)

The sodium hydrogencarbonate solution replaces the carbon dioxide taken up by the green

plant in photosynthesis. (1)

The amount of carbon dioxide inside the bell jar is kept constant. (1)

As a result, the air pressure inside the bell jar increases and the liquid level in side A falls. (1)

(c) The liquid level in side A will fall more slowly. (1)

This is because the green glasses only allow green light to pass through. (1)

Chlorophyll in the green leaves reflect most green light and therefore the rate of


photosynthesis decreases. (1)



1.D 2.B 3.D 4.C 5.B 6.A

Q1 Oxygen is consumed in oxidative phosphorylation.

Q5 If the mitochondria cannot function normally, Krebs cycle would be affected as it occurs

in the mitochondria. Glycolysis and the rest of the steps in anaerobic respiration occur in

the cytoplasm. Thus the two processes would not be affected.

Section B

7.

(a) It occurs in cytoplasm. (1)

(b) Pyruvate (1)

(c) The hydrogen is accepted by NAD to form NADH. (1)

Then, electrons from the hydrogen atoms pass through a series of electron carriers in the

electron transport chain. (1)

Finally, the electrons, together with the hydrogen ions released from NADH are accepted by

oxygen to form water. (1)

8.

(a) During strenuous exercise, skeletal muscle cells respire anaerobically. (1)

Glucose is first oxidized to pyruvate in glycolysis. (1)

In the absence of oxygen, pyruvate accepts hydrogen from NADH and is reduced to lactic

acid. (1)

(b) Some lactic acid breaks down into carbon dioxide and water. (1)

Some of it converts to glycogen in the liver. (1)

9.

(a) The glucose solution was boiled to expel air from the solution. (1)

The yeast-glucose mixture was covered with a layer of paraffin oil to prevent the entry of

atmospheric oxygen. (1)

(b) The reading rose gradually (1)

because heat was released in anaerobic respiration in yeast. (1)

(c) The hydrogencarbonate indicator changed from red to yellow (1)

because carbon dioxide was produced in anaerobic respiration in yeast. (1)

(d) In bread-making, yeast is used to break down the sugars in the dough. (1)

Carbon dioxide formed by alcoholic fermentation in yeast helps raise the dough. (1)

During baking, the carbon dioxide trapped in the dough expands, giving the bread a spongy

texture. (1)



1.B 2.C 3.C 4.D 5.A 6.D

Q1 The pathogens may be killed by the body defence mechanisms before they cause

any damage to the tissues.

Q2 Human swine influenza is thought to spread mainly through coughing or sneezing.

People may also become infected by touching objects contaminated with human swine

influenza virus. The virus is not transmitted by food.

Section B

7. Viruses — E (1)

Fungi — A (1)

Protists — C (1)

8.

(a) To see if there were any patterns in the outbreak. (1)

(b) It stopped people from drinking contaminated water from the pump. (1)

(c) There is proper treatment and disposal of sewage. /

There are well-developed water treatment systems to sterilize drinking water. /

There is more public awareness of water hygiene.

(any 2 or other reasonable answers) @(1) x2

(d) The history demonstrates the use of scientific method in scientific inquiry. The major steps in

scientific method include: observation, asking questions, proposing hypotheses, making

predictions and testing the hypotheses and drawing conclusions. (1)

John Snow observed that the cholera cases were centred around a particular water pump. He

then proposed that cholera was caused by contaminated water from the pump. He stopped

people from getting water from it and finally the result supported his hypothesis. (1)

(or other correct answers)

9.

(a) S. albus (1)

Hand-writing of scientific names must be underlined. e.g. S. albus

This method is commonly used to test the sensitivity of bacteria to different

antibiotics. The larger the diameter of the clear zone, the more sensitive is the bacteria

to the antibiotic.

(b) 10 i.u. of antibiotic X is twice as effective as 1 i.u. in controlling E. coli. (1)

There is little difference in the effectiveness of 1 i.u. and 10 i.u. of antibiotic X in controlling

S. albus. (1)

(c) Antibiotic Y (1)

10 i.u. of antibiotic Y is twice as effective as 10 i.u. of antibiotic X in controlling B. subtilis.

