S.

6 Mock Examination 2021-2022

Biology
Marking scheme

Paper 1 Section A

Question Answer Question no. Answer Question no. Answer

no.
1 A 13 A 25 C
2 A 14 D 26 B
3 C 15 C 27 A
4 B 16 A 28 D
5 D 17 D 29 C
6 A 18 D 30 B
7 D 19 B 31 C
8 D 20 D 32 B
9 C 21 B 33 D
10 B 22 D 34 C
11 A 23 B 35 B
12 A 24 A 36 C

Paper 1 Section B
1 C 1m
A 1m
D 1m

2 a X: Epidermis* Z: spongy mesophyll 1m, 1m

(tissues are NOT cells)
b The cells of tissue Y are closely packed / contain high desity of chloroplasts. 1m
This captures more light energy / intensity for photosynthesis. 1m

3 a Maltose* 1m
b i The pancreatic juice secreted into the duodenum contains pancreatic amylase which catalyses the
breakdown of starch into maltose. 1m
Also, maltase in the cell membranes of certain epithelial cells of the small intestine catalyses the
breakdown of maltose into glucose. 1m
Therefore, the concentration of sugar increases. 1m
ii Glucose is absorbed from the lumen of the small intestine into the blood in the capillaries by diffusion
and active transport. 1m
Therefore, the concentration of sugar decreases. 1m

4 a Cytoplasm 1m
b Amino acid U 1m
c P: Cys 1m
V: Gln 1m
d Mutation on mRNA leads to different codons on the mRNA strand 1m
Different tRNA / anticodon with different amino acids would match with the mutated sequence 1m
A difference amino acid chain leads to different folding / structure of the resulting protein. 1m

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 (change of protein structure leads to functional change)
5 a Due to the formation of the Isthmus of Panama, the ancestral population was separated into two groups
and the two groups became isolated from each other / could not meet to interbreed and gene flow stopped.
1m
Each isolated group was subjected to a different set of environmental conditions. 1m
The two groups evolved differently due to natural selection. / Different characters favourable to those
particular environmental conditions were selected by natural selection. 1m
Over time, their genetic compositions became so different that they could no longer interbreed to produce
fertile offspring even if they met again. They became two different species. 1m
b 1 a Without a vertical black stripe through the eye .………………………………………….. 2
1 b With a vertical black stripe through the eye .……………………………………………... 3
2 a Without two vertical black stripes behind its head (or other reasonable answers) ..… Fish P
2 b With two vertical black stripes behind its head (or other reasonable answers) ...…… Fish Q
3 a Body with horizontal stripes ……………………………………………………….... Fish R
3 b Body with vertical stripes …………………………………………………………… Fish S
3m

6 a To remove any substance present on the seeds that may inhibit germination / may interfere with the
results of the experiment. 1m
b Oxygen should be present so that the seeds can carry out respiration to release energy for growth. 1m
A warm environment should be provided so that the enzymes in the seeds can work efficiently. 1m
(NOT water, beasue there were water from step 3 already.)
c Repeat step 4 to prepare an identical set-up except that distilled water would be used instead of the extract
of tomato fruit wall 1m
and carry out the experiment under the same environmental conditions. 1m
This ensures any inhibition on seed germination in the experimental set-up is due to the extract of tomato
fruit wall only. 1m
d This ensures that seed germination will only occur after the fruit is eaten by animals or decomposed /
allows enough time for seed dispersal. 1m

7 a The water tank absorb heat / prevents heat emitted by the table lamp that heats up the dilute sodium
hydrogencarbonate solution. 1m
This helps ensure that temperature is kept constant during the experiment (and the difference in the rate of
photosynthesis is due to the independent variable only.) 1m
b By using dilute sodium hydrogencarbonate solutions of different concentrations. 1m
(or other reasonable answers)
c This allows time for the aquatic plant to equilibrate / adjust to a new concentration of carbon dioxide.
1m
d As the carbon dioxide concentration increases from 0% to 0.4%, the rate of photosynthesis increases
1m
because there is more carbon dioxide available for the Calvin cycle. 1m
As the carbon dioxide concentration increases beyond 0.4%, the rate of photosynthesis levels off 1m
because it is limited by other factors (e.g. light intensity or temperature). 1m

