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Biology 2
Suggested Answers
Chapter 9
Part 1
1.
B
2.
B
3.
C
4.
A
5.
C
6.
B
7.
B
8.
C
9.
A
Multiple Choice Questions
Part 2
Structured Questions
1.
(i)
food / glucose + oxygen  carbon dioxide + water
(1 mark)
(ii)
to absorb carbon dioxide in the air from the bell jar
(1 mark)
(iii)
(1)
(1 mark)
The weight of flask A did not change
because carbon dioxide produced by the rat is absorbed by potassium
hydroxide
(1 mark)
The weight of carbon dioxide absorbed was equal to the weight of
water lost from the potassium hydroxide
(1 mark)
(2)
(iv)
(211.8 - 211.8) – (206.1 – 260.7)/30 x 60 gCO2 h-1
(1 mark)
= 1.2 gCO2 h-1 (0.02 gCO2 min-1)
(1 mark)
The bell jar has to be covered by a black cloth/paper
(1 mark)
The pot of soil has to be covered by a plastic bag
(1 mark)
Total: 9 marks
2.
(i)
The mealworms take in oxygen during respiration
(1 mark)
Any carbon dioxide produced by the mealworms is absorbed by the soda
lime
(1 mark)
This leads to a drop in air pressure inside the syringe and the water droplet
will be drawn downwards
(1 mark)
Thus the rate of movement of the water droplet indicates the rate of
respiration
(1 mark)
Effective communication (C)
(1 mark)
1
(ii)
(1)
The rate of respiration of mealworms increases with an increase in
temperature from 20℃ to 50℃
(1 mark)
(2)
With an increase in the external temperature,
the body temperature of mealworms rises
(1 mark)
Thus the enzymatic activity of the worms increases
(1 mark)
This leads to an increase in the metabolic rate/body activities/ energy
demand
(1 mark)
thus the respiration rate of the mealworms rises
(iii) To allow time for the air temperature inside the syringe to become equal to
the temperature of the water bath
(1 mark)
and the respiration rate of the mealworms to become adjusted to the new
temperature
(1 mark)
Total: 10+1 marks
3.
(i)
The muscles carry out anaerobic respiration
(1 mark)
to release additional energy for muscle contraction.
(1 mark)
As anaerobic respiration produces lactic acid, it will lead to an increase in
blood lactic acid concentration.
(1 mark)
(ii)
glucose + oxygen  carbon dioxide + water
(2 marks)
(iii) Because it lowers the pH of the blood / tissue fluid which adversely affects
cellular activities / it inactivates enzymes.
(1 mark)
(iv)
Slow jogging
(1 mark)
This is because slow jogging can maintain a relatively high rate of heart beat
/ blood flow / breathing,
(1 mark)
which increases the rate of oxygen supply to the body,
(1 mark)
thus enhances the breakdown of lactic acid / conversion of lactic acid to
glycogen.
(1 mark)
Effective communication (C)
(1 mark)
Total: 10+1 marks
Chapter 10
Part 1
1.
D
2.
B
3.
C
4.
C
5.
A
6.
D
7.
D
Multiple Choice Questions
2
8.
C
9.
D
10. C
Part 2
Structured Questions
1.
(1)
(i)
xylem
(1 mark)
(2)
Structural feature
The xylem vessels in X have no cell
content / no end walls / large lumen
The xylem vessels in X have thick
cell wall
Adaptation
This allows a free flow of water
inside
This can prevent the collapse of
the xylem vessels
(any one set) (1+1 marks)
(ii)
Region R
(iii)
(1)
(1 mark)
The support of the stem of plant B is mainly due to the turgidity of
cells in region S / thin-wall cells
(1 mark)
Under a hot and sunny condition,
the rate of transpiration of the plant becomes greater than the rate of
water absorption
(1 mark)
The cells in region S lose water
(1 mark)
and hence lose their turgidity/ become flaccid and thus causing the
bending of the stem
(1 mark)
Effective communication (C)
(2)
(1 mark)
The support of the stem of plant A is due to the presence of xylem /
independent of the water content of the plant
(1 mark)
because the xylem contains thick-wall cells
(1 mark)
Total: 10+1marks
2.
