1997 Biology Paper I Marking Scheme

CE BIO 1997 1.

(a)

Alex Lam

(i)

*Transpiration------------------------------------------------------------------------------------------------------1

(ii)

Trace the outline of the leaf on a graph paper-------------------------------------------------------------1 then count the number of (1 cm2) squares within the outline------------------------------------------1 (No mark if the method is not workable) (Deduct  mark multiply the area by 2)

(iii) Leaf

Rate of water loss (g cm-2h-1)

A

0.002

B

0.006

C

0.001

D

0.002

(9.2-9.0)/100 (9.4-8.8)/100 (9.5 - 9.5)/100 (9.1-8.9)/100

Correct results-----------------------------------------------------------------------------------------------1 or 0 Results presented in table form with proper headings and units -----------------------------------,  (iv) (i) (2) (3)

the upper surface------------------------------------------------------------------------------------------1 There are more stomata on the upper surface of the leaves

)

The cuticle on the upper surface is thinner

) any ONE-----------------1

The rate of water loss increased at higher light intensity----------------------------------------1 Reason The stomatal pore opened wider at higher light intensity--------------------1 so the rate of diffusion of water vapour became faster-----------------------1 OR---------------------------------------------------------------------------------------------any ONE set The temperature increased at higher light intensity---------------------------1 so the rate of diffusion of water vapour / evaporation became faster----1

1.

(a)

Total : 10 marks (i) Some candidates wrongly answered that respiration was the process by which the leaves lose water. Some mistook evaporation as a biological process. Spelling mistakes such as 'transparation' and 'transperation' were quite common. (ii) Most candidates knew to use graph paper in estimating the area of a leaf though many of them could not express their ideas clearly. (iii) Most candidates calculated the rate of water loss correctly. However, the heading of rate of water loss in the table was commonly written as 'decrease in mass per unit area per hour'. Many candidates did not put down the unit. Some candidates wrongly expressed the results in terms of per 100cm2 instead of per cm2. (iv) This part was poorly answered. (1) A large number of candidates misinterpreted the question by stating that it was 'leaf B' instead of 'the upper surface' that lost water at a faster rate. (2) Most candidates just gave the structural feature 'stomata' without emphasizing the greater number or wider aperture of the stomata. Many candidates did not answer by referring to the leaf in question. They just stated in general that the large surface area of the leaf would result in a faster rate of water loss. (3) Some candidates had the misconception that the rate of water loss increased at higher light intensity because more water was used up to make starch due to faster photosynthesis. However, the amount of water used in photosynthesis is comparatively

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CE BIO 1997

Alex Lam much smaller than the amount lost in transpiration. It is not a significant factor to account for the increase in the rate of water loss.

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CE BIO 1997 1.

(b)

(i)

(ii)

Alex Lam

(1)

dietary fibre--------------------------------------------------------------------------------------------------1

(2)

It adds bulk to food,

)

promotes peristalsis of the intestine

) any TWO-----------------------------------------1,1

and prevents constipaton

)

(1)

Milk chocolate contains large proportion of energy-rich food substances / carbohydrates / fat------------------------------------------------1 If the energy intake of the child is greater than the energy needed ,-------------------------1 the excess energy-rich food substances will be stored in the body as fat------------------1 which lends to overweight.

(2)

Milk chocolate contains high proportion of carbohydrates / It sticks easily to the tooth surface-------------------------------------------------------------------------------------------1 Sugars in the chocolate are broken down by the bacteria in the plaque--------------------1 to form acid which dissolves the enamel and causes tooth decay ---------------------------1 Effective communication (C)-------------------------------------------------------------------------1

(iii)

Foot from animals is rich in protein---------------------------------------------------------------------------1 which is necessary for the growth of the child-------------------------------------------------------------1 Total : 11 + 1 marks

1.

(b)

(i)

The performance in this part was quite good. (1)

Most candidates gave the correct answer. However, there were many spelling mistakes, such as 'diary' or 'daily' for 'dietary', and 'cellose' for 'cellulose'.

