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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
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Chapter 1
Exercise 1.1
1
2
3
Term
Description
nutrition
making more of
the same kind of
organism
respiration
removing waste
products of
metabolism
growth
a permanent
increase in size and
dry mass
excretion
taking in materials
for energy, growth
and development
reproduction
chemical reactions
that release energy
from nutrient
molecules
Growth – the plant makes new
cells so it increases in size.
Reproduction – gametes
are made in the flower, which
fuse together to produce a
zygote; this is sexual
reproduction. (Note: If
students have not previously
studied reproduction, accept
an answer that refers simply to
reproduction as making more
of the same species.)
1
Points that learners may make include:
Birds are living things because they are able
to carry out all seven characteristics: they can
move of their own accord; they can reproduce;
they respire; they are sensitive to changes in
their environment; they grow; they excrete;
they take in nutrients.
Aeroplanes are able to move, and they also
take in ‘nutrients’, in the form of fuel. They
combine oxygen with fuel to provide energy
for their movement, which is similar to
respiration, and this reaction produces waste
products removed in the exhaust, which is
similar to excretion. They have sensors that
can detect and respond to changes in their
environment – for example, they may have
lights that come on automatically when light
intensity in their surroundings falls below a
particular level. However, aeroplanes do not
grow, and they are not able to reproduce.
Because aeroplanes are not able to carry out
all seven characteristics, they are not alive.
Excretion – the plant makes oxygen
as a waste product of phosynthesis,
and loses it from its leaves.
Sensitivity – the plant senses the
direction from which light comes, and
the stem and leaves grow towards it.
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Chapter 1 continued
Exercise 1.2
4
5
6
Exercise 1.6
An organism is a living thing. A species is a
group of living organisms that can reproduce
with each other to produce fertile offspring.
Each species of organism has a two-word
name. This system of naming is called the
binomial system. The first of the two words
in the name tells us the genus that the species
belongs to.
a They both belong to the same genus,
Panthera.
b They have different binomials, Panthera
tigris and Panthera leo. They cannot
interbreed to produce fertile offspring.
The two-word name provides information
about the genus and species that the organism
belongs to, so scientists can tell whether two
species are related or not. The name is used by
scientists all over the world, no matter what
language they speak, so all scientists can be
sure they are referring to the same species.
Exercise 1.3
7
Note that students cannot write in italic, so
should underline the binomials instead.
A (given) 1b, 2a, 3a – Crocodylus niloticus
B 1a – Geochelone elephantopus
C 1b, 2b – Ophiophagus hannah
D 1b, 2a, 3b – Chamaeleo gracilis
Exercise 1.4 and Exercise 1.5
8 and 9 Look for these features in the keys:
• It is made up of pairs of contrasting
statements.
• The statements are stand-alone and
can be selected by looking at only the
organism being identified; they do
not require comparison with another
organism.
• The key has no more than four pairs of
statements.
• The key works.
2
10 Learners may suggest these points:
• It is larger.
• Label lines are straight.
• Label lines always touch the part they
are labelling.
• There is no shading.
• The lines are continuous, not broken
which means they are clearer.
11 a Look for the features listed above,
for question 10.
b It has cells that do not have cell walls.
It has cells that do not have chloroplasts.
Some learners may also mention that it
has cells that do not have large vacuoles
containing cell sap. It is able to move its
body from place to place.
12 a Look for the features listed above,
for question 10 (but note that no labels
are required here).
b Fungi have cells with cell walls not made
of cellulose. They do not have chlorophyll,
and do not feed by photosynthesis.
They are made of hyphae. They feed
by digesting waste organic material and
absorbing it.
Exercise 1.7
13 A – amphibian; B – mammal; C – bird;
D – fish; E – mammal
14 Any two features of mammals, such as: they
have mammary glands; the young develop in
a uterus, attached by a placenta; they have
different types of teeth; they have a pinna;
they have sweat glands; they have
a diaphragm.
15 Reptile
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Chapter 1 continued
Exercise 1.8
16 They have several pairs of jointed legs; they
have an exoskeleton.
17
Group
Number
of pairs
of legs
Number
of pairs of
antennae
Other
distinguishing
features, if any
arachnids
4
0
body
divided into
cephalothorax
and abdomen
insects
3
1
body divided
into head,
thorax and
abdomen;
usually have
wings, breathe
through tubes
called tracheae
myriapods
many
(more
than 4)
1
body made up
of many similar
segments
crustaceans more
than 4
2
Exercise 1.9
18 They have cells with walls made of cellulose;
their cells contain chlorophyll; they feed
by photosynthesis.
19 Ferns do not produce flowers. They reproduce
by producing spores on the underside of
their fronds.
20 For example:
3
Monocotyledons
Dicotyledons
seeds have one
cotyledon
seeds have two
cotyledons
roots grow directly
from the stem
usually have a main
root that branches
leaves have parallel
veins
leaves have a
network of veins
flower parts in
multiples of three
flower parts in
multiples of four or
five
vascular bundles
in stem arranged
randomly
vascular bundles in
stem arranged in a
ring
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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
Chapter 2
Exercise 2.1
Exercise 2.2
1
6
cell membrane
magnification =
nucleus
size of image
size of actual object
7
cytoplasm
magnification =
mitochondrion
8
2
chloroplast
vacuole
containing
cell sap
cell wall
3
4
5
4
93 mm
93
32
= 2.9 to 1 d.p.
cell membrane
cytoplasm
= ×2
60
magnification =
mitochondrion
membrane
around
vacuole
a
b
120
9
a
45 mm
(allow any value between 42 and 47 mm)
b
nucleus
45
magnification =
Cell membrane: it is partially permeable, and
controls what enters and leaves the cell.
Mitochondrion: where aerobic respiration
happens, which releases energy from glucose.
Chloroplast: contains the green pigment
chlorophyll, which absorbs energy
from sunlight, used for making food by
photosynthesis.
Cell wall: supports the cell, helps to stop the
cell bursting when it absorbs water.
Ribosome: where amino acids are combined
together to make proteins, using instructions
on the DNA.
Circular DNA: provides instructions for
making proteins.
Mitochondria are the parts of the cell where
aerobic respiration happens, which is how
energy is released from glucose. If more
energy is needed in a cell, there will be more
mitochondria. Ribosomes are where proteins
are made. If more protein is needed in a cell,
there will be more ribosomes.
105
= ×0.43
(allow correctly
calculated answers from
the value given in a)
to 2 d.p.
10
size of actual object =
size of image
magnification
= 25 ÷ 12 = 2 mm
(to the nearest whole number)
11 a Root hair cell; it absorbs water and
mineral ions from the soil.
b length of cell in the diagram = 65 mm
= 65 × 1000 µm
So magnification =
65 000 µm
100 µm
= ×650
(allow correct
calculations
from a different
measurement of the
length of the cell)
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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
Chapter 3
Exercise 3.1
Exercise 3.2
1
2
7
3
4
5
6
5
A = 3; B = 5; C = 10; D = 20
Yes. As temperature increased, the distance
the red colour diffused through the jelly
increased. As the dishes were all left for the
same period of time, this must mean the
colour was moving faster in the warmer
dishes. A doubling of the temperature caused
the distance diffused by the colour to
roughly double.
The four most important variables to be
controlled are: concentration of the solution
of red pigment; size of hole in the jelly; depth
of jelly in the dish; volume of solution placed
in the hole.
This allows the learner to spot an anomalous
result. A mean can be calculated. It improves
the trust you can have in your results.
The higher the temperature, the more kinetic
energy the dye particles have. This means that
they move faster, so diffusion happens
more quickly.
a and b Possible answers include:
• Moving the dishes from one place to
another, after the dye had been put into
the holes, makes it likely that some dye
would overflow onto the surface of the
jelly. It would be better to place the dishes
in their final places, and then add the dye
to the holes.
• The dye samples placed into the holes will
all be the same temperature to start with,
and will take different amounts of time to
reach the four different temperatures in
the experiment. It would be better to leave
some dye at each temperature for a while,
and then add the dye to the holes.
8
Percentage
concentration
of solution
Mass / g
Before
soaking
After
soaking
Change
A
0.0
5.2
5.5
+0.3
B
0.1
5.1
5.2
+0.1
C
0.2
4.9
4.9
0.0
D
0.5
5.0
5.3
+0.3
E
0.8
5.1
5.0
–0.1
F
1.0
5.2
5.0
–0.2
The change for solution D should be ringed.
(The mass of the potato piece soaking in 0.5%
solution (D) has increased, but it would be
expected to decrease. This does not follow
the pattern of the other results and so
is anomalous.)
