MODEL ANSWERS IN
SCIENCE
FOR
CSEC
BIOLOGY
CHEMISTRY
Navindra Ramsaroop
Bella Seejoor
Tameez Newaj
Genesis Books
PHYSICS
Copyright © Navindra Ramsaroop et al
All rights reserved. No part of this publication may be reproduced or transmitted in any
form by any means, electronic or mechanical, including photocopying, recording or
otherwise without the written permission of the publisher.
Published by
Genesis Book Publisher
37 Gordon Street
San Fernando Trinidad West Indies
868 653 5234
868 738 0536
genesisbooks01@gmail.com
© MODEL ANSWERS IN SCIENCE FOR CSEC BIOLOGY CHEMISTRY PHYSICS
JANUARY – JUNE 2013 - 2017 AND JANUARY 2018
ISBN 978 976 96081 1 5
First Published
April 2018
Printed in Trinidad and Tobago by
The Office Authority Limited
Typing and Layout
Shoba Alicia Pooran
Kamita Sooknanan (Physics)
Genesis Book Publisher
INTRODUCTION
‘Model answer in science for CSEC Biology, Chemistry, Physics’ is geared specifically for
students of Form 4 and Form 5 levels. Perhaps the first of its kind in the region, it is expected to
be both convenient and economical for the users of this book. Solutions are generally presented in
a simple but systematic way that aligns with CSEC standard and coverage of syllabus. It is also
acknowledged that there would be differences of opinions amongst teachers with respect to
solutions or routes leading to numerical answers. This is the nature of science.
We strongly suggest that students attempt the CSEC Past Paper questions at the end of the topic
rather than at the end of the syllabus. This allows the students to establish the CSEC standard at
an early stage.
The authors of this model answer book have vast experience in teaching the science subjects both
at the CSEC and CAPE levels. At the same time they are also involved in the practical and
industrial applications of the scientific principles in the various subject areas. It is hoped that this
book would be highly beneficial to its users and leads to excellent exam results in the sciences.
ABOUT THE AUTHORS
1) Navindra Ramsaroop (Chemistry)

BSc. Chemistry/Analytical Chemistry

18 years teaching experience

Author of two Environmental Science Books
2) Bella Seejoor (Biology)

BSc. Chemistry/Biochemistry

Dip Ed.

18 years teaching experience.
3) Tameez Newaj (Physics)

BSc. Physics, Chemistry, Food Tech.

Dip. Ed.

Cert. Tech. Cryogenic Refrigeration, Revco. (USA)

Cert. Tech. Wet Chemistry Instrumentation, Metrohm. (Switzerland)

Cert. Tech Metallographic Instrumentation, Buehler. (USA)

Cert. Tech Spectroscopy Instrumentation, Perkin Elmer (Puerto Rico)

Cert. Tech Engineering Teaching Equipment, TQ (UK)

Cert. Tech FDM 3D-Printing, Stratasys. (USA)

32 years teaching experience

25 years Instrumentation Specialist, Scalar Scientific (Part time)
TABLE OF CONTENTS
JANUARY 2010 PAPER 2……………………………………………………………………2
JUNE 2010 PAPER 2…………………………………………………………………………8
JANUARY 2011 PAPER 2………….……………………………………………………….17
2…………………………………………………………………………8
JUNE 2011 PAPER 2…………………………………………………………………….….24
2…………………………………………………………………………8
JANUARY 2012 PAPER 2 ………………………………………………………………….32
2…………………………………………………………………………8
JUNE 2012 PAPER 2…………………………………………………………………….….41
2…………………………………………………………………………8
JANUARY 2013 PAPER 2 ………………………………………………………………….49
2…………………………………………………………………………8
JUNE 2013 PAPER 2…………………………………………………………………….….58
2…………………………………………………………………………8
JANUARY 2014 PAPER 2 ………………………………………………………………….66
2…………………………………………………………………………8
JUNE 2014 PAPER 2…………………………………………………………………….….73
2…………………………………………………………………………8
JANUARY 2015 PAPER 2 ………………………………………………………………….81
2…………………………………………………………………………8
JUNE 2015 PAPER 2…………………………………………………………………….….90
2…………………………………………………………………………8
JANUARY 2016 PAPER 2 ………………………………………………………………….97
2…………………………………………………………………………8
JUNE 2016 PAPER 2…………………………………………………………….………..107
2…………………………………………………………………………8
JANUARY 2017 PAPER 2 ………………………………………………………………..114
2…………………………………………………………………………8
JUNE 2017 PAPER 2…………………………………………………………….………..123
2…………………………………………………………………………8
JANUARY 2018 PAPER 2…………………………………………….…………………..131
2…………………………………………………………………………8
JANUARY 2013 PAPER 2…………………………………………………………………142
JUNE 2013 PAPER 2………………………………………………………………………148
JANUARY 2014 PAPER 2…………………………………………………………………156
JUNE 2014 PAPER 2………………………………………………………………………163
JANUARY 2015 PAPER 2…………………………………………………………………170
JUNE 2015 PAPER 2………………………………………………………………………177
JANUARY 2016 PAPER 2…………………………………………………………………184
JUNE 2016 PAPER 2………………………………………………………………………190
JANUARY 2017 PAPER 2…………………………………………………………………197
JUNE 2017 PAPER 2………………………………………………………………………203
JANUARY 2018 PAPER 2…………………………………………………………………210
JANUARY 2010 PAPER 2…………………………………………………………………217
JUNE 2010 PAPER 2………………………………………………………………………223
JANUARY 2011 PAPER 2…………………………………………………………………229
JUNE 2011 PAPER 2………………………………………………………………………235
JANUARY 2012 PAPER 2…………………………………………………………………242
JUNE 2012 PAPER 2………………………………………………………………………247
JANUARY 2013 PAPER 2…………………………………………………………………253
JUNE 2013 PAPER 2………………………………………………………………………259
JANUARY 2014 PAPER 2…………………………………………………………………265
JUNE 2014 PAPER 2………………………………………………………………………272
JANUARY 2015 PAPER 2…………………………………………………………………278
JUNE 2015 PAPER 2………………………………………………………………………284
JANUARY 2016 PAPER 2…………………………………………………………………291
JUNE 2016 PAPER 2………………………………………………………………………297
JANUARY 2017 PAPER 2…………………………………………………………………302
JUNE 2017 PAPER 2………………………………………………………………………308
JANUARY 2018 PAPER 2…………………………………………………………………315
BIOLOGY
1
1. (a)
ARRANGEMENT OF APPARATUS TO INVESTIGATE THE RESPONSE OF
LARVAE TO LIGHT
(b)
(i)
The table should include:
Total
Percentage
Dark side Light side
40
20
13
7
(ii)
Yes, the results support the hypothesis.
More larvae were found in the dark side.
(iii)
Two factors are:
 Moisture
 Temperature
(iv)
Factor – moisture
Place a plastic wall under the wire gauze such that the area under the
gauze is separated into halves, A and B. Place a drying agent such as silica
gel under the gauze in Half A. Place an equal volume of water in Half B.
Cover the entire petri dish cover with the black paper.
Alternatively, vary the temperature by using warm and cold water in the
two halves.
(v)
It helps them to avoid detection by predators. It reduces their risk of
desiccation, since an area illuminated by sunlight is usually warm,
encouraging evaporation of the water in their bodies.
2
(c)
(d)
2. (a)
(b)
(c)
(i)
It allows maximum absorption of light by the leaves for photosynthesis,
since the mesophyll layer is always at an appropriate angle for light
absorption. The rate of photosynthesis is high and glucose is produced at a
high rate.
(ii)
Test:
1) Place the plant in a dark area until its leaves show a negative test
for starch.
2) Expose the plant to light for a few, e.g. six, hours.
3) Remove a leaf. Boil it in water until it is soft, boil it in ethanol
until it is decolourised, dip it in hot water until it is soft, and cover
its surface with iodine solution. If the iodine changes from brown
to blue-black, the leaf contains starch and was photosynthesising.
(iii)
Water and carbon dioxide.
Chlorophyll and a suitable temperature are acceptable.
(i)
It ensures uniform/even distribution of auxins in the root by ensuring that
all sides of the root are equally subjected to the effects of gravity.
(ii)
Roots respond negatively and grow downwards. They obtain water and
ions and also a means of anchorage from the soil. Shoots respond
negatively, growing upwards. They are exposed to sunlight and carbon
dioxide for photosynthesis.
(i)
P: wall of capillary
Q: erythrocyte
R: lining of moisture on alveolar wall
S: alveolar wall
(ii)
A: carbon dioxide
B: oxygen
(i)
Diffusion
(ii)
It is thin – the flat cells arranged in only one layer, allowing quick
diffusion
Breathing moves the gases in and out of the alveoli. Inhaled air contains a high
concentration of oxygen and a low concentration of carbon dioxide. This ensures
that there is a steep concentration gradient of oxygen from the alveolus to the
blood and a steep concentration gradient of carbon dioxide from the blood to the
alveolus.
3
(d)
An exercising person needs more energy and therefore has a higher rate of
respiration. This causes the production of more carbon dioxide and the need to
inhale more oxygen. Faster and deeper breathing will expel more carbon dioxide
and take in more oxygen.
(e)
Carcinogens in smoke stimulate development of lung cancer. Tar and soot line the
alveoli and stimulate excess mucus production, which accumulates since tar
paralyses cilia. Cancer growths and mucus reduce the space available for air
within the lungs. Tar forms a physical barrier between the alveolar air and the
alveolar wall, reducing the efficiency of diffusion of gases and therefore, of
gaseous exchange. Tar and mucus stimulate coughing, which may rupture the
alveolar walls, reducing the surface area available for gaseous exchange.
3. (a)
(b)
(c)
4. (a)
(i)
I: scar of point of attachment to receptacle
II: embryo
III: testa
IV: food store (cotyledon /endosperm for endospermic seed)
(ii)
It provides nutrients for development of the embryo.
(i)
Pea: self/explosive mechanism
Coconut: water
(ii)
There are large air spaces within the mesocarp and seed, allowing the fruit
to float on water. It has a large food store, allowing survival of the
embryo during long travels on water.
Alternative correct responses include the waterproof testa.
(i)
Self/explosive mechanism
(ii)
When the fruit splits along lines of dehiscence, the force generated is
insufficient to send the seeds far away. They fall close to the parent plant,
where they germinate and grow.
In the light-dependent stage, chlorophyll in chloroplasts absorbs light. Its energy
is used for photolysis of water – it is split into hydrogen and oxygen. In the lightindependent stage, carbon dioxide from the atmosphere diffuses through stomata,
intercellular spaces and into the mesophyll cells. It combines with hydrogen
produced by photolysis of water to form glucose. Some of it is used to form
starch. The oxygen from photolysis of water is released as a waste product. The
process is summarised as:
Light absorbed by chlorophyll
6CO2  6H2O 
 C6 H12O6  6O2
4
(b)
Mineral ions are absorbed from the soil by roots. Glucose is produced in
photosynthesis in the leaves. Some of it is used in the synthesis of sucrose, amino
acids (using the mineral ions), fatty acids and glycerol. These are loaded into the
phloem and translocated to the non-green parts of the plants (sinks). On arrival at
the sinks, the food is unloaded from the phloem.
(c)
Three factors are:
 The availability of materials and conditions needed for photosynthesis,
e.g. water, sunlight, a suitable temperature. (If materials are unavailable,
little photosynthesis is done and so little food is available for storage.)
 The stage of the life cycle the plant is in. (During the growing stage, plants
store little food since the rate of respiration is high and food is used
rapidly. After growth, when the plant reproduces, food is stored in fruits
and seeds.)
 The presence or absence of storage organs. (Plants with storage organs
tend to store more food than those without.)
(d)
-Pollution is reduced. Human activity (vehicle emissions, combustion in factories)
produces large volumes of carbon dioxide. Plants in green spaces would absorb
carbon dioxide for photosynthesis, reducing the carbon dioxide levels in the air.
This reduces the rate of global warming.
-Oxygen levels are increased. Oxygen is a waste product of photosynthesis and is
released in the air. Humans and other animals need plants to produce this oxygen
for respiration. A large number of persons need a large supply of oxygen.
-Animals’ habitats are preserved. Green spaces provide animals with a space for
living and provide them with their other requirements, e.g., food and water.
Preservation of habitats reduces the chance of loss of biodiversity.
Alternative correct responses include reduction in noise, temperature control and
ecotourism, etc.
5. (a)
DIAGRAM OF HUMAN DIGESTIVE SYSTEM
5
(b)
(i)
Alternative response – the function of the small intestine (secretion of
lipase, sucrase, etc.) could be described in place of one of the annotated
organs
(ii)
- Consume protein from a variety of plants, including legumes, to increase
their chance of eating all the essential amino acids.
- Consume lipids from a variety of plants to increase their chance of eating
all the essential fatty acids.
- Use food supplements such as vitamins, minerals, amino acids, etc. to
compensate for deficiency in diet.
Control of the blood glucose level would be affected.
The α cells of the pancreas are stimulated by low blood glucose levels (below 80
mg mL-1) to produce glucagon. This causes conversion of glycogen to glycogen
and stimulates cells to absorb glucose, lowering the blood glucose level until it is
at the norm. Removal of the pancreas would prevent this and hyperglycaemia and
later, diabetes, would result.
The β cells of the pancreas are stimulated by high blood glucose levels (above 120
mg mL-1) to produce insulin. This causes conversion of glucose to glucose,
raising the blood glucose level until it is at the norm. Removal of the pancreas
would prevent this and hypoglycaemia would result.
(c)
Plants absorb mineral and other ions from the soil. These are used in the
synthesis of compounds essential for survival, growth and development of the
plant, e.g.:


Nitrates and sulphates (from soil) are used in amino acid synthesis (needed
for growth, especially of leaves)
Calcium (from soil) is used in cell wall formation
Alternative examples include:
 Magnesium (from soil) is used in synthesis of chlorophyll
 Phosphorous (from soil) is needed for root development and ATP
synthesis
 Potassium (from soil) is needed for fruit development
If these ions are lacking from the soil, the plant is unable to synthesise the
substances it requires and show signs of deficiency.
6
6. (a)
DIAGRAMS OF UNSPECIALISED PLANT AND ANIMAL CELLS
(b)
Nerve cells:
 Have elongated axons, minimising the number of cells needed to transmit
an impulse over a long distance, thus minimising the number of delays
associated with synapses.
 Are insulated by myelin sheaths at intervals, reducing the time taken for
an impulse to reach the end of the nerve cell
 End in dendrites, allowing them to transmit impulses to and from
numerous other nerve cells at the same time
Alternative responses could refer to the presence of synapses and
neurotransmitters, etc.
(c)
(i)
Abiotic factors:
 Large volume of water/aqueous habitat
 Low availability of oxygen
Alternative responses may include nutrient content, light intensity,
temperature, etc.
(ii)
Large volume of water: Non-swimming animals living in a freshwater
pond must overcome the problem of sinking if they are suspended in still
water. Some animals such as diving beetles have fixed air bubbles on the
outside of their bodies and this helps them to stay buoyant. Fishes have a
swim bladder for this purpose, inside them.
Low availability of oxygen: Burrowing invertebrates such as crabs and
crayfish often have modified gills, such that parts of them resemble,
physiologically, the mammalian lung. When oxygen levels are low, they
come to the surface and extract oxygen from the air (aerial respiration).
They may also be able to reduce their metabolic rate to reduce their rate of
use of oxygen.
There are alternative correct responses, e.g. continuous urine production
by freshwater fish, etc.
7
1. (a)
(i)
TABLE SHOWING OBSERVATIONS MADE WHEN LEAVES OF A
PLANT WERE TESTED FOR STARCH, REDUCING SUGAR,
PROTEIN AND OIL
Light intensity 1
Plant Plant Plant
A
B
C
Light intensity 2
Plant Plant Plant
A
B
C
Light intensity 3
Plant Plant Plant
A
B
C
Starch
Reducing
sugar
Protein
Oil
(ii)
It allows one to find the average of the results. The average result reduces
the effect of any error which may have occurred during any of the trials of
the experiment. This improves accuracy and so, reliability, of results.
(b)
DIAGRAM OF APPARATUS USED TO DEMONSTRATE WATER UPTAKE
BY PART Z FOR USE BY PART Y
8
(c)
Liver
There are other/alternative correct answers such as skin, etc.
(d)
Since respiration is the same in both organisms, the requirements/reactants
(oxygen and carbohydrate) and the products (carbon dioxide, water and heat) are
the same. Perform a test on the plant and human to demonstrate that the
requirements or products are the same for respiration in both organisms, e.g.
carbon dioxide produced by both turns lime water milky.
(e)
(i)
LONGITUDINAL SECTION OF THE FLOWER OF THE PLANT
SPECIES
(ii)
The anthers are within the flower and are not within easy reach of the
wind. Insects have access to them when walking on the large petals, which
enclose both stigma and anthers.
Any other visible feature which suggests that it is insect-pollinated or it is
NOT wind-pollinated, is acceptable such as stigma not feathery or an
attractive corolla, etc.
9
(f)
Growth of offspring is asexual reproduction. This allows rapid production of large
numbers of offspring, which quickly colonise a habitat. Growth of flowers allows
sexual reproduction. This causes variation among the offspring, increasing the
chance of survival of at least some of the offspring, if the environment of the plant
changes.
2. (a)
GERMINATING SEEDLING AT DAY 5
Other labels, such as testa, root hair, hypocotyl or epicotyl are also correct.
(b)
1)
2)
3)
4)
(c)
(d)
The seed absorbs water through the micropyle.
Starch, protein and lipid stores are hydrolysed by activated enzymes.
The rate of respiration increases (using glucose).
The hydrolysed food (glucose, amino acids, fatty acids, glycerol) is used
to synthesise plant material (cell walls, etc.) resulting in growth of
plumule, radicle and root hairs.
(i)
Growth movement or tropism
(ii)
Unlike the earthworm, the seedling:
 Grows in different directions at the same time
 Grows only at meristems
Other responses are correct, e.g. do not involve locomotion, is
irreversible, involves growth, is slower, etc.
(i)
High light intensity
High temperature
There are other/alternative correct responses, e.g. low moisture, etc.
(ii)
It exposes the earthworm to higher moisture levels, reducing its chances of
desiccation. It renders the earthworm less visible/invisible to its predators,
decreasing its chances of being eaten.
10
3. (a)
Plant materials, leaves, fruits, seeds  grasshopper  flycatcher bird  hawk
Or
Plant materials, leaves, fruits, seeds  caterpillar  flycatcher bird  hawk
(b)
(c)
(i)
Any organism after the second arrow of any of the food chains in the web
is a predator (that is, the third, fourth, and higher organisms in the food
chain are predators). The organism from whom the arrow points is its
prey, e.g.
Predator is ocelot and prey is squirrel.
(ii)
The predator population is smaller in numbers.
The predator population has adaptations for hunting, e.g. well-developed
senses of smell and sight.
There are other/alternative correct responses, e.g. prey population has
adaptations to escape predators, etc.
(i)
Relationship – parasitism
A tapeworm (parasite) lives in the intestine of a pig and obtains benefits
such as nutrients and a habitat which provides protection from extremes of
environmental conditions. The pig (host) suffers harm such as the
deprivation of its nutrients.
Examples of mutualism or commensalism are alternative correct
responses.
(ii)
Predator-prey, herbivore-producer and parasite-host relationships serve as
population control measures of the prey, hosts and producers. The
predators and parasites are selection pressures, causing the surviving prey
and host populations to be well adapted to their environments.
There are alternative correct responses, e.g. it allows use of a range of
resources, it increases the chances of survival of certain species, etc.
11
4. (a)
(i)
THE MAIN STAGES IN THE LIFE HISTORY OF A MOSQUITO
Insect vector – mosquito
(ii)
Control measures:
Stage
Egg
Larva
Pupa
Adult
Control measure
 Removal of likely breeding places of mosquito, e.g.
stagnant water in ponds, vases, discarded tyres and other
garbage which may collect water, dense vegetation (close
to homes)which may collect water
 Use of chemicals to kill eggs
 Spraying of surface of water with chemical toxic to larva,
e.g. insecticides
 Spraying of surface of water with oil or other chemical
which prevents breathing of larva
 Introduction of organisms which feed on larva (biological
control)
 Same measures as were outlined for larva
 Spraying with chemical toxic to adults, e.g. insecticides
 Introduction of organisms which feed on adults
(biological control)
12
(b)
5. (a)
(i)
They are difficult to cure because the viruses are often difficult to detect
because:
 they incorporate themselves within the DNA of their hosts
 remain latent for a long time (long incubation period)
 often adversely affect the function of the cells of the immune
system, which are the cells responsible for their detection
 have a high mutation rate, making recognition of them difficult
(immune cells which were once complementary to them may no
longer be complementary)
(ii)
Control of HIV/AIDS is difficult because the following factors cause a
high rate of infection of persons:
 Numerous persons are uneducated about the method of
transmission of the virus. They therefore are more likely to engage
in risky sexual behaviour, e.g. multiple partners and sexual
intercourse without use of a physical barrier such as a condom to
prevent transfer of body fluids. Lack of knowledge may leave
persons unaware of the need for testing their blood for the presence
of the virus.
People are also unaware of the length of the incubation period of
the virus. They may assume themselves to be virus-free if years
have passed since their last risky sexual behaviour and they are yet
symptom-free. They may unwittingly transfer the virus to other
people.
 Infected persons may not be viewed favourably by their relatives
or society. This causes persons to fear testing their blood for the
presence of HIV. If they get tested and the result is positive, they
may keep it a secret. If they are infected and engage in sexual
intercourse, they may transfer the virus to their partners.
 Some persons have a feeling of confidence about their ability to
resist infection by the virus, even if they are educated about it.
They may cause them to engage in risky sexual behaviour.
There are alternative correct responses, such as the infection of
babies and foetuses during childbirth and pregnancy, etc.
(i)
The differences are due to environmental effects. The phenotype of an
organism is the expression of the genotype, modified by environmental
effects. These include varying availability of sunlight, water and ions. The
differences among the plants demonstrate the effects of varying
environmental factors on a given genotype.
13
(b)
(ii)
The genetic make-up under consideration is one which gives the plants
desirable traits. These may include higher yield and greater resistance to
disease or extreme environmental conditions such as drought or
waterlogging. Plants with identical genetic constitution may be uniform in
size, taste, texture, nutrient content and appearance. This uniformity is
desirable when selling produce on a large scale.
(iii)
If the environment changes, all members of the population will be equally
affected and so the population will not adapt to the environmental changes
by natural selection. This may cause elimination of the entire population.
In artificial selection, the traits which are deemed as desirable are determined by
humans. These traits are useful to humans and not necessarily of benefit to the
organisms undergoing artificial selection, e.g. dogs may be bred for extremely
large muscles in the jaw area but this may cause them to have too little room for
development of the trachea. The deliberate selection of organisms to be mated
causes the desired changes to be effected much faster in the population than if
random mating is done.
In natural selection, the traits which are deemed as desirable are those which
cause the organisms to be well adapted to their environment, e.g. high speed of
predators. Such traits are not necessarily of benefit to humans but they ensure that
the population undergoing natural selection remains adapted to the environment.
Random mating of organisms causes the changes in the population to occur much
slower than if the organisms with the desired traits were forced to breed with each
other, as happens in artificial selection.
(c)
(i)
Genetic engineering involves transfer of genes to members of the same or
different species, since this transfer does not involve mating or pollination
(as is done in artificial selection). Since gametes are not involved, DNA of
somatic cells may be used in genetic engineering. In artificial selection,
however, only DNA of gametes is transmitted to offspring, which are of
the same species as the parent.
(ii)
Two concerns:
 There may be unpredictable outcomes. Cross-pollination or mating
of a genetically modified organism with wild relatives may cause
the transgene to “escape” to populations other than its target
population. This may cause development of “superweeds” and
“superbugs” and this could upset the balance of ecosystems, since
these have a competitive advantage and will dominate their habitat.
The physiology of genetically modified organisms may also be
adversely affected in unforeseen ways.
 Some applications of genetic engineering may not be considered
ethical. Whether the transfer of genes across species, the ability to
determine the traits of an unborn organism and the development of
14
sterile organisms, are acceptable, are questions for which society
has not yet formulated answers. There is concern about whether
humans should have so much control over DNA, since it could
lead to abuse of persons who do not possess the technology, money
or education associated with genetic engineering.
6. (a)
(i)
DIAGRAM OF THE HUMAN HIND LIMB
Antagonistic pairs of muscles (flexor and extensor) are attached by
inelastic tendons to two bones (one of which is larger) on opposite sides of
them.
When the flexor muscle (e.g. the hamstring) contracts, the tendons pull on
the bones and the smaller bone moves towards the larger one such that the
lower leg is flexed (see diagram for location of hamstring). At the same
time, the extensor muscle (e.g. the quadriceps) is relaxed, allowing flexion
of the leg.
Conversely, when the extensor muscle (e.g. the quadriceps) contracts, the
tendons pull on the bones and the smaller bone moves towards the larger
bone such that the leg is extended (see diagram for location of
quadriceps). At the same time, the flexor muscle is relaxed, allowing
extension of the leg.
Other pairs of antagonistic muscles are attached to the pelvic bone and the
femur (allowing flexion and extension of the thigh) and to the pelvic bone,
femur and ankle bones (allowing lifting and lowering of the heel).
15
When the thigh, lower leg and ankle of one limb move, the hind leg can
move and locomotion is possible.
(b)
(ii)
Moment allows humans to:
 Find food and other requirements essential for survival
 Find a mate for reproduction
 Avoid unfavourable environments which may threaten their
survival
Other correct responses include: allows them to work and
socialise, etc.
(i)
Seed dispersal may be done by:
 Wind
 Water
Other correct responses are animals and self/explosive
mechanism, etc.
(ii)
The seed coat/testa or the cotyledon(s) may be adapted:
 The testa may be extended into a wing-like structure or fine hairs,
enabling the seed to move with the wind using a parachute
mechanism. It may also be waterproof, preventing water from
entering the seed. Water entry would have made the seed too heavy
to move with the wind, apart from causing it to decay.
 The cotyledon(s) may be dry, small and light, allowing the seed to
move with the wind.
16
1. (a)
(b)
(i)
Fruits and seeds
Alternative correct responses include roots, leaves, stems, etc.
(ii)
Carbohydrates, e.g. starch, and protein
Other responses such as fats, etc. are correct.
(i)
Food test
Crush the storage organ in water to form a
suspension.
Add an equal volume of Benedict’s solution and
mix.
Warm the mixture and observe colour change.
Crush the storage organ in water to form a
suspension.
Add an equal volume of NaOH and a few drops
of CuSO4 (e.g., 4 drops to 5mL of mixture) and
mix.
Observe colour change.
Crush the storage organ in water to form a
suspension.
Place a few drops of the suspension on white tile.
Add an equal number of drops of I2/KI (aq) and
observe colour change
(ii)
Conclusion
Reducing sugar
present
is
Protein is present
Starch is present
It contains carbohydrate (reducing sugar and starch), which provides a
significant amount of glucose (when digested). Glucose is the main
substrate for respiration, which provides energy. A teenage boy needs a lot
of energy for growth and development of secondary sexual characteristics.
It contains protein. Protein is needed for tissue repair and synthesis of new
cells and muscle, all of which occur at high rates in a growing boy.
(iii)
The storage organ contain does not contain all the essential amino acids.
It also lacks vitamins, minerals and lipids. The boy needs all of these and
so will show signs of deficiency. He may also eat too much of the storage
organ and develop obesity, which is also a sign of malnourishment.
17
(c)
Emulsion test:
Crush the storage organ in a small volume of water to form a suspension. Add an
equal volume of ethanol and mix the contents of the test tube. Add an equal
volume of water to the tube and mix the contents.
If fats are present, a solution (transparent) will be formed on addition of the
ethanol. A cloudy emulsion appears on addition of the water, which later settles as
two layers.
Grease spot test:
Rub a sample of the storage organ on grease-proof paper.
If fats are present, a translucent area appears on the paper.
(d)
2. (a)
Liver and Adipose tissue
Alternative correct responses include muscle and the subcutaneous layer beneath
the dermis of the skin, etc.
(i)
COMPONENTS OF PLANT AND ANIMAL CELLS
Component
Plant Cell Animal Cell
Cell membrane
Cell walls
present
absent
Nuclei
Mitochondria
present
present
Chloroplasts
present
(b)
(ii)
Mitochondrion: it is the site of aerobic respiration, providing ATP for
cells.
Chloroplast: it is the site of photosynthesis, converting light energy into
chemical energy.
(i)
Osmosis
(ii)
It allows water to enter cells, where it:
 Dissolves enzymes and reactants, acting as a medium for metabolic
reactions
 Absorbs heat, thus cooling the plant
 Causes the protoplasm to expand, causing cell turgidity
(iii)
Turgid cells prevent leaves and stems from drooping, allowing the plant to
stay upright and the leaves to have maximum surface area for light
absorption.
18
(iv)
3. (a)
(b)
Intake of water causes the protoplasm of cells to expand slightly. In plant
cells, this causes the cell membrane to push against the cell wall. Being
inelastic, the cell wall resists this push (turgor pressure) and prevents the
cell membrane from bursting. Animal cells have no cell wall to resist the
slight expansion of the protoplasm, so they increase in volume until they
burst.
Tertiary consumer: any one of hawk or blue jay (only if the spider is listed among
the secondary consumers)
Secondary consumer: any two of spider, blue jay, woodpecker
Primary consumer: any two of caterpillar, ant, termite, butterfly
(i)
Fungi and bacteria
(ii)
They allow recycling of materials. When organisms die or shed structures,
their complex organic compounds are broken down by decomposers to
simple inorganic ions, which plants roots’ can absorb. When animals eat
plants, the materials are transferred to other trophic levels and assimilated
into the bodies of organisms as complex organic compounds once more.
(c)
- Competition
- Reproduction
Alternative correct answers are: irritability, availability of habitat/shelter, etc.
(d)
- pH of soil
- availability of water
- temperature
Alternative responses include light intensity, wind direction, wind speed, etc.
19
4. (a)
(i)
DIAGRAM ILLUSTRATING THE CIRCULATORY SYSTEM
1) Oxygen diffuses into blood in the alveolar capillaries.
2) The oxygenated blood flows in the pulmonary vein to the left side
of the heart and enters the left atrium.
3) It flows through the bicuspid valve to the left ventricle.
4) It leaves the left ventricle and enters the aorta.
5) The aorta divides into several arteries, with one artery transporting
blood to each organ.
6) Within the organ, the blood flows through the artery, arterioles and
capillaries. The oxygenated blood is now close to organs and
tissues.
20
(ii)
1)
2)
3)
4)
Oxygenated blood is in the capillary bed of tissues.
Blood plasma leaves the capillaries, forming tissue fluid.
Oxyhaemoglobin (in red blood cells) dissociates, releasing oxygen.
Oxygen diffuses into the tissue fluid, creating a concentration
gradient from the tissue fluid to the respiring cells.
5) Oxygen diffuses across the cell membrane of the muscle cells,
where it is available for aerobic respiration.
(b)
(c)
5. (a)
(b)
(i)
- It requires oxygen
- A lot of energy is produced (32 ATP per glucose molecule)
Alternative response: Carbon dioxide is produced
(ii)
Less energy is produced as ATP. The rest of the chemical energy of
glucose is used in the formation of lactic acid. The energy of lactic acid is
unavailable to cells. Lactic acid also causes muscle cramp and fatigue.
Transport vessels in both plants and animals:
- Are hollow tubes. The lumen allows unimpeded flow of materials.
- Form an extensive network throughout the organism, ensuring that all cells are
close to the transport system.
Alternative responses include adaptations of the walls to extreme pressure, etc.
Disease: Acquired Immune Deficiency Syndrome (AIDS)
Pathogen: Human Immunodeficiency Virus (HIV)
(i)
Mode of action of vaccine:
1) A vaccine introduces weakened antigens into the blood of a
person.
2) The antigens bind to their complementary B-lymphocyte, them to
divide repeatedly.
3) Some B-lymphocytes secrete antibodies and eventually die.
Antibodies destroy the antigens or pathogens. Some Blymphocytes persist in the bloodstream (memory cells).
4) The memory cells are a population of B-lymphocytes which is
large enough to cause such a quick and large immune response
(antibody secretion) on subsequent infection of the individual with
the pathogen, that the person shows no signs or symptoms of the
disease. The person is immune to the disease.
(ii)
A vaccine may not be effective because:
 The pathogen may have a high mutation rate. Memory cells
produced as a result of vaccination may not be complementary to
the antigens of the mutated pathogens and would not bind to them.
21

The target of the pathogen may be the lymphocytes themselves. If
these are destroyed by the pathogens, they will be unable to
execute the immune response.
Alternative responses include the possibility of allergic reactions to the
vaccines, etc.
(c)
(i)
This is done if:
 The plants reproduce asexually. All the plants will have the same
genotype and will be equally susceptible to the disease. It is likely
that all the plants will be infected.
 No chemicals are available for destruction of the pathogen. If the
plants are destroyed, there is a possibility that the pathogens will
be destroyed inside them.
 The high reproduction rate of pathogens means that the
neighbouring fields may be later infected, since the plants in those
fields are stationary and unable to move away from the infected
ones. Destruction of a field may prevent infection of non-infected
fields.
(ii)
Social:
 Shortage of food, causing malnutrition in the population
 Increased likelihood of crime as persons may resort to theft or
violence to obtain food
Economic:
 Loss of income for farmers or the country
 Increased expenditure on measures needed to eradicate the disease
Alternative responses are acceptable such as famine and increased debts
for the country as they would have to seek food from elsewhere, etc.
6. (a)
A gene is a length of DNA, in a particular locus on a chromosome, which codes
for a specific polypeptide. It causes an organism to possess a particular trait, e.g.
blood group.
An allele is one of the alternative forms of a gene. It causes an organism to
possess a specific version or form of the trait, e.g. possession of IA alleles will
cause a person to be of blood group A. Alleles may be dominant or recessive.
22
(b)
(i)
PP or Pp
(ii)
Let P represent the allele for purple flowers
Let p represent the allele for white flowers
Parental phenotype
purple
X
purple
Parental genotype
Pp
Pp
Gametes
(P) and (p)
(P) and (p)
Random fertilisation
(P) (p)
(P) PP Pp
(p) Pp pp
Offspring genotype
PP
Offspring phenotype
purple purple
Offspring phenotypic ratio
Pp
3 purple
Pp
pp
purple
white
:
1 white
Each plant has a 1 in 4 chance of producing white flowers.
(c)
The phenotype is the physical expression of the genotype, modified by
environmental effects. Some environmental influences can be quite large and
cause large deviations from the genotype-determined phenotype. If organisms of
identical genotype are exposed to different environmental influences, their
resulting phenotypes may be different from each other. “Different environmental
influences” refer to variation in factors such as temperature, nutrient availability,
exposure to pathogens, etc.
23
1. (a)
(i)




Write a self-explanatory title, e.g. DISTANCE DYE TRAVELLED UP
CELERY STALKS A AND B WHEN EXPOSED TO VARYING
TEMPERATURE. Label each axis: distance on the y-axis and time on
the x-axis.
Include the units on each axis.
Plot the points accurately and visibly.
Use scales for the axes such that maximum use is made of the graph
paper.
DISTANCE DYE TRAVELLED UP CELERY STALKS A AND B WHEN
EXPOSED TO VARYING TEMPERATURES
24
(ii)
The rate of transpiration was slower in stalk A because it was exposed to
less wind and heat.
(b)
To investigate the effect of wind speed on the rate of transpiration.
OR
To investigate the effect of temperature/heat on the rate of transpiration.
(c)
The stalk in Flask A is the control.
(d)
The rate of transpiration increases with wind speed OR the rate of transpiration
increases with temperature.
(e)
Water enters the root hairs of the epidermis by osmosis (down a water potential
gradient and across the partially permeable cell membrane). It then moves by
osmosis across the cortex cells towards the middle of the root and enters the root
xylem by osmosis. Water is then pulled up the xylem from root through stem to
leaf by capillarity (adhesion to xylem walls and cohesion to other water
molecules) and due to the transpiration pull caused evaporation of water from
mesophyll cells into intercellular spaces. The evaporated water diffuses out of the
leaf through stomata, down a concentration gradient.
Note that the physical processes must be described, not merely the route of the
water.
(f)
The protoplasm and cross-walls of xylem are degenerated, forming hollow
continuous tubes for the unimpeded flow of water. Xylem vessels are narrow,
allowing water flow by capillarity.
Other valid points are acceptable, e.g. lignification of tubes, etc.
(g)
The water carried through the plant by transpiration contains dissolved minerals
which are needed by the plant. Evaporation of water from leaves removes heat,
cooling the plant. Transpiration causes water uptake, needed as a reagent in
photosynthesis and hydrolysis reactions.
Other valid points are acceptable, e.g. allows maintenance of turgidity for
support, etc.
(h)
Phloem tissue transports sucrose (made from glucose formed by photosynthesis).
(i)
Difference: The heart’s pumping action, muscles and valves in veins effect blood
flow in mammals. These are absent in plants. Other valid points are acceptable,
e.g. separate tubes exist in plants for transport of water and sucrose, etc.
Similarities: A network of tubes is present throughout the organism, containing
the transport medium. A force / mechanism operates to move the transport
medium (transpiration pull / pumping of heart).
Other valid points are acceptable, e.g. water is the solvent / medium for
transported materials, etc.
25
2. (a)
(b)
(c)
3. (a)
(b)
(i)
It is necessary for the oxidation of food with the release of energy usable
by the body’s cells/ATP.
(ii)
C6 H12O6  6O2 
 6CO2  6H2O  32 ATP
(iii)
Respiration
(iv)
Mitochondrion
(i)
His rate of breathing and blood circulation was insufficient to inhale and
deliver the increased amount of oxygen for the increased respiration rate
done by exercising muscles.
(ii)
Lactic acid
(iii)
It causes pain and fatigue, decreasing his rate of running. It may cause him
to stop running if lactic acid production is high enough.
(i)
Anaerobic respiration OR alcoholic fermentation
(ii)
Glucose is only partly oxidised, hence, a small amount of energy is
produced. No oxygen is used. Carbon dioxide and ethanol are produced.
They do photosynthesis: they trap solar energy and convert it to chemical energy
(a form of energy which can be used by living organisms).
(i)
A – Herbivores OR primary consumers
B – Carnivores OR secondary consumers
(ii)
Example: Ocelot
Role: It acts as an agent of natural selection of the herbivores, preventing
overpopulation and ensuring that the population remains “fit”. It transfers
energy from the second trophic level to the fourth.
(iii)
Example: Putrefying bacterium OR Putrefying fungus
Carbon dioxide: It uses the dead body/body part of an organism as its
substrate for respiration. This process releases carbon dioxide to the
atmosphere.
(iv)
Decomposers return materials from all trophic levels to the soil, allowing
their cyclic uptake by plants (they are then re-used). There is no process
for returning energy to lower trophic levels.
26
(c)
In the diagram, be sure to include the four points / stages, that is, AIR, SOIL /
WATER, PLANTS and ANIMALS. There must be at least one arrow going to
each stage (adding nitrogen to it) and one arrow leaving each stage (removing
nitrogen from it). The names of the processes adding and removing nitrogen
should be written on the arrows. The processes should include four of:
 Nitrification: NH3 or NH4+  NO2-  NO3- in Soil/Water
 Denitrification: NO3- in Soil/Water  N2 in Air
 Nitrogen fixation: N2 in Air  NH3 /NH4+/NO3– in Soil/Water
 Ammonification /decomposition: Plants or Animals —› NH3/NH4+ in
Soil/Water
 Ingestion: Plants  Animals
DIAGRAM OF THE NITROGEN CYCLE
(d)
4. (a)
Rhizobium does nitrogen fixation.
Other examples are acceptable e.g. Nitrosomonos and Nitrobacter does
nitrification, etc.
Numerous stomata are located adjacent to the intercellular spaces for inward
diffusion of carbon dioxide and outward diffusion of oxygen. An extensive
network of vascular bundles, containing xylem, adequately supplies mesophyll
cells with water.
27
(b)
(i)




Drawing must illustrate: lamina, petiole, main vein
Must be of a reasonable size (at least two-thirds of the allotted space in
length.)
Lines must be drawn to show green and non-green areas and these must
be labelled
Appropriate title e.g. : DRAWING OF DORSAL VIEW OF
VARIEGATED LEAF
DIAGRAM SHOWING THE DISTRIBUTION OF GREEN AND NONGREEN AREAS OF A LEAF
(ii)
Write “changes colour” next to the label for the green part of the leaf.
(iii)
Yellow-brown to blue-black
(c)
In photosynthesis, chlorophyll traps light energy. This light energy is used to
synthesise glucose molecules. The light energy is converted to the chemical
energy of glucose. Glucose is then converted to starch. The presence of starch in
a leaf therefore indicates that photosynthesis occurred. Starch was found to be
present only in the green areas of the leaf, that is, only in the areas containing
chlorophyll. This indicates that chlorophyll is necessary for photosynthesis.
(d)
Some glucose made in photosynthesis (in green areas of leaves) is converted to
sucrose. This is loaded into phloem and translocated to non-chlorophyll
containing parts, such as storage organs. The sucrose is unloaded from the phloem
and converted to starch in these parts.
5. (a)
Urea and sodium chloride. Other examples are acceptable such as salt and water.
28
(b)
The length of the drawing should occupy at least two-thirds of the allotted space
and should be in proportion. Labels should include: Glomerulus, Bowman’s
capsule, proximal convoluted tubule, Loop of Henle, Distal convoluted tubule and
collecting duct.
Label
Glomerulus
Bowman’s capsule
Annotation
in Ultrafiltration of blood causes small soluble substances
from blood plasma to enter the lumen of the nephron in
the capsule
Proximal convoluted Selective reabsorption of all glucose, amino acids,
tubule
vitamins and a lot of the water and sodium chloride (and
other salts)
Loop of Henle
Reabsorption of water
Distal
convoluted Reabsorption of water and salts if necessary (controlled
tubule
by ADH secretion)
DIAGRAM OF A KIDNEY TUBULE
(c)
A diabetic patient’s blood glucose level is so high that the glomerular filtrate will
contain very high levels of it. The surface area of the proximal convolution will
be insufficient to reabsorb all of the glucose. Other parts of the nephron are not
adapted to reabsorb glucose. The glucose which was not absorbed by the proximal
convolution will therefore be present in the urine.
The blood pressure of the glomerulus is high. It will be even higher in a patient
with hypertension. It may be so high that the glomerular capillaries may rupture.
Since ultrafiltration is done by these capillaries, if they are ruptured then large
substances such as blood cells will pass, unimpeded, into the glomerular filtrate.
29
The nephron has no mechanism to reabsorb blood cells and so they appear in the
urine.
Other valid points are acceptable, e.g. the presence of kidney stones, etc.
6. (a)
(i)
Category
Pathogenic
Nutritional deficiency
Hereditary
Physiological
Example
HIV / AIDS
Scurvy
Haemophilia
Diabetes
(ii)
Treatment of pathogenic disease
(HIV / AIDS)
Use of antiretroviral medication to
decrease activity of pathogen
Use of medication to alleviate signs
and symptoms, e.g. fever reducers
Education of patient about disease
Treatment of physiological
disease (Diabetes)
No pathogen involved
Control of pathogenic disease (HIV
/AIDS)
Education of public about disease
Preventative measures such as limited
number of sexual partners, testing of
persons for presence of HIV, use of
condoms during sexual intercourse
Attempts to develop vaccine against
pathogen
Control of physiological disease
(Diabetes)
Education of public about disease
Preventative measures such as
control of body weight and a diet
of high fibre and less
carbohydrate and fat
No vaccine involved
Use of medication to alleviate
signs and symptoms, e.g. insulin
Education of patient about
disease
No dietary modification may be Modification to diet, e.g. reduced
necessary
carbohydrate intake
(b)
(i)
Genetic engineering is manipulation of genes which involves their transfer
from one organism (donor) to another (genetically modified organism –
which is not necessarily of the same species as the donor), resulting in its
altered genotype and phenotype.
(ii)
Ethical concern – Members of society has different views of whether
humans should transfer genes across species, determine the features of
offspring (in particular, human offspring) or determine which features
should be deemed acceptable enough to perpetuate in organisms. The
concern is whether there are manipulative boundaries that humans should
not defy.
30
Social concern – Genetically engineered crops and animals are likely to be
expensive because of their superior characteristics. These may also be
engineered to be sterile, so that farmers must continue to purchase stock
from the research companies. Some farmers may not be financially able to
do so, especially in developing countries.
Ecological concern – Transferred genes may “escape” to wild relatives by
cross-pollination. These may develop the superior characteristic of the
gene and have a tremendous selective advantage. Their populations may
soar and they may outcompete other plant species as well, disrupting the
natural mechanisms of population control. This may decrease the
biodiversity of the ecosystem and so disrupt the balance which is
necessary for its self-sustainability.
Other valid points are acceptable, e.g. the unknown effects of genetic
modification on plant and animal health.
31
1. (a)
(b)
(i)
Apparatus and materials:
Potato, sugar solutions of concentration 0.5M, 1.0M, 1.5M, 2.0M and
0.0M (distilled water), blotting paper, forceps, knife, balance, measuring
cylinders or large syringes, 5 Petri dishes, stopwatch
(ii)
Method:
1) Measure equal volumes of sugar solution of each concentration
(including water only) into appropriately labelled Petri dishes.
2) Cut the potato into 20 strips of dimensions such that each weighs
2.0g. Weigh each strip.
3) Place 4 strips in each Petri dish such that they are submerged and
cover them.
4) Calculate the average weight of the strips in each Petri dish.
5) After 20 minutes, remove the strips, blot them rapidly, weigh them
and calculate the average weight of the strips in each dish.
TABLE 1: AVERAGE WEIGHT OF POTATO STRIPS AFTER IMMERSION
IN DIFFERENT CONCENTRATIONS OF SUGAR SOLUTION
Concentration of sugar
solution / M
0.0
0.5
1.0
1.5
2.0
(c)
Average weight
of strips / g
3.0
2.3
2.0
1.7
1.6
(i)
The weight increased because water moved from a region of higher water
concentration (external water) to an area of lower water concentration
(potato cells).
(ii)
Less than 1.0M: The weight increased because water moved from a region
of higher water concentration (external solution) to an area of lower water
concentration (potato cells).
More than 1.0M: The weight decreased because water moved from a
region of higher water concentration (potato cells) to an area of lower
water concentration (external solution).
32
(iii)
Osmosis
(iv)
The time allowed for osmosis may have varied among the strips since it
takes time to place 20 strips in the various dishes.
Alternative responses include variation in weight and inaccuracy of
concentration of sugar solutions during preparation, etc.
(d)
(i)
As the animal cell gains water, the internal pressure increases and
protoplasm expands until the cell membrane bursts. Unlike the animal cell,
the plant cell has a tough inelastic cell wall which resists and limits the
expansion of the protoplasm, preventing it from bursting.
(ii)
As both plant and animal cells lose water, their protoplasm shrink. The
cell membrane of the plant cell now becomes visible as it pulls away from
the cell wall. The protoplasm shrinks less than in the animal cell because
the cytoplasm is continuous with plasmodesmata.
(e)
Diffusion or Active transport
(f)
- To remove waste materials from cells for excretion
- To obtain materials for growth, respiration, assimilation or any other cellular
process
2. (a)
A: stomach
B: small intestine
C: large intestine/colon
D: liver
E: pancreas
(b)
- It is long: it provides a large surface area for digestion
- It contains numerous villi: they provide a large surface area for absorption of
food
- Its villi contain numerous capillaries: blood flow maintains a concentration
gradient for digested food
Alternative responses include compartmentalisation, thin walls of villi, etc.
(c)
- Teeth vary in size and shape: different ones are used for different food types.
- A wide variety of enzymes are secreted: digestion of different food types is
possible.
- The gut is longer and more complex than that of carnivores (meat is easily
digested) but shorter and simpler than that of herbivores (some plant matter
remains undigested – roughage).
Alternative responses include maintenance of different pH levels at different areas
of the gut, facilitating the action of a variety of enzymes, etc.
33
(d)
(i)
Liver: production of bile for emulsification of fat
Pancreas: production of digestive enzymes trypsin, lipase and pancreatic
amylase
(ii)
- Impaired digestion due to insufficient production of enzymes
- Impaired regulation of blood glucose level due to insufficient/too much
production of insulin and /or glucagon
3. (a)
Parental phenotypes
Male
Female
Parental genotypes (i) XY
Gametes genotypes (ii) (X) and (Y) (iii) (X)
Offspring genotypes (iv) XY and XX
Offspring phenotypes
(b)
(i)
DIAGRAM OF A FLOWER
(ii)
Meiosis:
 Homologous chromosomes pair up, allowing crossing over to
occur within the bivalents.
 The bivalents attach to spindle fibres and are separated, with one
member of each homologous pair going to a daughter cell of the
first division. Each daughter cell has the haploid number of
chromosomes.
 Chromosomes attach to spindle fibres and are separated, with each
sister chromatid going to a daughter cell of the second division.
These daughter cells are haploid gametes.
(iii)
Four chromosomes
34
(c)
- It maintains the chromosome number across all generations of a sexually
reproducing population.
- It causes variation among offspring, increasing the chances of survival of the
population when the environment changes.
(d)
Parental phenotype
red flower
Parental genotype
Gametes
X
white flower
Rr
rr
(R) and (r)
(r)
Random fertilisation
(R) (r)
(r) Rr rr
Offspring genotype
(e)
4. (a)
Rr
Offspring phenotype
red flower
Offspring phenotypic ratio
1 red
rr
white flower
:
1 white
The differences illustrate environmental variation – they occur because the plants
were grown in different environments, e.g. varying exposure to light, water,
nutrients or were in different soil types.
Photosynthesis occurs in two stages and is summarised as:
light absorbed by chlorophyll
6CO2  6H2O 
 C6 H12O6  6O2
First/light dependent stage:
 Chlorophyll in chloroplasts absorbs light.
 Photolysis of water (absorbed from soil) occurs, using light energy: it is
split into hydrogen and oxygen.
 The oxygen produced diffuses through stomata as a waste product.
Second/light independent stage:
 Carbon dioxide diffuses from the atmosphere through stomata, then
intercellular spaces, into mesophyll cells.
 Carbon dioxide combines with hydrogen to produce carbohydrate
(glucose). The light energy is converted to the chemical energy of glucose.
35
(b)
DIAGRAM OF EXTERNAL FEATURES OF A LEAF SHOWING
ADAPTATIONS FOR PHOTOSYNTHESIS
Only three annotations are necessary. Alternative correct responses include:
Green parts of leaf- contain chlorophyll for light absorption.
Veins- form a “skeleton” which allows the leaf to remain flat for light exposure.
Cuticle- transparent (allows penetration of light) and waterproof (reduces water
loss, conserves it for photosynthesis.)
(c)
(d)
(i)
Decomposers
(ii)
- Decomposers do heterotrophic nutrition (they obtain organic compounds
from other organisms) while photosynthesis is autotrophic nutrition
(organic compounds are synthesised by the plant for its use).
- Decomposition is the degradation of complex insoluble organic
compounds to simple, ions or molecules, while photosynthesis is the
synthesis of complex organic compounds from simple molecules.
Advantages:
 Some excretory products are stored in leaves. When these leaves fall, the
excretory products are removed from the plant.
 Leaf fall reduces the number of stomata available for transpiration.
Reduced transpiration rates conserve water.
Alternative responses include return of nutrients to soil (for subsequent use by
plants) after decomposition, reduction of rate of soil erosion, etc.
Disadvantages:
 There is a reduced surface area available for photosynthesis (reduced
chlorophyll, stomata for diffusion of gases, etc.)
 There is reduced surface area available (reduced number of stomata) for
gaseous exchange associated with respiration.
36
Alternative responses include reduced uptake of mineral salts associated with
reduced transpiration rate, etc.
5. (a)
DIAGRAM OF THE HEART’S STRUCTURE




(b)
(i)
During diastole, blood flows into the atria.
During atrial systole, the atria contract. The atrial pressure increases,
pushing open the bicuspid and tricuspid valves and blood enters the
ventricles. (Blood leaves the left atrium and enters the left ventricle via the
bicuspid valve).
During ventricular systole, the ventricles contract. The ventricular pressure
increases, shutting the bicuspid and tricuspid valves and pushing open the
semilunar valves at the start of the aorta and pulmonary artery. (Blood
leaves the left ventricle and enters the aorta via the semilunar valves.)
Blood flows along the pulmonary artery to the lungs. Blood flows along the
aorta, then along arteries supplying blood to the organs (aorta branches into
smaller arteries).
Vaccines contain weakened/altered or dead antigens of pathogens.
Vaccination is used to prevent the spread of pathogenic diseases by
stimulating production of memory cells which are effective (stimulate
production of antibodies) against a particular pathogen. They are
ineffective against organ malfunction unless the malfunction is caused by
an infection.
Diseases of the heart and blood vessels are usually not caused by the entry
of pathogens. They are caused by (or their likelihood is increased by) diets
which are high in salt or saturated fat, lack of exercise, obesity, smoking
and high levels of stress. Some defects are congenital (present at
birth).These factors may cause the heart and blood vessels to malfunction.
37
(ii)
A patient is susceptible to infections following surgery it involves a
significant amount of cutting of skin and tissue. Pathogens are likely to
enter the wound.
The immune system responds to pathogen entry (infection) by stimulating
production of white blood cells (phagocytes and B-lymphocytes).
Phagocytes engulf pathogens and digest them. B-lymphocytes are
stimulated to divide upon binding to their complementary antigens (of
pathogens). This eventually stimulates production of antibodies, which
destroy pathogens or neutralise their toxins.
These reactions of the immune response lower the probability of
proliferation of pathogens which enter at the site of the surgery.
6. (a)
(i)
A community is the collection of all the members of all the species living
in a particular ecosystem or habitat.
A habitat is the place where an organism lives and fulfils its ecological
niche, e.g. a pond.
(ii)
Sampling a forest:
 A quadrat is thrown randomly and all plants of the species of
interest which are found within it are counted / recorded. This is
done multiple times and the data is used to estimate the
density/species cover/frequency of the species.
 A line transect is pegged along the ground/soil in a straight line
and all plant species touching it are recorded/counted. This shows
how the abundance of the species of interest varies across the area.
Only one method should be described.
Sampling a pond:
 A pond net is swept or dragged under the surface of the water. The
organisms caught in the trap of the net are transferred to a bottle or
jar. The sweep and collection of organisms is repeated multiple
times at varying depths or areas of the pond. The number and type
of organisms collected are recorded.
38
(b)
(i)
Relationship
Parasitism
e.g. wild duck
and intestinal
worm
Predator-prey
Partner 1
Partner 2
Host suffers harm and gains Parasite gains
no benefit
suffers no harm
e.g. The duck is deprived of
its products of digestion
(nutrients) as the worm
absorbs some.
benefit
and
e.g. The worm (parasite) gains
benefits such as a supply of
nutrients and protection from
extreme
environmental
conditions,
e.g. lack of
moisture or high temperature.
It confers no benefit to the
duck.
Predator hunts, kills and eats Prey is eaten by the predator
its prey
and so, it is the means of
transfer of energy to the higher
e.g. The lizard kills and eats trophic level.
the earthworm, gaining its
nutrients from a lower trophic e.g. The earthworm is food for
level.
the lizard.
Examples of alternative feeding relationships may be described, such as
mutualism, etc.
(ii)
A food web:
(iii)
Keskidee
(iv)
Disruption may result from:
 Change in environmental conditions, e.g. reduced water supply in a
drought
This may cause some organisms to be almost or totally eliminated
from the ecosystem. This in turn, leads to increase in numbers of
their prey and a decrease in numbers of their predators. These
changes will cause further changes in the populations of other
species. If the numbers of a species decrease, their niches, e.g.
39

pollination, dispersal of seeds, may not be adequately fulfilled by
the few remaining members (if any remain).
Introduction of pathogens to the ecosystem
This causes the incidence of disease in certain species. The species
may be totally eliminated from the ecosystem. If some survive, it
would take time to replenish the original number of organisms.
During this time, their niches may not be adequately fulfilled by
other species. This may threaten the survival of yet other species.
Alternative correct responses include pollution, causing destruction of
microorganisms, threatening the cycling of nutrients or migration of
species, removing food supplies or predators (which are a means of
population control) from the ecosystem.
40
1. (a)
(i)
TEMPERATURE/0C RATE OF PRODUCTION OF
REDUCING SUGAR/mg min-1
15
20
25
35
35
45
40
35
TABLE SHOWING EFFECT OF TEMPERATURE ON RATE OF AMYLASE
ACTIVITY
(ii)
Enzyme activity increases steadily with temperature to a maximum of 45
mg min-1 of reducing sugar at 35°C. As temperature increased beyond
this, enzyme activity decreased sharply to less than 5 mg min-1 at 55°C.
(iii)
The enzyme is denatured. Being a protein, its three-dimensional structure
is altered at high temperatures. This distorts its active site, making it is no
longer complementary to its substrate.
(b)
Experiment:
1) Measure 10.0mL of buffer solutions of different pH values (e.g. 2, 5, 7, 9,
14) into appropriately labelled test tubes.
2) Add 10.0mL of starch suspension to each tube.
3) Add 4.0mL of amylase to each tube. Mix the contents and record the time.
4) At two-minute intervals, remove 1.0mL of the mixture from each tube.
Test this sample for the presence of reducing sugar (mix with 1.0mL of
Benedict’s solution, warm, record weight of precipitate).
5) Plot, on graph paper, weight of precipitate against pH. The gradient of the
curve shows how the rate of breakdown of starch varies with pH.
Alternatively, the rate of breakdown of starch could be measured.
(c)
(i)
Mouth and duodenum
(ii)
Amylase (swallowed in the bolus) is denatured by the acidic conditions of
the stomach. Amylase works under alkaline conditions.
(iii)
Pepsin
Alternative answer – rennin
41
(d)
2. (a)
(i)
Seed, fruit and leaf
Alternative answer – storage organ
(ii)
Starch is hydrolysed to maltose using amylase. The maltose is converted
to sucrose, which is easily transported in the phloem to other parts.
(iii)
It eliminates the need for continuous manufacture of food
(photosynthesis).
Alternative responses may refer to periods of unfavourable conditions for
photosynthesis, synthesis of structures associated with reproduction
(fruits, seeds, storage organs, etc.) or growth of an embryo during
germination.
(i)
Name of A: humerus
Function of A: allows movement of limb by acting as a lever
Name of B: ligament
Function of B: attaches humerus to ulna/radius, preventing dislocation
during movement
Name of C: synovial fluid
Function of C: provides lubrication during movement and absorbs shock
(ii)
DIAGRAM OF A HINGE JOINT
(b)
Loss of cartilage results in loss of absorption of shock. The force associated with
mechanical impact is transmitted directly to bones, causing joint pain. Loss of the
smooth cartilage results in the ends of the bones rubbing against each other,
eroding them painfully.
42
(c)
The triceps point of origin is at the scapula and is inserted at the ulna. The biceps
originates at the scapula and is inserted at the radius. The biceps and triceps are
antagonistic muscles. When the biceps contracts, the triceps relaxes and the ulna
and radius are raised as a lever, raising the lower arm. When the triceps contracts,
the biceps relaxes and the ulna and radius are lowered as a lever, lowering the
lower arm.
(d)
Blood cells are produced in the bone marrow of the limb bones and the ribs.
Replacement of bone marrow involves replacement of blood-making tissue. This
would eliminate the disease if the condition is related to the production of
malfunctioning blood cells.
3. (a)
Name of A: cell wall
Function of A: protects the cell from intake of excess water (and therefore, from
bursting)
Alternative response – maintains shape of cell by resisting turgor pressure
Name of B: chloroplast
Function of B: traps light energy and converts it to chemical energy in
photosynthesis
(b)
DIAGRAM OF PLANT CELL IN CONCENTRATED SALT SOLUTION
(c)
Plasmolysed guard cells cause stomata to close, preventing entry of carbon
dioxide and transpiration. Cessation of transpiration slows or stops uptake of
water. If hydrogen (from water) and carbon dioxide are not available to combine
to produce glucose, photosynthesis is reduced or stopped.
Alternative response - Plasmolysed cells cause wilting of the leaf, reducing the
surface available for light absorption.
43
(d)
(i), (ii), (iii)
Feature
Vacuole
Plant cell
Animal cell
large, central, permanent small, no fixed location,
temporary
Chloroplast present
absent
Cell wall
present
absent
4. (a)
(b)
(i)
A gene is a length of DNA which codes for production of a particular
polypeptide. An allele is one of the possible/alternative forms of a gene.
(ii)
A genotype is the combination of alleles which an organism possesses for
the genes it contains. The phenotype is the physical expression of the
genotype, sometimes modified by environmental effects.
(iii)
Dominant is a description of allele if it is expressed in the phenotype once
it is present in the genotype, regardless of the presence of recessive alleles
(homozygous or heterozygous). Recessive is a description of an allele
which is expressed in the phenotype provided it is the only type of allele
present in the genotype (homozygous).
Let B represent the allele for black fur
Let b represent the allele for brown fur
Parental phenotype
Parental genotype
Gametes
father - black
X
mother – brown
Bb
bb
(B) and (b)
(b)
Random fertilisation
(B) (b)
(b) Bb bb
Offspring genotype
Bb
bb
Offspring phenotype
black
brown
Offspring phenotypic ratio
1
:
1
A 1:1 ratio means that half of the offspring are black and half are brown.
44
(c)
Parental phenotype
father - normal
XHY
Parental genotype
Gametes
mother – normal
X
XHXh
(XH) and (Y)
(XH) and (Xh)
Random fertilisation
(XH) (Y)
(X ) XHXH XHY
(Xh) XHXh XhY
H
Offspring genotype
XHXH
XHXh
Offspring phenotype
normal
normal
Offspring phenotypic ratio
1 normal
XHY
normal
:
XhY
haemophiliac
1 haemophiliac
XhY is a haemophiliac child.
5. (a)
(b)
Insect vector – mosquito
The first stage is the Egg. The immature offspring develop inside the eggs. The
eggs hatch, releasing the Larvae, or second stage. These are aquatic, feeding and
growing at a rapid rate. Their outer covering then forms a capsule. This is the
third stage and they are called Pupae. Within the pupae, their bodies degenerate
and are reorganised to form adult mosquitoes. When this development is
complete, the fourth stages, or Imago/Adults, emerge from the pupal cases.
Physiological disease: diabetes
Pathogenic disease: AIDS
Diabetes is not transmitted from one person to another. One may inherit a genetic
predisposition for the disease, but this does not mean that it is certain that the
person will develop the disease. A person is more likely to develop diabetes from
lifestyle habits such as consumption of foods high in carbohydrates and lipids, a
sedentary lifestyle and other habits which may encourage obesity. Any factor
which causes malfunction of the pancreas (e.g. destruction by one’s own immune
system) can cause the development of diabetes.
AIDS is transmitted when the HIV pathogen leaves one person (infected person)
and enters another (new patient) by the introduction of body fluids such as blood
or semen. This may happen during sexual intercourse, use of contaminated
needles or unscreened blood transfusions.
45
(c)
Disease
Coronary heart
disease
(physiological)
Treatment
1-Insertion of stents or
other devices to increase
the diameter of blood
vessels (surgery)
2-Use of medication to
remove the plaque lining
blood vessels
The purpose of these
measures is to correct the
malfunction, if possible, or
alleviate the suffering of
the patient if correction is
not possible.
Malaria
(pathogenic)
1-Use of medication
containing quinine to kill
the malarial pathogen
2-Use of drugs to reduce
signs and symptoms, e.g.
fever reducers
The purpose of these
measures is to kill the
pathogen, if possible, or
alleviate the suffering of
the patient if it is not
possible.
Control
1-Education of the public about the
risk factors and nature of heart disease
2-Avoidance of food with a high
content of saturated fat
3-Engagement in regular exercise
4-Avoidance of food with a high salt
content
5-Activites/strategies for reduction of a
stressful lifestyle
6-Avoidance of smoking
The purpose of these lifestyle measures
is to reduce the chances of healthy
persons developing the condition.
1-Covering the surface of water with a
chemical toxic to eggs, larvae or
pupae, e.g. oil prevents their gaseous
exchange
2-Removal of breeding grounds of
mosquito, e.g. draining stagnant water
3-Use of insecticide sprays and traps to
kill adult mosquitoes
4-Use of clothing to cover arms, legs,
etc.
5-Use of prophylactic drugs, e.g.
doxycycline, to prevent malarial
infections
6-Education of the public about the
risk factors and nature of malaria
6-Prevention of infected persons
coming into contact with non-infected
ones
The purpose of these measures is to kill
the pathogens by using methods
appropriate for particular stages of the
life cycle, or adopt other habits to
prevent healthy persons from
contracting malaria.
46
(d)
- The money spent medication, hospital space, virus-testing and other
costs associated with hospitalisation is a significant amount.
- The labour force is reduced. The majority of AIDS patients are of the
income-earning age. When persons of this age are unable to work,
productivity at workplaces is lowered.
- There is loss of jobs and income within households. As AIDS patients
decline in health, they may be forced to stop working altogether. This
affects the quality of life within the household, since the patients are often
in the income-earning age.
Alternative responses include:
- Vulnerable persons may be uncared for. If the working-age persons die
of AIDS, children and the elderly persons in the household may be left
without caregivers.
- AIDS patients may be the subject of social discrimination because of the
negative attitude towards AIDS patients. This may cause them to be fearful
of disclosure of their illness. The long incubation period of the HIV may
allow them to “hide” their illness and they may infect others. Persons who
engage in AIDS-risky behaviour may be hesitant to test their blood for
presence of the virus and may unwittingly spread it to others.
- The high incidence of AIDS in the region causes governments to spend a
lot of money on AIDS awareness campaigns and research. This reduces
the amount of money available to other sectors such as education and
agriculture.
6. (a)
DIAGRAM OF THE HUMAN MALE REPRODUCTIVE SYSTEM
(b)
(i)
Use of a physical barrier such as a condom prevents ejaculated sperm from
coming into contact with the ovum. The sperms do not reach the oviduct
since they do not enter the uterus.
Alternative responses include vasectomy, tubal ligation, etc.
(ii)
Use of oral contraceptives (hormonal pills) -they contain high proportions
of oestrogen- and progesterone-like molecules which inhibit FSH. This
inhibition prevents development of follicles and ovulation. If sperms are
present in the oviduct, they will come in contact with no ovum.
47
(c)
Flowering plants
Pollen is transferred from anther to
stigma
Agent of pollination transfers male
gametes to female part of flower
Male gametes are produced by mitosis of
pollen grain nucleus during germination
of pollen tube
Male gametes are carried by growth of
pollen tube down style and into ovary
Gametes are brought together for
fertilisation in the embryo sac of the
ovary
Female gamete stays within the embryo
sac of the ovary
Humans
Sperms are transferred from
epididymis to vagina
Sexual intercourse transfers male
gametes to female system
Male gametes are produced before
the gamete transfer process starts,
and are stored in the epididymis
Sperms swim up vagina, uterus and
oviducts, aided by muscular
contractions of the female tract
Gametes are brought together for
fertilisation in the oviduct
Female gamete is released from the
ovary at ovulation
48
1. (a)
(i)
ATHLETE’S RED BLOOD CELL COUNT DURING SIX MONTHS OF
TRAINING AT 2500 m ABOVE SEA LEVEL
49
(b)
(c)
(d)
(ii)
Months 1-3: There is a gradual but steady increase in the number of red
blood cells, reaching a count of 5.1 million per unit of blood. Months 4-6:
The red blood cell count reaches a maximum of 5.3 million per unit of
blood in the fifth month and remains at that value.
(iii)
The concentration of oxygen is lower at higher altitudes. Inhaled air
therefore contains less oxygen. An increased number of red blood cells are
needed to deliver sufficient oxygen to tissues, since each red blood cell
may absorb less oxygen than it would at lower altitudes.
(iv)
Continuous training at higher altitudes increase one’s red blood cell count.
(v)
An athlete as similar as possible to the test subject (same sex, height,
weight, health, etc.) should undertake the same training programme (type,
duration, time of day of training, diet, hours and time of sleep, etc.) but at
sea level.
(i)
Katz’s red blood cells are sickle-shaped (crescent-moon shaped) or
irregular while normal red blood cells are biconcave discs.
(ii)
The haemoglobin of sickled red blood cells is insoluble, causing them to
absorb less oxygen. Less oxygen causes a reduced rate of respiration.
Less ATP is produced and hence, she feels tired easily.
(i)
It occurs when the body is exposed to an antigen, due to entry of a
pathogen (infection) or an allergen.
(ii)
- Phagocytes engulf and digest pathogens.
- B-lymphocytes produce
(iii)
The nucleus of a red blood cell degenerates during maturation. The cell
contains no DNA and is unable to produce protein or divide.
(i)
- The upper surface of a leaf is covered by a waterproof, thick, shiny
cuticle. This reduces the rate of evaporation of water from leaves and
reflects some of the light (and heat), so reducing the amount of heat
absorbed by the plant. This reduces the need for transpiration (water loss).
- Under conditions of low water supply, guard cells become flaccid and
close the stomata, reducing the rate of transpiration and so, water loss.
Alternative responses may refer to adaptations of xerophytes such as
sunken stomata, hairy leaves and reduced leaf surface area, etc.
(ii)
Autotrophic
(iii)
Chloroplast
50
2. (a)
(b)
(c)
(d)
(i)
vascular tissues (xylem)
(ii)
flower (stamen – anther, filament) and carpel (stigma, style, ovary)
(iii)
fruit (contains seed/fertilised ovule within ovary)
(iv)
root (region of root hairs)
(v)
leaves (any green part of plant body)
(i)
Method: animal
Reason: the fruit appears succulent and edible.
(ii)
The plants grown from the seeds are products of sexual reproduction.
 The gametes were produced by meiosis, which introduces variation
among them.
 Fertilisation of gametes cause the genotypes of the offspring to be
hybrids of both parents.
 The environment of the offspring may be different from that of the
parent plant and this may modify their phenotypes.
(i)
- Desirable parental traits will also be present in the offspring.
- Much larger numbers of offspring will be produced in a shorter time than
if the plant were reproducing sexually.
(ii)
- Resistance to disease
- Production of large number of fruits
Alternative responses include tolerance to unfavourable environmental
conditions, short growing period, etc.
Adverse side effects such as allergic reactions or development of cancer caused
by consumption of GM foods are as yet unknown, especially the long term
effects.
Alternative responses include alteration of properties of the original food such as
taste, etc.
51
3. (a)
(i), (ii), (iii)
DIAGRAM OF A PART OF THE HUMAN BODY
(b)
(c)
(i)
- Antidiuretic hormone: stimulates the distal convoluted tubule and
collecting duct to reabsorb water.
- Follicle stimulating hormone: stimulates development of follicles within
the ovary of females.
Alternative responses include growth hormone, thyroid stimulating
hormone, luteinising hormone, etc.
(ii)
- Reduced efficiency of osmoregulation, causing the water level of blood
to be outside of physiological limits.
- Irregularities of events of menstrual cycle such as ovulation.
Alternative responses include disruption of metabolic rate, etc.
(i)
AN EXPERIMENT INVOLVING A SHOOT
52
4. (a)
(b)
(ii)
Auxins increase the rate of growth in shoots. The shaded side contains
more auxins and grow faster than the illuminated side which contains less
auxins. The shoot therefore bends towards the light.
(iii)
Plants need light for photosynthesis. Growth of shoots towards light
exposes the leaves to light, allowing them to synthesise the organic
compounds they need.
(i)
Humans: alveoli of the lungs
Fish: lamellae of the gills
(ii)
- large surface area
- thin (one cell thick)
- in close proximity to an extensive blood supply
(i)
DIAGRAM ILLUSTRATING GASEOUS EXCHANGE ACROSS AN ALVEOLUS
(ii)
a - Diffusion
b - Plants photosynthesise during the day, using carbon dioxide and
producing oxygen. They respire during the day, using oxygen and
producing carbon dioxide. Some of the carbon dioxide produced in
respiration is retained and used in photosynthesis. Similarly, some of the
oxygen produced in photosynthesis is retained and used in respiration.
This reduces the amount of gaseous exchange necessary, since some of the
gas produced is used.
At night, plants respire but do not photosynthesise. During respiration,
they absorb oxygen from the atmosphere and release the carbon dioxide
53
produced, resulting in a larger amount of gases exchanged than during the
day.
(c)
- Cutting forests reduce the number of trees available for photosynthesis. This
decreases the rate of removal of carbon dioxide from the atmosphere. It therefore
accumulates in the atmosphere. Decreased photosynthesis decreases the rate of
production of oxygen and its release to the atmosphere.
- Burning forests greatly increase the rate of return of carbon (in the form of
carbon dioxide) from plants (and the burnt animals as well) to the atmosphere,
leading to its accumulation in the atmosphere. Carbon dioxide is a greenhouse
gas. Its accumulation leads to global warming, with consequences such as rising
sea level and disruptions in weather patterns.
- Cutting or burning forests remove the leaf litter from the surface of soil. The
roots of the removed trees decay after a while. Loss of roots and leaf litter
increases the rate of soil erosion. This leads to leaching of minerals from soil,
landslides and destruction of habitats for animals.
Alternative responses include a reduced food supply for animals, etc.
5. (a)
(i)
DIAGRAM OF THE STRUCTURE OF A MOTOR NEURON
54



(ii)
(b)
Both the cell body and the end of the axon terminate in dendrites,
which can form synapses with other neurones. This facilitates
communication among several neurones at the same time.
The axon is insulated at intervals by myelin sheaths. This causes
impulses to be generated only at the nodes of Ranvier, increasing
the rate of impulse transmission.
The axon is long. This reduces the number of neurones necessary
to transmit an impulse across long distances and so increases the
rate of impulse transmission. (There is a time delay associated with
transmission of an impulse from one neurone to another.)
The cerebrum is the centre for reasoning and association of experiences.
Memory is the result of association of past experiences with some other
stimulus. Alcohol destroys the cells of the cerebrum. Loss of these cells
results a lowered rate of transmission of impulses among the cerebral
cells. This results in gaps in the communication among nerve cells of the
cerebrum and the efficiency of association of experiences is diminished.
This is known as loss of “memory cells”.
Alcohol causes dehydration of nerve cells in the brain and in the long term,
destruction of their myelin sheaths. This reduces the rate of transmission of
impulses and therefore, it has the following depressant actions on the brain:
 It reduces the activity of the cerebrum. Loss of reasoning, memory and
cognitive skills occurs.
 It reduces the activity of the medulla. Regulation of breathing rate is
impaired and a loss of consciousness or death may result.
 It reduces the activity of the cerebellum. Loss of motor co-ordination,
balance and posture results.
Social consequences:
 The loss of inhibition, accompanied by loss of reasoning, may result in
risky behaviour. Violence, acts of vandalism and aggressive behaviour are
common.
 Impairment of motor co-ordination and judgment are often accompanied
by vehicular accidents. