Answers to Workbook exercises
Chapter 1
Exercise 1.1 Observing and drawing
organisms
a, b, c Use the self-assessment checklist to assess the
drawing and labelling.
d
Check the measurements and calculation against
the student’s drawing.
e
Look for clear, comparable points opposite each
other.
Exercise 1.2 Using keys
Note that students cannot write in italic, so should
underline the binomials instead.
a
A (given) 1b, 2a, 3a, Crocodylus niloticus
B 1a, Geochelone elephantopus
C 1b, 2b, Ophiophagus hannah
D 1b, 2a, 3b, Chamaeleo gracilis
b
i
binomial
ii The first part is the genus the organism
belongs to, and the second part is its species.
c
© Cambridge University Press 2014 IGCSE® Biology
They all have scales.
Chapter 1: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 2
Exercise 2.1 Animal and plant cells
a
cell membrane
nucleus
Exercise 2.2 Drawing cells and
calculating magnification
a
cell wall
chloroplast
position of
cell
membrane
cytoplasm
nucleus
b
nucleus
cell membrane
cell wall
Use the self-assessment checklists to assess the
drawing and labelling.
large
vacuole
containing
contain
cell sap
plasm
cytoplasm
b
i
real size of object =
25mm
2000
= 0.0125 mm
So: real width of plant cell =
membrane
memb
around
aroun
vacuole
vacuo
chloroplast
roplast
ii magnification –
c
i
d
The length of the plant cell in the diagram is 60 mm.
The width of the cell in the diagram is 50 mm.
ii So magnification = 50 = ×500
0.1
magnification =
size of drawing
real size of object
So: real size of object = size of drawing
magnification
68
So: real height of plant cell =
= 0.75 mm
70
© Cambridge University Press 2014 IGCSE® Biology
size of drawing
magnification
size of drawing
real size of object
So: magnification of student’s drawing
width of drawing in mm
=
0.0125 mm
Exercise 2.3 Organelles
a
nucleus
b
cell wall
c
cytoplasm
d
cell membrane
e
chloroplasts
f
vacuole
g
mitochondrion
h
ribosome
Chapter 2: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 3
Exercise 3.1 Diffusion experiment
a
Dish
b
Temperature / °C
Distance red colour diffused into jelly / mm
Hole 1
Hole 2
Hole 3
Hole 4
Mean (average)
A
10
2
3
2
3
3
B
20
5
5
6
4
5
C
40
9
11
8
10
10
D
80
19
21
18
23
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.
c
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.
d
This allowed for a mean to be calculated. It
improves the reliability of the results.
e
Measurement of the distance diffused, because the
‘edge’ between the colour and the uncoloured jelly
will not be very distinct. Some dye may have got
into the jelly before the dishes are transferred to
their final temperatures (especially as they were
carried). Time taken for the dye and jelly in each
dish to reach their final temperature – the dye
won’t have been at the correct temperature for the
entire duration of the experiment.
© Cambridge University Press 2014 IGCSE® Biology
Exercise 3.2 How plants take up
water
a
cell wall, large vacuole
b
Label line to the cell surface membrane, or to the
membrane around the vacuole.
c
Water molecules move randomly. There is a
greater concentration of them outside the cell than
inside, so more will (by chance) move into the
cell than out of it, through the partially permeable
cell surface membrane. The solutes in the cell
cannot get out through the partially permeable
membrane. (Some students may answer in terms
of water potential. The water potential of the
solution outside the cell is higher than that inside,
so water moves down its water potential gradient.)
d
This provides a large surface area, so more water
can pass across the surface at any one time.
Chapter 3: Answers to Workbook exercises
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Exercise 3.3 Osmosis and potatoes
a
The table should have rows or columns for the
percentage concentration of the solution, and rows
or columns for the mass of potato pieces, with
the unit g in the heading. Students should also
calculate the change in mass. The following is an
example of a suitable results table.
e
Have several pieces of potato in each solution, and
calculate a mean change in mass for each.
f
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.
mass / g
Percentage
concentration
of solution
Before
soaking
After
soaking
Change
0.0
5.2
5.5
+0.3
0.1
5.1
5.2
+0.1
0.2
4.9
4.9
0
0.5
5.0
5.3
+0.3
0.8
5.1
5.0
–0.1
1.0
5.2
5.0
–0.2
b
The mass of the potato piece soaking in 0.5%
solution has increased, but it would be expected
to decrease. This does not follow the pattern of the
other results and so is anomalous.
c
Look for the following features on the graph:
◆ ‘percentage concentration of solution’ on the
x-axis, and ‘change in mass / g’ on the y-axis
◆ suitable scales
◆ all points plotted correctly (allow 0.5 mm
tolerance) as crosses or as encircled dots
◆ either a best-fit line, drawn as a smooth curve
with equal numbers of points above and below
the line, or points joined with straight lines
drawn with a ruler; the anomalous result
should be ignored.
d
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 solutions with higher
concentrations than this had water potentials
lower than that of the potato cells, so water
moved out of the cells by osmosis and their mass
therefore decreased.
The 0 and 0.1% solutions had a 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 a water potential
equal to that of the potato cells, so there was no
© Cambridge University Press 2014 IGCSE® Biology
Exercise 3.4 Diffusion and active
transport
S
a
A, because the concentration is the same inside
and outside.
b
B, because the concentration inside the root cell is
greater than outside, so it must have been moved
in against its concentration gradient.
c
C, because the concentration outside the cell is
greater than inside, so it must have been moved
out against its concentration gradient.
d
The roots would not be able to respire, 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. There
would probably be no measurable effect in the soil
water, because this is a huge volume compared
with that of the root cells.
Chapter 3: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 4
Exercise 4.1 Carbohydrates
Exercise 4.2 Proteins
a
The facts that students give will very much depend
on what they find, and what they think of as being
important. Give credit for being very brief and packing
information into a few words; very long answers do not
meet the criteria set by the question. The following are
some suitable answers.
Look for a single ruled table with fully headed
rows and columns.