(i)
(1/2 + 1/2 mark)
(1)
x = 1.5
y = 1.2
(2)
x represents the amount of water absorbed by the plant
(1 mark)
y represents the amount of water transpired / lost by the plant (1 mark)
(3)
Value x is larger than value y
(1 mark)
This indicates that there is a net gain of water by the plant
(1 mark)
The water gained is essential for various life processes e.g. formation
of new cells, photosynthesis, support, cellular metabolism, etc.
(1 mark)
Effective communication (C)
(ii)
(1 mark)
Value x would decrease
(1 mark)
because vaseline blocks the stomata and the leaf surfaces
(1 mark)
so the rate of transpiration would drop
(1 mark)
3
As transpiration enhances the absorption of water less water would be
absorbed by the plant
(1 mark)
Total: 10+1 marks
3.
(i)
(ii)
(1)
stoma
(1 mark)
(2)
It allows carbon dioxide to enter the leaf
(1 mark)
for photosynthesis in cell B
(1 mark)
The moss leaf has no cuticle / no waterproof covering
(1 mark)
And it is one-cell thick
(1 mark)
so the surface area to volume ratio is large
(1 mark)
(1)
This would lead to a high rate of water loss from the plant / so the moss
would become dehydrated easily in dry environment
(1 mark)
(2)
Atmospheric oxygen dissolves into the water film on the moss leaf
(1 mark)
and then diffuses in
(1 mark)
through the entire surface of the leaf
(1 mark)
OR
At night, moss cells carry out respiration only and thus lower the
oxygen concentration in the cells
(1 mark)
Atmospheric oxygen dissolves into the water film on the moss leaf
(1 mark)
and diffuses into the leaf
(1 mark)
(any one set)
Total: 10 marks
4.
(a)
(i) water
(1 mark)
mineral salts
(1 mark)
(ii) gives mechanical support to the tree
(b)
(1 mark)
The leaves at the aerial parts synthesise carbohydrates, the food for the
plant, by photosynthesis.
(1 mark)
The food is then transported to all parts of the plant by phloem.
(1 mark)
If that part of phloem is removed, the roots cannot receive any food for
respiration and finally die.
(1 mark)
but the aerial parts can survive longer with the food they produce. (1 mark)
(c)
(i)
4
The effect of the percentage of oxygen in the solution on the
rate of potassium ion absorption
Rate of
absorption of
potassium
ions in
arbitrary units
Oxygen in the solution (%)
(4 marks)
(ii) The rate of potassium ion absorption increases with the increasing
oxygen percentage in the solution.
(1 mark)
It can be deduced that the root cells uptake the ions by active transport.
(1 mark)
Root cells require oxygen to respire to supply energy to the process.
(1 mark)
(iii) Waterlogged soil lacks oxygen, so the root cells cannot respire
aerobically.
(1 mark)
As a result, there is insufficient energy for active transport to absorb
minerals for proper growth.
(1 mark)
Total: 16 marks
Chapter 11
Part 1
1.
C
2.
A
3.
C
4.
A
5.
A
Multiple Choice Questions
5
6.
D
7.
A
8.
C
9.
B
10. A
11. B
Part 2
Structured Questions
1.
(1)
dietary fibre
(2)
- It adds bulk to food,
(i)
(1 mark)
- promotes peristalsis of the intestine
- and prevents constipation
(ii)
(1)
(any two) (1+1 marks)
Milk chocolate contains large proportion of energy-rich
food substance/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 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)
Effective communication (C)
(iii)
(1 mark)
Food from animals is rich in protein
(1 mark)
which is necessary for the growth of the child
(1 mark)
Total: 11+1 marks
2.
