Paper 1 > Biology 1996 Paper 1

HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 1996 BIOLOGY PAPER I 1.Attempt THREE questions only 2.Each question consists of three parts. 3.All questions carrry equal marks . 4.In each question, 2 additional marks will be awarded for effective communecation. 5.The diagrams in this paper are NOT necessarity draw to scale. 1. (a) The diagrams below show three kinds of plants:

(i) Pine and bread mould belong to the same major plant group and marine belongs to another group. (1) State one feature that distinguishes these two major plant groups. (1 mark) (2) Bread mould belongs to the sub-group fungi. Name the sub-group to which belongs. (1 mark) (ii) State the role of the following structures in the reproduction of the plants: (1) P(1 mark) (2) R(2 marks) (iii) State one feature that is common to structures Q and R which enables them to be carried by wind. (1 mark) (iv) Referring to the diagram, explain how bread mould is adapted for obtaining nutrients. (4 marks) 1. (b) The photographs below show the transverse sections of two blood vessels:

(i) What type of blood vessel is (1) A (2) B (2 marks) (ii) Referring to one feature observed in each photograph, explain how each vessel is adapted to its function. (4 marks) (iii) What is the functional significance of the shape of the red blood cell? (2 marks) (iv) Blood vessel A can be found in the heart wall. Explain the possible effect on the heart if the vessel is partly blocked. (3 marks) 1. (c) A boy was standing by the roadside and saw a bus moving towards him. The photographs below show what he saw when the bus was 10 m and 5m away him respectively:

(i) (1) What kind of eye defect did the boy probably have? (1 mark) (2) State two features of the eyeball that might have caused this eye defect. (2 marks) (ii) Draw a ray diagram to show the pathway of light entering the eye of the boy when he was looking at the bus 10m away from him. (3 marks) (iii) How could this eye defect be corrected? (1 mark) (iv) The boy had normal colour vision, but when it was getting dark, he found it difficult to distinguish the colour of the cars on the road. Explain this briefly. (2 marks) 2. (a) The diagram below shows a set-up used to compare the rate of water loss from a leafy shoot at different light intensities:

At each light intensity, the initial and final positions of the air bubble were recorded as shown below: Light intensity Initial position of bubble Position of bubble after 10 minutes (mm) (arbitrary unit) (mm) 10 0 15 20 2 22 30 1 26 40 3 33 (i) Explain why the air bubble moved during the experiment. (2 marks) (ii) Calculate the rate of water loss at different light intensities in terms of the distance traveled by the bubble per minute. Present your results in a table.(3 marks) (iii) Based on the experimental results, explain the effect of light intensity on the rate of water loss. (3 marks) (iv) How would you adjust the position of the bubble before taking a new set of reading? (1 mark) 2.(b) Explain the following: (i) Leguminous plants can still grow well in soil deficient in nitrates. (4 marks) (ii) Vaccination can protect our body against certain disease. (4 marks) (iii) Birth control may contribute to the conservation of the environment. (3 marks) 2. (c) The diagram below shows the alimentary canal and the associated glands of a man:

(i) What is the function of A? (1 mark) (ii) Name the process by which food is moved along B. (1 mark) (iii) Explain how the secretion from C helps the digestion of protein. (4 marks) (iv) In an operation, a large part of D of the man was removed. Explain what change will occur to his faeces. (2 marks) (v) State two functions of E that are not related to digestion. (2 marks) 3. (a) Tomato plants produce fruits of two different shapes, spherical and pear-shaped. The shape of the fruit is controlled by a pair of alleles. In a study, two separate crosses were performed and the results are shown below: Cross Parent plants with Number of daughter plants with Pear-shaped fruits Spherical fruits A Pear-shaped fruits x spherical fruits 48 42 B Pear-shaped fruits x spherical fruits 0 84 (i) Which fruit shape is controlled by the dominant allele? Explain how you arrived at your answer. (Marks will not be given for genetic diagrams.) (4 marks) (ii) State the genotype of the parent plant with spherical fruits in (1) cross A (2) cross B Define the symbols you use. (3 marks) (iii) The parent plant with pear-shaped fruits in cross A is self-crossed. Use a genetic diagram to show the result. (3 marks) (iv) Explain whether the self-cross mentioned in (iii) is a kind of sexual or asexual reproduction. (2 marks)

