CE BIO 1990 1.
(a)
(i)
Alex Lam
cells A and D -------------------------------------------------------------------------------------------------1 or 0 possession of cell wall, -----------------------------------------------------------------------------------------1 and large / central vacuole / peripheral nucleus ---------------------------------------------------------1 (1)
cell A is for regulating the size of the stomatal opening / provides a passage for gases / water vapour ----------------------------------------1 cell B is for transmitting nerve impulses / message / information ----------------------------1
(2)
cell A - thicker inner wall and thinner wall / kidney-shaped ------------------------------------1 cell B - long cellular extension / nerve fibre / branched endings ------------------------------1
(ii)
cell C bursts / swells ---------------------------------------------------------------------------------------------1 cell D becomes turgid / shows no change -----------------------------------------------------------------1 water potential of distilled water is greater than that of the cells (C and D) -----------------------1 water enters the cells by osmosis ----------------------------------------------------------------------------1 cell C expands continuously until it bursts -----------------------------------------------------------------1 but cell D does not burst because of the presence of the inelastic cell wall ----------------------1 Total : 13 Marks
1.
(a)
(i)&(ii)
Most candidates were able to answer the question correctly with reference to the features shown in the diagram. A small number of candidates, however, included features not shown in the diagram such as the presence of chloroplast in cell A or myelin sheath in cell B. This suggests they had answered by rote memory. Some candidates did not realise that vacuoles are present in both plant and animal cells, although the vacuoles in plant cells are usually larger and central in position.
(iii)
Many candidates could give a clear and accurate account of the changes in cells C and D using the concept of water potential. This shows that water potential is a useful concept in understanding and explaining osmosis in living cells.
Paper I Marking Scheme & Year Report
P.1
CE BIO 1990 1.
(b)
(i)
Alex Lam
from 2.0 s to 4.0 s ------------------------------------------------------------------------------------------------1 pressure in the lungs is greater than the atmospheric pressure -------------------------------------1
(ii)
- intercostal muscles relax
)
- to lower the ribs
)
- diaphragm relaxes
) any THREE -----------------------------------------------------1,1,1
- to curve upward
)
these actions decrease the volume of the thorax / increase the air pressure of the lungs -----------------------------------------------------------------------------------1 (iii)
15 breaths per minute (no unit, no mark) ----------------------------------------------------------------1
(iv) showing a higher frequency -----------------------------------------------------------------------------------1 showing a greater magnitude----------------------------------------------------------------------------------1 (no unit/reference point, no mark) Total : 9 Marks 1.
(b)
(i)
Quite a number of candidates stated wrongly that air moves out of the lungs when the pressure in the lungs is increasing. They failed to see that air is breathed out only when the lung pressure exceeds the atmospheric pressure.
(ii)
This part was well answered, showing that the candidates were familiar with the mechanism of breathing movements.
(iii)
Some candidates did not express the rate of breathing as number of breaths per unit time.
(iv) Many candidates failed to provide a useful reference to indicate that the curve sketched shows an increase in rate and depth of breathing during exercise in comparison with those at rest.
Paper I Marking Scheme & Year Report
P.2
CE BIO 1990 1.
(c)
(i)
(ii)
Alex Lam
(1)
yellow (18 ± 2); purple (53 ± 2) ------------------------------------------------------------------1 or 0
(2)
1 : 3 -----------------------------------------------------------------------------------------------------------1
since the purple and yellow grains occur in a 3:1 ratio -------------------------------------------------1 both parents must be heterozygous -------------------------------------------------------------------------1 purple grain colour is the dominant character ------------------------------------------------------------1 therefore the phenotype of both parents must be purple grain colour -----------------------------1
(iii)
all grains will be yellow ------------------------------------------------------------------------------------------1 because both parents are homozygous for yellow grain colour / homozygous recessive------------1 Total : 8 Marks
1.
(c)
(ii)
Many candidates had difficulty explaining their deductions in words instead of using a genetic diagram. This may be due to the fact that some candidates could not express their deduction logically while others may have learned and used genetic diagrams in a stereotyped manner without understanding the principles involved. A small number of candidates were still unable to define the allele symbols they used in their explanation.