(i)
The daily energy intake of Jane is 6620 kJ
(1 mark)
which is less than her daily energy requirement
(1 mark)
Thus she has to consume / mobilise the fat / food reserve in her body leading
to a loss in her body weight
(1 mark)
Effective communication (C)
(1 mark)
Amino acids are broken down in the liver
(1 mark)
to from carbohydrate / a part without nitrogen (and urea)
(1 mark)
which will be oxidized / used in respiration to release energy
(1 mark)
(iii)
anaemia
(1 mark)
(iv)
Rub a piece of ham on a filter paper
(1 mark)
(ii)
6
A translucent spot will remain after drying
(1 mark)
Immerse the paper in an organic solvent and the spot will disappear(1 mark)
OR
Add alcohol to a piece of ham and obtain a clear solution
(1 mark)
Add water to the clear solution
(1 mark)
The solution will turn milky
(1 mark)
(any one set) Total: 10+1 marks
3.
(a)
- A vegetarian diet has low fat content
This will reduce the risk of obesity / heart diseases
- A vegetarian diet has a high content of dietary fibre
This helps maintain normal peristalsis / prevent constipation / reduce the
risk of colon cancer / avoid overeating
- A vegetarian diet has a high vitamin C content
This helps the formation of connective tissue / preventing scurvy.
(any two) (2+2 marks)
(b)
beans / peas / nuts / mushroom (accept other reasonable answers) (1 mark)
(c)
In the first way, humans feed directly on the crops grown, while in the
second way humans feeds on cattle which in turn feeds on grass / In the first
way, humans obtain food through a shorter food chain / a smaller number of
trophic levels.
(1 mark)
Since energy is lost along the food chain
(1 mark)
through respiration, death, excretory products, faeces
(1 mark)
more energy in the form of organic matter will be available to humans in the
first way.
(1 mark)
Effective communication (C)
(1 mark)
Total: 9+1 marks
4.
(a)
(i) (206－118) x 3＝264 kJ
(3 marks)
(ii) Sports Drink contains more carbohydrates which can be respired to
release energy during the race.
(1 mark)
It also contains more sodium which can compensate for the loss of
sodium when the athlete sweat.
(1 mark)
(b)
Semi-skimmed milk
(1 mark)
It contains more calcium which is the main component of bones and teeth.
(1 mark)
Total: 7 marks
7
Chapter 12
Part 1
1.
B
2.
A
3.
C
4.
D
5.
D
6.
C
7.
A
8.
D
9.
B
Multiple Choice Questions
10. B
11. C
Part 2
Structured Questions
1.
(i)
to close the opening of the trachea / to prevent food from entering the
trachea during swallowing
(1 mark)
(ii)
peristalsis
(1 mark)
(iii)
It contains protease
(1 mark)
to digest protein into short peptides / amino acids / polypeptides
(1 mark)
It is alkaline
(1 mark)
to neutralize the acid from the stomach / to provide a suitable pH for the
functioning of protease
(1 mark)
(iv)
(v)
the faeces will become more watery
(1 mark)
because less water is absorbed if a large part of D is removed
(1 mark)
- breakdown of excess amino acids / formation of urea
- storage of iron / vitamin A / vitamin D / glycogen
- regulation of blood sugar level
(any two) (1+1 marks)
Total: 10 marks
2.
(i)
Bile produced in the liver cannot be stored in A
(1 mark)
When food enters the duodenum, insufficient amount of bile is released
(1 mark)
(ii)
for emulsifying fats in the food
(1 mark)
Thus the surface area for the action of lipase decreases
(1 mark)
Effective communication (C)
(1 mark)
When part of structure D is removed, digestion and absorption of food is
reduced
(1 mark)
8
Energy intake becomes less than the energy expenditure in the body
(1 mark)
This may lead to the use / mobilisation of fat / food reserves stored in the
body
(1 mark)
(iii)
(1)
insulin
(1 mark)
(2)
It cannot regulate the blood glucose level
(1 mark)
as blood glucose would not be converted into glycogen in the cells of B
(1 mark)
Total: 10+1marks
3.
(i)
(ii)
(1/2 mark)
(1)
C
(2)
A, B, D
(1/2+1/2+1/2 marks)
Drawing (D): (resemblance, large & clear drawing)
(1 mark)
Labels (L): enamel, dentine, pulp/pulp cavity, jaw bone, root, crown
(any five)
(21/2 marks)
(1/2 mark)
Title (T)
enamel
dentine
crown
pulp / pulp cavity
root
jaw bone
Structure of the molar
(iii) Food could not be ground or crushed into small pieces / there would be less
mechanical digestion
(1 mark)
The surface area for the action of digestive enzymes would not be large enough
(1 mark)
It would take longer time for the food to be broken down into simple soluble
forms
(1 mark)
Total: 9 marks
Chapter 13
Part 1
1.