3. (b) Patients with kidney failure will die if they do not receive proper treatment. One method of treatment is to use a kidney machine. The diagram below shows the workings of a kidney machine:

(i) Urea is found in the solution passing out of the kidney machine. Account for this. ( 2 marks) (ii) Give a reason for each of the following: (1) Each treatment takes a long time (about 6 hours) (1 mark) (2) The treatment needs to be repeated at regular intervals (about 2-3 times a week) (1 mark) (iii) Both the human kidney and the kidney machine are able to retain useful substances in the blood. Explain how this is achieved by (1) the human kidney(1 mark) (2) the kidney machine( 2 marks) (iv) Another method of treatment kidney failure is to transplant a healthy kidney into the patient . Suggest two reasons why only a small number of such patients can receive this treatment in Hong Kong. ( 2 marks) 3. (c) The diagram below shows a set-up used to measure the rate of respiration of grasshoppers:

(i) (1) Write a simple word equation of aerobic respiration. (1 mark) (2) The set-up measures the change in the amount of a substance in the word equation. What is this substance? (1 mark) (ii) What change will occur to the water level in the U-tube after 15 minutes? Explain your answer. (4 marks) (iii) What is the use of flask B? (1 mark) (iv) How will the result be different if the experiment is repeated at a higher room temperature? Explain your answer. (2 marks) 4. (a) In order to study the conditions for osmosis, a student prepared two set-up as shown in the diagrams below. The skin of each potato was removed and a cavity was made for holding the sucrose solution.

(i) Explain why the level of sucrose solution of set-up A rose after 12 hours.(3 marks) (ii) What can you conclude from the results of the two set-ups? (2 marks) (iii) Referring to set-up A, draw a labeled diagram to show the possible appearance of a complete potato cell which is in contact with (1) the sucrose solution (2) the distilled water(4 marks) (iv) At the end of the experiment, sucrose was found in the distilled water in set-up B.

Explain why this occurred. (2 marks) 4. (b) The diagram below shows some organisms in a food web in the South Pole region.

(i) Referring to the food web, state the relationship(s) between (1) the blue whale and the seal (1 mark) (2) the squid and the sea bird (1 mark) (ii) In the South Pole region, the period of daylight becomes much shorter in winter. Explain how this affects the algal population. (2 marks) (iii) Although there is no farming activity in the South Pole, a pesticide, DDT, is found accumulated in penguins. Explain why this occurs. (5 marks) 4. (c) An experiment was performed to study the effect of environmental temperature on the skin temperature and the body temperature of a person. During the experiment, the person was kept in a room with a constant relative humidity of 65%. He was allowed to eat and move around freely but he was not allowed to change the amount of clothing.

(i) (1) Describe the change in the skin temperature when the room temperature rose from 18℃ to 28℃. (1 mark) (2) What physiological change in the skin may cause this change in the skin temperature? (2 marks) (ii) When the room temperature rose from 38℃ to 40℃, (1) describe the change in the body temperature. (1 mark) (2) what was the main way by which body heat was lost to the environment? Explain how you arrived at your answer. (3 marks) (3) explain why the person’s life might be endangered if the relative humidity of the room rose to 95%.(3 marks) ANSWER 1.(a) (Total :10 +1 marks) (i) (1) The major plant group to which maize belongs produces flowers/fruits while the other group does not (1 mark) (2) gymnosperm(1 mark) (ii) (1) To form a new plant/to protect the embryo/for dispersal of the plant(1 mark) (2) To carry the male gamete(1 mark) To the female gamete for fertilization(1 mark) (iii) dry, small, light (any one) (1 mark) (iv) Bread mould has branching rhizoids/root-like structures(1 mark) to provided a large surface area. (1 mark) For secreting digestive enzymes/enzymes digest the organic food(1 mark) And then absorb the digested products(1 mark) (Effective communication (C)) (1 mark) 1. (b) (Total:11+1 marks) (i) (1)artery (1 mark) (2)capillary(1 mark) (ii) Blood vessel A has thick wall