Result
of test
with iodine
Result of
test with
Benedict’s
Food
A
brown
orange–red
contains
reducing
sugar but
not starch
Food
B
black
blue
contains
starch but
not reducing
sugar
Conclusion
Students might decide to have two separate
columns for the conclusions, one for starch and
one for reducing sugar, which would be fine.
b
Type of
carbohydrate
Example
monosaccharide
glucose
polysaccharide
Role in living
organisms
provides energy;
released by
respiration; also
the form in which
carbohydrates are
transported in
mammalian blood
sucrose
the form in which
carbohydrates are
transported in
plants
starch
the form in which
plants store energy
cellulose
forms cell walls of
plant cells
glycogen
the form in which
animals store energy
© Cambridge University Press 2014 IGCSE® Biology
a
Haemoglobin: a red pigment found in red blood
cells of mammals that contains iron. It combines
reversibly with oxygen, and so is used for the
transport of oxygen from lungs to respiring
tissues.
b
Keratin: a protein found in hair, nails and the
upper layers of skin. It is insoluble and forms long
fibres. It is a structural protein.
c
Collagen: a protein found in skin, bone and other
tissues. It forms long, stretchy fibres and so helps
to provide strength and elasticity. Vitamin C is
required to make it.
d
Antibodies: proteins secreted by white blood cells
(lymphocytes) in response to antigens. Specific
antibodies attach to specific antigens, and help to
destroy them.
Exercise 4.3 Testing a hypothesis
a
Add dilute sodium hydroxide (or potassium
hydroxide) and very dilute copper sulfate solution
to the milk. A purple colour indicates the presence
of protein. (Alternatively, biuret reagent could
be added.)
b
i
The variable to be changed is the type of
milk – cow’s milk and goat’s milk.
ii 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 reagents added to it, the time
left before the intensity of the colour is assessed.
Chapter 4: Answers to Workbook exercises
1
iii The quantity to be measured is the intensity of
the colour produced after the biuret test has
been carried out on the milk.
iv This could be measured by comparing the
colours visually.
v 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.
© Cambridge University Press 2014 IGCSE® Biology
Exercise 4.4 DNA
a
bases
b
Upper strand: A, G. Lower strand: C
c
chromosomes, nucleus
Chapter 4: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 5
Exercise 5.1 Writing enzyme
questions
h
40 °C is certainly the temperature at which the
enzyme worked fastest in this experiment, but the
optimum could actually be somewhere either side
of this – either a bit below or anywhere between
40 °C and 100 °C.
i
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 – say 35 °C,
45 °C, 50 °C and so on. Once these results have
been found, the temperature range can be
narrowed down even more to keep moving in
closer and closer to the optimum temperature.
j
Take equal volumes of cow’s and goat’s milk. Add
equal volumes of lipase to both samples. Keep the
tubes at 40ºC for five minutes.
Look for questions that are very clear, biologically
correct and that have unambiguous answers.
Exercise 5.2 Lipase experiment
a
fats (lipids)
b
fatty acids and glycerol
c
Fatty acids are produced, which are acids and
therefore have pH below 7.
d
Tube
1
2
3
4
5
Temp / °C
20
20
0
40
100
Milk added?
no
yes
yes
yes
yes
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
6.55
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
e
There was no milk, so no fat, so no fatty acids
were made.
f
The high temperature denatured the lipase
molecules, so there was no digestion of fats and no
fatty acids were made.
g
These tubes differed only in their temperature.
Lipase acts more rapidly at 20 °C than at 0 °C
because its molecules (and those of its substrate)
are moving round faster and therefore collisions
between enzyme and substrate molecules happen
more frequently and with more energy. This means
the rate of reaction is faster at 20 °C than at 0 °C.
© Cambridge University Press 2014 IGCSE® Biology
Measure the pH every two minutes.
Repeat the experiment three times, and calculate
the mean pH for cow’s milk and mean pH for
goat’s milk at each time interval.
The milk in which the pH drops faster is the one
that contained most fat.
Exercise 5.3 Finding the optimum
pH for amylase
a
pH
b
1 to 14 (a narrower range would be acceptable)
c
Using buffer solutions. Tubes could be set up using
buffers for pH 1, 2 and so on.
d
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
Chapter 5: Answers to Workbook exercises
1
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 too need to be
kept constant – use water baths.
e
f
g
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 1, 3, 5, 7, 9 and 11, 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.
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 times 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
then be colours.
© Cambridge University Press 2014 IGCSE® Biology
h
The sketch 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.
Exercise 5.4 How enzymes work
S
a
b
c
i
Each type of enzyme has an active site of a
specific shape. It can only bind with a substrate
whose shape is complementary to this.
ii The molecules of the enzyme and the substrate
move faster at higher temperatures, so collide
with each other more frequently.
iii Above its optimum temperature, an enzyme
molecule loses its shape (becomes denatured),
so the substrate cannot fit into its active site.
Chapter 5: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 6
Exercise 6.1 How a palisade cell
obtains its requirements
c
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.
◆
by diffusion into the air spaces then out of a
stoma into the air.
stored as starch in the chloroplast, or changed to
sucrose and transported away in the phloem.
b
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 the cells
except those in the upper and lower epidermis; but
guard cells should also contain green spots.
© Cambridge University Press 2014 IGCSE® Biology
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 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 Sun and shade leaves
a
cuticle
e
Carbohydrates:
◆
Shade leaf
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.
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:
Sun leaf
d
Water:
◆
Part of leaf
Exercise 6.3 Limiting factors
a
S
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’.
Chapter 6: Answers to Workbook exercises
1
S
b
0.04%
c
53 arbitrary units
d
0.12% (Note that if students 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.)
e
light intensity
f
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.
g
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.
Exercise 6.4 Effect of increased CO2 and temperature on tree growth
a, b
Group
CO2
concentration
Temperature
A
normal
B
Increase in diameter / mm
Year 1
Year 2
Year 3
Mean
normal
2.0
4.8
3.8
3.5
increased
normal
5.0
5.8
5.9
5.6
C
normal
increased
4.9
4.2
3.9
4.3
D
increased
increased
6.0
6.1
5.9
6.0
Look for a chart that is clear and easy to
understand. It could be orientated as in the
example above, or students could construct a chart
in which the quantities are arranged down the side
rather than along the top. All columns (or rows)
should be fully headed, including units. All values
should be entered correctly. The mean should be
correctly calculated, and given to one decimal
place only (as for all the individual values).
c
For the bar chart, look for:
◆ group on the x-axis
◆ increase in diameter / mm on the y-axis, with
a suitable scale
◆ each bar accurately and cleanly plotted, with
bars not touching.
d
Group B had increased carbon dioxide but normal
temperature, and this grew by an average of 2.1 mm
per year more than Group A which also had
normal temperature but did not have increased
carbon dioxide.