D
2.
D
3.
C
4.
C
5.
D
Multiple Choice Questions
9
Part 2
Structured Questions
1.
Rate of breathing = 12 breaths per minute
(1 mark)
Depth of breathing = 500 cm3
(1 mark)
(ii)
between 0 and 2.5 second
(1 mark)
(iii)
Intercostal muscles relaxed, so that the ribs moved downward and inward
(1 mark)
(i)
Diaphragm muscles relaxed, so that the diaphragm recoiled to the dome-shape
(1 mark)
(iv)
The thoracic / lung volume decreased
(1 mark)
leading to an increase in air pressure in the lungs
(1 mark)
Effective communication (C)
(1 mark)
(1)
The breathing movements would become faster
(1 mark)
(2)
In the left lung, air flow would decrease
(1 mark)
In the right lung, air flow would increase
(1 mark)
Total: 10+1 marks
(a)
backbone
2.
(3 marks)
(b)
intercostal muscle
(1 mark)
(c)
the volume increases
(1 mark)
the pressure decreases
(1 mark)
During inhalation, the diaphragm muscles contract,
(1 mark)
the diaphragm is flattened
(1 mark)
(d)
to increase the volume of the thorax and decrease the pressure inside.
(1 mark)
Total: 9 marks
10
Chapter 14
Part 1
1.
A
2.
C
3.
D
4.
C
5.
C
6.
C
7.
B
8.
C
9.
A
Multiple Choice Questions
10. C
11. C
Part 2
Structured Questions
1.
(1)
Blood capillary
(2)
Its wall is one-cell thick / the diameter of the lumen is slightly greater
than or similar to that of a red blood cell
(1 mark)
(i)
(ii)
(1 mark)
The pressure of blood in A is higher than the pressure of the fluid surrounding
the muscle cells
(1 mark)
This forces the plasma except the plasma protein out of A
(iii)
(1 mark)
Drawing (D): clear and accurate diagram, double line for capillary wall,
biconcave RBC
(11/2 marks)
Labels (L): any three of the following labels and title; 1/2 marks each
(11/2 marks)
nucleus
cell of capillary wall / capillary wall
red blood cell
lumen / blood plasma / blood
A section along line BC
(iv)
(small intestine)  hepatic portal vein  liver  hepatic vein  vena cava 
heart  pulmonary artery  (lung)
(1/2 marks for each term)
(3 marks)
Deduct 1/2 marks if there is no arrow sign.
Total: 10 marks
11
2.
(i)
The barbecued pork contains fat
(1 mark)
The digested products are absorbed into the lacteal in the form of fat / oil
droplets
(1 mark)
As a result, there will be numerous fat / oil droplets in the lymph in X
(1 mark)
Effective communication (C)
(ii)
(1 mark)
Contraction of skeletal muscles around vessel Y forces the lymph to flow
upward
(1 mark)
The valves in vessel Y help to prevent the backflow of the lymph (1 mark)
(iii) As the lymph vessels are blocked, tissue fluid in the leg cannot be
transported away
(1 mark)
Meanwhile it is continuously formed in the leg thus tissue field accumulates
(1 mark)
(iv)
The pathogen that causes sore throat
(1 mark)
stimulates the proliferation of lymphocytes / white blood cells in structure Z
thus structure Z becomes enlarged
(1 mark)
Total: 9+1 marks
3.