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8 a Mucus secreted by mucus-secreting cells on the inner wall of the respiratory tract may trap
M. tuberculosis in inhaled air. 1m
Cilia of ciliated epithelial cells beat to sweep the mucus with trapped M. tuberculosis towards to pharynx
for coughing or swallowing. 1m
b The tubercles replace some air sacs. The surface area for diffusion of gases is reduced. 1m
Hence, the rate of diffusion of oxygen into blood decreases. The patients experience shortness of breath
due to low blood oxygen levels. 1m
c The capillaries in the lungs rupture / are damaged. 1m
d i This ensures that bacteria which are resistant to a particular antibiotic can be killed by another
antibiotic that they are sensitive to. 1m
ii Complete the whole course of treatment even symptoms have disappeared. 1m
Take the right dose of antibiotics at the right time. 1m
(or other reasonable answers)
e i Cover our nose and mouth with tissue paper when coughing or sneezing and dispose of nasal and
mouth discharge properly. /
Wear a face mask if we have any respiratory symptoms and seek treatment promptly. /
Wash our hands with liquid soap and water, especially after coughing or sneezing. /
Receive vaccination against TB according to immunization schedule. 1m
(or other reasonable answers)
ii Establish and implement immunization programmes for TB. /
Carry out disease surveillance for TB. /
Educate the public about the importance of taking preventive measures against TB. 1m
(or other reasonable answers)

9 a i A transect was placed from the edge of the woodland towards the centre of the woodland. 1m
Quadrats were placed at 5-m intervals on one side of the transect. 1m
The positions of the quadrats and the areas covered by each species in the quadrats were recorded.
1m
The percentage cover of each species was calculated by dividing the area covered by the species in
the quadrat by the total area of the quadrat. 1m
ii Correct title 1m
Choice of axes 1m
Correct plotting and joining of lines 1m
Axes with labels and units 1m
(see garph below)

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 Abundance of plant species X and Y
from the edge towards the centre of a woodland

100 Key:
species X
90 species Y

80
abundance (percentage cover)

70

60

50

40

30

20

10

0
0 5 10 15 20 25 30

distance from the edge of the woodland (m)

b Individual plants were not shown clearly from the topic view . 1m

10 a Amino acid X is constantly formed as the proteins in food are broken down by proteases. 1m
However, as no functional enzyme P is produced in the patients, amino acid X cannot be converted to
amino acid Y / Therefore, amino acid X accumulates to a high level, 1m
while the level of amino acid Y in blood becomes low as amino acid Y is used for the synthesis of
pigments and cannot be formed from amino acid X. 1m
As a result, the ratio of amino acid X to amino acid Y in blood in the patients becomes significantly
higher than that in healthy people. 1m
b The patients can still obtain a certain amount of amino acid Y from their diet as the proteins in food are
broken down. 1m
However, as the patients do not have functional enzyme P for converting amino acid X into amino acid Y,
amino acid Y cannot be formed in their bodies. 1m
Therefore, the supply of amino acid Y for the synthesis of pigments is reduced 1m
and most patients have a lighter skin and hair colour than healthy people.
(Any two above)

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 c i Cindy has the genetic disease. She must be homozygous recessive. 1m
She must have received one recessive allele for the disease from Mr Wong and one from Mrs Wong.
1m
Mr Wong is normal. He must have at least one normal allele. 1m
Therefore, Mr Wong is heterozygous. 1m
ii Partially agree/ do not agree
The probability of Elaine being normal is 50% because she could have inherited a normal allele or a
recessive allele from Mr Wong. 1m
However, the probability of Elaine having the disease or not is not determined by her siblings’
phenotype. 1m

11 How water is transported in humans: (max.3)

⚫ Water is absorbed from the contents of alimentary canal into the blood and then transported to the heart.
1m
⚫ The heart acts as a pump to drive the flow of blood in the arteries. Blood eventually reaches the capillaries
in different body tissues. 1m
⚫ At the arteriole end of the capillary bed, the pressure of blood / hydrostatic pressure in the capillaries is
higher than that of tissue fluid. This forces some components of the plasma, including water, out of the
capillary walls to form tissue fluid. 1m
⚫ Water in tissue fluid then enters body cells by osmosis. 1m
How water is transported in terrestrial plants: (max. 2)
⚫ Water is lost from leaves through transpiration. (Transpiration pull is created.) 1m
⚫ Water is drawn up the xylem vessels by transpiration pull as a continuous stream from the roots to the
leaves. 1m
⚫ Water in the soil is absorbed into the roots by osmosis. 1m
The role of water in the transport of materials in humans: (max. 2)
⚫ Carbon dioxide (produced by body cells during respiration) is transported in plasma, in the forms of
dissolved gas and hydrogencarbonate ions, (to the air sacs of the lungs for removal.) 1m
⚫ Water-soluble nutrients (e.g. glucose, amino acids and vitamin C) are absorbed from the alimentary canal
into the blood and are transported by blood to body cells. 1m
(or other reasonable answers)
The role of water in the transport of materials in terrestrial plants: (max. 1)
⚫ Minerals absorbed into the roots dissolve in water and are transported up the plant along with the water to
other parts of the plant. 1m
(or other reasonable answers)
Communication 3m