© Cambridge University Press 2014 IGCSE® Biology
S
We can also compare Groups C and D, in which
both had increased temperature, but only Group
D has increased carbon dioxide. Again, the
increase in carbon dioxide resulted in an increase
in growth, this time by an average of 1.7 mm
per year.
e
Here we can compare Groups A and C, where
both had normal carbon dioxide but only
Group C had increased temperature. The higher
temperature resulted in a higher growth rate, by an
average of 0.8 mm per year. We can also compare
Groups B and D, where both had increased carbon
dioxide but only D had increased temperature.
Again, this resulted in a higher growth rate, by an
average of 0.4 mm per year.
f
It is possible that some other factor (e.g.
availability of nitrate ions – accept other
suggestions) is limiting the rate of growth when
both temperature and carbon dioxide levels
are raised.
Chapter 6: Answers to Workbook exercises
2
Answers to Workbook exercises
Chapter 7
Exercise 7.1 Diet
a
They all come from plants.
b
Scrambled egg contains a large quantity of fat,
which contains more energy per gram than any
other nutrient.
c
Spinach, because the total mass of the listed
nutrients in 100 g of food is least, and therefore
the remaining mass, which is mostly water, is
greatest.
d
Perhaps children who have a low standard of
living do not have as much calcium in their diet.
Perhaps they do not clean their teeth or use
fluoride toothpaste. Perhaps they eat more sweets
or drink more carbonated drinks.
c
This decreases the number of decayed teeth. We
can see the difference between the results for town
A and town B, where fluoride was added to the
water. This roughly halved the number of decayed
teeth in five-year-olds at any particular standard of
living.
d
Perhaps there is more fluoride in the naturally
fluoridated water than was added to the water in
town B. Perhaps the fluoride in town B has only
been added recently, so the children didn’t have
fluoride in the water when they were younger.
Pepsin … in stomach
Mastication … in mouth
Amylase … in mouth and in duodenum
Exercise 7.4 Cholera patterns in
Bangladesh
Pancreatic juice … in duodenum
a
The incidence fluctuates greatly. There are peaks
approximately every year, generally towards the
end of the year. In some years, e.g. 1987 and
1988, there are several peaks and troughs in just
one year.
b
There seems to be a correlation between peaks in
sea surface temperature and especially high peaks
in cholera incidence. For example, in 1983, sea
surface temperature peaked at more than 40 °C,
and this coincided with a peak in cholera cases
at just under 40%. This happened again in
1987–1988, when the surface sea water
temperature was about 35 °C, and cholera
incidence rose to over 40%. (Other points on the
graphs could be chosen to illustrate this.)
Gastric juice … in stomach
Lipase … in duodenum
Bile salts … in duodenum
Sodium hydrogencarbonate … in duodenum
Saliva … in mouth.
Exercise 7.3 Tooth decay data
analysis
a
b
Egg and spinach, as these have the highest
concentrations of iron. Iron is needed to make
haemoglobin. Anaemia is caused by a lack of
haemoglobin.
Exercise 7.2 Functions of the
digestive system
◆
◆
◆
◆
◆
◆
◆
◆
◆
decayed tooth), but where standard of living was
lowest they had a mean score of 4.0 (on average,
each child had four decayed teeth).
The incidence of tooth decay increases as standard
of living decreases. Where the standard of living
was highest, five-year-olds had a mean tooth decay
score of 1.0 (that is, on average each child had one
© Cambridge University Press 2014 IGCSE® Biology
Chapter 7: Answers to Workbook exercises
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1
c
d
S
Cholera is caused by a bacterium. It leaves the
body in faeces, and is transmitted to another
person if they drink water or eat food that has
been contaminated by these faeces.
Bacteria reproduce more rapidly in warm
temperatures, so the populations of bacteria
in contaminated food or water might be larger
than when it is cooler. People drink more when
it is warm, so may be more likely to drink
contaminated water. They may swim more, to try
to keep cool, increasing the likelihood of bacteria
getting from an infected person into the water, or
from the water into an uninfected person. (There
are other possible answers that you might think of.)
Exercise 7.5 Vitamin D absorption
a
It rose very rapidly over the first 12 hours, from
0 to just over 140 arbitrary units. After peaking
at 12 hours, it fell less rapidly, reaching 60 a.u. at
© Cambridge University Press 2014 IGCSE® Biology
48 hours. It then continued to fall but now very
slowly, reaching 56 a.u. at 72 hours.
S
b
Small intestine / ileum.
c
glucose, amino acids, fatty acids, glycerol, water,
any other vitamin, any mineral (e.g. calcium)
d
It is long, so food is in contact with its walls for a
long time. It is covered with villi, which increase its
surface area. It is folded, which also increases surface
area. The walls of the villi are thin, and there is a
good blood supply, so it is easy for digested nutrients
to diffuse through the walls and into the blood.
e
Its molecules are already small enough to be
absorbed.
f
Vitamin D is made in the skin when sun shines
onto it. If this had happened, we would not know
how much of the vitamin D in the blood had come
from this source, and how much from the vitamin
D that was ingested.
Chapter 7: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 8
Exercise 8.1 A transpiration
experiment
a
The results chart could look like this:
Condition
b
c
Still air
time / min
0
2
4
distance / cm
0
2.8 6.1 10.0
Moving air
6
8
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.
Still air: meniscus moved 16.2 − 0 = 16.2 cm in
10 minutes. So, mean rate was 1.62 cm per minute.
Moving air: meniscus moved 44.9 − 16.2 = 28.7 cm
in 10 mins. So, mean rate was 2.87 cm per minute.
d
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.
e
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.
Exercise 8.2 Tissues in a root
a
cortex
xylem
© Cambridge University Press 2014 IGCSE® Biology
phloem
endodermis
Chapter 8: Answers to Workbook exercises
1
b
Diameter in diagram = 10 mm
allow any measurement between 10 and 12
d
i
e
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 roots,
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 roots to turn into
starch.
So real diameter = 10 ÷ 200 = 0.05 mm
S
c
Transporting water; transporting mineral ions;
support.
d
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.
e
Water enters the root hairs by osmosis, down
a water potential gradient from the soil to the
cytoplasm, through the partially permeable cell
membrane. It moves across the cortex by osmosis,
and finally into the xylem vessels.
Exercise 8.3 Sources and sinks
a
sucrose
b
starch
c
i
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.
ii Leaves will be sources in summer. They
photosynthesise, producing sugars that can be
converted to sucrose and transported to other
parts of the plant.
iii 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.
© Cambridge University Press 2014 IGCSE® Biology
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.
ii 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.
iii 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.