(2)

Some candidates answered wrongly that dietary fibre helps in digestion. There were many wrong spellings for 'peristalsis' such as 'peristalsis', 'peristalisis' etc.

(ii)

(1)

Very few candidates could relate the weight problem of the child to the excessive energy intake in food. Many candidates answered wrongly that the increase in weight was due to the conversion of excess carbohydrates into glycogen which was stored in the liver and the muscles. In fact, tile increase in weight was due to the storage of fat, not glycogen. Some candidates had the misconception that fat was digested into fatty acid and glycerol which were stored in the body; others answered wrongly that fat in food was stored under the skin. In fact, fat in food is digested into fatty acid and glycerol first before absorption. These are resynthesised into fat in the epithelial cells of villi before being transported to the body organs for storage.

(2)

Most candidates understood the concept but some could not express themselves clearly using appropriate wording. Many candidates wrongly said that bacteria in the plaque digested / fed on / reacted with sugars in the chocolates to form acid. In fact, the conversion of sugars to acid by the metabolism of bacteria should not be viewed as a feeding or digestion process. It is also wrong to consider the bacteria as a reactant.

(iii)

Many candidates just listed all food substances that can be found in pork and fish without realizing that the main difference between these two types of food and the others was the

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Alex Lam

high proportion of protein they contained.

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CE BIO 1997 1.

(c)

Alex Lam

(i)

homozygous recessive------------------------------------------------------------------------------------------1

(ii)

Peter is heterozygous--------------------------------------------------------------------------------------------1 because John must receive a recessive allele from each of his parents---------------------------1 As Peter is normal for this character, he must have at least one dominant allele----------------1 Effective Communication (C)-------------------------------------------------------------------------------1

(iii)

May is heterozygous, so the genotypes of Paul and June are also heterozygous--------------1 There is a chance for them to have a diseased child----------------------------------------------------1 The occurrence of a diseased child is random------------------------------------------------------------1

(iv) to predict the risk of getting a diseased child--------------------------------------------------------------1 (v)

lack of iron in the diet--------------------------------------------------------------------------------------------1 Total : 9 + 1 marks

1.

(c)

(i)

Most candidates answered this part correctly. Some candidates confused the terms 'homozygous' and 'homologous' while some mistook characters for alleles. Some candidates used symbols which were not defined.

(ii)

A large number of candidates answered that 'Peter is normal but he has a diseased child, so Peter is 'heterozygous.' or 'If Peter is not heterozygous, it is not possible for him to have a diseased child'. It was emphasised repeatedly in the previous examiner's reports that genotypes should be logically deduced by means of the concept of alleles, not by referring only to the phenotypes of offspring and parents.

(iii)

This part was poorly answered. Only a few candidates pointed out the genotypes of Paul and June. A fair proportion of candidates realized that each fertilization process was random so that it was possible for Paul and June to have all normal children though theoretically they had a  chance of having a diseased child. However, some erroneously related the phenomenon to the random assortment of chromosomes in meiosis.

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CE BIO 1997 2.

(a)

Alex Lam

(i)

air sac----------------------------------------------------------------------------------------------------------------1

(ii)

The intercostal muscles contract to raise the rib cage--------------------------------------------------1 The diaphragm muscles contract to flatten the diaphragm--------------------------------------------1 The volume of the thoracic cavity increases,--------------------------------------------------------------1 and hence the pressure inside decreases------------------------------------------------------------------1 so air rushes into structure A Effective Communication (C)-------------------------------------------------------------------------------1

(iii)

The number / surface area of structure A is reduced / the surface of structure A of the smoker is less folded--------------------------------------------------1 which greatly reduces the rate of gaseous exchange---------------------------------------------------1

(iv) (1)

lung cancer--------------------------------------------------------------------------------------------------1

(2)

Large and accurate diagram (D)-----------------------------------------------------------------------1 Label and title (L) : *cigaittte, filter *pump, *cotton wool-------------------------------------4 ×  (No mark if the set-up is not workable) filter pump cigarette cotton wool cotton wool plastic bottle

OR

A set-up to show the pressure of tar in cigarette smoke Total : 11 + 1 marks 2.