(i)
The blood pressure in the vein is much lower than that in the artery (1 mark)
Reasons:
-
the blood in the artery in directly under the pumping action of the heart,
while that in the vein is not
-
the blood in the vein has overcome great friction / resistance after
travelling over a long distance
-
there is a loss of fluid from the blood during the formation of tissue fluid
(any two) (1+1 marks)
(ii)
To allow more time
(1 mark)
for the exchange of materials between the blood and the tissue cells(1 mark)
(iii)
The volume of blood flow through each section of the circulation per unit
time is the same
(1 mark)
From the capillary to the vein, the total cross-sectional area decreases, so the
velocity of blood flow increases
(1 mark)
OR
Contraction of skeletal muscle adjacent to the vein helps to force the blood
to flow / inspiration helps to draw blood toward the thorax
(1 mark)
At the same time, valves are present in the veins to prevent the backflow of
the blood
(1 mark)
(1/2 marks)
(iv) Title (T)
Shape of the curve showing the drop in O2 content at the capillary (S)
(11/2 marks)
12
Correct axis labels (A): oxygen content, heart-heart / artery-vein / aorta-vena
cava / arteriole-venule
(2 x 1/2 marks)
Oxygen
content
artery
capillary
vein
Change in oxygen content of the blood in its circulation between the heart and the leg
Total: 10 marks
4.
(i)
To withstand the high blood pressure.
(1 mark)
To regulate the blood flow to the organ / control the diameter of the vessel
lumen.
(1 mark)
(ii)
The blood pressure in vessel type II is lower.
(1 mark)
The larger lumen would have a smaller resistance, thus facilitating the blood
flow.
(1 mark)
(iii)
Lung
(1 mark)
(iv)
(1) Vessel type II of organ A has a lower carbon dioxide content than
vessel I.
(1 mark)
As blood flows from vessel type I to the lung, then to vessel type II,
(1 mark)
carbon dioxide diffuses out of the blood to the air sac in the lungs.
(1 mark)
(2) Blood in vessel I of organ A comes from the veins / the right side of the
heart,
(1 mark)
which collect blood from the body tissues
(1 mark)
where oxygen is consumed in respiration.
(1 mark)
Total: 11 marks
5.
(a)
(1/2 mark)
(i) Q: pulmonary artery
(1/2 mark)
R: vena cava
(ii) X: left ventricle of the heart
Y: right ventricle of the heart
(b)
13
(1/2 mark)
(1/2 mark)
Blood vessel R
Blood capillary
large lumen
small lumen
one-cell thick wall
thin, less
elastic wall
(2 marks)
(c)
(d)
-
branch repeatedly to reach all cells of the body
-
thin-cell thick wall to give a short pathway for quick diffusion
-
provide a large surface area for rapid diffusion of food and oxygen from
blood into the tissue cells.
(any two points, 1+1 mark)
(i) adrenaline
(1 mark)
(ii) lactic acid
(1 mark)
Total: 8 marks
6.
(a)
artery
(1 mark)
(b)
P: fatty deposits
(1/2 mark)
Q: muscular wall
(1/2 mark)
The fatty deposits would harden the artery,
(1 mark)
increasing the risk of blood clotting in the artery.
(1 mark)
It also constricts the artery
(1 mark)
and so causes an increase in blood pressure.
(1 mark)
His diet is rich in animal fats and cholesterol.
(1 mark)
(c)
(d)
These substances increase the deposition of fats on the wall of artery.
(1
mark)
Total: 8 marks
7.
(a)
(b)
(c)
(i) pulmonary vein
(1 mark)
(ii) vein
(1 mark)
F
(1 mark)
A
(1 mark)
E
(1 mark)
H
(1 mark)
F
(1 mark)
(i)
14
(1 mark)
a white blood cell
(1 mark)
(ii) haemoglobin
(1 mark)
Total: 10 marks
8.
(a)
M: right auricle
(1 mark)
N: aorta
(1 mark)
(b)
Air entering at P has more oxygen and less carbon dioxide, while air
entering at Q has less oxygen and more carbon dioxide
(1 mark)
(c)
alveoli
(d)
(i)
(1 mark)
to increase the surface area for the diffusion of oxygen and carbon
dioxide
(1 mark)
(ii) only small molecules, oxygen and carbon dioxide molecules, can pass
through the pores
(1 mark)
(e)
(i)
white blood cells
(1 mark)
(ii) to produce antibodies
(1 mark)
to kill invading bacteria
(1 mark)
Total: 9 marks
Chapter 15
Part 1
1.
D
2.
B
3.
B
4.
B
5.
A
6.
D
7.
A
Part 2
1.