Paper I Marking Scheme & Year Report
P.3
CE BIO 1990 2.
(a)
Alex Lam
(i)
the shorter the time, the faster the rate of enzyme activity --------------------------------------------1
(ii)
pH 2 ------------------------------------------------------------------------------------------------------------------1
(iii)
- incubate the tubes at a higher temperature but not exceeding 40°C
)
3
- cut the 1 cm cube into smaller cubes / using smaller cubes of egg white ) any TWO---1,1 - using a more concentrated enzyme solution / using more enzyme
)
(iv) to measure the changes in weight / size of the egg white cubes within a fixed period of time ---------------------------------------------------------1+1 (v)
stomach -------------------------------------------------------------------------------------------------------------1
(vi) replace the enzyme solution with boiled enzyme solution / distilled water-------------------------1 and repeat the whole experiment ----------------------------------------------------------------------------1 Total : 9 Marks 2.
(a)
(iii)
Many candidates could apply their knowledge on enzyme activity in answering this question.
(iv) This question assesses candidates’ ability in experimental design. The performance on this question was not satisfactory. A common mistake is to use the Biuret test to check the disappearance of the egg white cube rather than the presence of digested proteins in the solution. (vi) Most candidates were able to point out that a control can be set up by replacing the enzyme solution with boiled enzyme solution or distilled water. However, many failed to realise that, for this particular experiment, a control should be set up for each of the given pH values.
Paper I Marking Scheme & Year Report
P.4
CE BIO 1990 2.
(b)
(i)
Alex Lam
to compare / growth at different regions of the radicle -----------------------------------------------1,1 e.g. to find out which region of the radicle grows faster
(ii)
(1)
3 2 1 growth in length mainly takes place in the region of elongation (starting from region 2 or region 3) -------------------------------------------------1 radicle bends downwards -------------------------------------------------------------------------------1 (2)
enable the roots to grow deeper into the soil ------------------------------------------------------1 for better anchorage --------------------------------------------------------------------------------------1 to ensure the root to absorb enough water / mineral nutrients (dissolved mineral salts) / to reach a new source of water (dissolved mineral salts) ------------------------------------------------------------------------1
(iii)
the radicle grows in length but does not bend ------------------------------------------------------------1 (N.B. diagram acceptable) because no gravity acting on it -------------------------------------------------------------------------------1 Total : 9 Marks
2.
(b)
(ii)
(1)
Many failed to show in their diagrams that bending of the radicle takes place at the region of elongation.
(iii)
Many candidates could give a reasonable prediction on the possible changes on the radicle in a space shuttle. Some candidates, however, considered wrongly that *no geotropism* had the same meaning as "no gravity".
Paper I Marking Scheme & Year Report
P.5
CE BIO 1990 2.
(c)
(i)
Alex Lam
A - * vena cava ----------------------------------------------------------------------------------------------------1 for transporting blood / wastes to the heart ----------------------------------------------------------------1 B - urinary * bladder ---------------------------------------------------------------------------------------------1 for storage of urine -----------------------------------------------------------------------------------------------1
(ii)
(1)
X from B / D -------------------------------------------------------------------------------------------------1 Y from C (also accept A) --------------------------------------------------------------------------------1
(2)
(I)
protein is present in Y but not in X because - proteins are too large ---------------------------------------------------------------------------1 - to pass through the glomerulus / capillaries ----------------------------------------------1 (to pass into bowman’s capsule)
(II)
glucose is present in Y but not in X because -glucose passes into the nephron -------------------------------------------------------------1 - is reabsorbed from the glomerular filtrate -------------------------------------------------1 (absorbed from the kidney)
(III) the concentration of urea in X is higher than that in Y because - urea passes into the nephron / urea in nephron is not reabsorbed -----------------1 - water is reabsorbed form the glomerular filtrate -----------------------------------------1 Total : 12 Marks 2.
(c)
(i)
Many candidates had mistaken “structure A”. the vena cava. as the renal vein.