(a)

(i)

There were many spelling mistakes for 'alveolus', such as 'alvolus', 'alveoi' etc.

(ii)

Most candidates answered correctly, though some of them on seeing the alveolus in the photographs erroneously described the process of gaseous exchange in the alveoli instead of the inhalation process.

(iii)

Some candidates described the inhibitory effect of the smoke particles on the action of cilia which was not shown in the photographs.

(iv) (2)

The quality of drawing was poor. Many drew small diagrams while some used shaky discontinuous lines. Most of the diagrams did not indicate the pumping action. These candidates did not realize that the apparatus would not work if there was nothing to set up an air current to draw in the cigarette smoke through the cotton wool. Most

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Alex Lam candidates did not give a title to the diagram drawn.

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CE BIO 1997 2.

(b)

(i)

Alex Lam

(1)

insect----------------------------------------------------------------------------------------------------------1

(2)

A is club-shaped / has a broad tip

)

to receive pollen grains on the insect body

)

OR Presence of nectary / D

)

which produces sugary secretion / nectar to attract insects ) any ONE set---------1 + 1 OR C is relatively large in size

)

to attract insects (ii)

)

(1)

Animal dispersal--------------------------------------------------------------------------------------------1

(2)

The fruit has fleshy portion / E is fleshy and juicy

)

OR The fruit has a good smell / taste

) any ONE------------------------------1

OR The outer skin of the fruit is in bright colour

)

to attract animals to eat tile fruit -----------------------------------------------------------------------1 (iii)

Because cells in tissue E are formed from cells / ovary wall of the mother plant only----------1 white cells in tissue F are developed from the zygote---------------------------------------------------1 with contains a combination of genes from two parent plants-----------------------------------------1 Effective Communication (C)-------------------------------------------------------------------------------1 Total : 9 + 1 marks

2.

(b)

(i)

(2)

When answering this part, many candidates did not refer to the features shown in the diagram. They simply answered from rote memory, citing features of insect-pollinated Bowers such as sticky stigma or brightly coloured petals, which were not indicated in the diagram of the question.

(ii)

(2)

Instead of answering how the fruit attracted the animals to disperse the seeds, some candidates described how the seeds were adapted to dispersal by animals.

(iii)

Most candidates did not give a satisfactory explanation. They could not tell clearly that cells in tissue E came from the parent plant while cells in tissue F were formed from the zygote which contained the combination of genetic materials from two different parents. Some had the misconception that E was formed by mitosis so the cells were diploid, and F was formed by meiosis so the cells were haploid. In fact, cells in E and F were both diploid. Most candidates just ascribed the terms mitosis or meiosis, fertilized or not fertilized to the two tissues. Some even attributed the difference in genetic composition between E and F to their functional difference.

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CE BIO 1997 2.

(c)

Alex Lam

(i)

left ventricle---------------------------------------------------------------------------------------------------------1

(ii)

The pressure increases sharply-------------------------------------------------------------------------------1 because the muscles of chamber Q is contracting-------------------------------------------------------1

(iii)

(1)

closed---------------------------------------------------------------------------------------------------------1

(2)

The pressure in the left ventricle (Q) is greater than that in the left atrium (P)------------1 thus closing valve A The heart tendons prevent valve A from turning into the left atrium (P)---------------------1

(iv) The unidirectional flow of the blood through the heart cannot be maintained---------------------1 Thus, less blood is pumped out of the heart---------------------------------------------------------------1 (v)

(1)

chamber P---------------------------------------------------------------------------------------------------1

(2)

chamber S---------------------------------------------------------------------------------------------------1 Total : 10 marks

2.

(c)

(ii)

Many of candidates misinterpreted the question and wrongly explained the importance of the rapid increase in pressure of chamber Q instead of explaining the cause of the increase in pressure.

(iii)

(2)

This part was poorly answered. Many candidates failed to explain that the closure of valve A was caused by the difference in pressure between chamber Q and chamber P. Most did not mention the function of the heart tendon in preventing valve A from turning into chamber P and thus keeping the valve closed Many candidates only gave the importance of closing valve A instead of the mechanism that led to its closure.