Paper 2
1 a i Proteins are too large to pass through the walls of the glomerulus and the Bowman’s capsule.
1m
ii (1) No glucose is present in the fluid sample taken from point Z 1m
because all glucose in the glomerular filtrate has been reabsorbed in the first coiled tubule.
1m
(2) The concentration of urea is the highest in the fluid sample collected from point Z
1m
because a large amount of water is reabsorbed from the glomerular filtrate as the filtrate flows
along the kidney tubule. 1m
(NOT urea being reabsorbed)
iii As the release of ADH is inhibited, the walls of the second coiled tubule and collecting duct becomes
less permeable to water, 1m
a smaller proportion of water is reabsorbed from the glomerular filtrate. 1m
A larger volume of dilute urine is produced. As a result, the body loses a larger amount of water.
1m

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 iv Increased thirst / frequent urination / a large volume of urine / very dilute urine is produced / salt
imbalance (any 2) 1m x 2

1 b i The LH level remains fairly constant / There is no LH surge over the period. 1m
Ovulation is not triggered. 1m
(Surge or peak of LH is NOT the same as high LH level; the LH level was not shown anyway.)
ii As the menstrual cycle length becomes much longer than normal, 1m
the uterine lining becomes excessively thick. 1m
Heavy menstrual bleeding results when this thickened uterine lining sheds. 1m
iii High levels of oestrogen and progesterone inhibit FSH secretion 1m
by the pituitary gland. 1m
Due to the low level of FSH in blood, follicle development stops and cysts do not develop. 1m
iv Mary still have mature follicle that can be fertilized. 1m
Could recommend her to conduct IVF by extracting and putting the zygote back to her uterus.
/ intake of LH to stimulate the ovulation. 1m

2 a i Dissolved oxygen in the water is consumed by microorganisms 1m

in the decomposition of organic matter. 1m
A high BOD value reflects that a large amount of organic matter is present and a large amount of
oxygen is required for microbial decomposition. 1m
ii Excess nitrate and phosphate in the water may lead to an algal bloom. 1m
The algae use up the oxygen in the water for respiration at night, causing suffocation of aquatic
organisms. 1m
The decomposition of algae releases toxic substances, which may kill aquatic organisms. 1m
iii Organic nitrogen in the sewage is converted to ammonium compounds / ammonia through
decomposition by microorganisms / decomposers / putrefying bacteria in the artificial wetland. 1m
The ammonium compounds / ammonia can be converted to nitrates through nitrification by nitrifying
bacteria. 1m
The nitrates are absorbed by the wetland plants for growth. / The nitrates are converted to nitrogen
gas through denitrification by denitrifying bacteria. 1m
All these processes lead to a decrease in the total nitrogen content in the effluent compared with the
influent. 1m
iv Perennial plants can grow for many years without harvest / without the need of replanting every year.
1m
This allows the accumulation/storage of pollutants in the plant bodies over a long period of time. /
This can reduce the maintenance cost of the wetland system. 1m

2 b i Starch-based plastics are made from starch which is renewable, while conventional plastics are made
from non-renewable petroleum / fossil fuels. 1m
ii Carbon dioxide 1m
Water 1m
iii Reduce the use of disposable plastic items / Reuse plastic items / Bring you own bags / Separate
plastic waste for recycling 1m
(Accept other reasonable answers.)
iv (1) At pH 7.0, increasing the temperature from 30 °C to 40 °C increases the amount of the plastic
being broken down. (1)
At pH 9.0, increasing the temperature from 30 °C to 40 °C decreases the amount of the plastic
being broken down. (1)
(2) Any two of the following: (2)
 Size/thickness of the plastic sheets
 Concentration of the enzyme solution
 Time of immersion