(ii)
(2)
Many candidates failed to provide a complete explanation for this question. They missed some points although they seemed to understand the mechanisms involved. A common misconception is that protein can pass into the Bowman’s capsule and be reabsorbed later. Some candidates failed to point out that the increase in urea concentration is due to the reabsorption of a large amount of water.
Paper I Marking Scheme & Year Report
P.6
CE BIO 1990 3.
(a)
(i)
(ii)
Alex Lam
(1)
* reptile - possession of dry scales -----------------------------------------------------------------1,1
(2)
* bird - possession of feathers / wing / beak -----------------------------------------------------1,1
(1)
toadstools - saprophytic / heterotrophic tree - autotrophic ,
(2)
------------------------------------------------------------------1 or 0
toadstools - no formation of flowers / fruits / seeds tree - formation of flowers / fruits / seeds ;
(iii)
---------------------------------------------1 or 0
(1)
centipede, mite --------------------------------------------------------------------------------------1 or 0
(2)
crow, snake -------------------------------------------------------------------------------------------1 or 0
(iv) (1)
* predation / prey and predator ------------------------------------------------------------------------1
(2)
* competition ------------------------------------------------------------------------------------------------1 Total : 10 Marks
3.
(a)
(i)
Some candidates did not point out that dry scale is a characteristic feature of reptiles. It is surprising to note that quite a number of candidates wrongly classified crows as mammals. This reflects a very weak knowledge of the general classification of living organisms in the study of Biology.
(ii)
Many candidates answered wrongly that the toadstool carries out asexual reproduction while the fruit tree carries out sexual reproduction. They did not realise that both the toadstool and the fruit tree, like most plants, can carry out both asexual and sexual reproduction.
(iii)
Many candidates failed to examine the food web diagram carefully and made mistakes in the identification of carnivores and omnivores.
Paper I Marking Scheme & Year Report
P.7
CE BIO 1990 3.
(b)
(i)
Y
Alex Lam ------------------------------------------------------------------------------------------------------------------1
Y causes a greater drop in weight of soil animals -------------------------------------------------------1 (ii)
less mites in area B / chemical X kills a lot of mites -----------------------------------------------------1 therefore predation on springtails is reduced -------------------------------------------------------------1 springtails multiply more rapidly / reproduce---------------------------------------------------------------1
(iii)
crop plants absorb DDT at low concentration -------------------------------------------------------------1 - pest feed on a lot of crop -------------------------------------------------------------------------------------1 - birds feed on a lot of pests -----------------------------------------------------------------------------------1 N.B.
mention feeding relationship ----------1 mention a lot of ---------------------------1
DDT accumulates in the body/cannot be removed or excreted / not biodegradable -----------1 therefore DDT is concentrated along the food chain Total : 9 Marks 3.
(b)
(ii)
Although most candidates pointed out that there is less predation on the springtails, very few realised that the rapid multiplication or reproduction of the springtails is also a contributing factor to the subsequent increase in their number.
(iii)
This part is quite demanding in that it requires the candidates to organise their ecological concepts in order to give a clear and accurate account of the concentration effect of DDT along the food chain. Many candidates, for example, did not state the correct feeding relationship of the food chain. The performance on this part was rather unsatisfactory.
Paper I Marking Scheme & Year Report
P.8
CE BIO 1990 3.
(c)
(i)
(1)
Alex Lam children ------------------------------------------------------------------------------------------------------1 by calculation : ---------------------------------------------------------------------------------------------1 378 children 7560 20
(2)
(ii)
>
162 men 11340 70
,
153 women 8400 55
,
149 pregnant women 9660 65
children need to maintain a higher metabolic rate in order to -------------------------------1 - support greater growth/muscular activities
)
- compensate for the higher rate of heat loss
) any ONE ----------------1
men require less vitamin D than children ------------------------------------------------------------------1 because they have already well developed bones / teeth ---------------------------------------------1
(iii)
pregnant women --------------------------------------------------------------------------------------------------1 for the growth and development of the foetus / embryo ------------------------------------------------1
(iv) women require more iron than men -------------------------------------------------------------------------1 for the formation of red blood cells in the blood ----------------------------------------------------------1 which is lost periodically (during menstruation) ----------------------------------------------------------1 Total : 11 Marks 3.
