New Senior Secondary Mastering Biology (Third Edition)
Book 1A
Suggested answers to Exercise,
Reading to learn and Cross-topic
exercise
MS Word file is available in Teaching Resource Centre:
https://trc.oupchina.com.hk/biology
The overseas examination boards bear no responsibility for the suggested answers contained in
this publication. Answers for HKDSE, HKCEE and HKALE questions are not available due to
copyright restrictions.
Ch 1 Introducing biology
Exercise
Section 1.1
Level 1 (p. 1-19)
1
C
Section 1.2
Level 1 (p. 1-19)
2
C
Level 2 (p. 1-19)
3
a
b
No
The death rates of childbed fever were different in the two clinics. /
If fear of hospitalization caused childbed fever, the death rates would have been
similar in the two clinics.
1
1
i
The death rate of childbed fever in a clinic in which doctors handle deliveries
straight after dissecting dead bodies.
1
ii
The death rate of childbed fever would be lower in the clinic in which doctors
wash their hands before handling deliveries.
1
Section 1.3
Level 2 (p. 1-19)
4
a
To prevent the exchange of any materials between the soil and the environment.
b
i
ii
1
Plants grow by absorbing materials from the soil.
1
The increase in mass of the tree would be equal to the decrease in mass of the
soil.
1
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iii
c
The hypothesis is falsified
1
because the increase in mass of the tree was much higher than the decrease in
mass of the soil.
1
Science advances through reasonable skepticism.
1
Van Helmont questioned the general belief that plants grew by absorbing materials
from soil.
1
(or other reasonable answers)
Reading to learn (p. 1-20)
1
If the S-shaped neck of a flask was broken to allow microorganisms on dust particles to
reach the boiled broth, then microorganisms would appear in the boiled broth.
1
2
It was used to show that the result of the experiment was due to the presence of
microorganisms in the air.
1
3
Nature of science
Science is affected by the
technology and the types of
equipment available at the time.
Scientists have to explore using
different techniques and methods
to find out a good solution to a
problem.
Scientific knowledge is tentative
and subject to change.
Elaboration
The invention of microscopes
helped the discovery of
microorganisms in the air.
1
Pasteur used flasks with an
S-shaped neck to allow air but not
microorganisms to enter the
flasks.
1
Scientists no longer think
microorganisms arise from
non-living things by spontaneous
generation.
1
Ch 2 The cell as the basic unit of life
Exercise
Section 2.1
Level 1 (p. 2-38)
1
D
Level 2 (p. 2-38)
2
B
Section 2.3
Level 1 (p. 2-38)
3
B
4
D
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Level 2 (p. 2-38)
5
A
6
7
a
C
Length of the cell in the photomicrograph (L) = 0.7 cm
Magnification = 40
length of the image
Magnification =
length of the object
1
0.7 cm
actual length of the cell
Actual length of the cell = 0.0175 cm
40 =
b
c
= 175 μm
1
Move the slide until the cell is located in the centre of the field of view.
Rotate the nosepiece to select an objective of higher magnification.
1
1
Focus by turning the fine adjustment knob.
1
He should use a 4X objective.
1
At low-power observation, the area of the leaf epidermis observed is larger.
1
The results will be more accurate.
1
Section 2.4
Level 1 (p. 2-39)
8
C
9
10
D
CE Bio 2008 I Q4c
Level 2 (p. 2-40)
11 D
12
B
13
A
14
Rough endoplasmic reticulum / ribosomes are involved in the synthesis of proteins. /
Mitochondria convert chemical energy in food into energy that the cell can use for protein
synthesis.
2
15
AL Bio 2011 IA Q3a, b
16
This cell is a eukaryotic cell.
1
Mitochondria, which are bounded by a double membrane, can be observed in the electron
micrograph.
1
17
a
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Title
1
Resemblance of drawing
1
Labels: cell wall / cell membrane / nucleus / cytoplasm / chloroplast (any 2)
b
1×2
The magnification and resolution of the light microscope may be not high enough.
1
c
18
Organelle X captures light energy and converts it into the chemical energy in food
during photosynthesis.
