1997 Biology Paper1

HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 1997 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) An experiment was carried out to study the rate of water loss from the leaves of a plant under different conditions. Four similar leaves A, B, C and D were detached. For each leaf, the area of one surface was estimated to be 100 am2. The four leaves were then smeared with Vaseline and put under different light conditions as shown in the diagrams below: (To illustrate the treatments, only the transverse sections of the leaves are shown )

The initial mass of the leaves and their mass after 1 hour of light treatment were measured and the results are shown in the table below: Leaf Initial mass of leaf (g) Mass of leaf after 1 hour (g) A 9.2 9.0 B 9.4 8.8 C 9.5 9.4 D 9.1 8.9 (i) Name the biological process by which the leaves lose water. (1 mark) (ii) Describe how the area of one surface of a leaf can be estimated. (2 marks)

(iii)Calculate the rate of water loss of each leaf in terms of the decrease in mass per unit area hour. Present your results in a table. (2 marks) (iv) (1) Based on your calculations in (iii) above, state which leaf surface lost water at a faster rate in bright light. (1 mark) (2) Suggest one structural feature of this leaf surface which would result in a faster rate of water loss. (1 mark) (3) Explain the effect of light intensity on the rate of water loss of the leaf. (3 marks) 1. (b) The bar chart below shows the composition of food eaten by an overweight child for lunch:

(i) (1) What is food substance Q? (1 mark) (2) Explain its importance to the body. (2 marks) (ii) The child often eats milk chocolate as a snack. (1) Explain how it may contribute to his weight problem. (3 marks) (2) Explain how it may cause tooth decay. (3 marks) (iii) Explain why food from animals, such as pork and fish, is an important part of a child’s diet. (2 marks) 1. (c) Mediterranean anaemia is a human blood disease. Whether a person has this disease or not is determined by a pair of alleles. The allele for the normal character is dominant, which the allele for the disease is recessive. The pedigree below shows the inheritance of this disease in a family: 2. (a) The photographs below show the sections of the lung tissues of a cigarette smoke and a non-smoker observed under the microscope with the same magnification:

(i) What is structure A? (1 mark) (ii) Describe and explain the mechanism by which air from the atmosphere is drawn into structure A. (4 marks) (iii) With reference to the photographs, explain how the function of structure A is a affected by cigarette smoking. (2 marks) (iv) (1) State a disease which may be caused by tar in cigarette smoke. (1 mark) (2) Draw a labeled diagram of a set-up used to show the presence of tar in cigarette smoke. (3 marks) (i) State the genotype of John. (1 mark) (ii) Deduce, with reasons, the genotype of Peter. (3 marks) (Marks will not be awarded to genetic diagrams.) (iii) the genotypes of Paul and June are the same as that of their daughter May. Explain why they do not have a diseased child which May does. (3 marks) (iv) About 7% of the Hong Kong population is heterozygous for this character. Suggest a reason why it is preferable for a couple to have a test to find out their genotypes for Mediterranean anaemia before marriage. (1 mark) (v) State one other cause of anaemia. (1 mark) 2. (b) Pollen grains from a flower of an orange tree were transferred to a flower of another orange tree. After some time, a fruit was formed. The diagrams below show the structure of an orange flower and the transverse section of the fruit:

(i) With reference to the diagram, (1) State the agent involved in the transfer of pollen grains between the orange flowers under natural conditions. (1 mark) (2) Explain one way in which the structure of the flower is adapted to this mode of pollination. (2 marks) (ii) (1) State the dispersal mechanism of orange seeds. (1 mark) (2) Explain one way in which the fruit helps in this dispersal mechanism. (2 marks) (iii) Explain why the genetic composition of cells in tissue E is different from that in tissue F. (3 marks) 2. (c) Figure 1 shows a section of the human heart. Figure 2 shows the change in pressure in chamber P and chamber Q during a cardiac cycle:

(i) What is chamber Q? (1 mark) (ii) Describe and explain the change in pressure in chamber Q from 0.1 second to 0.2 second. (2 marks) (iii) (1) State the condition of valve A at 0.2 second. (1 mark) (2) Explain your answer in (1) with reference to figure 1 and figure 2. (2 marks) (iv) Some patients may suffer from a kind of heart defect in which valve A cannot close completely. Explain the probable effect of this defect on the function of the heart. (2 marks) (v) Using the letters in the diagram, state the heart chamber which first receives the following substances after their entry into our body: (1) carbon monoxide in exhaust fumes (1 mark) (2) vitamin C in food. (1 mark) 3. (a) In humans, the normal range of glucose concentration in the blood plasma is

between 70 and 100 cm3. As the concentration glucose in the blood plasma changes, both the rate of filtering glucose from the glomerulus and the rate of excreting glucose in the urine also change. The table below shows the variation of the two rates with the glucose concentration in the blood plasma: Concentration of glucose in Rate of filtering glucose Rate of excreting glucose in blood plasma (mg per from glomerulus (mg min-1) urine (mg min-1) 100cm3) 70 95 0 100 150 200 250 300 350 400