Chapter 8: Answers to Workbook exercises
2
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Answers to Workbook exercises
Chapter 9
Exercise 9.1 Risk of heart attack
a
S
She has a 13% (13 in 100) chance of having a heart
attack in the next five years.
b
She should stop smoking. This will reduce the risk
from 13% to 7%. 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%.
c
Health records have been kept for large numbers
of women over long periods of time. The records
have been grouped into women in a particular age
group, 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.
Exercise 9.2 The heart in a fetus
a
O in the left atrium.
b
OF in the right atrium.
c
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.
d
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.
© Cambridge University Press 2014 IGCSE® Biology
Exercise 9.3 Double and single
circulatory systems
S
a
b
human (accept any mammal or bird)
c
fish (accept any named fish)
d
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 faster.
Exercise 9.4 Changes in the blood
system at high altitude
a
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
Chapter 9: Answers to Workbook exercises
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◆
S
◆
◆
◆
◆
b
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 a 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.
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 at low altitude
© Cambridge University Press 2014 IGCSE® Biology
◆
◆
S
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.
c
Oxygen transport.
d
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.
e
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.
Chapter 9: Answers to Workbook exercises
2
Answers to Workbook exercises
Chapter 10
Exercise 10.1 Food poisoning in
the USA
Exercise 10.2 Waste disposal
in Australia
a
An organism that causes disease.
a
b
Look for:
◆ pathogen on the x-axis, and number of cases
or percentage of cases or both on the y-axis; if
both are plotted, then two axes will be needed,
one on the left and one on the right
◆ suitable scale or scales on the y-axis or y-axes,
fully labelled
◆ bars plotted accurately
◆ if only one of number or percentage is plotted,
the bars should not touch; if both are plotted,
then the two bars for one organism can touch
◆ if both number and percentage are plotted,
there should be a key or label to make clear
what each bar refers to.
The amount of solid waste that was recycled
increased from 15 000 000 tonnes (1.5 × 107) to
23 000 000 tonnes, an increase of 8000 000 tonnes.
The amount of solid waste that was deposited as
landfill also increased, from 17 000 000 tonnes to
21 000 000 tonnes, an increase of 4000 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
2006–7, unlike 2002–3, the amount of waste that
was recycled was greater than the amount of waste
deposited as landfill.
b
Answers could include some of these ideas.
◆ 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 which 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.
c
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.
d
Most people would not bother to go to a doctor
when they have food poisoning, so there will be
many unrecorded cases.
e
For example: keep food cool (in a fridge); 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.
© Cambridge University Press 2014 IGCSE® Biology
Chapter 10: Answers to Workbook exercises
1
◆
◆
c
S
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.
2002–3: 15 000 000 + 17 000 000 = 32 000 000
tonnes
2006–7: 23 000 000 + 21 000 000 = 44 000 000
tonnes
ii 44 000 000 – 32 000 000 = 12 000 000 tonnes
iii (12 000 000 ÷ 32 000 000) × 100 = 37 %
c
Immunisation coverage increased sharply from
1980 to 1991, from about 22% to 76%. 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.
d
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.
e
The antigens on the polio viruses would be
recognised as foreign by lymphocytes. These
lymphocytes would multiply to form a clone,
which would then make antibodies against the
viruses. Some of them would remain in the blood
as memory cells. If the polio virus is encountered
again, these memory cells will rapidly make
antibodies to destroy them.
f
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.
i
Exercise 10.3 Eradicating polio
a
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.
b
Look for some of these ideas. (For some of the
points, accept other years to be quoted.)
◆ 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 1983
or 1984.
◆ 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.
© Cambridge University Press 2014 IGCSE® Biology
Chapter 10: Answers to Workbook exercises
2
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Answers to Workbook exercises
Chapter 11
S
Exercise 11.1 Effect of temperature
on the rate of respiration
Look for the following points being made somewhere
in the plan:
◆ temperature varied, over a stated range (say,
0–50 °C)
◆ how the temperature is varied (for example,
placing in fridge, warm incubator, or standing
in a water bath)
◆ important variables controlled – type and age
of seeds, mass or number of seeds, length of
time seeds are soaked before placing in a flask
or other container, size and insulation of flask
◆ details of how the dependent variable – e.g.
carbon dioxide concentration – will be
measured
◆ outline results chart.
b
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.
c
Respiration would continue, but photosynthesis
would not. The indicator would therefore go
yellow in tubes A, B and C, and remain unchanged
in D.
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.
Exercise 11.2 The effect of animals
and plants on the carbon dioxide
concentration in water
a
In tube A, the animal respired, giving out carbon
dioxide.
The results chart could look like this:
Tube
A
B
C
D
Contents
animal
plant
animal
and
plant
no
animal
or
plant
orange
orange
Colour of
indicator
at start
orange
Colour of
indicator
at end
yellow
orange
deep
red
orange
orange
Exercise 11.3 A simple
respirometer
S
a
Towards the container. As the woodlice use
oxygen, this reduces the volume of air around
them. The carbon dioxide they give out is
absorbed by the soda lime.
b
Look for ‘time / minutes’ on the x-axis, and
‘distance travelled / cm’ on the y-axis; both axes
fully labelled with units, and with suitable scales;
points accurately plotted with crosses or encircled
dots; two clean best-fit lines drawn and labelled.
Students might also want to include a row stating
the conclusions that can be made.
© Cambridge University Press 2014 IGCSE® Biology
Chapter 11: Answers to Workbook exercises
1
S
c
Distance travelled in 8 mins is 2.4 cm. So, mean
distance travelled in 1 min = 0.3 cm
d
It was probably because the soda lime took up the
small amount of carbon dioxide already present
in the air. It could also be caused by a change in
temperature – if temperature fell, then the volume
of gas would be decreased.
e
Repeat the experiment twice more, and calculate
mean rates of movement of the oil drop. Place
more animals in the apparatus, so that they respire
faster and it is easier to measure the distance
moved by the drop. Use a longer tube so results
can be collected over a longer period of time.
Exercise 11.4 Gas exchange
surfaces in rats
a
Look for:
◆ ‘age / days’ on the x-axis
◆ ‘ratio of alveolar surface to body mass / cm2
per gram’ on 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.
© Cambridge University Press 2014 IGCSE® Biology
b
The individual rats may have differed in size, 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.
c
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.
d
When pregnant, the female’s alveolar surface has
to supply the growing embryo with oxygen, as well
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.)
Chapter 11: Answers to Workbook exercises
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2
Answers to Workbook exercises
Chapter 12
Exercise 12.1 The human excretory
system
a, b
c
S
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.
b
This increases the surface area across which
diffusion can take place, speeding up the process.
c
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.
d
The concentration of protein will remain
unchanged. Protein molecules are too large to get
through the holes in the dialysis tubing, so they
will all stay in the blood.
e
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 dialysis
membrane.