1
Mitochondria convert the chemical energy in food into energy in usable form.
1
Staining cells can increase contrast so that cell structures can be observed more
clearly.
1
b
Muscle cells are long in shape, while white blood cells are round in shape.
The size of the nucleus is larger relative to the cell size in the white blood cell.
1
1
c
i
a
Z is likely to be the muscle cell.
1
It has a large amount of mitochondria to meet the energy requirement of muscle
contraction.
1
Y is likely to be the white blood cell.
1
It has a large amount of rough ER for producing antibodies which are made up
of proteins.
1
ii
Yes, I agree.
According to the table, W contains no mitochondria and rough ER. Both are
membrane-bound organelles.
1
A bacterial cell, which is a prokaryotic cell, does not contain membrane-bound
organelles.
1
19
Similarities:
Both of them are enclosed by a cell membrane. /
Both of them have a nucleus that contains DNA. /
Both of them have cytoplasm. /
Both of them have rough ER which is a site for the synthesis of proteins. /
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Both of them have smooth ER which is a site for the synthesis of lipids. /
Both of them have ribosomes which are involved in the synthesis of proteins. /
Both of them may have vacuoles. (any 3)
1×3
Differences:
Plant cells are generally larger than animal cells. /
Plant cells generally have a more regular shape than animal cells. /
Plant cells have a cell wall but animal cells do not. /
Green plant cells have chloroplasts but animal cells do not. /
Plant cells often have a large central vacuole but most animal cells have only a few small
vacuoles or do not have any vacuole. (any 4)
1×4
Communication
3
Level 3 (p. 2-42)
20 A
21
a
X: mitochondrion
Y: rough endoplasmic reticulum
1
1
b
c
Glycogen acts as an energy reserve in humans.
1
i
Liver cells have a high level of metabolic activities. A lot of energy is required
to support their activities.
1
They need a larger number of X to meet the energy requirement.
ii
1
Cell Z has a larger total membrane surface area of organelle Y. This means it has
a larger amount of Y.
1
Y is the site for the synthesis of proteins. A larger amount of Y indicates cell Z
may produce more proteins.
1
Section 2.5
Level 1 (p. 2-42)
22 C
Reading to learn (p. 2-43)
1
Umbrella alga is a eukaryote.
It has a true nucleus.
1
1
2
Nucleus of the umbrella algae is located at their foot.
1
The part that is cut off can grow back only when the nucleus is present in the remaining
part, indicating that genetic information which instructs the growth of cells is likely to be
contained in the nucleus.
1
3
Doing science requires creativity and imagination.
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Hammerling designed this experiment to investigate whether genetic information is
contained in the nucleus. The process required creativity and imagination. /
The success of scientific investigation is the result of dedication, ingenuity and luck.
Hammerling chose the umbrella alga to carry out his investigation.
2
Ch 3 Movement of substances across cell membrane
Exercise
Section 3.1
Level 2 (p. 3-33)
1
a
Similarity:
A phospholipid bilayer is present in both models.
1
Difference:
In the ‘sandwich’ model, protein molecules are located at the surfaces of the
phospholipid bilayer. In the fluid mosaic model, protein molecules are interspersed
among the phospholipid molecules.
1
b
c
2
i
Hydrophobic protein molecules are repelled by water.
1
They are not likely to be located at the surfaces of the membrane, where they are
in contact with water.
1
ii
Polar molecules are repelled by the phospholipid bilayer and cannot move
through it.
1
Channel proteins and carrier proteins which transport them across the membrane
are absent in the ‘sandwich’ model.
1
Scientific knowledge is tentative and subject to change.
1
DSE Bio 2015 IB Q6
Section 3.2
Level 1 (p. 3-34)
3
A
4
6
C
5
B
a
The core of the phospholipid bilayer is hydrophobic.
1
It is permeable to non-polar substances but impermeable to polar substances and ions.
1
Fatty acids are small and non-polar. They dissolve in the phospholipid bilayer and
move across the membrane.
1
On the other hand, amino acids are small but polar. They are repelled by the
phospholipid bilayer and cannot move through it.
1
b
They are transported by channel proteins or carrier proteins.