(1)

(d) There is no evidence that antibiotic Y is more effective at a concentration higher than 10 i.u.

(1)

The antibiotic at high concentrations may cause serious side effects. (1)

Indiscriminate use of the antibiotic may speed up the development of antibiotic resistance in

large populations of B. subtilis (1)


(e) It inhibits the formation of bacterial cell walls. The bacteria therefore lyse when they divide. /

It damages the cell membranes of bacteria. The cell contents leak out and the bacteria die. /

It inhibits the synthesis of nucleic acids so that the bacteria cannot carry out cell division. /

It binds to bacterial ribosomes and inhibits protein synthesis in bacteria. As a result, the

bacteria cannot function and grow.

(any 2) @(1) x2



1.D 2.B 3.C 4.C 5.D

Q2 [(1113 + 417) / (7831 + 5245)] x 100 %

= 11.7%

Section B

6.

(a) Cardiovascular (1)

(b) oxygen / nutrients (1)

(c) nutrients / oxygen (1)

(d) coronary arteries (1)

(e) cholesterol (1)

(f) carbon monoxide (1)

7.

(a) Increase in urine production / excessive thirst / weight loss / fatigue / blurred vision

(any 2) @(1) x2

(b) The insulin lowers the blood glucose level (1)

by stimulating liver cells to convert excess glucose to glycogen / stimulating body cells to

take in more glucose from the blood. (1)

As patients with type I diabetes cannot produce enough insulin, their blood glucose

level becomes very high, especially after a meal. An external supply of insulin helps

reduce the blood glucose level to the normal level.

(c) Small or regular meals to prevent a rush of glucose into the blood. /

Meals with low sugar or carbohydrate content to reduce the need for insulin.

(any 2) @ (1) x 2

8.

(a) They knew that tar was a carcinogen. (1)

(b) To reduce individual variation so that the data obtained are more reliable. (1)

(c) 205 / 9 (1)

=23 times (1)

(d) Tar and other chemical carcinogens in cigarette smoke (1)

cause mutations in the segments of DNA which control cell division of lung cells. (1)

The lung cells divide uncontrollably. (1)

This results in excessive growth of the lung cells and the formation of a malignant tumour. (1)



1.D 2.A 3.B 4.D 5.C 6.C

Q3 Not all antigens come from outside of the body.

Some antigens are produced by the body itself.

Q6 Active immunity involves the production of antibodies by our own plasma cells. It can

be acquired naturally when a person recovers from an infection, or acquired artificially

by vaccination.

Section B

7. In infection by pathogen X7 B cells were activated. (1)

The activated B cells multiplied and differentiated into plasma cells and memory B cells. (1)

The plasma cells produced antibodies to act against the pathogen. (1)

In infection by pathogen Yr T cells were activated. (1)

The activated T cells multiplied and differentiated into killer T cells and memory T cells. (1)

The killer T cells bound to the infected cells and killed them directly by making holes in their cell

membranes. (1)

B cells are activated when they come into contact with pathogens that have not entered

host cells. Helper T cells are activated when they come into contact with infected cells.

8. A vaccine gives active immunity / induces the body to produce antibodies and killer T cells to

destroy the pathogen and develop memory for the pathogen (1)

while serum containing antibodies gives passive immunity / is introduced into the body to fight

against the pathogen. (1)

Immunity given by a vaccine starts relatively slow as time is needed for the body to produce

antibodies, killer T cells and memory cells (1)

while immunity given by serum starts immediately once the antibodies enter the body. (1)

Immunity given by a vaccine is more long-lasting (1)

while immunity given by serum is lost when the antibodies break down after a short period of

time. (1)

9.

(a) The immune systems of young children have not fully developed so they have a lower

immunity against malaria. (1)

(b) The drugs were used on a large scale. (1)

The drugs were always used singly in a treatment. (1)

(c) The vaccine contains an antigen from the malarial parasite. (1)

The antigen stimulates the body to produce a primary response / memory cells that

'remember' the type of antigen. (1)

When the body is later invaded by the malarial parasite, the body will produce a secondary

response / a larger amount of antibodies and killer T cells in a shorter time to destroy the

pathogen. (1)

(d) Sleep under protective bed nets treated with insecticides. /

Spray indoor areas with appropriate insecticides.