S
Exercise 12.2 Dialysis
a
Blood in the artery might be at too high a
pressure, and would be pulsing. It would be better
for the blood to flow smoothly from the patient
to the pump. It might also be dangerous to the
patient to have blood taken from an artery.
© Cambridge University Press 2014 IGCSE® Biology
Chapter 12: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 13
Exercise 13.1 Caffeine and reaction
time
Look for these points being made somewhere in
the plan:
◆ caffeine intake varied (for example, drinking
coffee or cola and drinking water); some
students may wish to use a range of caffeine
concentrations
◆ important variables controlled: volume and
concentration of caffeine in the liquid drunk;
time between drinking and doing the reaction
time test; time of day; age and sex of person;
what the person has done just before the test
is carried out; how many times the person has
done a reaction time test before (in practice, it
is impossible to control all of these variables)
◆ reaction time measured, using a stated method
(for example, using a test on the Internet, or
catching a dropped ruler)
◆ repeats done (probably using different people,
as any one person will improve as they do
more tests, up to a limit)
◆ outline results chart drawn, and sketch graph
showing predicted results if the hypothesis
is correct.
S
Exercise 13.2 Accommodation in
the eye
a
© Cambridge University Press 2014 IGCSE® Biology
b
The ciliary muscles contract, which narrows the
diameter of the circle of muscles. This loosens
the tension on the suspensory ligaments, which
allows the lens to revert to its natural, rounded
shape. The lens now refracts light rays more
strongly, bringing the diverging rays from the
nearby object to a focus on the retina.
c
i A fast, automatic response to a stimulus.
ii A blurred image on the retina.
d
They are less able to focus on objects at different
distances. They may be able to see clearly at a
particular distance, but vision will be blurred at
other distances.
S
Exercise 13.3 Auxin and tropism
a
A response in which part of a plant grows away
from the direction in which it is pulled by gravity.
b
i
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.
◆
◆
◆
◆
ii There was more auxin on the lower surface
than on the upper surface. This made the cells
on the lower surface get longer than those
on the upper surface, so the shoot curved
upwards.
Chapter 13: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 14
Exercise 14.1 Endotherms and
ectotherms
a
b
c
d
Endothermic animals: cat, rabbit, burrowing
bettong;
Ectothermic animals: alligator, gopher snake,
cyclodus lizard
The cat, rabbit and burrowing bettong use
respiration to provide heat to keep their bodies
warm when the environmental temperature is
below 37 °C. This requires fuel, which is in the
form of carbohydrates, fats or proteins. The other
three animals don’t use food to produce heat
energy, so they need much less.
At 5 °C, the cat has a core body temperature of
about 38 °C, so its metabolic reactions will be
taking place quickly, and it will be active. The
lizard has a body temperature of about 5 °C, so its
reactions will be taking place slowly and it will be
inactive.
As they have a constant core temperature, cats are
able to be active in winter and summer, at night
and in the daytime. This means they can hunt in
all seasons and at all times of day. Rabbits, too, can
be active at all of these times, so they are able to
flee from predators no matter what the external
temperature.
© Cambridge University Press 2014 IGCSE® Biology
Exercise 14.2 Diabetes
S
a
When blood glucose levels rise higher than
normal.
b
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.
c
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.
d
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.
Chapter 14: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 15
Exercise 15.1 Alcohol and traffic
accidents
a
b
c
Young drivers have most accidents, with more
than five fatal collisions per 10 000 licensed
drivers, for 16-year-olds. The number of accidents
gradually decreases as the drivers get older,
reaching a minimum of about 1.3 fatal collisions
per 10 000 licensed drivers in the 61–70 age group.
The number of accidents then increases again, so
that drivers aged 81 and over have about the same
number of collisions per 10 000 licensed drivers as
the 21–30 age group.
Young drivers probably have so many accidents
because they are not very experienced, and are
not aware of the circumstances that may cause
accidents. They may have a tendency to drive
faster than older drivers, and with less caution
and understanding of road conditions and the
likely behaviour of other drivers. As drivers get
older, they gain experience and become better
drivers. After the age of 70, however, this increase
in experience is outweighed by factors relating
to ageing, such as poor eyesight or slower
reaction times.
The highest number of fatal accidents involving
drivers drinking alcohol was in the 18–20 age
group. The highest proportion of fatal accidents
involving drivers drinking alcohol was in the
21–30 age group.
© Cambridge University Press 2014 IGCSE® Biology
d
Alcohol slows down reactions, increasing reaction
time and therefore braking time – the time taken
to bring a car to a stop after seeing a danger and
responding to it by braking. Alcohol can also
increase a person’s self-confidence, so that they
don’t realise they have been affected by alcohol,
and may drive more recklessly than they
normally would.
Exercise 15.2 Smoking and life
expectancy
a
i 100
ii Just below 80
b
i 100
ii Just below 50
c
Look for:
a general statement – for example, the more
cigarettes a person smokes, the less long they
are likely to live
◆ a reference to specific survival rates at one or
more ages, comparing at least two different
rates of smoking (or not smoking)
◆ a recognition that at least some people who
smoke a lot of cigarettes survive into their 90s
◆ a recognition that at least some people who do
not smoke at all die in their 40s.
◆
d
Answers could mention any smoking-related
disease, such as lung cancer, chronic bronchitis,
emphysema.
Chapter 15: Answers to Workbook exercises
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Answers to Workbook exercises
Chapter 16
Exercise 16.1 Grass pollen
a
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 falling sharply to 15 hours,
then remaining low during the late afternoon
and evening.
b
dull or no petals; anthers dangling outside flower;
feathery stigma outside flower; large quantities
of pollen
c
i
pathogen: an organism that causes disease
immune system: the white blood cells
(lymphocytes) that defend the body against
disease; lymphocytes produce specific
antibodies to destroy antigens
ii There is much more of it, and it is more easily
carried in the air, so it can be breathed in.
Exercise 16.2 Pollination in forests
of different shapes and sizes
a
The most fruits per flower were produced in Area
A, the set of patches of forest that were connected
to each other by corridors. Here, there was an
average of 0.5 fruits per flower. The least fruits per
flower developed in the set of unconnected forest
patches, Area B, and the set of smaller patches of
forest came in between, Area C.
b
Pollination leaves the pollen grains on the stigma.