-6-
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Level 2 (p. 3-34)
7
8
a
Oxygen molecules are small and non-polar.
1
They can dissolve in the phospholipid bilayer. Therefore, they can move across both
membranes at a similar rate.
1
b
Sodium ions are charged. They are repelled by the phospholipid bilayer.
1
They cannot move through the phospholipid bilayer.
1
In the cell membrane, they can be transported across the membrane by channel
proteins and carrier proteins.
1
As channel proteins and carrier proteins are absent in the artificial membrane, sodium
ions cannot move across the membrane.
1
Structure of the cell membrane:
The cell membrane is mainly made up of phospholipids and proteins. /
The phospholipid molecules are arranged in a bilayer. /
Their hydrophilic heads point outwards and hydrophobic tails point inwards. /
The protein molecules are interspersed among the phospholipid molecules. (any 3) 1 × 3
How the structure contributes to its differential permeability:
Since the core of the phospholipid bilayer is hydrophobic, it is permeable to non-polar
substances but impermeable to polar substances and ions.
1
Only small, non-polar molecules can dissolve in the phospholipid bilayer and move across
the membrane.
1
Polar substances and ions are repelled by the phospholipid bilayer and cannot move
through it.
1
Channel proteins and carrier proteins transport these substances across the membrane. 1
Communication
3
Section 3.3
Level 1 (p. 3-35)
9
A
10
D
11
C
Level 2 (p. 3-36)
13 B
14
C
15
D
16
DSE Bio 2018 IB Q2
17
DSE Bio 2017 IB Q2
18
DSE Bio 2014 IB Q7
12
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A
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19
CE Bio 2009 IA Q4
Level 3 (p. 3-38)
20 D
21
23
a
b
A
22
B
Chloride ions are charged. They are repelled by the phospholipid bilayer.
1
Therefore, they cannot move through it.
1
i
As chloride ions move out of the cells, solute concentration in the cells
decreases.
1
Thus, water potential of the cells increases. The water potential of the cytoplasm
becomes higher than that of the fluid in the lumen of the respiratory tract.
1
Therefore, water moves out of the cells by osmosis.
ii
1
Since the defective channel proteins transport less chloride ions from the cells to
the lumen,
1
less water moves out of the cells to the lumen.
1
As a result, the water content of the fluid decreases, thus it becomes thick.
1
Reading to learn (p. 3-39)
1
The concentrated sugar solution in the bag has a lower water potential than the seawater.
1
There is a net water movement from the seawater (high water potential) to the
concentrated sugar solution (lower water potential).
1
2
The emergency water filter bags are lighter in weight.
The sugary drink produced in the emergency water filter bags can provide energy.
1
1
Ch 4 Enzymes and metabolism
Exercise
Section 4.1
Level 1 (p. 4-27)
1
B
Level 2 (p. 4-27)
2
A
3
B
Section 4.2
Level 1 (p. 4-28)
4
DSE Bio 2019 IB Q2
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Level 2 (p. 4-28)
5
C
6
8
B
7
A
Functions:
Enzymes are biological catalysts.
1
They speed up metabolic reactions in our body by lowering the activation energy.
1
Importance of the shape of the enzymes in relation to their functions:
Enzymes bind with substrate molecules to form enzyme-substrate complexes during
reaction.
1
Each enzyme has an active site with a specific shape.
1
An enzyme only acts on a substrate that can fit into its active site.
1
Therefore, each enzyme catalyses one type of reaction only.
1
Factors like high temperatures or extreme pH which can cause a change in shape of the
active site can affect the activity of enzymes.
1
Section 4.3
Level 1 (p. 4-29)
9
B
Level 2 (p. 4-29)
10 B
11
12
a
B
Plant tissues may be damaged by mechanical force upon dropping, the vacuole
membrane and cell membrane rupture.
The ruptured vacuole membrane allows the enzyme PPO in the cytoplasm and
polyphenolic compounds in the vacuole to come together.
1
1
The ruptured cell membrane exposes polyphenolic compounds to another substrate
oxygen, producing brown pigment.
1
b
i
Boiling / High temperatures cause a change in shape of the active site of PPO.