135 203 270 338 405 473 540

0 0 7 18 40 75 130

(i) Using the same x and y axes, present the above data in the form of a graph. (4 marks) (ii) (1) When the concentration of glucose in the plasma is 90 mg per 100cm3 (I) what is the rate of filtering glucose from the glomerulus? (II) what is the rate of excreting glucose in the urine? (1 mark) (2) With reference to the function of the kidney tubule, explain your answer in (II) above. (2 marks) (iii) Explain one situation which leads to the excretion of glucose in the urine of a healthy person. (3 marks) 3. (b) The diagrams below show the different stages of development of a germinating mung bean seed:

(i) (1) With reference to stages 1 and 2, what tropic response is demonstrated by the radicle? (1 mark) (2) What is the significance of this tropic response to the plant? (2 marks) (ii) With reference to the diagrams, deduce, with reasons, two functions of the cotyledons. (4 marks) (iii) If an excessive amount of chemical fertilizer is added to the soil at stage 7, the seeding will wilt after several hours. Explain this phenomenon. (4 marks) 3. (c) The diagram below shows part of a food web in a gei wai at Mai po:

(i) Describe how the energy from the sun becomes available to the gei wai shrimp in this web. (4 marks) (ii) Explain what will happen to the yield of the mouthbrooder after all the shrimps have been harvested from the gei wai. (2 marks) (iii) Two of the animals in the food web are classified into the same vertebrate group. (1) Name the group. (1 mark) (2) State one external feature used for classifying the two animals into this group. (1 mark) (iv) Give one reason why the Mai Po reserve area is ecologically valuable. (1 mark) 4. (a) To study the conditions required for photosynthesis, a destarched plant with variegated leaves was put under sunlight for 4 hours as shown in the diagram below:

(i) Explain why the plant can be destarched by keeping it in darkness for 48 hours. (2 marks) (ii) What is the use of soda lime in the set-up? (1 mark) (iii) After 4 hours, both leaf A and leaf B were detached and tested to see whether photosynthesis had taken place. Describe how the test should be carried out. (iv) After the test in (iii) , what is the observed result of (4 marks) (1) leaf A (2) leaf B? (2 marks) (v) What conclusion, if any, can be drawn by comparing the results of (1) the green part and the non-green part of leaf A only, (2) the green part of leaf A and the green part of leaf B only, (3) the green part of leaf A and the non-green part of leaf B only? Give an explanation if no conclusion can be drawn. (4 marks) 4. (b) Yeast is commonly used in the making of bread. The diagrams below show the steps in bread-making:

(i) Explain what happens to the volume of the dough after keeping it at 30℃ for 1 hour. (3 marks) (ii) Give a reason why the volume of the dough will not change any more after it has been kept in the oven for 10 minutes. (1 mark) (iii) Suggest another industrial application of yeast. (1 mark) (iv) If the bread is left in a warm and humid place for several days, black dots will be found on the bread surface. Make a labelled drawing to show some of these black dots and their associated structures when observed under a microscope. (3 marks) 4. (c) The photographs below show two postures of a woman doing sit-up exercises:

(i) Describe how the movement of the head can be detected by the semi-circular canals of the ear when the woman changes her posture from 1 to 2. (5 marks) (ii) Referring to photograph 2, which muscles in her arms, biceps or triceps, are in a

contracted state so that she can touch her knees? (1 mark) (iii) Give two structural features of the backbone which allow it to bend to a smooth and curved shape as shown in photograph 2. (2 marks) (iv) Suggest one advantage of doing regular exercise. (1 mark) ANSWER 1. (a) (Total : 10 marks) (i) *Transpiration (1 mark) (ii) Trace the outline of the leaf on a graph paper(1 mark) then count the number of (1 cm2) squares within the outline(1 mark) (no mark if the method is not workable) (Deduct 0.5 mark if multiply the area by 2) (iii) Leaf Rate of water loss (g cm-2h-1) A 0.002 B C D