The grains then grow pollen tubes down through
the style. The male nuclei (gametes) travel down
the tubes into the ovule, where a male nucleus
© Cambridge University Press 2014 IGCSE® Biology
fuses with a female nucleus to produce a zygote.
This develops into an embryo plant, inside a seed.
The ovary becomes a fruit.
c
Fruits will only develop after a flower has been
pollinated. This is done by butterflies that prefer the
edges of forests. So flowers near the edges of forests
were more likely to produce fruits than ones deep
inside. The small patches of forests had a larger edge
(surface) to volume ratio than the large patches, and
the patches joined by corridors had even more edges.
d
There are many different suggestions students
could put forward. For example, the researchers
could make different patches of forest that were all
identical in volume, but had different lengths of
edges, and compare the mean number of fruits per
flower in each one.
e
There are many possible answers to this question,
and students are likely to put forward a range of
ideas. In this particular case, it does appear that
many small patches of forest are ‘better’ than a
few big ones, but this is unusual because these
particular butterflies happen to need forest edges.
There will be many more animals and plants that
need large areas of deep forest to survive, and
they will do better in large patches, preferably
connected. Some animals need large territories in
which to hunt. Some only need small areas, but
there needs to be a large population to be sure
they will not become extinct. Students may also
refer to the importance of forests in the carbon
cycle and in the production of oxygen.
Chapter 16: Answers to Workbook exercises
1
Answers to Workbook exercises
Chapter 17
Exercise 17.1 Gametes
a
b
d
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.
i
ii The energy required for active transport is
released from glucose molecules by respiration.
Aerobic respiration requires oxygen.
Exercise 17.2 Gas exchange in the
placenta and lungs
Exercise 17.3 Breast-feeding
statistics
a
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
b
i
c
From the air spaces inside the alveoli, to the
interior of the red blood cells in the capillaries.
ii 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 air 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
© Cambridge University Press 2014 IGCSE® Biology
S
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.
Look for:
◆ country on the x-axis and percentage on the
y-axis
◆ suitable scale on the y-axis, with even intervals
ranging from 0 to at least 72 (probably 75)
◆ bars drawn correctly and neatly
◆ bars not touching
◆ key to identify the bars for exclusive breastfeeding and exclusive bottle feeding.
The bar chart could look like this:
100
80
Percentage
S
Black or blue labels: cell membrane, cytoplasm,
nucleus.
away, maintaining a steeper diffusion gradient
down which oxygen will diffuse more rapidly.
60
40
20
0
ia
liv
Bo
il
bia
az
Br
m
olo
C
la
an
ma
nic
mi ublic uate
o
G
D ep
R
iti
Ha
ay
gu
ra
Pa
ru
Pe
Exclusive breast-feeding up to 4 months
Exclusive bottle feeding up to 4 months
Chapter 17: Answers to Workbook exercises
1
b
This means that the baby was being only breastfed – it was not bottle fed at all.
c
Some babies were partly breast-fed and partly
bottle fed.
d
Peru
e
Breast-feeding has many advantages, including the
following:
◆ it provides the right nutrients for a growing
baby
◆ it contains antibodies that help the baby to
fight infectious diseases
◆ it is unlikely to contain infectious
microorganisms, which may contaminate
bottle milk
◆ it is free
◆ it helps the mother and baby to form an
emotional bond.
Exercise 17.4 Birth control data
a
There are various ways in which these data could
usefully be displayed, for example a bar chart or
a pie chart. Some students may think of more
original methods. Give credit for anything that
is easy to understand and correct. Note that a
© Cambridge University Press 2014 IGCSE® Biology
line graph is not correct, because the different
methods are discrete, and not a continuous
variable.
b
Of women who used a diaphragm, 40% became
pregnant, compared with only 14% of women
whose partner used a condom. Perhaps the
diaphragm was not correctly fitted, so that sperm
could get around the edge of it and swim to the
oviducts. Perhaps the women did not remember
to put the diaphragm into place before having sex.
Perhaps they took the diaphragm out too early,
while there were still sperm in the vagina.
c
The pill prevents eggs being released from the
ovaries. If taken regularly, this works very well, so
there are no eggs present to be fertilised.
d
Of women who used the diaphragm, 42%
continued for more than one year, compared with
86% of women who used the pill. This could be
because women did not like using the diaphragm,
which requires them to remember to insert it
before having sex, whereas taking a pill is much
easier – you simply take it once a day. It could be
because they became pregnant when using the
diaphragm, so stopped using it. Students may
think of other reasons.
Chapter 17: Answers to Workbook exercises
2
Answers to Workbook exercises
Chapter 18
Exercise 18.1 Fruit fly inheritance
a
Body divided into head, thorax and abdomen; six
jointed legs; one pair of antennae; wings attached
to thorax.
b
Drosophila
c
d
b
NN
normal wings
Nn
normal wings
nn
vestigial wings
gametes from
black stallion
vestigial
wings
Nn
nn
N and n
all n
N
Nn
normal wings
n
nn
vestigial wings
About half would have each phenotype, so about
41 with normal wings and 41 with vestigial wings.
Exercise 18.2 Black and chestnut
horses
© Cambridge University Press 2014 IGCSE® Biology
all
e
E
Ee
black
e
ee
chestnut
So there was about a 1 in 2 chance that the foal
would be chestnut.
iii Exactly the same: 1 in 2. Each time they have a
foal, half of the stallion’s sperm will be carrying
an E allele and half carrying an e allele, so
the chance of a sperm carrying an e allele
fertilising the egg is 1 in 2.
n
i E
ii Ee
iii chestnut
e
e
gametes from vestigial
winged fly
a
E and
ee
gametes from chestnut mare
phenotypes of parents normal wings
gametes
Ee
gametes
Phenotype
genotypes of parents
e
Ee The foal must have had two e alleles, one
from each of its parents. The mare was ee. The
stallion was black, so he must have had one
black allele and one chestnut allele.
ii phenotypes of parents black
chestnut
genotypes of parents
Genotype
gametes
from normal
winged fly
i
Exercise 18.3 Pedigree
a
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.
Chapter 18: Answers to Workbook exercises
1
b
c
d
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.
Person 1 could be either QQ or Qq.
Person 2 could also be either QQ or Qq.
Person 3 must be Qq, as they don’t 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.
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 good
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 she marries someone
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.
© Cambridge University Press 2014 IGCSE® Biology
Exercise 18.4 Sex linkage in
fruit flies
a
For example, XR and Xr.
b
phenotypes
of parents
white-eyed
male
XrY
genotypes
of parents
red-eyed
female
XR Xr
Xr and Y
gametes
S
XR and Xr
gametes from red-eyed female
gametes
from
white-eyed
male
Xr
Y
XR
Xr
Xr XR
red-eyed
female
Xr Xr
white-eyed
female
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.
Chapter 18: Answers to Workbook exercises
2
Answers to Workbook exercises
Chapter 19
S
Exercise 19.1 Water hyacinth
experiment
Exercise 19.2 Big-horn sheep
a
hair
a
5 μm
b
b
guard cell measures 12 mm in length = 12 000 μm
i nucleus
ii All of them. (All the body cells have a complete
set of genes, but each type of cell only uses a
particular number of them.)
c
There has been selection against the sheep with
the largest horns. Sheep with smaller horns are
most likely to survive and reproduce. The alleles
for smaller horns are therefore passed on to the
next generations more often than the alleles for
larger horns. Over time, more and more of the
big-horn sheep population have small horns, and
the mean horn length therefore deceases.
d
i
magnification = length in diagram ÷ real 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
of course vary from that obtained here. Check that
the method of calculation is correct.
c
d
The leaves have many stomata on the
upper surface. This is not usually found in landliving 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 don’t need to conserve water
and having stomata on the upper surface allows
them to absorb carbon dioxide easily from
the air.
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 on the
upper surfaces of the leaves is the same in clean
and polluted water. The mean number of stomata
on the lower surfaces is a little higher in the plants
grown in clean water than in those grown in
polluted water.
© Cambridge University Press 2014 IGCSE® Biology
As the temperature rises, the sweat glands
secrete more sweat onto the surface of the skin.
The water in the sweat evaporates, taking heat
with it and cooling the skin surface.
ii Vasoconstriction is the narrowing of the
arterioles that supply blood to the skin
capillaries. This reduces the amount of blood
flowing close to the skin surface, and therefore
reduces the rate of heat loss from the blood
to the air (by radiation). Instead, the blood
is diverted to flow through deeper vessels,
separated from the air by adipose tissue, which
insulates the body and decreases heat loss.
Exercise 19.3 Selective breeding
for high milk yield
a
i
Value in 1990 = 11.0, value in 1965 = 7.2, so
the change is an increase of 3.8 kg per cow.
ii Value in 1990 = 5.8, value in 1965 = 7.2, so the
change is a decrease of 1.4 kg per cow.
Chapter 19: Answers to Workbook exercises
1
S
b
c
Only the cows that gave the highest milk yield
would have been allowed to breed. They would be
bred with bulls whose female family members also
gave high milk yields. This would be done over
several generations, each time only choosing the
animals giving the highest milk yield to breed.
We can only guess – there is no evidence to tell us
why the milk yield fell. In this group of cows, all
the cows were equally likely to breed, so perhaps
it is just chance that the mean milk yield fell over
time. However, perhaps there is a disadvantage in
© Cambridge University Press 2014 IGCSE® Biology
having a high milk yield – for example, perhaps
these cows were less healthy in other respects so
they are actually less likely to have offspring.
d
i
The selected line were the cows with high
milk yields. The large amounts of milk in their
udders may have increased the incidence of
inflammation, and the heavy weight of milk
they have to carry around may have increased
the degree of lameness.
ii They would need more food, to supply the
materials needed to produce the extra milk.
Chapter 19: Answers to Workbook exercises
2
Answers to Workbook exercises
Chapter 20
S
Exercise 20.1 Energy transfer in a
food chain
a
The position in a food chain at which an
organisms feeds.
b
In sequence: producer, primary consumer,
secondary consumer, tertiary consumer.
c
20 × 100 = 0.1%
20810
ii Much of it is lost as heat to the environment,
through respiration. Some goes to the
decomposer food chain.
d
f
a
b
i
ii
c
i
There is not enough energy available at
the higher trophic levels to support large
populations.
ii By the time you reach a fifth or sixth link,
there is not enough energy to support any
organisms at all.
Exercise 20.2 Fish tank
b
protein
c
The dead phytoplankton were decomposed by the
bacteria in the water. These 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.
d
The ammonia came from the breakdown of
nitrogen-containing substances, such as proteins,
in the cells of the dead phytoplankton.
e
Nitrate first appeared in December – that is, about
one month after the start of the experiment. The
quantity of nitrate increased sharply in April.
© Cambridge University Press 2014 IGCSE® Biology
i
ii
i
They had no light, so they could not
photosynthesise.
S
Exercise 20.3 Goats on an island
i
a
The nitrate was produced from ammonia, by
nitrifying bacteria.
The curve should be the classic S-shape,
beginning fairly flat, then rising steeply before
levelling off.
Limiting factors begin to cut in at the point
where the curve begins to flatten out.
A change in a gene or chromosome.
The long-hair allele, a, is recessive, so a goat
needs two copies (one from each parent)
in order to have long hair. Only goat P
can pass on an allele for long hair, not the
females, so none of its offspring could have
long hair.
iii Some of the offspring from goat P would
have inherited one copy of the a allele. If
these bred with each other (or with goat P),
then there would be a 1 in 4 chance of each
offspring having the genotype aa and having
long hair.
ii
The long-haired goats did not lose as much
heat from their well-insulated bodies, and
so needed to generate less heat through
respiration. They therefore needed less
glucose (or other nutrients) to use as fuel in
respiration.
The goats with long hair would have been
at a selective advantage – they would be
more likely to survive and breed than the
shorthaired goats. In each generation, there
would therefore be more chance of the alleles
for long hair being passed on than the alleles
for short hair.
Chapter 20: Answers to Workbook exercises
1
Answers to Workbook exercises
Chapter 21
Exercise 21.1 Pectinase
a
e
Pectinase breaks down the pectin that holds
together the cells in the fruit. This makes it
faster and easier to extract the juice, and also
increases the total yield of juice from a given
mass of fruit. Pectinase also helps to make
cloudy fruit juices clear.
Any three from: they are very small and therefore
easy to grow in large numbers; there are no ethical
issues associated with their use; they have the
same type of genetic material as other organisms
(including humans); they have plasmids, which
can be used to move genes from one organism to
another; accept other valid reasons.
Exercise 21.2 Yoghurt
b
Fungi
a
c
the antibiotic, penicillin
d
i
Yoghurt is made from milk. Bacteria convert the
lactose in the milk to lactic acid, which is an acid
and therefore has a pH below 7.
b
Perhaps the bacteria were continuing to change
more lactose to lactic acid.
c
The changes in pH are all very small. The greatest
change is in the yoghurt that did not contain
stabilisers, where the pH fell by 0.5 (from 4.1 to
3.6). The changes in the other three experiments
were even smaller.
d
In the past, yoghurt was consumed soon after
it was made, so it did not have time to separate.