to withstand a high blood pressure (1+1 marks) OR Blood vessel A has elastic wall To withstand/maintain a high blood pressure. (1+1 marks) OR Blood vessel A has muscular wall To control the blood flow/the diameter for the blood vessel(1+1 marks)(any one set) The wall of blood vessel B has one-cell thick(1 mark) So as to facilitate the exchange of substances between the blood and the tissue cells. (1 mark) (iii) To provided a large surface area to volume ratio(1 mark) To facilitate the diffusion of gases(1 mark) (iv) The blood supply to the heart muscle id reduced. (1 mark) So there is less food and oxygen supply to the heart muscle cells(1 mark) The heart muscle cells would die/heart attack would occur. (1 mark) (Effective communication (C)) (1 mark) 1. (c) (Total: 9 marks) (i) (1) short sight (1 mark) (2) The eyeball might be too long(1 mark) The lens might be too thick(1 mark) (ii) Large, accurate diagram with smooth lines and parallel incident rays(D) (1 mark) Title(0.5 mark) Presence of an arrow sign. (0.5 mark) Focus in front of retina. (0.5 mark) Light rays continue to retina. (0.5 mark)

(iii) Wear a concave lens/have an operation to decrease the curvature of the cornea (1 mark) (iv) Cones, which are responsible for colour vision (1 mark) can’t function properly in dim light. (1 mark) 2. (a) (Total :9+1 marks) (i) As the plant lost water by transpiration/evaporation, (1 mark) it absorbed water from the set-up(1 mark) so the air bubble moved towards the left.

(ii) Light intensity(arbitrary unit) 10

Rate of water loss (mm min-1) 1.5

20 30 40

2.0 2.5 3.0

Rate of water loss at different light intensities Results present in a table with proper headings and units(T) (1 mark) Correct results (4 x 0.5 mark) (iii) The rate of water loss increased with an increase in light intensity (1 mark) Reason: The temperature increased at higher light intensity So the rate of diffusion of water vapour/evaporation became faster(1+1 marks) OR The stomatal pore increased in size at higher light intensity So the rate diffusion of water vapour became faster (1+1 marks) (any one set) (Effective communication (C)) (1 mark) (iv) open the tap of the reservoir until the moved to the desired position(1 mark) 2. (b) (Total :11+1 marks) (i) Leguminous plants contain nitrogen-fixing bacteria (1 mark) Which can change atmosphere nitrogen (1 mark) Into nitrogenous compounds (1 mark) For protein synthesis of the plants (1 mark) (Effective communication (C)) (1 mark) (ii) Vaccination involves the introduction of a weakened of killed pathogen/antigen into the body (1 mark) It stimulates certain white blood cells which will develop a memory for the antigen (1 mark) When the body is attacked by similar pathogen/antigen(1 mark) It will produce a large amount of antibodies in a short time, so as to destroy to pathogen(1 mark) (iii) Birth control serves to limit the increase in population. (1 mark) Too large a population will lead to faster exhaustion of natural resources,) Destruction of natural habitats ) And production of large amount of waste/more serious pollution )any two 1+1 2. (c) (Total: 10 marks) (i) to close the opening of the trachea/to prevent food from entering the trachea during swallowing (1 mark)

(ii) peristalsis(1 mark) (iii) it contains protease(1 mark) to digest protein into short peptides/amino acids/polypeptides(1 mark) it is alkaline(1 mark) to neutralize the acid from the stomach/to provide a suitable pH for the functioning of protease. (1 mark) (iv) The faeces will become more watery(1 mark) because less water is absorbed if a large part of D is removed. (1 mark) (v) –breakdown of excess amino acids/formation of urea.) -storage of iron/vitamin A/ vitamin D/glycogen ) -regulation of blood sugar level )any two 1+1 3. (a) (Total: 12+1 marks) (i) Spherical fruits (1 mark) In cross B, some offspring must have received an allele for spherical fruits and an allele for pear-shaped fruits from their parents/heterozygous(1 mark) All daughter plants have spherical fruits. (1 mark) Therefore , the allele for pear-shaped fruits must be masked by the allele for spherical fruits. (1 mark) Effective communication (C) (1 mark) (ii) Let F represents the allele for spherical fruits f represents the allele for pear-shaped fruits (1 or 0 mark) (1) Ff (1 mark) (2) FF (1 mark)