9
0.4
0.3
0.2
Mean change
in mass / g
0.1
0.0
−0.1
−0.2
−0.3
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Percentage concentration of solution
0.9
1.0
10 The 0% and 0.1% solutions had a lower
concentration (higher water potential) than
inside the potato cells, so water moved in by
osmosis and made the cells increase in mass.
The 0.2% solution had the same concentration
(water potential) as the potato cells, so there
was no net movement of water into or out of
the cells (the same amount went in as came
out) so there was no change in mass. The 0.8%
and 1.0% solutions had higher concentrations
(lower water potential) than that of the potato
cells, so water moved out of the cells by
osmosis and their mass therefore decreased.
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Chapter 3 continued
11 The most useful improvement would be to
have several pieces of potato in each solution,
and calculate a mean change in mass.
12 Yes, this would have been better because the
original masses of the potato pieces were
not identical. Calculating percentage change
would give a fairer comparison between the
pieces – it would avoid discrepancies caused by
this uncontrolled variable.
13 He could look at the graph, and determine
the concentration of solution where there
would be no change in mass (i.e. the value
on the x-axis where the line intercepts the 0
on the y-axis). (It would be helpful to repeat
the experiment using more concentrations
between 0.1% and 0.5%, to narrow down
this value.)
Exercise 3.3
14
6
Term
Description
diffusion
movement of
particles through
a cell membrane,
against a
concentration
gradient
concentration
gradient
a difference in
concentration
between two places
osmosis
the diffusion of
water through a
partially permeable
membrane
active
transport
the net movement
of particles down
a concentration
gradient
15 a Ion A
b Diffusion
16 B, because the concentration inside the root
cell is greater than outside, so it must have
been moved in against its concentration
gradient.
17 The roots would not be able to respire
aerobically, so they would not be able to
release energy to use in active transport. This
would have no effect on the concentration
of A, as these ions are moving passively by
diffusion. Active transport of B and C would
stop, so they would now move by diffusion
alone and their concentrations in the soil and
cells would become equal. For ion B, this
would mean that the concentration inside the
cells would decrease and for ion C, it
would increase.
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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
Chapter 4
4
Exercise 4.1
1
Look for a single table, with ruled and fully
headed rows and columns.
For example:
Food
Result of
test with
iodine
Result of
test with
Benedict’s
Conclusion
A
remained
brown
changed
from blue
to orangered
contains
reducing
sugar but
not starch
B
changed
from
brown to
black
remained
blue
contains
starch
but not
reducing
sugar
5
There are several other ways in which the table
could be organised. For example, learners
could decide to have two separate columns for
the conclusions, one for the starch test and
one for the reducing sugar test.
2
Example of
carbohydrate
Function in organisms
glucose
provides energy; the form
in which carbohydrates are
transported in mammalian
blood
starch
the form in which plants store
energy
cellulose
makes up the cell walls of
plants
glycogen
used to store energy in
animal cells
Exercise 4.2
3
7
Cut up the substance into very small pieces
and mix with water. Add biuret solution.
(Note: no heating is required.) If the mixture
remains blue, there is no protein. A purple
colour indicates the presence of protein.
a
b
c
Ribosomes
Amino acids
Each protein is made from a particular
sequence of amino acids. A different
sequence would make a different protein,
which would have a different function (or
no function at all).
Answers will depend on the learner’s choice
of protein, and the information that they find.
Look for answers that appear to be written in
the learner’s own words, rather than copied
from the internet or other source.
Exercise 4.3
6
The variable to be changed is the type of milk
– cow’s milk and goat’s milk.
The most important variables to be controlled
are: the volume of milk, the age of the milk,
the temperature of the milk, the volume and
concentration of biuret reagent added to it
and the time left before the intensity of the
colour is assessed.
The variable to be measured is the intensity of
the colour produced after the biuret test has
been carried out on the milk. This could be
measured by comparing the colours visually.
The apparatus that students choose will
depend on their choice of method, but should
include a way of measuring volume (e.g. a
measuring cylinder or syringe), a timer and
a thermometer.
If the hypothesis is correct, the purple colour
formed in the cow’s milk will be more intense
than the colour in the goat’s milk.
Check that the results table has been drawn
with ruled lines, the independent variable
placed in the first column and the dependent
variable placed in the remaining columns
(with a column for a mean if repeats have been
included) and that, if appropriate, there are
units in the headings.
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Chapter 4 continued
Exercise 4.4
7
8
9
8
In an animal or plant cell, DNA is found
in the nucleus. It forms long thread-like
structures called chromosomes.
In bacterial cells, there is no nucleus. Instead,
the DNA is free in the cytoplasm. It is in
the form of a circle. These cells also contain
smaller circles of DNA, called plasmids.
a Bases
b Upper strand: A, G; lower strand: C.
A DNA molecule is made of two strands,
coiled around each other to form a
double helix.
Cross-links between the bases hold the two
strands together.
The sequence of bases in a DNA molecule
determines the sequence of amino acids that
are used to make protein molecules.
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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
Chapter 5
9
Exercise 5.1
Tube
1
2
3
4
5
Temp / °C
20
20
0
40
100
enzyme
Milk added?
no
yes
yes
yes
yes
substrate
pH at:
0 min
7.0
7.0
7.0
7.0
7.0
2 min
7.0
6.8
7.0
6.7
7.0
4 min
7.0
6.7
7.0
6.5
7.0
6 min
7.0
6.6
7.0
6.3
7.0
8 min
7.0
6.6
6.9
6.2
7.0
10 min
7.0
6.5
6.9
6.2
7.0
1
active site
of enzyme
2
3
The active site of the enzyme is complementary
in shape to the substrate. Only a molecule
with the shape of the substrate can form an
enzyme–substrate complex with the enzyme.
Exercise 5.2
4,5 Look for questions that are very clear,
biologically correct and that have
unambiguous answers.
Exercise 5.3
6
7
8
9
lipids (fats and oils)
Fatty acids and glycerol
Fatty acids are produced, which are acids and
therefore lower the pH.
10 There was no milk, so no fat, so no fatty acids
were made.
11 The high temperature denatured the lipase
molecules, so there was no digestion of fats
and no fatty acids were made.
12 These tubes differed only in their temperature.
Lipase acts more rapidly at 20 °C than at 0 °C.
Students studying the supplement should also
refer to the lipase molecules moving around
faster and therefore collisions between enzyme
and substrate molecules happening more
frequently and with more energy. This means the
rate of reaction is faster at 20 °C than at 0 °C.
13 40 °C is the temperature at which the enzyme
worked fastest in this experiment, but the
optimum could be somewhere either side of
this – either a bit below or anywhere between
40 °C and 100 °C.
14 To find a more reliable value of the optimum
temperature, the experiment could be
repeated, to obtain another set of results,
to see if these matched the first ones.
Alternatively (or as well), three tubes could
be set up for each temperature, and a mean
calculated. To find a more precise value of
the optimum temperature, more temperatures
need to be tested on either side of 40 °C – for
example 35 °C, 45 °C, 50 °C. Once these results
have been found, the temperature range can
be narrowed down even more to keep getting
closer to the
optimum temperature.
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Chapter 5 continued
g
Exercise 5.4
15 a
b
c
d
e
f
10
pH
From pH 2 to pH 10 (a different range
would be acceptable).
Using buffer solutions. Tubes could be set
up using buffers for pH 2, 4, and so on.
The volume and concentration of starch
solution used should be kept constant.
Do this by making up one lot of starch
solution, keeping it well mixed, and
measuring volumes using a syringe or
other calibrated instrument. The volume
and concentration of amylase solution
should also be kept constant – do this as
for the starch solution. The temperatures
of all solutions need to be kept constant –
use water baths.
The time taken for the starch to disappear
should be measured. Take samples from
the mixtures of amylase and starch
at timed intervals (for example, every
minute); place them on a tile and add
iodine solution. Record the colour. The
time at which the sample does not go
black with iodine solution is the time
to record.
Measure equal volumes of starch solution
into six tubes. Add equal volumes of
different buffer solutions, for pH 2, 4, 6,
8 and 10 to each tube. Stand the tubes in
a water bath at a known temperature (for
example, 30 °C). Measure equal volumes
of amylase solution and add them to the
starch mixtures. Use a clean glass rod to
take samples from each tube (a different
glass rod for each, wiped clean between
samples) and place them on a tile. Add
iodine solution and record the colour
obtained.
h
Look for columns or rows for the pH and
the time taken for the brown colour to
disappear. In this case, the values written
in the table would be time in minutes.
Students may also like to show the colour
each time a sample was tested, in which
case the results table should also have
columns or rows with headings for the
time intervals. The results written in the
table would be colours.
The graph should have an x-axis labelled
‘pH’, and a y-axis labelled ‘Time taken for
starch to disappear / minutes’. The line
should begin high at the lowest pHs, drop
down to pH 7.5 and then rise again.
