Marking Scheme > Biology 2002 Paper 1 Marking Scheme

2002 CE Biology Paper I Marking Scheme 1. (a) (i) The plant tissue cannot carry out photosynthesis / produce its own sugar, so it needs an external supply for sugar for respiration to release energy and as raw material for growth Effective communication (C) (ii) The genetic make-up of the daughter plants was the same as that of the parent plant because the daughter plants were formed by mitosis of the parent tissues cells. (iii) For plant B, the sterilized soil had no microorganisms for recycling the fallen leaves into minerals After several weeks, the minerals in the soil became exhausted, so insufficient chlorophyll was made. (iv) This method is a faster / surer way of producing daughter plants. The desirable characteristics of the parent can be retained in the daughter plants. 1.

(b) (i)

(1) Z Deficiency of vitamin D will lead to poor bone growth. As Z is not strong enough, the body weight exerting on it will cause it to bend. (2) The body produces it own vitamin D under sunlight / UV light.

(ii) Because when producing movements, one muscle contracts, the other relaxes / the contraction of M bends the arm and that of N extends the arm. (iii) Title (T) Position of load, effort, fulcrum correct. Direction of arrows correct. The weight of the forearm forms the load. The force generated by the contraction of muscle M forms the effort. The elbow joint forms the fulcrum.

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1.

(c) (i)

(1) 40 mg per 100cm3 of blood (2) 65 mg per 100cm3 of blood

(ii) In the healthy person, the initial rise in blood glucose level stimulates the secretion of insulin by the pancreas while there is no / less insulin secretion in the diabetic person. Insulin stimulates the conversion of glucose into glycogen in the liver / uptake of glucose by body cells, So the increase of blood glucose level in the healthy person is smaller. Effective communication (C) (iii) Since the 36 th minute, the blood glucose level of the diabetic person is higher than the upper limit for complete reabsorption of glucose, so glucose is present in the filtrate / urine in the collecting duct. The water potential of the filtrate / urine is lowered by the glucose present, Thus the reabsorption of water is reduced and a larger volume of urine would be produced.

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2.

(a) (i) It forms a physical barrier to prevent sperms from meeting the ovum. (ii) AIDS / hepatitis B, C or E / gonorrhea / syphilis. (iii) Menstruation will still occur because the operation does not affect the production of the female sex hormones by the ovary and the transport of the hormone is by blood. Under the influence of the female sex hormone, the uterine lining will get thicker and shed off later. Effective communication (C) (iv) Title. Drawing including cell outline. Two chromosomes shown, i.e. haploid, chromosomes as single thread.

2.

(b) (i) (1) stoma (2) It allows carbon dioxide to enter the leaf for photosynthesis in cell B. (ii) (1) The moss leaf has no cuticle / no waterproof covering. And it is one-cell thick, So the surface area to volume ratio is large. This would lead to a high rate of water loss from the plant / so the moss would become dehydrated easily in dry environment.

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(2) Atmospheric oxygen dissolves into the water film on the moss leaf and then diffuses in through the entire surface of the leaf. OR At night, moss cells carry out respiration only and thus lower the oxygen concentration in the cells. Atmospheric oxygen dissolves into the water film on the moss leaf and diffuses into the leaf. 2.

(c) (i) July – September because the temperatures in these months were the highest so the growth / activities of bacteria was the fastest. (ii) Because the patient suffers from excessive loss of minerals / water / dehydration (iii) (1) Biotoxins (2) The harmful substance in microscopic algae is passed to shellfish through feeding. As a shellfish feeds on many algae and the substance can neither be excreted nor broken down by shellfish, The substance would be accumulated to a high level in the shellfish. Effective communication (C) (iv) Cooking the food thoroughly is to kill the microorganisms in the food. The shorter the time for the food to be left at room temperature, the less the microorganisms can grow in the food.

3.

(a) (i) (1) x = 1.5 y = 1.2 (2) x represents the amount of water absorbed by the plant. y represents the amount of water transpired / lost by the plant.