1
It can no longer bind to the substrate molecules. Thus the reaction does not
occur.
1
ii
At low temperature, PPO becomes inactive.
1
The chance of PPO and its substrate molecules colliding with each other is low,
thus the reaction occurs only at low rate.
1
iii
Oxygen is not available in a vacuum pack.
1
Oxygen is one of the substrates of the reaction. Without oxygen, the enzymatic
reaction cannot occur.
1
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Level 3 (p. 4-30)
13 B
14
A
15
DSE Bio 2015 IB Q7
16
a
X and Y
1
b
The rate at the 2nd minute is higher than that at the 20th minute.
The concentration of substrate at the 2nd minute is higher than that at the 20th
minute.
Therefore, the enzyme molecules collide with the substrate molecules more
frequently at the 2nd minute.
The chance of forming enzyme-substrate complexes is higher, thus the rate of
reaction is higher.
1
17
1
1
1
c
X
1
The sum of the substrate concentration in tube Y and the product concentration in
tube W is always 100%, indicating that the experimental conditions of these two
tubes are the same.
1
d
Lower temperature
(or other reasonable answers)
1
a
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Title
1
Choice of axes
1
Plotting and joining of line
1
Labels and units
1
b
i
ii
pH 6
1
When pH increases from 4 to 6, the area of the remaining jelly block decreases.
This shows the activity of the protease increases as pH increases from 4 to 6. 1
When pH increases from 6 to 10, the area of the remaining jelly block increases.
This shows the activity of the protease decreases as pH increases from 6 to 10.
1
The protease denatures when pH is too low or too high. The substrate can no
longer fit into the active site of the protease to form the enzyme-substrate
complex.
1
c
Put two or more jelly blocks in each Petri dish. /
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Measure the mass of each of the remaining jelly blocks instead of the areas of the
upper face of the jelly blocks. /
Mix the protease solution with the buffer solution before adding the jelly blocks.
(any 2 or other reasonable answers)
1×2
Section 4.4
Level 1 (p. 4-32)
18
a
b
The beakers were the control set-ups with no enzymes added.
Water was added to these beakers to keep the total volume the same, so that the
results of different beakers can be compared.
1
1
Type of fruit / time of filtering / temperature
(any 2 or other reasonable answers)
1×2
c
Compared with pH 7–8, the enzyme works better at pH 3–4.
1
At pH 7–8, the amount of juice collected was much greater in the beaker with enzyme
added compared with the breaker with no enzyme, while the amount of juice
collected showed no significant difference in both beakers at pH 3–4.
1
d
At unsuitable pH, the activity of the enzyme decreases.
1
This is because unsuitable pH causes denaturation of the enzyme.
The substrates can no longer fit into the active site of the enzyme to form the
enzyme-substrate complex. The enzyme loses its catalytic ability permanently.
1
1
Level 2 (p. 4-32)
19
DSE Bio Sample paper IB Q9
20
a
b
i
Enzymes X, Y and Z are all moderately active at 35 °C.
1
ii
45 °C
45 °C is the optimum temperature of enzyme Z, which is the enzyme for
removal of egg stains.
1
1
At a temperature higher than 60 °C, the activity of all the enzymes in the washing
powder becomes 0.
1
This is because the high temperature causes denaturation of the enzymes.
The substrates can no longer fit into the active site of the enzymes to form
enzyme-substrate complexes. The enzymes lose catalytic ability permanently.
1
1
Level 3 (p. 4-33)
21
a
The active site of the enzyme glucose oxidase has a specific shape.
1
Only glucose can fit into the active site of glucose oxidase to form an
enzyme-substrate complex and produce hydrogen peroxide, which causes the colour
change.
1
b
The products of the reaction catalysed by glucose oxidase are invisible.
- 12 -
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Peroxidase catalyses the reaction between the product and a colourless compound,
forming a coloured compound. Thus the result becomes visible.
1
c
Enzymes become inactive at low temperatures.
1
The strip may give a false negative result if the refrigerated sample is tested directly.
1
Reading to learn (p. 4-34)
1
2
The inhibitor decreases the activity of ALDH.