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3 a i (1) The mouth of the flask should be flamed each time it is opened and closed. 1m
The cap of the flask should be held between the little finger and the palm after it is removed
from the flask. 1m
(or other reasonable answers)
(2) To help avoid contamination of the cultures with unwanted microorganisms from handlers and
the environment. 1m
To help avoid contamination of handlers and the environment with E. coli. 1m
(3) Using OD to measure bacterial growth is easy and quick. 1m
ii Bacteriophage P1 together with P2 1m
The OD or E. coli level of this treatment drops to a very low level / close to zero 1m
and remains low at the end of the investigation. 1m
iii It takes some time for the bacteriophages to bind to E. coli / multiply in the bacteria and cause the
bacteria to burst. 1m
iv The binding proteins on bacteriophages cannot bind to the receptors on V. cholerae. / The shape of
the binding proteins on bacteriophages does not match the receptors on V. cholerae. 1m
3 b i Food-borne infection 1m
ii (1) 7.5 seconds 1m
(2) Enzyme B 1m
From the graph, the denaturation of enzyme B requires a longer duration than the killing of
M. bovis under the same temperature. / enzyme B has a higher tolerance to heat than M. bovis.
1m
If no activity of enzyme B is detected in a pasteurized milk sample, there should be no living
M. bovis bacteria in the milk. 1m
iii No. 1m
Raw milk is not healthier as it may contain M. bovis which causes peritoneal TB or some other
pathogens that can cause serious diseases or even death. /
Pasteurized milk is as nutritious as raw milk as the temperature used in pasteurization is not high
enough to destroy most nutrients in milk. /
Raw milk is not a better choice for people with lactose intolerance because it still contains a large
amount of lactose / does not contain lactase to catalyze the breakdown of lactose. 2m
(any two or other reasonable answers)
iv Ultra high temperature treatment 1m
o
Milk is heated to 135 C for 3–5 seconds to kill all the spoilage microorganisms. 1m
(or other reasonable answers)

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4 a i (1) 7 1m
(2) CAACAGCCGCCACCGCCG 1m
ii Band B 1m
The DNA fragments with a larger number of CAG repeats are longer and thus larger in size. 1m
As DNA is negatively charged and will move towards the positive pole during gel electrophoresis,
1m
the fragments with a larger number of (or 39) CAG repeats move through the pores of the gel 1m
slower than those with a smaller number of (or 33) CAG repeats.
(Therefore, in a fixed period of time, the DNA fragments with a larger number of CAG repeats travel
a shorter distance.)
iii Individual 3 is homozygous for the HTT gene. /
Both alleles of the HTT gene have the same number of CAG repeats. 1m
iv (1) Individual 4 has a high risk of Huntington’s disease and the onset would likely occur during
adulthood. 1m
(2) Yes. 1m
Since Huntington’s disease can be inherited, the test result of individual 4 is relevant to the
health of his daughter / may influence the couple’s decision on having another child or not
in the future. 1m
OR
No. 1m
Individual 4 has the ownership of his test result and only he can decide whether to disclose
the test result to his family members or not. 1m
(or other reasonable answers)

4 b i These cells are able to undergo unlimited mitotic cell division /

able to differentiate into different types of plant cells. 1m
ii (1) The transformed maize cells have the plasmid with the kanamycin resistance gene, which
allows the cells to grow well and develop into a plantlet on the agar plate with kanamycin.
1m
The non-transformed maize cells do not have the resistance gene and do not grow well on the
agar plate / eventually die. 1m
(2) Sugar / glucose 1m
The maize cells cannot carry out photosynthesis / cannot produce their own sugar so they need
an external supply of sugar for respiration to release energy and as raw materials for growth.
1m
Plant hormones / auxins 1m
This is to promote the growth and development of shoots and roots of maize plants. 1m
(or other reasonable answers)
iii (1) The higher the concentration of Bt toxins in the maize leaves, the higher the death rate of the
caddisfly larvae. /
The longer the period of feeding maize leaves with Bt toxins to the caddisfly larvae, the higher
the death rate of the caddisfly larvae. 1m
(2) In nature, the caddisfly larvae live underwater and fallen leaves from Bt maize plants may
not be their only food source. /
Some of the Bt toxin concentrations used in the experiment may be significantly higher than
the concentrations to which the caddisfly larvae are exposed in nature. /
Caddisfly is only one of the insect species in the river. Comprehensive field studies should be
carried out to determine the (long term) ecological effects of growing Bt maize plants on nearby
freshwater communities. 2m
(Any two or other reasonable answers)

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