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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
Chapter 6
7
Exercise 6.1
1
2
3
4
5
6
11
carbon dioxide + water ➞ glucose + oxygen
Students may also include reference to
sunlight and chlorophyll.
Light energy from sunlight, which passes
through transparent epidermis cells to reach
the chlorophyll in the chloroplasts.
Carbon dioxide from the air, by diffusion
through a stoma and then the air spaces in the
spongy mesophyll.
Water from the soil, by osmosis into the root
hair cells, then up through the stem in the
xylem vessels, then by osmosis out of the
xylem and into the palisade cell.
Oxygen by diffusion into the air spaces then
out of a stoma into the air.
Carbohydrates stored as starch, or changed to
sucrose and transported away in the phloem;
made into cellulose for cells walls; made into
nectar to attract pollinators; made into amino
acids for growth; made into chlorophyll.
The sequence of labels runs from upper
epidermis at the top, then palisade mesophyll,
then spongy mesophyll, and finally lower
epidermis at the bottom of the diagram.
Green spots should be put inside all palisade
mesophyll cells, spongy mesophyll cells and
guard cells.
Part of leaf
Sun leaf
Shade leaf
cuticle
relatively thick
relatively thin
palisade
mesophyll
two layers
one layer
spongy
mesophyll
more loosely
packed; larger
cells and more
air spaces
quite tightly
packed; small
cells and small
air spaces
The cuticle helps to prevent water loss from
the leaf. The sun leaf will be hotter, so would
tend to lose more water by evaporation, so the
thicker cuticle helps to prevent this. The shade
leaf has a thin cuticle so more of the limited
amount of sunlight can get through it and
reach the palisade cells.
The sun leaf is exposed to much more
sunlight, so having more palisade cells
enables it to make more use of this light
and photosynthesise more. There can be
two layers of cells because at least some
sunlight will penetrate through the top layer
and reach those underneath. The shade leaf
has much less light, so only very little would
pass through the top layer of cells to reach a
second layer, so there is no point in having a
second layer of palisade cells.
Exercise 6.2
8 and 10 Look for: ‘Percentage concentration
of carbon dioxide’ on the x-axis, and ‘Rate of
photosynthesis / arbitrary units’ on the y-axis;
suitable scales; points plotted accurately, as
crosses or encircled dots; best-fit lines drawn
(though you could allow points joined with
ruled lines); the two lines labelled ‘low light
intensity’ and ‘high light intensity’.
9 Carbon dioxide is one of the raw materials for
photosynthesis.
10 See 8 above.
11 0.04%
12 53 Arbitrary units
13 0.12% (Note that if learners have drawn a
best-fit line, their line may flatten a little before
or after this value; if so, take the reading from
their graph.)
14 Light intensity
15 Carbon dioxide is often a limiting factor for
photosynthesis, so adding more will make
photosynthesis take place faster. This enables
the plant to make more carbohydrates and
grow faster, therefore producing higher yields.
16 Around 0.08 to 0.10%. Above this, the
increase in rate of photosynthesis is quite
small (the graph is flattening off) so the extra
yield is likely to be small.
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CAMBRIDGE IGCSE™ BIOLOGY: WORKBOOK
Chapter 7
Exercise 7.1
1
2
3
4
5
6
7
8
Egg, scrambled
Fat and carbohydrate (also protein to some
extent)
Calcium and iron
12.5 times as much
The carbohydrates come from plants.
It contains the most fat. Fat contains more
energy per gram than other nutrients.
Spinach, because it contains the most iron.
Anaemia results from a lack of haemoglobin,
and therefore a lack of red blood cells.
Iron is needed to make haemoglobin.
Spinach. Add up the total mass of the
contents listed in the table; what is left over
from 100 g is water.
Apple: 0.2 + 9.0 = 9.2 g
100 − 9.2 = 89.8 g of water
Chicken: 25 + 5 = 30 g
100 − 30 = 70 g of water
Egg: 10 + 23 = 23 g
100 − 23 = 77 g of water
Rice: 2 + 0.3 + 30 = 32.3 g
100 − 32.3 = 67.7 g of water
Spinach: 5 + 0.5 + 1.5 = 7 g
100 − 7 = 93 g of water
(Calcium, Iron, Vitamin C and Vitamin D
values are negligible)
lipasean enzyme that breaks down its
substrate to fatty acids
and glycerol
stomachan organ that secretes a juice
containing hydrochloric acid
digestionthe breakdown of food into small
molecules so that they can move
from the intestine into the blood
Exercise 7.2
9
12
Terms
Descriptions
pancreasan organ that produces enzymes
that digest starch, protein and fat
absorptionthe movement of nutrient
molecules and ions through the
wall of the intestine into
the blood
enamelthe outer, very hard layer of
a tooth
duodenumthe part of the alimentary canal
into which bile and pancreatic
juice flow
amylasean enzyme that digests starch to
reducing sugar
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Chapter 7 continued
Exercise 7.3
10
amylase acts on starch
physical digestion increases
the surface area of food
amylase is secreted
secretes a liquid with a low pH
which kills bacteria
protease is secreted
the gall bladder
amylase acts on starch
lipase breaks down fats
absorption happens
Exercise 7.4
11 The concentration rises and then falls. It
rises to a maximum of 142 a.u. at 12 hours.
The most rapid rise is between 0 hours and
6 hours. It falls quite steadily from 12 hours to
48 hours, from 142 a.u. to 60 a.u, a change of
82 a.u. It then falls more slowly from 48 hours
to 72 hours.
12 142 – 60 = 82
82
× 100 = 58%
142
13 Ileum
14 It has villi, which increase the surface area
so that absorption can happen more quickly.
The villi have microvilli, which further
increase surface area. The villi contain blood
capillaries and lacteals, into which absorbed
nutrients can pass. Some students may also
mention that the epithelium is only one cell
thick, minimising the distance that nutrients
have to travel.
13
15 Its molecules are already small enough to be
absorbed.
16 a Bile emulsifies fats, breaking large drops
into small droplets. This ensures that more
vitamin D is exposed on the surface of a
small droplet, so that it can more easily
move out of the droplets and be absorbed.
b Lacteals, as this is where fats are
absorbed.
17 The skin can make vitamin D when sunlight
falls onto it. This would confuse the results, as
the researchers would not know whether the
vitamin D in the volunteer’s blood came from
what he had absorbed or what he had made in
his skin.
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Chapter 8
Exercise 8.1
1
Xylem is a tissue that transports water and mineral ions from the roots of a plant to its leaves.
Phloem transports amino acids and sucrose from the leaves to other parts of the plant.
2
cortex
xylem
phloem
A
3
4
Diameter in diagram = 10 mm (allow any measurement between 10 and 12)
Therefore, real diameter = 10 ÷ 200 = 0.05 mm
They are hollow and empty so water containing dissolved mineral ions can easily flow through them.
They have no end walls, so they can fit end to end to form continuous tubes. Their walls contain lignin,
which is very strong, to provide support.
Exercise 8.2
5
6
7
8
9
14
The results table could look like this:
Condition
still air
Time / min
0
2
4
Distance / cm
0
2.8
6.1
moving air
6
8
10.0
12.9
10
12
14
16
18
20
16.2
21.8
27.9
31.1
39.5
44.9
Look for:
• ‘time’ on the x-axis and ‘distance’ on the y-axis, both with units and sensible scales
• points plotted accurately either as crosses or encircled dots
• ruled straight best-fit lines drawn, with change in gradient sharp and clear at time 10 mins.
a Still air: meniscus moved 16.2 − 0 = 16.2 cm in 10 minutes. So, mean rate was 1.62 cm
per minute.
b Moving air: meniscus moved 44.9 − 16.2 = 28.7 cm in 10 minutes. So, mean rate was 2.87 cm per
minute.
Yes. The mean rate per minute of movement of the meniscus is much higher in moving air than still air.
This means that the shoot was taking up water faster in the moving air. The rate at which it takes up
water is determined by the rate at which transpiration is taking place within the leaves.
It is likely that the temperature was not controlled – it could have been warmer or colder in the moving
air than in the still air. It is possible that light intensity was not controlled. The student was actually
measuring the rate at which water was taken up, rather than the rate at which it was lost – but we can
assume that they are very similar to each other, if not identical.
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Chapter 8 continued
Exercise 8.3
10 Sucrose
11 Starch
12 There is plenty of light in summer, but not
enough in winter. It is warmer in summer than
in winter. Liquid water may be in short supply
in winter if the ground is frozen.
13 Leaves will be sources in spring and summer.
They photosynthesise, producing sugars that
can be converted to sucrose and transported
to other parts of the plant.
14 Leaves will be sinks in winter. They cannot
photosynthesise, so they need to obtain sugars
from other parts of the plant, such as
storage organs.