Acetaldehyde is broken down at a lower rate.
1
1
As a result, acetaldehyde accumulates and causes unpleasant symptoms.
1
The shape of their active sites may be different.
The inactive form may have an active site that is less likely to bind to substrates. This
decreases the chance of forming enzyme-substrate complexes, causing a decrease in
activity.
1
1
Cross-topic exercise 1
Multiple-choice questions (p. 4-36)
1
5
9
A
C
B
2
6
10
B
A
C
3
7
D
B
4
8
D
D
Short questions (p. 4-38)
11
a
Magnesium is a component of chlorophyll / activates some enzymes.
Nitrate is a source of nitrogen for the synthesis of proteins.
b
The concentrations of the two ions in the root hair cells are higher than those in the
soil.
1
This indicates that the cells take up these ions against concentration gradients.
1
Addition of chemical fertilizers lowers the water potential of the soil water.
The water potential of the soil water becomes lower than that in the cytoplasm.
1
1
Water cannot enter the root hair cells by osmosis.
1
a
Chloroplast
Cell wall
1
1
b
i
c
12
1
1
ii
The water potential of the water in the pond is higher than that of the cytoplasm.
1
Water enters the cells by osmosis.
1
The cells may burst.
1
Cell wall is rigid.
1
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As water enters the cells causing cytoplasm to swell, pressure that builds up on
the cell wall prevents further entry of water. This prevents the cells from
bursting.
1
Structured questions (p. 4-39)
13
CE Bio 2000 I Q3c
14
a
b
It contains a large number of mitochondria.
Mitochondria release energy for active transport.
1
1
The cell membrane of the cell is highly folded.
1
This increases the surface area for diffusion.
1
i
A bacterial cell has no true nucleus while a eukaryotic cell has a true nucleus. /
DNA is lying free in the cytoplasm in a bacterial cell, but DNA is enclosed in the
nucleus in a eukaryotic cell. /
A bacterial cell has no membrane-bounded organelles, while a eukaryotic cell
has membrane-bound organelles. /
Ribosomes are lying free in the cytoplasm in a bacterial cell. In a eukaryotic cell,
some ribosomes are attached to endoplasmic reticulum and some are lying free
in the cytoplasm.
(any 2 or other reasonable answers)
ii
15
a
i
1×2
I
The secretion of chloride ions into the lumen would decrease the water
potential of the contents in the lumen.
1
II
The water potential of the cells becomes higher than that of the contents in
the lumen.
1
There will be a net water movement from the cells to the lumen by osmosis.
1
Keep the same size / thickness / surface area / surface area to volume ratio /
diameter of discs. /
Use the same variety / part of potato. /
Remove potato skin. /
Remove excess water before weighing, e.g. by blotting. /
Keep the same number of discs in each solution. /
Keep the same volume of sucrose solution. /
Keep the same temperature. /
Cover the tubes. (any 2)
ii
1×2
When the water potential of sucrose solution equals the water potential of potato
tissue, the mass of potato discs remains unchanged.
1
There is no change in mass of potato discs when the concentration of sucrose
1
solution lies at a certain point between 0.2 and 0.3 mol dm−3.
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The student may plot a graph of the change in mass of potato discs against
concentration of sucrose solution to determine the exact concentration at which
the mass of potato is unchanged. / Carry out the experiment again with more
sucrose concentration intervals between 0.2 and 0.3 mol dm−3 to determine the
exact concentration at which the mass of potato is unchanged.
1
b
i
ii
c
X: cell wall
Y: cell membrane
1
1
Z: vacuole membrane
1
Sucrose solution
1
The root hair cells absorb ions / salts / solutes by active transport. / The root hair cells
store ions / salts / solutes.
1
Therefore, the water potential is lower inside the root hair cells.
1
Essay (p. 4-40)
16
Comparison between animal cells and plant cells:
In terms of size, animal cells are generally smaller than plant cells.
1
In terms of shape, plant cells generally have a more regular shape.
1
In terms of structure, animal cells have no cell wall while plant cells have a cell wall. /
Animal cells have small or no vacuoles while plant cells often have a large central vacuole.