0.006 0.001 0.00

Correct results (1 or 0) Results presented in table from with proper heading and units (0.5 + 0.5) (iv) (1) The upper surface(1 mark) (2) There are more stomata on the upper surface of the leaves) The cuticle on the upper surface is thinner )any one (1 mark) (3) The rate of water loss increased at higher light intensity Reasons: The stomatal pore opened wider at higher light intensity So the rate diffusion of water vapour became faster(1+1 mark) OR The temperature increased at higher light intensity So the rate of diffusion of water vapour/evaporation became faster(1+1 mark) Any one set 1. (b) (Total :11+1 marks) (i) (1) dietary fibre (1 mark) (2) It adds bulk to food ) promotes peristalsis of the intestine ) and prevents constipation ) any two 1+1 marks (ii) (1) Milk chocolate contains large proportion of energy-rich food substances/carbohydrates/fat(1 mark)

If the energy intake of the child is greater than the energy needed. (1 mark) The excess energy-rich food substances will be stored in the body as fat Which leads to overweight(1 mark) (2) Milk chocolate contains high proportion of carbohydrates/It sticks easily to the tooth surface(1 mark) Sugars in the chocolate are broken down by the bacteria in the plaque(1 mark) To form acid which dissolves the enamel and causes tooth decay(1 mark) Communication Skill (C) (1 mark) (iii) Food from animals is rich in protein(1 mark) Which is necessary for the growth of the child(1 mark) 1. (c) (Total 9+1 marks) (i) homozygous recessive(1 mark) (ii) Peter is heterozygous(1 mark) because John must receive a recessive allele from each of his parents(1 mark) As Peter is normal for this character, he must have at least one dominant allele (1 mark) Communication Skill (C) (1 mark) (iii) May is heterozygous, so genotypes of Paul and June are also heterozygous (1 mark) There is a chance for them to have a diseased child(1 mark) The occurrence of a disease child I random(1 mark) (iv) to predict the risk of getting a diseased child(1 mark) (v) lack of iron in the diet(1 mark) 2. (a) ( Total : 11+1 marks) (i) air sac (1 mark) (ii) The intercostals muscles contract to raise the rib cage(1 mark) The diaphragm muscles contract to flatten the diaphragm(1 mark) The volume of the thoracic cavity increases(1 mark) And hence the pressure inside decreases(1 mark) So air rushes into structure A Communication Skill (C) (1 mark) (iii) the number/surface area of structure A is reduced/the surface of structure A of the smoker is less folded(1 mark) which greatly reduces the rate of gaseous exchange(1 mark) (iv) (1) lung cancer(1 mark) (2) large and accurate diagram (D) (1 mark) Label and title (L): *cigarette, filter *pump, *cotton wool (4 X 0.5 mark) (No mark if the set-up is not workable)

2. (b) (Total :9+1 marks) (i) (1)insect (1 mark) (2) A is club-shaped/has a broad tip to receive pollen grains on the insect body (1 +1 marks) OR Presence of nectary/D Which produces grains on the insect body(1 +1 marks) OR C is relatively large in size To attract insects(1 +1 marks) (any one set) (ii) (1) Animal dispersal (1 mark) (2) The fruit has fleshy portion/E is fleshy and juicy ) OR The fruit has a good smell/taste ) OR The outer skin of the fruit is in bright colour ) any one 1 mark To attract animals to eat the fruit (1 mark) (iii) Because cells in tissue E are formed from cells/ovary wall of the mother plant only(1 mark) while cells in tissue F are developed from the zygote(1 mark) which contains a combination of genes from two parent plants(1 mark) Communication Skill (C) (1 mark) 2. (c) ( Total : 10 marks) (i) left ventricle (1 mark) (ii) The pressure increases sharply(1 mark) because the muscles of chamber Q is contracting(1 mark) (iii) (1) closed(1 mark) (2) the pressure in the left ventricle (Q) is greater than that in the left atrium (P) (1 mark) thus closing valve A

the heart tendons prevent valve A from turning into the left atrium (P) (1 mark) (iv) The unidirectional flow of the blood through the heart cannot be maintained (1 mark) Thus, less blood is pumped out of the heart(1 mark) (v) (1) chamber P(1 mark) (2) chamber S(1 mark) 3. (a) (Total : 10 marks) (i) Title (T) (0.5 mark) Correct choice of axes (A) (0.5 mark) Correct labeling of axes with units (L) (0.5 +0.5 mark) Correct plotting and joining of all points (P) (0.5 +0.5 mark) Labelling of the two lines (B) (0.5 +0.5 mark)