(iv) Some candidates answered wrongly that there would be a mixing of oxygenated and deoxygenated blood and thus the oxygen supply to body cells would be insufficient. They confused this heart defect with the defect that is caused by the presence of a hole between the left and right heart chambers.

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CE BIO 1997 3.

(a)

(i)

Alex Lam

Title(T)----------------------------------------------------------------------------------------------------------------- Correct choice of axes (A)--------------------------------------------------------------------------------------- Correct labelling of axes with units (L)---------------------------------------------------------------------,  Correct plotting and joining of all points (P) --------------------------------------------------------------,  Labelling of the two lines (B)----------------------------------------------------------------------------------,  The variation in the rate of filtering glucose and the rate of excreting glucose with glucose concentration in blood plasma

600

Rate of filtering glucose / -3 Rate of excreting glucose (mg min )

500

400

300

200

(1) (2)

excreting glucose

100

0

(ii)

filtering glucose

50

100 150 200 250 300 350 Concentration of glucose in blood plasma 3 (mg per 100 cm )

400

(I)

120 ± 5 mg min-1-------------------------------------------------------------------------------------

(II)

0 mg min-1 (no unit, no mark)-------------------------------------------------------------------

because all glucose---------------------------------------------------------------------------------------1 is reabsorbed from the glomerular filtrate-----------------------------------------------------------1

(iii)

After excessive intake of sugary food / glucose-----------------------------------------------------------1 the digested sugar / glucose is absorbed into the blood------------------------------------------------1 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,--------------------------------------------------------------------------------1 glucose will be excreted in urine.

3.

(a)

(i)

(ii)

(iii)

Total : 10 marks

A large number of candidates did not give a title to the graph while many chose the wrong axes. It seemed that many of them failed to realise that the x-axis is for the independent variable while the y-axis is for the dependent variable. Some did not use a proper scale for the graph A large number of candidates wrongly joined the graph to the origin. The labelling of the axes was often incomplete and units were sometimes left out. (2) Many candidates failed to point out that all glucose is reabsorbed from the glomerular filtrate. Some candidates explained wrongly that since 90 mg glucose per 100 cm plasma fell between the normal range of glucose concentration in the blood plasma, there was no need to excrete glucose in the urine. Candidates seemed to think that the excretion of glucose was a means of regulating the blood glucose concentration ! This part was poorly answered. Some candidates were careless in reading the question.

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Alex Lam They overlooked the words 'healthy person' in the question. Answers involving liver damage, deficiency of insulin, diabetes or kidney failure were not accepted.

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CE BIO 1997 3.

(b)

(i)

(ii)

Alex Lam

(1)

Geotropism--------------------------------------------------------------------------------------------------1

(2)

To enable the root to penetrate into the soil

)

for better anchorage

) any TWO-----------------------------------1,1

for absorbing more water

)

They protect the delicate plumule

)

because they cover up the plumule before it emerges from the soil

)

OR They provide food for the germination of the seed

) any

because they decrease in size while other structures are developing

) TWO --1+1

OR They provide more food for the development of the seedling by photosynthesis ) set because they turn green after emerging from the soil (iii)

)

The water potential of the soil water becomes lower than that of the cells of the root----------1 As a result, the cells of the root lose water by osmosis-------------------------------------------------1 and hence the cells of lie seedling become flaccid-------------------------------------------------------1 Since there is little mechanical tissue in the seedling / the seedling is mainly supported by cell turgidity---------------------------------------------------------------------------------------1 the seedling cannot support its own weight and wilt Effective Communication (C)-------------------------------------------------------------------------------1 Total : 11 + 1 marks

3.

(b)

(i)

(1)

Most candidates answered correctly though there were many spelling mistakes, such as, 'geotrophism', or 'gravitropism' etc.

(ii)

Most candidates deduced the functions of the cotyledons correctly but they failed to explain their deductions by referring to the information provided in the diagrams. Some candidates failed to recognise that cotyledons are part of the seed and wrongly answered that they protected the seed.