15 The concentration of starch in the leaves
increases slightly, by 0.6% of their dry mass,
between spring and summer, reaching a peak
of 15.6% of dry mass. It then falls to only
4.9% of dry mass in the autumn.
16 The concentration of starch in the roots
increases from 2.6% to 3.1% of dry mass
between spring and summer, and then
continues to increase to reach 4.1% of dry
mass by autumn.
17 In spring and summer, leaves make more
glucose than they need by photosynthesis,
and store some of this as starch. In autumn,
they are photosynthesising much less and may
be using up their starch stores. Also, some of
the sugars will have been transported to other
parts of the plant – such as the roots – for
storage. This can explain the increase in starch
content of the roots in the autumn.
18 Removing the buds had no effect on the
amount of starch in the leaves. This is because
removing the buds did not affect the rate
at which the leaves could photosynthesise.
Removing the leaves reduced the amount of
starch in the buds, from 7.1% to 6.5% of dry
mass. This could be because there was less
sugar being made now that the leaves had
been removed, so there was less sucrose to
transport to the buds to turn into starch.
15
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Chapter 9
Exercise 9.1
1
2
Exercise 9.2
The circulatory system is a system of blood
vessels in which blood is transported. The
heart acts as a pump to move the blood. There
are valves in the circulatory system, which
ensure a one-way flow of blood.
Blue shading on the left of the diagram, and
red shading on the right.
6
3
4
5
Fish (accept any named fish)
In a double circulatory system, blood
is returned to heart after it has become
oxygenated. The heart then pumps it at high
pressure to the rest of the body. In a single
circulatory system, the blood moves directly
from the oxygenating organ (gills, lungs) to the
rest of the body, at a relatively low pressure.
A double system is therefore able to supply
oxygen more quickly to respiring body cells,
which allows metabolic rate to be higher.
Letter Function
aorta
H
transports oxygenated
blood to body cells
right
ventricle
D
pumps deoxygenated
blood into the
pulmonary artery
left atrium
B
receives oxygenated
blood from the
pulmonary vein
left
ventricle
C
pumps oxygenated
blood into the aorta
right atrium
E
receives deoxygenated
blood from the vena
cava
pulmonary
artery
G
transports
deoxygenated blood to
the lungs
pulmonary
vein
A
transports oxygenated
blood from the lungs
vena cava
F
transports
deoxygenated blood
from the body cells
Exercise 9.3
7
8
9
16
Structure
She has a 13% (13 in 100) or greater chance of
having a heart attack in the next five years.
She should stop smoking. This will reduce the
risk from 13% to 7% (or greater). She cannot
do anything about her diabetes. If she carries
on smoking as she gets older, the risk of heart
attack will rise to 22% when she reaches her
60s. If she stops smoking, it will only be 12%.
Health records have been kept for large
numbers of women over long periods of time.
The records have been grouped into women
of a particular age, and into smokers and
non-smokers, people with diabetes and people
without. The percentage of people in each group
having heart attacks has been worked out.
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Chapter 9 continued
Exercise 9.4
Exercise 9.6
10 O in the left atrium.
11 OF in the right atrium.
12 It allows oxygenated blood to flow directly
from the right atrium to the left atrium. This
oxygenated blood then leaves the heart in the
aorta, to deliver oxygen to respiring tissues all
over the fetus’s body.
13 This prevents oxygenated blood in the left
atrium mixing with deoxygenated blood in the
right atrium. If they mixed, then there would
be less oxygen in the blood in the aorta, so
body tissues would not get as much oxygen
delivered to them and would not be able to
respire as fast. The tissues might run short
of energy.
16 Use the peer assessment checklist to mark this
exercise.
Exercise 9.5
14
15
17
Feature
The answer was written in full sentences.
The answer was written in a sensible
sequence, so that it was easy to follow.
The answer referred to the thickness of
the walls of arteries and veins.
The answer correctly explained why
arteries and veins have walls with
different thicknesses.
The answer referred to the quantity of
elastic tissue in the walls of arteries and
veins.
The answer correctly explained why
arteries and veins have walls with
different amounts of elastic tissue.
The answer referred to valves in veins,
and explained why veins need valves and
arteries do not.
Arteries
Veins
Capillaries
contain valves
✗
✓
✗
wall is one cell
thick
✗
✗
✓
Exercise 9.7
carry blood at
high pressure
✓
✗
✗
have a wide
lumen
✗
✓
✗
17 Look for some or all of the following ideas:
• the correct data being described – that is,
the lighter grey bars
• reference to the overall trend – that is,
pulse rate increases at high altitude
• reference to the fall during the period at
high altitude
• reference to the initial fall and then rise
when returning to low altitude
• some comparison of time scales – for
example, the slow fall in pulse rate over
the almost two years at high altitude,
compared with the very rapid fall in just
two weeks at low altitude
• reference to the slightly lower pulse rate
at low altitude after having been at high
altitude, compared with before travelling
to high altitude
• at least two sets of figures quoted, stating
both time and the value for pulse rate,
including units.
Component
Function
red blood cell
transport nutrients
plasma
destroy pathogens
white blood
cell
clotting
platelet
transport oxygen
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Chapter 9 continued
18 Look for some or all of the following ideas:
• the correct data being described – that is,
the dark grey bars
• reference to the overall trend – that is, red
blood cell concentration increases at high
altitude but falls with time, then decreases
again when back at low altitude
• reference to the slightly lower
concentration six weeks after having
returned to low altitude, compared with
before travelling to high altitude
• at least two sets of figures quoted, stating
both time and the value for red blood cell
concentration, including units.
19 Oxygen transport.
20 There is less oxygen available in the air at high
altitude, so less diffuses into the blood. The
person adapted to this by producing more red
blood cells, to help to increase the amount of
oxygen that could be absorbed into the blood
and transported to body cells for respiration.
21 A person who has trained at high altitude
will have a faster pulse rate and more red
blood cells. This will increase the rate at which
oxygen can be supplied to muscles, making it
possible for them to work faster because they
can respire faster.
18
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Chapter 10
Exercise 10.1
Exercise 10.3
1
2
9
3
Viruses
Yes; it is caused by a pathogen and can be
passed from one person to another.
a The acid kills pathogens in food that
we eat.
b Phagocytosis: some white blood cells
engulf pathogens and digest them.
Producing antibodies: some white blood
cells secrete antibodies, which stick to
pathogens and help to destroy them.
Exercise 10.2
4
5
6
7
8
19
A microorganism that causes disease.
Look for these criteria on the bar chart:
• pathogen on the x-axis, and percentage of
cases on the y-axis
• suitable scale on the y-axis, fully labelled
• bars plotted accurately.
You could use the Self-assessment table to
assess the bar chart.
Perhaps there were other pathogens causing
food poisoning; perhaps not all cases of food
poisoning were able to be identified as being
caused by a particular pathogen.
Most people would not bother to go to a
doctor when they have food poisoning, so
there will be many unrecorded cases.
For example: keep food cool (in a fridge);
wash fresh foods, such as fruits and vegetables,
in clean water before eating; wash hands and
cooking implements carefully before allowing
them to come into contact with food; cook
food thoroughly and either eat while hot, or
cool rapidly; keep raw meat and other food
that may carry pathogens away from food that
is to be eaten cold.
The mass of solid waste that was recycled
increased from 15 000 000 tonnes (1.5 × 107)
to 23 000 000 tonnes, an increase of 8 000 000
tonnes.
The mass of solid waste that was deposited as
landfill also increased, from 17 000 000 tonnes
to 21 000 000 tonnes, an increase of 4 000 000
tonnes.
The total increase in all solid waste was
therefore 12 000 000 tonnes.
The increase in recycled waste was twice the
increase of landfill waste. This means that in
2006–2007, unlike 2002–2003, the mass of
waste that was recycled was greater than the
amount of waste deposited as landfill.
10 Answers could include some of these ideas.
Note, however, that learners are not likely to
have studied recycling yet.
• Landfill sites can cause pollution, if they
are not well constructed and maintained.
For example, run-off from them can carry
pollutants (such as heavy metals or other
named substances) into nearby waterways,
where they can harm aquatic animals or
humans coming into contact with
the water.
• Uncovered landfill sites can be a magnet
for houseflies, rats and other pests, which
can then carry pathogens to human
habitations.
• Landfill sites take up space that could be
habitats for plants and animals.
• Non-biodegradable plastics on landfill
sites can harm animals that may eat them
or get trapped in them.
• Recycling means that less landfill has to
be used.
• Recycling reduces the need to mine
resources such as metals, fossil fuels (used
for making plastics) and sand (used for
making glass), and so reduces the damage
to habitats and the pollution that can be
caused by these activities.