Multiple Choice Questions
Structured Questions
(i)
(1 mark)
(ii)
Light-sensitive cells on the retina were stimulated
(1 mark)
Nerve impulses were set up
(1 mark)
15
(iii)
(iv)
and transmitted along the optic nerve
(1 mark)
to the cerebral cortex for interpretation into vision
(1 mark)
Effective communication (C)
(1 mark)
Ciliary muscles would contract
(1 mark)
Suspensory ligaments would slacken
(1 mark)
Lens would become more convex
(1 mark)
- Convex lens
- Lens of the eye
- Nearby object (diverging rays)
- Converging rays
- Arrow sign
- Image on retina
- Title
(any six) (3 marks)
Path of light rays from a nearby object after correction
Total: 11+1 marks
2.
(i)
From 1 to 5 units of light intensity, the size of the pupil decreases with
increasing light intensity
(1 mark)
However, when the light intensity is above 5 units, the size of the pupil will
remain unchanged with increasing light intensity
(1 mark)
(ii)
This response helps to prevent excessive light entering the eye
(1 mark)
so that the light-sensitive cells of the retina may not be damaged
(1 mark)
(iii) light sensitive cells / photoreceptors  sensory neurone / optic nerve
 association neurone in brain  motor neurone / nerve  iris muscle
(1/2 mark for each term)
Deduct 1/2 mark if there is no arrow sign.
(iv)
(3 marks)
Because all the light entering the eye is absorbed by the retina and the
choroids
(1 mark)
As a result no light is reflected out of the eye through the pupil
(1 mark)
Total: 9 marks
3.
(i)
To supply nutrients / oxygen to the eyeball
(1 mark)
as it is richly supplied with blood / capillaries
(1 mark)
16
OR
To prevent reflection of light within the eye
(1 mark)
as it contains dark pigments
(1 mark)
(any one set)
(ii)
(iii)
The image formed on C can be seen, while that formed on B cannot be seen
(1 mark)
because there are light-sensitive cells at C, but none at B
(1 mark)
Structure D would become thinner / less convex
(1 mark)
This is caused by the relaxation of the ciliary muscle
(1 mark)
which leads to an increase in tension in the suspensory ligaments and, in turn,
stretches the lens
(1 mark)
(iv)
Effective communication (C)
(1 mark)
Drawing (D): large, clear and accurate diagram
(1 mark)
(accuracy includes outline of eyeball, lens of eye, straight lines for light rays)
Label (L): Parallel rays from object (with arrow sign), focus in front of
retina, rays extended to retina
(3 x 1/2 marks)
(1/2 mark)
Title (T)
Ray diagram showing light rays entering a short-sighted eye
Total: 10+1 marks
4.
(i)
(1) To allow the transmission of light to the retina without obstruction
(1 mark)
(ii)
(2) aqueous humour / choroid
(1 mark)
(1) It helps to reduce the amount of light entering the eyes,
(1 mark)
so as to prevent over-stimulation / damage of the light-sensitive cells.
(1 mark)
(2) reflex action
(1 mark)
(3)
Constriction of B
Putting on sunglasses
Does not involve the cerebrum Involves the cerebrum
Inborn
Learned action
Faster in action
Slower in action
Stereotyped response i.e. same Variable responses to the same
stimulus always evokes the
stimulus
17
same response
(any two) (1+1 marks)
(iii)
Dark-coloured sunglasses reduce the light intensity entering the eye,
(1 mark)
so the pupil will not constrict / constrict to a smaller extent in bright
sunlight.
(1 mark)
As a result, more UV light can enter the eye and cause damage.
(1 mark)
Total: 10 marks
5.
(a)
U: ciliary muscle
(1/2 mark)
V: cornea
(1/2 mark)
W: iris
(1/2 mark)
X: sclera
(1/2 mark)
(b)
light rays
lens
(2 marks)
(c)
(i) lens B
(1 mark)
(ii) When the eye focuses on a near object, the ciliary muscles contract
(1 mark)
to reduce the tension in the suspensory ligaments.
(1 mark)
Since the lens is elastic
(1 mark)
the lens becomes more convex.
(1 mark)
Total: 9 marks
18