(ii) (1) (I) 120+5 mg min-1(0.5 mark) (II) 0 mg min-1(0.5 mark) (no unit, no mark) (2) because all glucose (1 mark) is reabsorbed from the glomerular filtrate(1 mark) (iii) After excessive intake of sugary food/glucose(1 mark) the digested sugar/glucose is absorbed into the blood(1 mark) when the glucose concentration in the blood plasma is so high

(e.g. exceed 150 mg per 100 cm3) that the kidneys cannot reabsorb all the glucose from the filtrate, glucose will be excreted in urine (1 mark) 3. (b) (Total : 11+1 marks) (i) (1) Geotropism (1 mark) (2) To enable the root to penetration into the soil) for better anchorage ) for absorbing more water )(any two 1+1 marks) (ii) They protect the delicate plumule because they cover up the plumule before it emerges from the soil(1+1 marks) OR They provided food for the germination of the seed Because they decrease in size while other structures are developing(1+1 marks) OR They provided more food for the development of the seedling by photosynthesis Because they turn green after emerging from the soil(1+1 marks)any two set (iii) The water potential of the soil water becomes lower than that of the cells of the root (1 mark) As a result, the cells of the root lose water by osmosis (1 mark) And hence the cells of the seedling become flaccid (1 mark) Since there is little mechanical tissue in the seedling/the seedling is mainly supported by cell turgidity (1 mark) The seedling cannot support its own weight and wilt Communication Skill (C) (1 mark) 3. (c) (Total : 9+1 marks) (i) The mangrove plants and/or the reeds can undergo photosynthesis (1 mark) converting the solar energy into chemical energy (1 mark) the chemical energy in the plant fragments (1 mark) is then passed to gei wai shrimps via the worms by feeding (1 mark) Communication Skill (C) (1 mark) (ii) The yield of the mouthbrooder increases (1 mark) because there is less competition for food/more food is available to the mouthbrooder (1 mark) (iii) Group External feature *fish Presence of fins/lateral lines/slimy scales/gills *bird Skin covered with feathers (Any one 1 + 1 marks) (iv) It provides a habitat for the wildlife/protects the wildlife

)

It provides a resting place for migratory birds (accept other reasonable answers)

)any one 1 mark

4. (a) (Total :13 marks) (i) In darkness, starch in the leaves is converted to sugars (1 mark) which are transported away from the leaves/oxidized in respiration (1 mark) (ii) To absorb carbon dioxide in the plastic bag (1 mark) (iii) Put the leaf in boiling water (1 mark) Immerse the leaf in hot alcohol (1 mark) Immerse it in water (1 mark) Add iodine solution onto the leaf (1 mark) (iv) (1) In leaf A, the green part turned to dark blue and the non-green part became brown in colour after the iodine test (1 or 0) (2) In leaf B, both the green part and the non-green part became brown after the iodine test (1 mark) (v) (1) Chlorophyll is necessary for photosynthesis (1 mark) (2) Carbon dioxide is necessary for photosynthesis (1 mark) (3) No conclusion can be drawn (1 mark) because the non-green part of leaf B differs from the green part of leaf A by the absence of chlorophyll and carbon dioxide/by two variables (1 mark) 4. (b) (Total : 8+1 marks) (i) The volume of the dought increases (1 mark) because the yeast carries out anaerobic respiration/alcoholic fermentation(1 mark) which produces carbon dioxide is produced(1 mark) Communication Skill (C) (1 mark) (ii) The yeast are killed/enzymes are denatured under high temperature(1 mark) thus no more carbon dioxide is produced (iii) brewing of beer/wine(1 mark) (iv) Large, accurate drawing (D) (1 mark) Labels and title (L) (any four): * sporangium, *sporangiophore, *spore, *hypha, *rhizoid, * columella (optional) (4 X 0.5 mark)

4. (c) (Total :9+1 marks) (i) The movement of the head results in the movement of the endolymph/the gelatinous structure of the semi-circular canal (1 mark) in the direction opposite to the head movement (1 mark) This stimulates the sensory hair cells (1 mark) And nerve impulses are generated and (1 mark) Carried to the cerebrum for interpretation (1 mark) Communication Skill (C) (1 mark) (ii) triceps (1 mark) (iii) The backbone is made up of many vertebrae/small bones ) which are articulated by joints ) there are compressible cartilage discs between the vertebrae ) (any two 1+1 marks) (v) Improve the function of the lungs/the heart ) Help to reduce body weight ) Improve musculature/improve strength of muscles ) Improve/maintain the flexibility of joints ) Reduce stress/tension )any one 1 mark (accept other correct answers)