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Chapter 10 continued
11 a
b
c
Total waste in 2002–2003: 15 000 000 +
17 000 000 = 32 000 000 tonnes
Total waste in 2006–2007: 23 000 000 +
21 000 000 = 44 000 000 tonnes
44 000 000 – 32 000 000 = 12 000 000
tonnes
12 000 000
32 000 000
× 100 = 38%
Exercise 10.4
12 For example: children are more likely to put
their hands to their mouths without washing
them first; they are more likely to play in
contaminated water.
13 Look for some of these ideas. (For some of
the points, accept years other than those
quoted below.)
• The number of polio cases has fallen from
about 53 000 in 1980 to just over 3000 in
2005.
• The highest number of cases was in 1981,
when 66 000 cases were recorded.
• The steepest fall was from 1981 to 1982 or
1983.
• Numbers of cases fluctuated between
1982 and 1988, remaining roughly
constant at just below 40 000 cases per
year.
• Numbers fell fairly steadily from 1987 to
1995 or 1996.
• Numbers remained very low, fluctuating
only slightly, between 2001 and 2005.
14 Immunisation coverage increased sharply
from 1980 to 1991, from about 22% to 75%.
This coincided with a sharp decrease in
the number of polio cases. Immunisation
coverage remained high from 1991 onwards,
increasing slightly to 78%. This coincided
with a steady fall, and then constant low level,
in the number of polio cases. This could be
explained if immunisation does reduce the
number of cases. However, it is not impossible
that some other factor is causing the fall in
cases, as a correlation does not prove cause.
20
15 The antigens in the vaccine would be digested
by enzymes, or broken down by stomach acid,
in the alimentary canal, before they could be
absorbed into the blood.
16 The antigens on the polio viruses in the
vaccine would be recognised as foreign
by lymphocytes that are able to produce
complementary antibodies. These lymphocytes
would multiply to form a clone, which would
then make antibodies against the antigens of
the virus. These lymphocytes would also make
memory cells. If the polio virus is encountered
again, these memory cells will rapidly make
antibodies to destroy them.
17 The sequence of the bases in the virus’s
DNA codes for the sequence of amino acids
in proteins that are made. If the bases are
different, the amino acid sequence in the
proteins will also be different, so the protein
will not work in the same way as usual. If
this protein is needed to help the virus to
reproduce, then it will not be able to do so.
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Chapter 11
Exercise 11.1
Exercise 11.2
1
2
For each statement, there are many possible
sentences that could be written. Look for
evidence that the learner has identified the
mistakes described below and rewritten a
correct sentence without the mistake.
a Every cell uses energy to help it to respire.
Cells do not use energy for respiration –
respiration releases energy.
b Aerobic respiration produces energy by
combining nutrient molecules, such as
glucose, with oxygen.
Respiration does not ‘produce’ energy.
Respiration releases energy that is already
present in glucose molecules.
c Anaerobic respiration happens in
mitochondria.
Anaerobic respiration happens in
cytoplasm, not mitochondria. Aerobic
respiration happens in mitochondria.
d In human muscle, both aerobic respiration
and anaerobic respiration produce carbon
dioxide.
Anaerobic respiration in humans does not
produce carbon dioxide.
e Anaerobic respiration releases much more
energy from each glucose molecule than
aerobic respiration does.
Anaerobic respiration releases less energy
than aerobic respiration.
Tube
A
B
C
D
Contents
animal
plant
animal
and
plant
no
animal,
no
plant
Colour
of
indicator
at start
orange
orange
orange
orange
Colour
of
indicator
at end
yellow
deep
red
orange
orange
3
4
5
21
The results table could look like this:
Students might also want to include a row
stating the conclusions that can be made.
In tube A, the animal respired, giving out
carbon dioxide.
In tube B, the plant photosynthesised (faster
than it respired), taking in carbon dioxide.
In tube C, the carbon dioxide given out
by the respiring animal was used by the
photosynthesising plant, so there was no
change in the carbon dioxide concentration in
the water.
In tube D, neither photosynthesis nor
respiration took place.
Respiration would continue, but
photosynthesis would not. The indicator
would therefore go yellow in tubes A, B and
C, and remain unchanged in D.
During the day, aquatic plants take in carbon
dioxide (and give out oxygen), which helps
the animals in the tank. At night, the plants
use oxygen and give out carbon dioxide, so
this could mean less oxygen for animals for
respiration, and a higher concentration of
carbon dioxide in the water.
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Chapter 11 continued
Exercise 11.3
6
You could use the self-assessment table to assess the plan.
Statement
Notes
I stated clearly what the independent
variable is, and described how I would
change it.
The independent variable is temperature. It could be
varied using water baths.
I suggested five different temperatures that I
would use.
Values should include 0 °C, and a reasonably high
temperature such as 60, 70 or 80 °C. (Note that seeds are
being used, so we do not have to worry about harming
animals with the higher temperatures.)
I stated clearly what the dependent variable
is and described how I would measure it.
The dependent variable is the rate of respiration
of the peas. Students might think about using
hydrogencarbonate indicator, as this has been used in the
previous exercise, and placing some in a container with
the germinating peas. They could time how long it takes
for the indicator to change to yellow.
I identified at least two important variables
that should be kept the same and described
how I would do this.
Important variables to standardise include the type of
peas, the age of the pea seeds, how long they have been
soaked in water (to stimulate germination) and the mass
of peas. There may be others, depending on the method
chosen by the student.
I made a list, or gave a description, of all the
apparatus and materials I would use.
This will depend on the method chosen.
I described how I would keep myself and
others safe as I worked.
There should be reference to using hot water in the water
baths, and how risk will be reduced – for example, by not
sitting down when using hot water; by not carrying hot
water around the laboratory.
I drew an outline results table with headings.
This will depend on the method chosen. It is likely to have
columns or rows headed: ‘Time taken for indicator to
change to yellow / minutes’, and columns or rows headed:
‘Temperature / °C’.
I sketched a graph, with the axes labelled, to The graph should have temperature on the x-axis, and
show what I predict the results would be if
rate of respiration or time taken for indicator to change
the hypothesis is correct.
to yellow on the y-axis. The line should rise to whatever
the student predicts will be the optimum temperature and
then fall.
Exercise 11.4
7
8
22
Look for:
• age / days on the x-axis
• ratio of alveolar surface to body mass / cm2 per gram on the y-axis
• both axes with suitable scales with equal intervals (not the intervals in the first column of the
results chart)
• points accurately plotted as neat crosses or encircled dots
• two separate lines drawn
• a key or labelling to show which line is for females and which for males.
40 × 23.1 = 924 cm2
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Chapter 11 continued
9
The individual rats may have differed in
mass, so comparing the alveolar surface area
for a small rat with that of a big rat would
introduce another variable. The important
feature is the ratio between surface area and
mass or volume, as this gives information
about how effectively the body cells (mass) can
be provided with oxygen by the gas exchange
surface (area).
10 At 21 days, males have a higher ratio of
surface area to body mass than females; the
difference is 1.5 cm2 per gram. However, from
33 days onwards, females always have a higher
ratio than males. The greatest difference is
at 95 days, when females have a ratio that is
4.0 cm2 per gram higher than males.
11 When pregnant, the female’s alveolar surface
has to supply the growing embryo with
oxygen, as well as her own cells. She therefore
needs a larger surface area in order to obtain
this extra oxygen. This could explain why the
female rats’ ratio of alveolar surface area to
body mass is higher than the males’ ratio at
60 days (when pregnancy can first occur) and
95 days. (However, it does not explain why the
ratio is actually at its highest at age 21 days,
and then falls to age 45 days. This pattern
is the same for both males and females, so
perhaps this is related to the rate of growth of
the rats at those stages in their development.)
23
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Chapter 12
•
Exercise 12.1
1
The human nervous system is made of
specialised cells called neurones. These cells
have a long thread of cytoplasm called an
axon. They can transmit electrical impulses
very quickly.
The brain and spinal cord make up the central
nervous system. The nerves outside the brain
and spinal cord form the peripheral nervous
system.
•
Exercise 12.4
5
Exercise 12.2
2
3
24
Letter
Name
Function
A
retina
contains receptor cells that
are sensitive to light
B
optic
nerve
transmits electrical impulses
from the receptor cells in the
retina to the brain
C
iris
controls how much light
passes through the pupil
and into the eye
Letter
Name
A
spinal cord
B
relay neurone
C
motor neurone
D
pupil
allows light to pass through
D
muscle / effector
E
cornea
E
receptor
refracts light as it enters the
eye
F
sensory neurone
G
synapse
Exercise 12.5
Arrows towards spinal cord on neurone F,
then towards C in neurone B, then towards the
muscle in neurone C.
