hkcee - biology - 1999 - paper i - a.pdf

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Biology 1999

1. (a)

(b)

(c)

(i) Cell C contains a diploid set of chromosomes because its nucleus

comes from cell A which is a body cell

(ii) Because the cytoplasm of cell D is formed by the repeated divisions

(mitosis) of the cytoplasm of cell C

(iii) Sheep X

The body characteristics of Dolly is determined by its genetic material

which is derived from and as same as the body cell of Sheep X

(iv) No

It does not involve the fertilization of gametes

(i) During the fitness test, the heart beats more rapidly to supply more

blood to the skeletal muscle.This provides an abundant supply of food

and oxygen to the muscles.

Hence, a large amount of energy can be released in respiration for thevigorous muscular activities. Meanwhile, more blood supply to the

muscles also helps to remove carbon dioxide and waste products

produced by the muscles

(ii) Peter’s heart

Since the energy demand of Peter and David was the same. But Peter’s

heart is able to supply the same amount of blood to the skeletal muscle

at a lower heart rate

(iii) (1) glucose → lactic acid + energy

(2)

To provide additional energy for the skeletal muscle cells, so that

the skeletal muscle can contract more strongly

(i) Gelatin in jelly solution B was digested because of the presence of

fresh pineapple juice

(ii) Gelatin was present in solution C but not in solution B. The result

revealed that pineapple juice contains enzyme that digests gelatin.

However, the enzyme was denatured at high temperature

(iii) Pineapple juice contains protease which could digested beef.

Moreover, allow the beef and pineapple mixture to stand for half an

hour before cooking, in order to provide enough time for the enzyme

to work

2. (a) (i) (1) - mass decrease by 2.4 g

(2) - mass decrease by 0.6 g

(ii) To study the effect of direct illumination on the lower or upper

epidermis with the rate of transpiration of the leaf

(iii) The decrease in mass in treatment I is greater than that in




Biology 1999

(b)

(c)

treatment II. This suggests that the transpiration rate in

treatment I is higher.

Under the direct illumination by the spotlight, the temperature

of the lower epidermis in treatment I is higher than that in

treatment II. Thus, the evaporation of water is faster.

OR Under the direct illumination by the spotlight, stomata of the

lower epidermis open more widely in treatment I than in

treatment II. As the surface area for the diffusion of water

vapour is greater, the diffusion of water vapour becomes faster

and hence the transpiration rate is higher in treatment I.

(iv) Peel off the upper and lower epidermis. Then, observe under

the microscope whether stomata are only present on lower

epidermis

(i) Bile produced in the liver can no longer be stored in structure

A. When food enters the duodenum, there are insufficient

amount of bile is released to emulsify fats in the food. Hence,

the surface area for the action of lipase decreases and cannot

digest fatty food properly.

(ii) The removal of part of structure D will reduce the digestion

and absorption of food. Then, the energy intake may becomes

less than the energy expenditure of the body. Therefore, it lead

to the use of food reserves stored in the body

(iii) (1) Insulin

(2)

The blood glucose level cannot be regulated as blood

glucose would not be converted into glycogen in the cells

of B

(i) On Day 1, as the air temperature was higher than the body

temperature. So, heat from the body could not be lost to the

environment by conduction, convection and radiation. However,

heat would be lost through the evaporation of sweat which

absorbs heat from the body surface

(ii)

On Day 2, at the higher relative humidity, sweat could not

evaporate efficiently so less heat was lost from the body through

sweating.

Undergo vigorous exercise would generate a large amount of

heat, so the body temperature would rise steadily. As a result, it

lead to the breaking down of normal metabolism

3. (a) (i) Mass of the algae in which the water has been completely removed

(ii) dry mass of the algae at hour 24 was greater than that at hour 0 because




Biology 1999

(b)

(c)

algae have carried out photosynthesis and had made carbohydrates. In

addition, the rate of photosynthesis is greater than the rate of respiration

(iii) (1) if exhaust gas is supplied, it will enhance the growth of algae

(2) exhaust gas has high composition of carbon dioxide which can increase

the rate of photosynthesis of algae

(iv) carbon dioxide will be reduced which lead to a decrease in the greenhouse

effect. Thus, it helps to slow down global warming

(i) 12

(ii) (1) 10-5 ppm, 10-4 ppm, 10-3 ppm

(2) 10-2 ppm, 10-1 ppm

(iii) (1) As cells in region A are capable of rapid growth while cells

in region B are not

(2) (i) Mark the whole length of the root with lines at regular intervals(ii) compare the distance between markings after few days .

It will find out that the markings at region A would be further apart

OR

(i) Use sections from region B

(ii) Repeat the experiment

It will show that these sections would show little elongation

(iv) Two days later, the root would bend towards the side with auxins. It is

because the auxins concentration on this side is almost 10-2 ppm which

inhibited the growth of root. However, on the other side, the auxins

concentration is about 0 ppm.

(i) In order to push the baby out of the uterus, the muscular wall of uterus will

contract rhythmically and powerfully

(ii) Period X

Ovulation

Higher chance of fertilization might occur after sexual intercourse.

Moreover, uterine lining is ready for the implantation of the embryo during

this period.

(iii) The duration of the menstrual cycle may vary and the day of ovulation may

vary from cycle to cycle as well.

4. (a) (i) (1) A and B

(2) A is semi-circular canals which detect the direction of head

movement

B is cochlea which detect sound vibration

(ii)

On reaching the 50th floor, the atmospheric pressure is lower than that

in the middle ear. So, the eardrum was pressed outwards and could notconvert sound waves into vibrations freely.




Biology 1999

(iii) germs in the throat. They have spread along the Eustachian tube to the

middle ear and cause infection

(b) (i) The mealworms absorbed oxygen during respiration.

Any carbon dioxide produced by the mealworms is absorbed by thesoda line. There will be a drop in air pressure in the syringe and the

water droplet will be moved downwards. Thus the rate of movement of

the water droplet indicates the rate of respiration.

(ii) (1) The rate of respiration of mealworms increases as the temperature

increase from 20oC to 50oc

(2) The body temperature of mealworms will rise with an increase in

the external temperature. Thus the enzymatic activity of the

worms increases which leads to an increase in the metabolic rate.

Thus the respiration rate increases.

(iii) To provide enough time for the air temperature inside the syringe to

reach the temperature of the water bath and allow time for the

respiratory rate of the mealworms to become adjusted to the new

temperature

(c) (i) During daytime, the rate of photosynthesis of the algae exceeds

the rate of respiration. There is sufficient oxygen supply to the

fish for respiration.

At night, the large population of algae undergoes respiration. So,

most of the dissolved oxygen content in seawater is greatly

reduced. Therefore, the fish die of suffocation.

(ii) Small fish eat a large quantity of microscopic algae and large

marine fish eats many small fish.Since the toxins cannot be

excreted or broken down by the fish. The toxins of the

microscopic algae are passes along the food chain to the large

fish. As a result, large fish accumulate a higher concentration of

toxins.

hkcee - biology - 1999 - paper i - a.pdf

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