Format (F) (1 or 0 mark) (iv) sexual reproduction(1 mark) because the self-cross involves the fusion of two gametes/fertilization(1 mark) 3. (b) (Total: 9 marks)

(i) Urea is present in the blood but not in the solution (1 mark) It diffuses into the surrounding solution(1 mark) (ii) (1) Because only a small volume of blood enters the kidneys ) machine per unit time ) to allow time to remove most of the urea from the body ) any one 1 mark (2) To remove urea which is continually produced/harmful ) at high concentration ) to remove excess salts taken up in the diet ) any one 1 mark (iii)(1) All useful substances in the glomerular filtrate are reabsorbed back into the blood along the kidney tubule (1 mark) (2)As the solution contains the same concentration of glucose, amino acids and minerals as the normal plasma(1 mark) there is mot net movement of these substances from the blood into the solution (iv) Not many people are willing to donate their kidneys after their death (1 mark) The kidney transplanted must match with the tissue of the patient(1 mark) 3. (c) (Total 9+1 marks) (i) (1) food + oxygen → carbon dioxide + water (+ energy) (1 or 0) (2) oxygen (ii) The water level in limb X would rise Reasons: The live grasshoppers taken up oxygen during respiration (1 mark) And give out carbon dioxide which is absorbed by the sodium hydroxide solution (1 mark) As a result, the air pressure inside flask a becomes lower than that in flask B (1 mark) Effective communication (C) (1 mark) (iii) To eliminate the error caused by changes in environmental temperature/pressure (1 mark) (iv) The rise in the water level is faster/the water level rises higher (1 mark) because the rate respiration of grasshoppers is faster at a higher temperature (1 mark) 4. (a) (Total :11+1 marks) (i) As the water potential of the sucrose solution is lower than that of the distilled water (1 mark) water passes through the potato(1 mark) into sucrose solution(1 mark) causing the level to rise Effective communication (C) (1 mark)

(ii) Osmosis occurs when two solutions of different concentrations are separated. By living tissue(1 mark) Which serves as a selectively permeable membrane(1 mark) (iii) Large, clear drawing with smooth lines (D) (1 mark) *Labels (any three) : cell wall, cell membrane, cytoplasm (0.5 X 3 marks) (optional: starch grains, nucleus, vacuole) Title (0.5 mark) Signs of plasmolysis and turgidity (S) (1 mark)

(iv) The cell membrane of potato cells became freely permeable/was destroyed after boiling(1 mark) so sucrose can diffuse out to the distilled water(1 mark) 4. (b) ( Total :9+1 marks) (i) (1) competition (1 mark) (2) predation and competition (1 or 0) (ii) The shorted period of daylight would result in a lower photosynthetic activity (1 mark) so the algal population would decrease(1 mark) (iii) DDT is leached into the sea from farming areas in other regions(1 mark) It is absorbed by microscopic algae(1 mark) DDT or microscopic algae are carried by oceanic currents to the South Pole region(1 mark) As penguins feeds on squids, while squids feed on shrimps, and shrimps feed on algae(1 mark) And DDT cannot be metabolized or excreted by the organisms(1 mark) So it accumulates up the food chain into the body of penguin(1 mark) Effective communication (C) (1 mark) 4. (c) (Total : 10 marks) (i)(1) The skin temperature increased (1 mark) (2) Vasodilation occurs in the arterioles of the skin(1 mark) so that more blood bringing heat flows to the skin(1 mark) (ii) (1) The body temperature remains constant(1 mark) (2) Sweating(1 mark)

As the room temperature was higher than the body temperature(1 mark) heat could not be lost by conduction/convection/radiation(1 mark) so sweating was the main way of heat loss. (3) Evaporation of sweat is too slow to lose heat effectively(1 mark) while the body absorbs heat from the environment(1 mark) Thus the body becomes overheated(1 mark)