6
Exercise 12.3
4
an outline results table, with headed
columns including units (distance / mm
or cm)
a description of the expected results if the
hypothesis is supported.
Look for:
• changing the independent variable by
testing reaction time in someone who has
drunk caffeine (e.g. cola drinks, coffee)
and someone who has not; learners may
wish to have a range of the independent
variable, by testing people who have
drunk different quantities of caffeine
• measuring the dependent variable by
recording the distance reading on the ruler
• standardising other important variables –
examples could include: doing the
experiment in a quiet room; giving all
subjects the same volume of drink (either
with or without caffeine); ensuring that
subjects have not drunk anything else before
the experiment is done; keeping the position
from the which the ruler is dropped and
caught the same; using the same ruler
• doing replicates – testing each subject
several times
As you move into darkness, the intensity of
light falling onto the eye decreases. This is
sensed by cells in the retina of the eye. They
send an electrical impulse along the optic
nerve to the brain.
The brain then sends an impulse to the muscles
in the iris of the eye. The radial muscles
contract and the circular muscles relax, which
makes the diameter of the pupil increase.
This is an example of a reflex action.
Exercise 12.6
7
The thick lens bends
the light rays greatly.
light focused
on the retina
light rays
diverging
greatly
The cornea
bends the
light rays
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Chapter 12 continued
8
9
The ciliary muscles contract, which loosens
the tension on the suspensory ligaments. This
allows the lens to revert to its natural, more
rounded shape. The lens now refracts light
rays more strongly, bringing the diverging
rays from the nearby object to a focus on
the retina.
They are less able to adjust focus for looking
at objects at different distances. They may be
able to see clearly at a particular distance, but
vision is blurred at other distances.
Exercise 12.7
10
Component
Function
hormone
a hormone secreted by
the testes
target organ
a chemical substance
produced by an
endocrine gland and
carried in the blood
insulin
a part of the body
that is affected by a
hormone
ovaries
organs that secrete
oestrogen
testosterone
a hormone secreted by
the pancreas
Exercise 12.8
14 A response of a plant, in which the direction
of the growth is away from the direction in
which gravity is acting.
15 We are told that the plant was kept in a place
with light coming equally from all sides, so
the plant could not grow towards or away
from light.
16 Look for:
• ‘time / minutes’ on the x-axis
• ‘percentage increase in length’ on the
y-axis
• suitable scales on both axes
• accurately plotted points using small
crosses or encircled dots
• neat best-fit lines
• a key or labels to identify the two lines.
17 The data show that there was more auxin on
the lower surface than on the upper surface.
Auxin makes cells elongate. The greater
quantity of auxin on the lower surface made
the cells on the lower surface get longer than
those on the upper surface, so the shoot
curved upward.
11 Adrenal glands
12 It prepares the body for fight or flight. It
increases breathing rate and heart rate, and
increases pupil diameter. Learners studying
the Supplement should also know that it
increases blood glucose concentration.
13 Feature
Control by
Control by
25
nerves
hormones
how
information
is transmitted
between
different parts
of the body
as electrical
impulses,
which are
transmitted
along
neurones
to specific
effectors
as chemicals,
which travel
in the blood
to all parts of
the body, but
affect only
their target
organs
speed of
action
relatively fast
relatively slow
duration of
effect
relatively short relatively long
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Chapter 13
Exercise 13.1
Exercise 13.2
1 and 2
6
7
vena cava
renal
artery
aorta
kidney
renal
vein
ureter
bladder
urethra
3
4
5
26
The liquid contained in the ureter does not
contain red blood cells, white blood cells
or platelets. It contains more urea and less
oxygen than the liquid in the renal artery.
Students could also state it contains less
oxygen than the renal artery.
This increases the surface area across which
diffusion can take place, speeding up
the process.
a The concentration of glucose will remain
unchanged, as there is no diffusion
gradient for it. The number of glucose
molecules moving in each direction will be
roughly equal.
b The concentration of protein will remain
unchanged. Protein molecules are too
large to get through the holes in the
partially permeable membrane, so they
will all stay in the blood.
c The concentration of urea in the blood
will fall (but it will not become 0). There
is a higher concentration of urea in
the blood than in the dialysis fluid, so
it will diffuse down its concentration
gradient, through the partially permeable
membrane.
Pancreas
When blood glucose levels rise higher than
normal.
8 The starch is digested by amylase (in saliva
and pancreatic juice) to produce maltose.
Maltose is digested by maltase to produce
glucose. Glucose is absorbed into the blood
capillaries in the villi in the small intestine.
9 Person A. The blood glucose level rose higher
after eating the starch and stayed high for
longer. In person B, insulin was secreted from
the pancreas when the glucose rose above
normal. This caused the liver to take some
of the glucose out of the blood and change
it into glycogen and store it. This did not
happen in person A.
10 If blood glucose concentration is too high,
water is drawn out of the blood cells and body
cells by osmosis. This means that metabolic
reactions cannot take place normally in their
cytoplasm. If blood glucose concentration
is too low, cells do not get enough glucose
to be able to carry out respiration, which is
essential to supply them with energy for active
transport and other processes.
11 The set point is the required or normal
concentration of glucose in the blood.
In practice, this is a range rather than a
‘point’ value. Negative feedback is the
process by which action is taken to bring
the concentration back to this set point if
it drifts away from it. The change in blood
glucose concentration is detected by receptors
in the pancreas cells. If it is too high, insulin
is secreted, and if it is too low, glucagon is
secreted. Insulin causes the liver to decrease
the blood glucose concentration, while
glucagon causes the liver to increase it.
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Chapter 14
Exercise 14.1
1
2
3
4
5
Exercise 14.3
Asexual
reproduction
Sexual
reproduction
always only
one parent
✓
✗
offspring are
genetically
identical
✓
✗
gametes are
involved
✗
✓
a zygote is
produced
✗
✓
Feature
The nuclei in pollen grains (male) and in the
ovules (female).
Sperm (male) and egg cell / ovum (female)
There is only one parent, which is growing
new plants from itself. There are no gametes,
fertilisation or zygotes involved.
Haploid means containing only a single set
of chromosomes. When two gametes fuse, the
single sets of chromosomes from each one join
together to form two sets in the zygote, so that it
is diploid. The zygote can then divide to produce
all of the cells in the new organism, which need
to be diploid.
Exercise 14.2
6
7
8
27
9 Coffea
10 They are insect-pollinated. They have flowers
with white petals and scent to attract insects.
11 Asexual
12 The trees belong to different species, so they
would probably not be able to reproduce with
each other. If they did, their offspring would
not be fertile.
13 Pollination; pollen grains from an anther are
transferred to a stigma. The pollen grain then
grows a tube, ready for the male nucleus to
travel down it.
14 There are various possibilities.
• Use the resistant trees to produce more
genetically identical ones, using asexual
reproduction.
• Use self-pollination of the naturally
resistant trees; the seeds are likely to
produce similar but not identical trees,
which will mostly be resistant to the fungus
and may have new good features such as
better-quality beans.
• Use cross-pollination between the naturally
resistant trees and others, to produce a
wide range of very different seedlings from
which you can select the best.
Dull or no petals; anthers dangling outside
flower; feathery stigma outside flower; large
quantities of pollen.
Little or no pollen is emitted at night, between
about 22 and 7 hours. Pollen emission rises
sharply during the morning, peaking at
around 11 hours, then falls sharply to 15
hours, then remains low during the late
afternoon and evening.
The pollen grain grows a tube down through
the style and the ovary, into an ovule. A
nucleus in the pollen grain travels down the
tube and into the ovule. It fuses with a nucleus
in the ovule, producing a diploid zygote. This
develops into an embryo plant, while the ovule
develops into a seed.
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Chapter 15
Exercise 15.1
Exercise 15.2
1
7
2
3
4
5
6
28
Black or blue labels: cell membrane,
cytoplasm, nucleus.
Red or other colour labels: look for about five
labels altogether, each of which includes an
explanation of the function of the feature.
For example:
egg cell: haploid nucleus that will become a
diploid nucleus when it fuses with the sperm
nucleus
sperm cell: long tail to help it to swim to
the egg.
The lungs are made up of millions of tiny
alveoli. Although each of these is very small,
there are so many of them that their total
surface area is huge.
a From the air spaces inside the alveoli, to
the interior of the red blood cells in
the capillaries.
b There is a lower concentration of oxygen
in the red blood cells than in the alveoli,
because the blood has travelled past
respiring cells that have taken oxygen
from it and made it deoxygenated. There
is a high concentration of oxygen in the
alveoli because fresh air is drawn in by
breathing movements. Oxygen therefore
moves by diffusion, down its diffusion
gradient.
The lungs have a surface area that is more
than three times greater than the placenta,
so more oxygen can diffuse across at any one
moment in time.