(c)
(i)
(1)
Many candidates did not present their calculations as required by the question.
(iv) Many candidates could relate the greater demand for iron by women to the loss of blood during menstruation. However, some failed to pinpoint that iron is required for the formation of red blood cells or haemoglobin.
Paper I Marking Scheme & Year Report
P.9
CE BIO 1990 4.
Alex Lam
(a) 14.0
(i)
as shown in the graph correct choice of axes ---------------------------1
12.0
correct labelling of axes and giving a key to each curve ----------------1 correct plotting of 5 points
10.0
for each curve ---------------------------------1@ 8.0
leaf
(ii)
(1)
(10.2 ± 0.2) × 10 mg = 102 ± 2 mg--------------------------------1
6.0
(N.B. No unit, no mark) (2)
Net release of CO2 Net uptake of CO2 (mg/cm2/h2)
[ (10.2 ± 0.2) + 1.0] × 10 mg ------------1 = 112 ± 2 mg---------------------------------1
4.0
(N.B. No unit, no mark) stem
2.0
0
2
4
6
8
10
-2.0 Light intensity (arbitrary units)
(iii)
(1)
the leaf -------------------------------------------------------------------------------------------------------1 because the carbon dioxide uptake by the leaf is always higher than that by the stem -----------------------------------------------------------------1
(2)
- more choloplast (per unit area) / closely packed palisade cells
)
- more stomata / numerous air space
) any TWO -------1,1
- thin / flat
) Total : 11 Marks
4.
(a)
(i)
Graph plotting was satisfactory in general but some candidates failed to label the axes with correct units. Some candidates plotted the data on wrong axes while a small number of candidates plotted their graphs on the answer book instead of on graph paper.
(ii)
This part was poorly answered as many candidates gave the net carbon dioxide uptake by a leaf of 1 cm2 area. They failed to read the question carefully which states that the leaf has an area of 10 cm2. For the second Part of this question, many failed to consider the effect of respiration when calculating the net carbon dioxide uptake by the leaf.
(iii)
(2)
Some candidates failed to compare the leaf with the stem and did not stress the point that the leaf contains more chloroplasts and stomata than the stem.
Paper I Marking Scheme & Year Report
P.10
CE BIO 1990 4.
(b)
(i)
Alex Lam
(1)
skeletal muscle and skin --------------------------------------------------------------------------1 or 0
(2)
skeletal muscle : to carry more nutrients /oxygen -----------------------------------------------------------------------1 for muscle contraction -----------------------------------------------------------------------------------1 skin : to carry more blood to surface to increase heat loss --------------------------------------------1 to help maintaining a constant body temperature / to prevent overheating of the body -------------------------------------------------------------------1
(3)
(ii)
increased heart beat rate
)
stronger heart beat
) any TWO -------------------------------------1,1
vasodilation in the organs concerned
)
increased blood pressure
)
the brain the brain will be damaged when there is a shortage of blood supply -------------------------------1 (or other reasonable answers) Total : 9 Marks
no year report
Paper I Marking Scheme & Year Report
P.11
CE BIO 1990 4.
(c)
(i)
Alex Lam
production of eggs -----------------------------------------------------------------------------------------------1 secretion of hormones ------------------------------------------------------------------------------------------1
(ii)
(iii)
(1)
the fusion / union of male and female gametes ---------------------------------------------------1
(2)
B ---------------------------------------------------------------------------------------------------------------1
(1)
the egg and the sperms cannot reach each other ------------------------------------------------1
(2)
- to ensure that the fertilized eggs
)
- to ensure that the embryo is ready for implantation
) any ONE -------------------1
- to ensure that the uterine wall is ready for implantation
)
(3)
to allow the formation of placenta / for implantation of embryo -------------------------------1 for gaseous exchange / nutrition / excretion of the embryo ------------------------------------1 and for protection of the embryo ----------------------------------------------------------------------1
OR (As a shock absorber / stable environment) (4)
the baby is, in fact, developing inside the uterus of the mother for most of the time --------------------------------------------------------------------1 Total : 10 Marks
4.
(c)
(iii)
Many candidates were able to apply their knowledge of mammalian reproduction to explain the process of “in vitro fertilization” although this topic is not explicitly mentioned in the syllabus.
Paper I Marking Scheme & Year Report
P.12
CE BIO 1990 5.
(a)
(i)
Alex Lam
set-up C -------------------------------------------------------------------------------------------------------------1 set-up D -------------------------------------------------------------------------------------------------------------1 no transpiration occurs in both set-ups ---------------------------------------------------------------------1 because in C, the Vaseline blocks the stomata / prevents transpiration to take place ---------1 because in D, there is no leaf surface for transpiration ------------------------------------------------1
(ii)
(1)
curve II : set-up B -----------------------------------------------------------------------------------------1 curve III : set-up A -----------------------------------------------------------------------------------------1
(2)
Deduction : more stomata on the lower epidermis ----------------------------------------------1 Reason : If the lower surface is smeared with Vaseline, the rate of water loss is reduced by more than half / greatly reduced when compared with untreated leaves ----------------------------------------------------------------------------1
(iii)
(1)
the rate of weight loss would be reduced -----------------------------------------------------------1 because of the higher humidity ------------------------------------------------------------------------1
(2)
the rate of weight loss would be reduced -----------------------------------------------------------1 because stomata close in darkness / temperature is lower ------------------------------------1 Total : 13 Marks
5.
(a)
(i)& (iii)
Some candidates wrongly related the changes in weight to photosynthesis instead of transpiration.
(ii)
The question demands clarity of thought as it involves the comparison of conditions in four different set ups.
Paper I Marking Scheme & Year Report
P.13
CE BIO 1990 5.
(b)
Alex Lam
(i)
fingertip -------------------------------------------------------------------------------------------------------------1
(ii)
because the pin (stimulus) is not applied directly onto a touch receptor / nerve ending ------1 (or other reasonable answers)
(iii)
touch receptor stimulated --------------------------------------------------------------------------------------1 nerve impulses pass along the sensory neurone --------------------------------------------------------1 and via the relay neurone / assoication neurone / intermediate nueorne to the brain where the sensation of touch is produced -------------------------------------------------1 and then nerve impulses sent from the brain via the motor neurone to the muscles responsible for speech -------------------------------------------------1 (N.B. accept flowchart)
(iv) (1)
* reflex --------------------------------------------------------------------------------------------------------1
(2)
avoid danger immediately / provide immediate protection -------------------------------------1
(3)
inborn / learning not required
)
involuntary / not controlled by will ) any ONE ---------------------------------------------------1 stereotype / fixed response
) Total : 9 Marks
5.
(b)
(iii)
While this part was well attempted, some candidates did not present a complete nervous pathway to indicate the coordinating role of the brain in sensation and in voluntary action.
Paper I Marking Scheme & Year Report
P.14
CE BIO 1990 5.
(c)
(i)
dehydration -
Alex Lam milk powder, dried
)
mushroom, spaghetti,
)
baked bean
)
ice cream, cheese, butter,
) any TWO pairs ----------------------------1+1
fresh milk
)---------------------------------------------------1+1
canning -
baked bean, milk powder
)
adding -
baked bean, cheese, butter,
)
preservative
ice cream, milk powder
)
refrigeration -
(N.B. 2 examples of food required for each method) (ii)
dehydration refrigeration -
canning -
preservative (iii)
without water, microorganisms
)
cannot grow
)
low temperature inactivates
)
the microorganisms/lower
)
enzyme activity in
)
microorganisms
) any TWO ---------------------1,1
bacteria are killed by heating /
)
sealing the can in vacuum prevents
)
entry of microorganism OR depletes
)
the can of oxygen for further
)
growth of bacteria
)
prevent growth of microorganism
)
change in quality of food (smell, taste, texture, appearance, etc.) ----------------------------------1 danger of food poisoning / harmful to the body ----------------------------------------------------------1 Total : 8 Marks
5.
(c)
For part (iii), however, while many candidates stated the effect of bacteria in causing diseases, very few pointed out the effect of microorganisms on the quality of food.
Paper I Marking Scheme & Year Report
P.15