(or other correct answers) (1)



1.C 2.D 3.B 4.B 5.C

Q1 Each body cell has a complete set of genes in its nucleus (except mature red blood cells

that do not have nuclei).

Q4 In option B, the normal father must carry a dominant allele. This dominant allele would

pass on to his daughters who would receive an X chromosome from each of their parents.

Hence, all the daughters of a normal father must carry at least one dominant allele.

Therefore, it is not possible for the daughters to have haemophilia.

Section B

6.

(a)

Percentage of bases

A C T G

Dog 29 21 29 21

Cat 28 22 28 22

As C is paired with G, the amounts of C and G should be equal. The bases remaining

would be A and T, which should also be equal in amount. Therefore, the percentage of

base G in the DNA of the dog is 21, and the percentage of bases A or T would be (100 -

21 - 21)/2 = 29.

(b) phosphate group (1)

deoxyribose (1)

7.

(a)

Phenotype normal normal normal muscular (female) (male) female dystrophy (carrier) (male)

(b) Genotype of the normal son: XDY (1)

Chance of having a normal son = 1/4 (1)

8.

(a) The disease is caused by a dominant allele. (1)

Individual 4 does not have disease H. She must have at least one normal allele that is

inherited from her parents. (1)

Both parents (individuals 1 and 2) have the disease. They must have at least one allele for the


disease. (1)

Therefore, at least one of the parents is heterozygous (1)

with the dominant allele being expressed. (1)

(b) If disease H is X-linked, individual 1 who has the disease would carry a disease allele on his

X chromosome. (1)

This dominant disease allele would pass on to individual 4 and she would have the disease.

(1)

However, individual 4 does not have the disease. Therefore, the allele for disease H is not

X-linked. (1)

(c) Individual 8: hh, individual 9: Hh (1)



1.B 2.D 3.C 4.C 5.B

Q1 Three bases code for one amino acid and stop signal does not code for any amino acid.

Thus, the polypeptide consists of (1800 / 3) - 1 = 599 amino acids.

Section B

6.

(a) ser - asp - gly - val - leu (1)

(b) Some amino acids are coded for by more than one triplet codes, e.g. both triplet codes AAT

and GAT (mRNA codon UUA and CUA) specify the amino acid leu. (2)

7.

(a) 24 (1)

(b) 47 (1)

(c) chromosomal (1)

(d) Down syndrome (1)

(e) age (1)

8.

(a) X: messenger RNA (mRNA) (1)

Y: transfer RNA (tRNA) (1)

(b) 3: U (uracil) (1)

9： G (guanine) (1)

12： T (thymine) (1)

(c)

(i) The synthesis of molecule X (mRNA) requires a DNA strand which acts as the template.

(1)

DNA is located in the nucleus. (1)

(ii) The formation of a polypeptide molecule requires ribosomes. (1)

Ribosomes are present in the cytoplasm.

(d) A missing base shifts the reading frame. (1)

The whole amino acid sequence is altered. (1)

The conformation of the protein molecule is changed. So the protein becomes non-functional.

(1)



1.D 2.D 3.B 4.B 5.B

Section B

6.

(a) Human insulin / vaccine (or other correct answers) (1) x2

(b) Bacteria are relatively easy to manipulate genetically. /

Bacteria can take up recombinant plasmids easily. /

Bacteria can divide / replicate rapidly. /

The space needed to culture bacteria is small. /

The cost of culturing bacteria is relatively small. /

Bacteria can be cultured in any part of the world. /

The use of bacteria to produce proteins raises fewer ethical issues than the use of animals.

(any 4) @(1) x4

7.

(a) Blood / semen / saliva / hair with hair root (any 2) @(1) x2

(b) DNA fragments are negatively charged. (1)

Under an electric field, they will move towards the positive terminal. (1)

There are many molecular spaces in the gel slab. (1)

Shorter DNA fragments can travel further in a fixed period of time. (1)

8.