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(3) Value x is larger than value y. This indicates that there is a net gain of water by the plant. The water gained is essential for various life processes, e.g. formation of new cells, photosynthesis, support, cellular metabolism, etc. Effective communication (C) (ii) Value x would decrease because vaseline blocks the stomata and the leaf surfaces so the rate of transpiration would drop. As transpiration enhances the absorption of water, less water would be absorbed by the plant. 3.

(b) (i) (1) Mammal (2) Body covered with hair (ii) Insects / invertebrates (iii) Bat B feeds on fruits. Some seeds escape the chewing action of the teeth / the seed coat protects the seeds from being digested in the alimentary canal. Thus, intact seeds are egested in the faeces. (iv) Bat B produces a larger amount of faeces per day than bat C because bat B feeds on plants which contains more indigestible material / bat C feeds on blood which contains less indigestible material. (v) Different types of bats have different diets. Thus, the competition for food among the bats would be less / more resources would be available to the bats / the risk of extinction of bats due to the disappearance of a certain food source would be smaller.

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3.

(c) (i) (1) Mutation is a sudden change in the genetic materials (2) A mutation may lead to the synthesis of new protein / failure to synthesis a certain protein. Since proteins may be enzymes or may have other function in the cell, absence / change of the original protein will result in a change in the metabolic activities of cells. (ii) Individual 3 / 6 is normal, so she / he must have received at least one allele for normal fingers from either of her / his parents (individual 1 or 2). Individual 1 and 2 have short fingers, so each of them must carry at least one allele for short fingers. Thus, at least one of the individual 1 and 2 is heterozygous. In the heterozygous condition, only the dominant character is shown. Thus, short finger is the dominant character. Effective communication (C) (iii)

4.

Individual 6 Parents ff

x

Female homozygous for normal fingers ff

Gamete

f

f

Offsprings

ff All offsprings have normal fingers

(a) (i) The loudness of the sound heard would be smaller because A is used for collecting sound waves into the external ear canal / less sound waves are collected. (ii) C, E and F (iii) (1) Cochlea (2) They serve as receptors to detect sound vibrations / They are stimulated by sound vibrations and they produce nerve impulse upon receiving the vibrations. (iv) Due to the frequent loud noises, the sensory hair cells are over-stimulated 6

so they are damaged / become fewer in number. (v) Hearing would be lost because D, being damaged, cannot transmit nerve impulses to the cerebrum where the nerve impulses are interpreted into hearing. Effective communication (C)

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(b) (i)

The blood pressure in the vein is much lower than that in the artery. Reasons: The blood in the artery is directly under the pumping action of the heart, while that in the vein is not. The blood in the vein has overcome great friction / resistance after traveling over a long distance. There is a loss of fluid from the blood during the formation of tissue fluid.

(ii) To allow more time for the exchange of materials between the blood and the tissue cells. (iii) The volume of blood flow through each section of the circulation per unit time is the same. From the capillary to the vein, the total cross-sectional area decreases, so the velocity of blood flow increases. OR Contraction of skeletal muscle adjacent to the vein helps to force the blood to flow / inspiration helps to draw blood toward the thorax. At the same time, valves are present in the veins to prevent the backflow of the blood. (iv) Title. Shape of the curve showing the drop in O2 content at the capillary. Correct axis labels: oxygen content, heart-heart / artery–vein / aorta– vena cava / arteriole-venule

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Change in oxygen content of the blood in its circulation between the heart and the leg 4. (c) (i) From 20oC to 60oC, the rate of coagulation of milk by ginger juice increases with increasing temperature. From 80oC to 100oC, the rate of coagulation decreases as temperature increases. The rate of coagulation is highest at 60oC – 80oC / is very low at 20oC and 10oC. Effective communication (C) (ii) Fresh ginger juice contains an enzyme / a heat-sensitive substance that converts soluble milk protein into the insoluble form as the coagulating effect of fresh ginger juice is lost upon boiling / deceases when the temperature of milk increases beyond 80oC. (iii) In the stomach. Coagulated milk protein, being in semi-solid form, can be retained in the stomach for a longer period of time for the protease to digest it.

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