Nowadays, it is transported long distances so
it is not consumed as soon after manufacture.
The table shows that untreated yoghurt has
separated to a value of 7.9 by only five days after
manufacture, and to a value of 21.1 after 15 days.
Adding stabilisers slows down separation, so the
yoghurt is probably more pleasant to eat after
being transported.
e
Cornstarch worked best. The yoghurt to which
cornstarch had been added only separated to
a level of 6.9 after 15 days, which is lower than
untreated yoghurt and the yoghurts to which
pectin or carrageenan were added.
f
The volume of stabiliser added to a fixed volume
of yoghurt; the temperature at which the yoghurt
was kept; the batch of yoghurt that was used (so
the milk had exactly the same composition).
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 + 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.
ii If the waste materials are not used, then they
have to be disposed of. 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, then
other plant material would have to be used to
make the pectinase, which means more land
would be used for growing plants that might
otherwise be used for growing food, or for
habitat for wildlife.
© Cambridge University Press 2014 IGCSE® Biology
Chapter 21: Answers to Workbook exercises
1
S
Exercise 21.3 Golden Rice
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 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 daffodils and Erwinia into Agrobacterium.
© Cambridge University Press 2014 IGCSE® Biology
e
Agrobacterium naturally infects plant cells, so
it was able to carry the plasmids carrying the
genes from the daffodils and Erwinia into the
rice cells.
f
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 beta
carotene, so there was no real starting point
for selective breeding. Instead, genes that were
already present in other species had to
be used.
Chapter 21: Answers to Workbook exercises
S
2
Answers to Workbook exercises
Chapter 22
S
Exercise 22.1 Acid rain and wildlife
in Canadian lakes
a
b
c
d
The low pH could be caused by acid rain. This
could be produced when fossil fuels are burnt,
releasing sulfur dioxide into the atmosphere. This
dissolves in water droplets in clouds and reacts
to form sulfuric acid, which falls to the ground as
rain with a lower pH than normal. (Students may
also suggest that it could be caused by the kind
of rocks on which the lake is lying and the type
of soil around it. Peaty soils, for example, tend to
contain acids, which are carried into the lake as
water drains through them.)
The fish-eating birds are most common in habitat
with lakes with a pH of around 6.5. They are not
common in habitat where lakes with low pH,
and only very small amounts (less than 10%)
of the available habitat contain lakes with a pH
of less than 5. This is because fish are unable to
live in such acidic water. There are often large
concentrations of aluminium ions in acidified
lakes, washed out of the nearby soils by the acid
rain. Aluminium ions affect the functioning of the
fish gills, and may kill young fish.
The divers are more common in lakes with a low
pH than in lakes with a higher pH. This could
be because there is more food for them in these
lakes, perhaps because the plants and invertebrates
that they eat do well in these waters. Another
possibility is that they might compete for nesting
sites with fish-eating birds; these are less common
in habitat with lakes with low pH so there would
be more nest sites available for the divers. Students
might also come up with other answers.
The answer will depend on the student’s
suggestion in c above. Look for an outline of
© Cambridge University Press 2014 IGCSE® Biology
an experiment that makes clear what variable
is changing and what is being measured,
perhaps also with some mention of the control
of other variables.
S
Exercise 22.2 Eutrophication
a
Plants need nitrogen-containing ions (such
as ammonium or nitrate) to make proteins.
They need proteins to build new cells, and
therefore for growth. The soil in the field may
be deficient in nitrogen-containing ions, which
would limit the growth of the crop. The farmer
therefore gets higher yields by adding these ions
to the soil.
b
i
Through their root hairs, by active transport.
Energy, provided by respiration in the root
hair cells, would be used to move the ions into
the cell against the concentration gradient.
ii xylem
c
i
The population of algae rises rapidly just
downstream of where the fertiliser flowed into
the river. This is because algae can use the
nutrients (nitrates and ammonium ions) in
the fertiliser for growth. Further downstream,
there are fewer nutrients because they have
been used by algae upstream. The population
size therefore decreases with the distance
downstream.
ii Many of the algae die. The population of
bacteria rises because they can feed on the
increased quantity of dead algae. These
bacteria use up the dissolved oxygen in
the water in respiration. This decreases the
quantity of oxygen. Fish need oxygen to
respire, so in the area where oxygen levels are
low they either die or move away.
Chapter 22: Answers to Workbook exercises
1
Exercise 22.3 Fertiliser experiment
Look for the following features in the plan.
◆ A clear statement that amount of fertiliser (the
independent variable) will be changed, with at
least five different values being used, including
a control with no fertiliser.
◆ A clear statement that crop yield (the
dependent variable) will be measured; some
idea of exactly how this will be done (e.g.
measure the mass of grain or fruit produced per
plant, or per unit area of the field) and when it
will be done.
◆ A clear statement about all the variables that
will be controlled, and how this will be done
(e.g. do the experiment in a laboratory and
keep temperature, water availability, light, etc.
the same for all the plants, with an outline
of how these will be kept constant; or do the
experiment in the field and grow all in the same
type of soil with the same amount of shade or
shelter).
◆ A clear statement about how many repeats
will be done, and how mean results will be
calculated.
◆ An outline result chart, with headings, showing
how results will be recorded.
b
Flight allows birds to escape predators. Predators
are therefore a selection pressure which gives
birds that can fly a greater chance of survival
and reproduction. However, where there are no
predators, there are fewer advantages in being
able to fly, and there may be disadvantages. For
example, birds that can fly use more energy than
birds that cannot, and need to grow strong flight
muscles, and may need more food. Birds that do
not have wings, do not have these energy costs
and may be better able to survive and reproduce,
passing on their alleles for winglessness to their
offspring. Over time, this process of natural
selection will result in the whole population
having no wings.
c
In all three areas, the number of seedlings in the
areas where rats were trapped were greater than
in the areas where they were not trapped. (Credit
reference to comparative numbers of seedlings
in any trapped and untrapped area.) Removing
rats would therefore be expected to increase the
population, as more seedlings would survive and
grow into adult palms.
S
Exercise 22.4 Introduced species in
New Zealand
a
They have wings, so could fly across the oceans
separating New Zealand from the nearest land.
© Cambridge University Press 2014 IGCSE® Biology
Chapter 22: Answers to Workbook exercises
2