/
Animal cells have no chloroplasts while green plant cells have chloroplasts. (any 2) 1 × 2
Responses when placed in a hypotonic solution:
Water enters the cell by osmosis because the water potential of the hypotonic solution is
higher than that of the cytoplasm.
1
Animal cells do not have a cell wall. They will swell and may finally burst.
1
In plant cells, the cytoplasm swells. The swelling is resisted by the rigid cell wall. They
may finally become turgid.
1
Communication
3
Ch 5 Food and humans
Exercise
Section 5.1
Level 1 (p. 5-35)
1
B
2
A
Level 2 (p. 5-35)
3
D
4
A
5
A
6
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Level 3 (p. 5-36)
7
a
Vitamin C content of the fresh cabbage juice = (0.1 × 15) / 50
= 0.03%
1
1
b
The vitamin C content of sample X is higher than that of samples Y and Z.
1
This indicates that boiling causes a decrease in the vitamin C content of cabbage. 1
The vitamin C content of sample Z is higher than that of sample Y.
1
This indicates that compared with adding cabbage to cold water and boiling, adding
cabbage to boiling water and boiling results in less vitamin C loss.
1
8
a
i
ii
The mass of group A falls slightly
then rises.
1
1
The mass of group B rises
1
then falls.
1
Milk is needed for growth.
The effect is delayed.
1
1
b
This is to ensure the difference in mass is due to the milk but not other differences
between the groups of rats.
1
c
Environmental factors (e.g. temperature / light / activity) controlled by keeping the
rats in cages of the same settings. /
Genetic factors controlled by using same species / strain / gender of rats.
2
d
Vitamin
A
C
D
Function
night vision
skin and gum
development
bone / teeth
development
Source
carrot
kiwi fruit / bell pepper
/ guava
fatty fish / liver / egg
yolk
2
Section 5.2
Level 1 (p. 5-37)
9
A
Level 2 (p. 5-37)
10 D
11
12
a
i
ii
C
The amount of protein in salad X
= 0.5 × 4 = 2 g
1
The amount of protein in salad Y
= 10 × 2 = 20 g
b
1
Proteins are important for growth and repair of body tissues. /
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Proteins act as enzymes, hormones, antibodies and haemoglobin. /
Proteins act as energy source if the carbohydrates and fat stored in our body are used
up. (any 2)
1×2
c
13
If Jane only eats a box of salad X for each of her main meals, her daily energy and
protein intakes will be 120 kcal and 6 g respectively,
1
which is far less than her requirements. She may become thin and weak.
1
d
Heart disease / stroke
1
a
The energy requirement is the highest for people aged 35–54, and is lower for
children and elderly.
1
Children are growing actively. They have highest metabolic rate because they have
highest growth rate and highest rate of heat loss.
1
But as their body mass is small, the recommended daily energy intake is smaller than
adults.
1
Elderly have lower energy requirement than adults. This is due to their lower
metabolic rate.
1
b
Level of activity
1
The energy requirement of a very active male aged 35–54 is higher than that of a
moderately active male of the same age.
1
c
Females generally need less energy.
1
This is because they have a lower metabolic rate due to their smaller body size / being
less muscular / lower rate of heat loss.
1
Level 3 (p. 5-38)
14 a
The ‘light option’ sandwich has a lower energy value. Thus, it is less likely to cause
weight gain / obesity. /
It is lower in fat / lower in saturated fat. Thus, it is less likely to cause heart disease /
obesity. /
b
c
15
It is lower in salt. Thus, it is less likely to cause high blood pressure.
2
The amount of saturated fat in the guideline daily amount:
6g
× 100
26
= 23 g
1
Elderly have lower energy requirement than adults.
This is due to their lower metabolic rate.
1
1
The boy needs more energy per unit body mass than the adult woman. /
It is because he has a higher growth rate. /
He also has a higher rate of heat loss due to his high surface area to volume ratio. /
Furthermore, he has a higher level of activity, more energy is needed for muscular
activities. /
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The boy needs more proteins and calcium than the adult woman as he is growing actively.
/
Proteins are needed to build body tissues like muscles. /
Calcium is needed for building bones and teeth. /
Both the boy and the woman need more iron. /
The boy needs iron to produce new blood cells as he is actively growing. /
The woman needs iron to replace the loss of iron during menstruation. (any 7)
Communication
1×7
3
Reading to learn (p. 5-39)
1
Heart disease and stroke (or other reasonable answers)
1×2
2
Vegetables and fruits are high in dietary fibre.
If a person does not eat enough of them, constipation may result.
1
1
(or other reasonable answers)
3
Plan the diet carefully to avoid excess energy intake.
Do more exercise.
1
1
Ch 6 Nutrition in humans
Exercise
Section 6.3
Level 1 (p. 6-37)
1
B
2
A
3
C
4
C
9
B
10
A
Section 6.4
Level 1 (p. 6-37)
5
A
6
A
Level 2 (p. 6-38)
7
C
8
C
11
DSE Bio 2012 IB Q10
12
DSE Bio 2018 IB Q8
13
a
i
Hydrochloric acid
1
ii
The cardiac sphincter at the junction of the oesophagus and the stomach
prevents the contents of the stomach from flowing back into the oesophagus. 1
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iii
b
i
ii
Alkaline sodium hydrogencarbonate is present in bile and pancreatic juice.
It neutralizes the acidic chyme to protect the small intestine from being
damaged.
1
Pancreas is the main organ that produces lipase.
When the pancreatic duct is blocked, pancreatic lipase cannot reach the
duodenum.
Therefore, lipids cannot be broken down and thus cannot be absorbed.
Undigested lipids are egested in faeces.
1
1
1
1
When the pancreatic duct is blocked, pancreatic juice accumulates in the
pancreas.
1
Proteases in the pancreatic juice may digest the pancreatic tissues.
1
Level 3 (p. 6-40)
14 B
15
A
16
C
Section 6.5
Level 1 (p. 6-40)
17
DSE Bio 2013 IB Q3
Level 2 (p. 6-40)
18 B
Level 3 (p. 6-41)
19
a
b
i
Mitochondrion
1
ii
Food molecules are absorbed by active transport into the epithelial cells.
1
Mitochondria provide energy for the active transport of food molecules.
1
i
Shorter microvilli in the epithelial cell of the infected person.
Fewer microvilli in the epithelial cell of the infected person.
1
1
ii
The surface area for absorption of food molecules is reduced.
1
iii
The large amount of unabsorbed food decreases the water potential of the
contents in the lumen.
The water potential gradient between the contents in the lumen and the
cytoplasm of epithelial cells is reduced.
Less water is absorbed, and this results in the formation of watery faeces.
1
1
1
Section 6.6
Level 2 (p. 6-41)
20 C
21
D
22
D
Level 3 (p. 6-41)
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23
How the food is digested in the person’s body:
In the mouth cavity, food is chewed by teeth into small pieces and mixed with saliva
containing salivary amylase, which catalyses the breakdown of starch into maltose.
1
In the stomach, muscles in the stomach wall contract to churn the bolus and mix it with
gastric juice, which contains pepsin to digest proteins into peptides.
1
In the duodenum, pancreatic juice is present. It contains pancreatic amylase catalysing the
breakdown of the remaining starch into maltose, and proteases catalysing the breakdown
of some proteins into peptides, and some peptides into amino acids.
1
Enzymes embedded in the cell membranes of the specialized cells in the epithelium of the
small intestine also help digestion. Carbohydrases catalyse the breakdown of
disaccharides into monosaccharides, and protease catalyses the breakdown of some
peptides into amino acids.
1
How the person’s body uses the products digested:
Glucose is broken down by respiration in cells for releasing energy.
1
Amino acids are used by cells to make different types of proteins for growth and repair. 1
24
They are also the raw materials for making enzymes, antibodies and some hormones.
1
Communication
3
The concentration of oxygen in blood leaving the liver is lowered as oxygen is consumed
in respiration to supply energy for metabolic activities. /
The concentration of carbon dioxide in blood leaving the liver becomes higher as carbon
dioxide is produced when the liver cells carry out respiration. /
When the blood glucose level is high, the concentration of glucose level in blood leaving
the liver is lowered as excess glucose in blood is converted to glycogen. /
When the blood glucose level is low, the concentration of glucose in blood leaving the
liver becomes higher as stored glycogen is converted back to glucose and released into the
blood. /
The amount of amino acids in blood leaving the liver is reduced when amino acid is taken
in in excess as the excess amino acids are broken down by deamination. /
The amount of toxins in blood leaving the liver is reduced as they are broken down into
milder toxic substances by detoxification. (any 4)
2×4
Communication
3
Reading to learn (p. 6-42)
1
2
After the surgery, the patients feel full sooner and the feeling lasts longer.
1
They eat less food as a result.
1
Large pieces of food may not be able to pass through the narrow channel between the two
portions of the stomach.
1
They may remain in the upper portion for a long time. This may cause discomfort.
- 20 -
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3
Liquid can pass through the narrow channel quickly without giving the feeling of fullness.
1
The surgery cannot help reduce the intake of these high-energy foods. The patients are not
able to lose weight if they continue to take large amounts of these foods.
1
Cross-topic exercise 2
Multiple-choice questions (p. 6-44)
1
5
9
B
A
A
2
6
10
B
B
D
3
7
C
B
4
8
D
D
Short questions (p. 6-46)
11
a
The rate of lactose digestion is limited as lactase is insufficient or even absent.
1
The large amount of lactose decreases the water potential of the contents in the lumen
of the small intestine.
1
The water potential gradient between the contents in the lumen of the intestines and
the cytoplasm of epithelial cells is reduced. Less water is absorbed, and this results in
the formation of watery faeces.
1
b
At low temperature, lactase is inactive.
1
Thus, passing the milk through the column several times can increase the time for
lactase to work.
1
This allows more lactose to be broken down.
12
1
a
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b
The colour of the iodine drop remained brown.
1
The temperature of the mixture rose after it was removed from the 0 °C water bath,
1
amylase became active again and catalysed the hydrolysis of starch.
c
i
ii
1
Mouth / small intestine
1
Addition of amylase helps the breakdown of large starch molecules in the food
into smaller maltose molecules.
1
Maltose can be further digested into glucose, which can be absorbed by the
babies.
1
Structured questions (p. 6-47)
13
a
b
c
This tube remains pink throughout the experiment.
At high temperatures / Upon boiling, lipase is denatured.
1
1
Lipids can no longer fit into the active site of lipase to form enzyme-substrate
complex. Boiled lipase loses catalytic ability.
1
The presence of bile salts causes the emulsification of lipids into small droplets. 1
This facilitates chemical digestion by increasing the surface area of lipids for lipase to
act on.
1
Fatty acids are released more quickly and an acidic pH is attained more quickly.
1
Enzymes are sensitive to pH and work best at their optimum pH.
1
At unsuitable pH, enzymes denature and lose their catalytic ability.
Sodium hydrogencarbonate provides an alkaline medium for the action of the
enzymes in the small intestine.
1
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14
d
Bile contains brown bile pigments which are waste products formed from the
breakdown of haemoglobin from red blood cells.
1
When the bile duct is blocked, these pigments cannot enter the intestine and thus the
faeces appears pale in colour.
1
a
To show the effect of the inhibitor / drug
To show that yoghurt does not affect blood glucose concentration on its own
b
Food affects blood glucose concentration. / Different food contains different amount
of starch / glucose / sugar / carbohydrate.
1
All of the mice were given the same food each day to keep the starch / dietary fibre
intake the same / similar.
1
c
When inhibitor is added, fewer enzyme-substrate complexes are formed.
Less / No starch is digested into maltose and subsequently into glucose.
1
1
Therefore, less glucose is absorbed in the small intestine.
1
d
1
1
The sample size is not large enough. The results might not be representative. /
The investigation only lasted for 20 days. The long-term effects are still unknown. /
The decrease in blood glucose level is small. Mice with inhibitor still have a large
increase in blood glucose level. / The difference is not significant.
(or other reasonable answers)
2
Essay (p.6-48)
15
DSE Bio 2014 IB Q11
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