The lungs have a thinner barrier than the
placenta, so the diffusion distance is much
smaller, and diffusion takes less time.
The rate of blood flow in the lungs is 10 times
that in the placenta, so the oxygen is quickly
taken away, maintaining a steeper diffusion
gradient down which oxygen will diffuse
more rapidly.
Active transport moves substances up their
concentration gradient, whereas in diffusion
substances move down their concentration
gradient. Active transport requires input of
energy from the cell, whereas diffusion does not
require the cell to use energy.
8
9
Look for:
• x-axis labelled ‘year’, and the four years
shown
• y-axis labelled ‘percentage of people living
with HIV who knew they had the virus’,
and a suitable scale (note that this does not
need to start at 0)
• bars accurately plotted
Although these data could be shown as a
histogram, the instructions are for a bar chart
so the bars should not touch.
If people do not know that they have HIV,
then probably no one else knows either, so it is
not possible to be certain that these numbers
are correct. Estimates can be made from the
proportion who are tested for HIV who did
not think that they had the virus, but who are
found to have it. Researchers can then use this
proportion to estimate how many people in the
general population would be found – if tested –
to have HIV.
The third part of the target – suppression of the
virus – has already been met. The first part –
percentage of people living with HIV who
know they have it – looks unlikely to be met,
as the increase between 2015 and 2018 is very
small and looks to be levelling off. If this trend
continues, by 2020 the percentage would be
predicted to be no higher than 65% or 66%. The
second part – percentage of people diagnosed
with HIV who are receiving antiretroviral drugs
– is increasing steadily, and it is possible that the
90% target will be reached by 2020. There has
been an increase of 5% to 6% of people each
year, so if that continues then perhaps it might
reach 91% in 2020.
Students might like to look on the UNAIDS
website to check what actually happened
by 2020.
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Chapter 16
Exercise 16.1
Exercise 16.3
1
2
3
4
12 Neither of the parents of the two people with
PKU have PKU. If the allele was dominant,
then at least one of the parents would have it
and would therefore show the condition. (This
situation can only be explained if both parents
are heterozygous, with one copy of the normal
allele and one of the recessive PKU allele.
Two of their children must have received the
recessive allele from both parents.)
13 Both of person 4’s copies of this gene are the
recessive allele. It is virtually impossible for
the same mutation to have occurred in both of
them, as mutation is a random event.
14 Person 1 could be either QQ or Qq.
Person 2 could also be either QQ or Qq.
Person 3 must be Qq, as they do not show the
condition but do pass on a q allele to a child.
Person 4 is qq.
Person 5 could be QQ or Qq, as both of her
parents have the genotypes Qq.
15 The only way person 5 could have a child with
PKU is if she has the genotype Qq, and her
partner has this genotype as well. There is a
1 in 2 chance that she does not have the q allele
(in other words that she is QQ) and it is likely
that her partner will also be QQ. However, if she
does have the genotype Qq, and if her partner
is from a family in which some members have
PKU, then there is a risk that he could also be
Qq, in which case there is a one in four risk of
them having a child with PKU.
5
6
Nucleus
DNA
A length of DNA that codes for a protein.
a Diploid
b 16
a 32
b Genetically identical
c Three of: growth, repair of damaged
tissues, replacement of cells, asexual
reproduction.
Four cells produced instead of two; cells are
haploid and not diploid; cells are genetically
different, not genetically identical.
Exercise 16.2
7
8
9
Body divided into head, thorax and abdomen;
three pairs of jointed legs; one pair of
antennae; one pair of wings.
Drosophila
Genotype
Phenotype
NN
normal wings
Nn
normal wings
nn
vestigial wings
10 phenotypes of parents normal wings vestigial
wings
genotypes of parents
gametes
Nn
nn
N and n
all n
gametes from
vestigial-winged fly
n
gametes
from
normalwinged fly
N
Nn
normal wings
n
nn
vestigial wings
11 About half would have this phenotype, so
about 41 with vestigial wings.
29
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Chapter 16 continued
21 The genes in a stem cell are exactly the same
as the genes in every other cell.
22 Different combinations are expressed in
each type of specialised cell. Some genes are
‘switched on’ while others are ‘switched off’.
This means that each type of cell synthesises
only the proteins that are required to carry
out its function. For example, a cell in the
skin might express the gene to make keratin,
whereas a cell in the pancreas would express
the gene to make insulin.
Exercise 16.4
16 For example, XR and Xr.
17 phenotypes of parents
white-eyed male red-eyed female
genotypes of parents XrY XR Xr
XR Xr
gametes Xr Y
gametes from red-eyed
female
gametes
from
white-eyed
male
XR
Xr
Xr
Xr XR
red-eyed
female
Xr Xr
white-eyed
female
Y
XRY
red-eyed
male
XrY
white-eyed
male
The predicted ratio is therefore 1 red-eyed
female : 1 white-eyed female : 1 red-eyed
male : 1 white-eyed male.
Exercise 16.5
18 A gene is a length of DNA that codes for the
production of a protein. The sequence of
bases in a gene determines the sequence of
amino acids in the protein that is made.
Proteins are synthesised on the ribosomes in the
cytoplasm of a cell. A copy of the gene is carried
to the cytoplasm by a molecule called mRNA.
19 For example: enzymes, antibodies, receptors.
20 A change in the base sequence of the DNA
would result in a change in the amino acid
sequence in the protein that is made. This
affects the shape of the protein, which affects
its function. For example, if the shape of an
enzyme is altered, then its active site may
no longer be a complementary shape to its
substrate, so it cannot form enzyme–substrate
complexes and therefore cannot catalyse
the reaction.
30
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Chapter 17
Exercise 17.1
1
Continuous variation. There are no definite weight categories that a plant must fit into. The mass of
the plants can be any value between the smallest (0.5 kg) and largest (3.4 kg) values.
2
9
8
7
6
Number
of plants
5
4
3
2
1
0
3
4
5
31
0.5–0.9
1.0–1.4
1.5–1.9
2.0–2.4
Mass range / kg
2.5–2.9
3.0–3.4
5 μm
a 12 mm in length; accept figures between 11 and 13 mm
b 12 mm = 12 000 μm (Accept other appropriate answers depending on the guard cell measured.)
c magnification = length in diagram ÷ actual length
= 12 000 μm ÷ 5 μm
= ×2400
If the student has measured a different guard cell in the diagram, and arrived at a slightly different
length value, the magnification value obtained will be different from that obtained here. Check that
the method of calculation is correct.
The leaves have many stomata on the upper surface. This is not usually found in land-living plants,
where most stomata are on the lower surface to reduce the rate at which water vapour is lost through
them – the lower surface is out of direct sunlight and therefore cooler, reducing the rate of evaporation
and diffusion. The water hyacinth leaves are at the surface of the water, so they do not need to conserve
water and having stomata on the upper surface allows them to absorb carbon dioxide easily from
the air.
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Chapter 17 continued
6
The stomatal pores of the plants growing in
polluted water are 1 μm smaller than those
in clean water. The guard cells of the plants
growing in polluted water are 2 μm shorter
than those in clean water. The mean number of
stomata in the upper epidermis is the same in
clean and polluted water. The mean number of
stomata in the lower epidermis is a little higher
in the plants grown in clean water than in those
grown in polluted water.
Exercise 17.2
7
Natural selection depends on the fact that
there is variation within populations.
In most populations, far more young are
produced than will live long enough to be able to
reproduce. The organisms in the population have
to compete for scarce resources.
As a result, only the individuals that are best
adapted to their environment are likely to have
offspring. Their alleles are the ones that are most
likely to be passed on to the next generation.
Exercise 17.3
8
9
32
There has been a steady increase in frequency
of resistance to the insecticide over the whole
period. It has changed from 30% to 64%, so it
has more than doubled.
A random mutation occurred in a cotton
bollworm, producing an allele that conferred
resistance to the insecticide. This resulted in
natural variation in the population, with some
bollworms having resistance and others having
no resistance.
When the insecticide was used on the cotton
plants, it provided a selection pressure. The
individual insects that were resistant to
the insecticide were more likely to survive
and reproduce. They passed on their alleles
for resistance to the next generation.
This continued in each generation, so the
percentage of individuals with the alleles for
resistance increased.
Exercise 17.4
10 The milk yield in the selected population
has increased, but it decreased in the control
population. There are fluctuations in both.
The two populations began with very similar
(but not identical) yields, with the selected
population having a yield of 8000 kg and the
control population having a yield of 7700 kg.
The cows that were born in 1990, however,
had a yield of 10 800 kg in the selected line, but
only 5600 kg in the control line – a difference
of 5200 kg.
11 In the selected line, cows would only have
been selected for breeding if they had a high
milk yield. They would have been bred with
bulls whose female relatives also had a high
milk yield. This continued in each generation.
Alleles for high milk yield would therefore
have been passed on from the selected parents
to their offspring, increasing the frequency of
these alleles in each generation.
In the control line, any cow was allowed
to breed with any bull. We cannot be sure
why the milk yield fell, but it is possible that
cows that have high milk yields are not as
well adapted in other respects, so they may
not breed as successfully, or have as many
offspring, if they are not selected for. For
example, cows with high milk yields may
be more likely to suffer from diseases or
lameness. Therefore, with no selection for high
milk yield by humans, the high-milk-yield
cows are at a selective disadvantage, and are
less likely to reproduce and pass their alleles
on to the next generation.
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Chapter 18
Exercise 18.1
1
2
Community
birds
spiders
snakes
grasshoppers
mice
grass
3
4
a
b
a
b
foxes
voles
other plants
The position in a food chain at which an organisms feeds.
In sequence: producer, primary consumer, secondary consumer, tertiary consumer.
20
× 100 = 0.1%
20 810
The percentage of energy transferred from the first to the fourth trophic level is 0.1%.
Much of the energy is lost as heat to the environment, through respiration. Some goes to the
decomposer food chain.
Exercise 18.2
carbon dioxide
in the air
combustion
combustion
photosynthesis
carbon compounds
in plants
respiration
respiration
feeding
carbon compounds
in animals
feeding
carbon in fossil fuels
respiration
feeding
carbon compounds
in decomposers
6
7
8
They had no light, so they could not photosynthesise.
Protein and DNA
The dead phytoplankton were decomposed by the bacteria in the water. The bacteria secreted enzymes
that digested the proteins and other compounds in the cells of the dead phytoplankton. The digested
products were absorbed into the decomposers’ cells.
9 The ammonia came from the breakdown of nitrogen-containing substances, such as proteins, in the
cells of the dead phytoplankton.
10 Nitrate first appeared in December – that is, about one month after the start of the experiment.
The quantity of nitrate increased sharply in April.
11 The nitrate was produced from ammonia, by nitrifying bacteria.
33
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Chapter 18 continued
Exercise 18.3
12 A group of organisms of one species, living in the same area at the same time.
13
50
stationary
phase
40
Number of
adult beetles
30
log (exponential)
phase
20
10
0
death phase
lag phase
0
100
200
300
400
500
Time / days
600
700
800
900
14 a Death phase
b Log (exponential) phase
c Lag phase and stationary phase
15 The beetles were running out of flour to eat.
16 a The curve should show the population following a similar sigmoid shape, but not rising to such a
high number. The curve could rise and fall at similar times or each change could happen earlier.
b The explanation should match the predictions. Students should mention that there is competition
between the two types of beetle for food, and that food supplies run out earlier. Food supply
becomes a limiting factor.
34
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Chapter 19
9
Exercise 19.1
1
2
3
4
5
Tethering
Most chickens were free range, whereas no
goats were free range.
Most goats were tethered, whereas no chickens
were tethered.
Both goats and chickens were fed in an
enclosure, but this was more common for
chickens than for goats.
Method A. The number of animals kept in
a certain area is greater than for the other
methods. Also, this method involves bringing
in food for the animals from elsewhere – there
are higher inputs.
Chemical fertilisers are used to add mineral
ions, such as nitrate ions, to a soil, where the
soil does not have enough of these to support
good growth of plants. Manure from the
animals can be collected and added to the soil.
This releases nitrate ions and other minerals
required by plants.
a This could be done to improve the
quality of the offspring, through selective
breeding. For example, the farmer might
want to increase the quantity of milk
produced by her cows.
The increase in milk produced might
increase the money she earns from selling
the milk, more than the costs of using the
non-native bull.
b The native breeds have always been
exposed to these parasitic worms, so they
have developed adaptive features that
enable them to resist infection with them.
Non-native breeds have not been exposed
to this selection pressure, so most of them
have not developed resistance.
Each species has developed adaptive features,
through natural selection, that enable it to
survive and reproduce in its habitat. If that
habitat changes, these adaptive features may no
longer be as useful, so individuals may be more
likely to die before they can reproduce.
10 Chinchilla lanigera is still threatened by loss
of habitat. The population cannot increase
significantly if there is not sufficient suitable
habitat where it can live.
11 When an individual is first exposed to a new
pathogen, it takes time for its white blood cells
to respond to the pathogen. Lymphocytes
that can produce antibodies that have a
complementary shape to the antigens on
the pathogen divide to form a clone of cells,
which then all secrete the antibody. While this
is happening, the animal may become ill and
perhaps die. If it survives, memory cells retain
the ability to reproduce rapidly and secrete large
quantities of antibody if the same pathogen is
encountered again. Pudu puda would not have
encountered the pathogens brought in by the
introduced species before, so may not be able to
develop immunity to them quickly enough
to survive.
12 The isolated populations cannot exchange
genes with one another. Genetic variations
within each of the small, isolated populations
will remain low. This makes it more likely that
recessive alleles will come together in offspring,
making them less able to survive. It also means
that the populations are less able to respond,
through natural selection, to changes in
their environment.
Exercise 19.2
6
7
8
35
The first word is the genus that the species
belongs to, and the second word tells us
the species.
It is a species at high risk of becoming extinct.
Humans using ground for growing food crops
or livestock production; for building houses; for
extracting resources from the ground; humans
causing pollution of land or water; climate
change may cause changes in plant species that
can live in an area.
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Chapter 20
Oxygen concentration: to provide enough
oxygen for the microorganism to respire
aerobically; air is bubbled into the
fermenter to supply oxygen.
Nutrient concentration: to provide
carbohydrates for energy release in
respiration, and a nitrogen source for the
synthesis of amino acids and proteins;
nutrients are provided through an inlet
into the fermenter.
Exercise 20.1
1
2
3
4
5
6
36
A protein that functions as a biological catalyst;
enzymes are involved in all metabolic reactions.
Pectin
It digests the pectin that holds cells together
in fruits, allowing more juice to be extracted.
It can also be used to clarify the juice, by
breaking down insoluble particles to produce
soluble products.
The most likely choice of display is a bar
chart. (A line graph would not be suitable,
because the substrate is a discontinuous
variable.) ‘Substrate’ should be on the x-axis.
‘Production of pectinase / arbitrary units’
should be on the y-axis, with a suitable scale
with equally spaced intervals, ranging from 0
to 1500. Students may have six equally spaced
bars for the six types of substrate, but a better
choice would be to have the bars for wheat
bran and wheat bran with sugar cane bagasse
next to each other (these could be touching),
and the same for the other two pairs. Each bar
could be separately labelled or one of each of
the pairs could be shaded to indicate that it
includes sugar cane bagasse, and a key given
to explain what the shading means.
If the waste materials are not used, they have to
be disposed of in some way. They might pollute
waterways, causing eutrophication (as they are
likely to contain nutrients that could be used
by bacteria, which would then use up a lot of
oxygen as their increased populations respire).
Also, if waste materials are not used, other
plant material would have to be used to make
the pectinase. This means more land would be
used for growing plants as a substrate for the
bacteria instead of being used for growing food
or for habitat for wildlife.
a A container in which microorganisms can
be grown in a liquid medium.
b pH: to maintain a suitable pH for enzyme
activity; buffers can be used to maintain
the pH at a particular level, or controlled
quantities of acid or alkali added if the
pH probe gives a reading that deviates
from the required value.
Temperature: to maintain the optimum
temperature for enzymes activity; water at
a chosen temperature is passed through a
jacket that surrounds the fermenter.
Exercise 20.2
7
Changing the genetic material of a cell or an
organism by removing, changing or inserting
individual genes.
8 By inserting genes to confer resistance to
herbicides; by inserting genes to confer
resistance to insect pests; by inserting genes to
improve nutritional qualities.
For each of these, accept a specific example
instead of the general statement.
9 a Restriction enzymes would be used at
step 1.
b i
Sticky ends are unpaired lengths of
DNA (i.e. just a single strand).
ii The unpaired bases on the sticky
ends of two pieces of DNA will pair
up with each other, as long as the
unpaired bases are complementary to
each other. This enables the joining of
the DNA of the plant genes with the
DNA of the plasmids.
c DNA ligase would be used at step 2.
d The plasmids are used to transfer the
genes from the original bacteria into
Agrobacterium tumefaciens.
10 Agrobacterium tumefaciens naturally infects
plant cells, so it was able to carry the plasmids
carrying the required gene into the rice cells.
11 With selective breeding, you can only build
on variation that is already there, by selecting
organisms with the best features for breeding.
There is no natural variant of rice that has
genes for making large amounts of carotene in
its grains, so there was no real starting point
for selective breeding. Instead, genes that were
already present in other species were used.
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