(iii)

This part was poorly answered. Many candidates did not point out that the seedling was mainly supported by cell turgidity. Some candidates suggested that the roots absorbed the excessive chemical fertilisers which were toxic to the plant, and thus the root cells were killed. Many candidates had the misconception that the plant became flaccid or plasmolysed. Actually the terms 'flaccid' or 'plasmolysed' should only be used to describe the condition of the cells, not to describe the plant. Some candidates answered that the soil water became hypertonic to plant cells so that water moved out of the cells. It should be emphasised that tonicity of the solution, which is affected by the amount of solutes in it, only contributes to the factor of osmotic potential. The movement of water is determined by the difference in water potential between the root cells and the soil water. Osmotic potential forms only a part of the water potential, thus it is inaccurate to use the term 'hypertonic' when accounting for the water movement from the cells to the soil water. In addition, there was a general lack of communicative skills shown by the candidates in answering this

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question.

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CE BIO 1997 3.

(c)

(i)

Alex Lam

The mangrove plants and / or the reeds can undergo photosynthesis-----------------------------1 convert the solar energy into chemical energy------------------------------------------------------------1 The chemical energy in the plant fragments---------------------------------------------------------------1 is then passed to gei wai shrimps via the worms by feeding------------------------------------------1 Effective Communication (C)-------------------------------------------------------------------------------1

(ii)

The yield of mouthbrooder increases------------------------------------------------------------------------1 because there is less competition for food / more food is available to the mouthbrooder-----1

(iii) Group

External feature

*fish

presence of fins / lateral lines / slimy scales / gills

*bird

skin covered with feathers

(iv) It provides a habitat for the widelife / protects the wildlife It provides a resting place for migratory birds

) ) any ONE---------------------1+1

) ) any ONE----------------------------1

(accept other reasonable answers) Total : 9 + 1 marks 3.

(c)

(i)

The performance in this part was satisfactory. However, many candidates did not mention the conversion of solar energy to chemical energy by the plants. Many answered wrongly that the worm fed on the plants instead of on the plant fragments

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CE BIO 1997 4.

(a)

(i)

Alex Lam

In darkness, starch in the leaves is converted to sugars-----------------------------------------------1 which are transported away from the leaves / oxidized in respiration-------------------------------1

(ii)

To absorb carbon dioxide in the plastic bag----------------------------------------------------------------1

(iii)

Put the leaf in boiling water-------------------------------------------------------------------------------------1 Immerse the leaf in hot alcohol--------------------------------------------------------------------------------1 Immerse it in water------------------------------------------------------------------------------------------------1 Add iodine solution onto the leaf------------------------------------------------------------------------------1

(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

(v)

(2)

In leaf B, both the green part and non-green part became brown after the iodine test--------------1

(1)

Chlorophyll is necessary for photosynthesis--------------------------------------------------------1

(2)

Carbon dioxide is necessary for photosynthesis--------------------------------------------------1

(3)

No conclusion can be drawn----------------------------------------------------------------------------1 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 Total : 13 marks

4.

(a)

(i)

Most candidates answered incompletely that starch was used up in darkness because there was no photosynthesis. However, they failed to mention that starch was changed to glucose first before it was oxidised in respiration. Many candidates gave the purpose of destarching the plant instead of the mechanism by which the plant was destarched.

(iii)

Many candidates did not realise that the alcohol used to extract chlorophyll from the leaf should be hot or warm, and that iodine solution instead of iodine should be used.

(v)

Most candidates answered this part correctly. But still there was a fair proportion of candidates who did not know how to draw the conclusion. They simply described the results or explained the results instead of drawing conclusions. For part (3), some candidates concluded wrongly that both chlorophyll and carbon dioxide were necessary for photosynthesis.

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CE BIO 1997 4.

(b)

(i)

Alex Lam

The volume of the dough increases--------------------------------------------------------------------------1 because the yeast carries out anaerobic respiration / alcoholic fermentation--------------------1 which produces carbon dioxide that raises the dough--------------------------------------------------1 Effective Communication (C)-------------------------------------------------------------------------------1

(ii)

The yeast are killed / enzymes are denatured under high temperature----------------------------1 thus no more carbon dioxide is produced

(iii)

brewing of beer / wine-------------------------------------------------------------------------------------------1

(iv) Large, accurate drawing (D)------------------------------------------------------------------------------------1 Labels and title (L) (any four) *sporangium, *sporangiophore, *spore, *hypha, *rhizoid, *columella(optional)------------------------------------------------------------------------------------------4 × 

Diagram of bread mould observed under the microscope Total : 8 + 1 marks 4.

(b)

(ii)

Many candidates answered wrongly that yeast was denatured or used up when the dough was kept in the oven. As yeast is not an enzyme, it is inappropriate to say that yeast is denatured by high temperature.

(iv) The standard of drawing was poor and very few candidates drew a clear and accurate diagram. Some drew single lines for hyphae or drew disproportionately thick hyphae. Most candidates did not give an appropriate title to their drawing. Many confused the terms 'sporangium' and 'sporangiophore', while many others mixed up 'spore' and 'sporangium' when labelling the structures. There were a lot of spelling mistakes on the labels such as 'rhizod' or 'rizhoid' for 'rhizoid', 'sporangiospore' or 'sporangiore' for 'sproangiophore', 'sporagium' for 'sporangium' and 'hypae' or 'hyphea' for 'hyphae'.

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CE BIO 1997 4.

(c)

(i)

Alex Lam

The movement of the head results in the movement of the endolymph / the gelatinous structure of the semi-circular canal---------------------------------------------------------------------------1 in the direction opposite to the head movement----------------------------------------------------------1 This stimulates the sensory hair cells------------------------------------------------------------------------1 and nerve impulses are generated and---------------------------------------------------------------------1 carried to the cerebrum for interpretation-------------------------------------------------------------------1 Effective Communication (C)-------------------------------------------------------------------------------1

(ii)

triceps----------------------------------------------------------------------------------------------------------------1

(iii)

The backbone is made up of many vertebrae / small bones

)

which are articulated by joints

) any TWO----------------1,1

There are compressable cartilage discs between the vertebrae

)

(iv) Improve the functioning the lungs / the heart

)

Help to reduce body weight

)

Improve musculature / improve strength of muscles

) any ONE-------------------1

Improve / maintain the flexibility of joints

)

Reduce stress / tension

)

(accept other correct answers) Total : 9 + 1 marks 4.

(c)

(i)

Many candidates answered wrongly that the ampulla would move in the opposite direction to that of the head movement. In fat, ampulla is a swollen part at the base of the semicircular canal and it cannot move. Many answered that the lamph / perilymph inside the semi-circular canals moved. Actually, lymph is the fluid in lymph vessels while perilymph is the fluid outside the middle canal in the bony cavity of the inner ear. Many candidates answered wrongly that the head movement stimulated the sensory hair / sensory neurone, and nerve impulses were then sent to cerebellum for interpretation. In fact, it is the sensory hair cells, not the hairs or the nerve fibres, that are stimulated. The cerebellum, though, receives information on the balance of the body from the ear and the muscles; it is the centre of muscular coordination and is not responsible for the detection of body movement. The site of interpretation is the cerebrum. Quite a number of candidates inaccurately mentioned that the impulses were sent to the brain. (iii) Many candidates just said that the presence of cartilage between the vertebrae allowed the bending of the backbone into a smooth curve. They failed to stress that it was the compressible nature of the cartilage that allowed such bending. Very few candidates answered that the presence of joints also played a part in allowing the backbone to bend smoothly. (iv) A considerable number of answers were ambiguous or inaccurate e.g. increase blood circulation / heartbeat / respiratory rate / metabolism, make the bones soft / flexible, for body defence etc. In fat, the advantage of doing regular exercise is to improve the function of the lungs / heart, not to increase the respiratory rate or heart beat rate. Besides, doing regular exercise will make the joints (not the bones) more flexible.

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