(a) Obtain the DNA fragment containing the insecticide-producing gene from the bacteria. Cut

the DNA fragment with a restriction enzyme. (1)

Obtain a plasmid from a bacterial cell and cut open it using the same restriction enzyme. (1)

Insert the insecticide-producing gene into a plasmid with the help of a DNA ligase. (1)

Introduce the recombinant plasmid into a maize cell. (1)

(b) The crop yield increases. /

Reduced loss of crops due to insect pests. /

Less money is spent on insecticides. /

The risk of environmental pollution by insecticides is reduced. /

There is no need for farmers to handle toxic insecticides.

(any 2 or other correct answers) @(1)x2

(c) Golden Rice (1)

Its grains are rich in p-carotene. (or other correct answers) (1)



1.A 2.D 3.D 4.B 5.A 6.B

Q1 Fossil records, homologous structures and comparative biochemistry provide

evidence for evolution.

Q5 The fewer the differences in the amino acid sequence in the same type of protein of

different species, the closer the phylogenetic relationship between the species.

Q6 Pigs and sperm whales share the same amino acid sequence in that part of the a-chain.

Section B

7.

(a) Fish B is probably more primitive than fish A. (1)

(b) Changing environment on earth leads to the extinction of fish B. (1)

(c) The fossil record provides information about the time of existence of organisms and (1)

the structures of organisms for comparison. (1)

(d) Soft-bodied organisms cannot be fossilized. /

The bodies of dead organisms may be eaten by other organisms or decay before fossilization

occurs. /

Most fossils are incomplete and may be damaged. /

Some fossils are located in inaccessible areas.

(any 2) @(1) x2

8.

(a) base (1)

(b) mutations (1)

(c) proteins (1)

(d) larger (1)

9.

(a)

(b) The higher the % of similarities of the base sequence of DNA, the closer the phylogenetic

relationship with humans. (2)

(c) Mutation leads to changes of the base sequence of DNA. (1)

Organisms with closer phylogenetic relationship would have fewer mutations or more similar

base sequence. (1)

(d) Protein (1)



1.B 2.B 3.D 4.C 5.C 6.C 7.D

Section B

8.

(a) Percentage of white peppered moth recaptured:

(253 / 460) x 100% = 55% (1)

Percentage of black peppered moth recaptured:

(129 /430) x 100% = 30% (1)

(b) An unpolluted woodland (1)

A higher percentage of white peppered moths were recaptured. (1)

This indicated that the white peppered moths were better camouflaged / fewer white peppered

moths were eaten by predators. (1)

The white peppered moths on light-coloured tree trunks were hard for birds

(predators) to see.

9.

Variation in copper tolerance existed in individuals of the grass species. (1)

The copper-tolerant individuals had a higher chance to survive in the area of mine waste. (1)

They reproduced and passed the character of copper-tolerance on to their offspring. (1)

The proportion of copper-tolerant individuals in the population increased. (1)

10.

(a) Large and long ears provide a larger surface area for dissipating body heat, (1)

thus helping lower the body temperature in the hot climate. (1)

(b) One species (1)

because they can interbreed to produce fertile offspring (i.e. all the four populations are of the

same species). (1)

(c) Three species may be present. (1)

There is no gene flow between populations W and Y, and populations Y and Z. (1)

Mutations take place in each population independently to produce different genetic variations.

(1)

Since the isolated populations live under different environmental conditions, natural selection

acts on them in different ways. (1)

The three populations can no longer interbreed to produce fertile offspring as a result of the

increasing genetic differences. (1)


Chapter E1 1.

(a) The arterioles that supply blood to the capillaries near the surface of John's skin would have

dilated. (1)

(b) Evaporation of sweat. (1)

At 1300, the air temperature was higher than John's body temperature. (1)

Heat could not be lost from the skin into the surroundings effectively by conduction,

convection or radiation. However, evaporation could still occur. (1)

(c)

(i) A smaller volume of concentrated urine would be produced. (1)

Since the water loss from John's body was not replaced, the water potential of John's

blood dropped below normal. (1)

The hypothalamus detected this change and stimulated the pituitary gland to secrete more

ADH. (1)

ADH increased the permeability of the wall of the collecting duct to water. (1)

Thus a greater proportion of water was reabsorbed from the glomerular filtrate. (1)

(ii) John might suffer from dehydration and his body would fail to lose heat through the

evaporation of sweat. (1)

Therefore, his body temperature would rise. This could lead to heatstroke and even death.

(1)

2.

(a) The rate of breathing stays nearly constant as the carbon dioxide concentration increases from

0.03% to 3%. (1)

It increases sharply when the carbon dioxide concentration reaches 4%. (1)

(b) Percentage change

%10014670

14670201850×

×

×−×= (1)

%294= (1)

(c) When the carbon dioxide concentration of inhaled air increased to 4%r the carbon dioxide

content of the blood rose, leading to a fall in blood pH. (1)

This was detected by the chemoreceptors in the respiratory centre and in the aortic and carotid

bodies. (1)

The receptors sent nerve impulses to the respiratory centre. (1)

After processing, the respiratory centre sent nerve impulses to the intercostal muscles and

diaphragm muscles to cause them to contract faster and more strongly. (1)

(d) The concentration of carbon dioxide of exhaled air (4%) is high enough to stimulate the

respiratory centre to trigger the person's breathing to start up again. (1)


Chapter E2 1.

(a) The number of cases of red tides in Tolo Harbour increased from 1984 to 1988. (1)

It decreased from 1988 to 2008. (1)

(b) Detergents used in homes contain phosphates, nitrates and ammonium compounds which are

nutrients for algae. (1)

Also, keeping livestock in rural villages produced animal waste including dead bodies and

manure. The waste contained large amounts of organic matter and bacteria. (1)

The sewers collected nutrient-rich sewage. This prevented the sewage from being discharged

directly into Tolo Harbour. (1)

(c)

(i) The rate of photosynthesis of the algae is greater than the rate of respiration during the

day. (1)

There is a net release of oxygen from the algae. The oxygen is used by the fish for

respiration and the fish is not affected. (1)

At night, algae carry out respiration only. (1)

They use up the dissolved oxygen in the water. The fish dies of suffocation. (1)

(ii) They could pump air into the sea water at night. (1)

(d) Some algae produce toxins. (1)

These toxins can accumulate in the tissues of shellfish when the shellfish feeds on the algae.

(1)

If we consume shellfish contaminated with these toxins, food poisoning may result.

2.

(a) The high temperature kills the harmful pathogens in the compost. (1)

It provides an optimum temperature for decomposition by bacteria. (1)

(b)

(i) This provides a suitable temperature for bacteria to convert decomposed food waste in the

compost to humus. (1)

(ii) Humus allows soil particles to be held together better. This improves the water- holding

capacity of soil. (1)

(c) Using compost can replace the humus in soil so that soil structure is improved. (1)

(d)

(i) Methane (1)

(ii) Methane is renewable. /

The use of methane reduces the use of fossil fuels. /

Burning methane produces less sulphur dioxide.

(any 2) @(1) x2


Chapter E4 1.

(a)

(i) restriction enzyme (1)

(ii) DNA ligase (1)

(b) Plasmids can pick up foreign DNA easily. (1)

Plasmids can be picked up by host cells because of their small size. (1)

(or other reasonable answers)

(c) Yes (1)

The bacteria which have taken up the plasmid are antibiotic resistant. They can survive and

divide to form colonies on the agar plate containing an antibiotic. (1)

The bacteria which have not taken up the plasmid are not resistant to the antibiotic and will be

killed. (1)

(d) Culture the bacteria in a fermenter. (1)

Carry out further processes (e.g. extraction and purification). (1)

(e) vaccines / human growth hormone (1)

(f) The insulin produced does not cause an immune response. /

The insulin produced contains fewer impurities. /

The product yield is higher.


2.

(a)

(i) Virus (1)

(ii) Viruses may cause severe immune responses. (1)

Viruses may regain the ability to cause diseases during gene therapy. (1)

(b) The cells lining the lungs and airways are constantly being replaced with new cells. (1)

(c) Introduce the normal gene into stem cells instead of cells lining the lungs and airways. (1)

(d) It may be difficult to decide when gene therapy should be used. /

Gene therapy may be misused to create 'designer babies'. /

It raises the question: are we playing the role of God? /

Gene therapy may cause unforeseeable harmful effects on future generations. /

Gene therapy may further widen the social class gap. /

It is difficult to obtain informed consent from seriously ill patients.


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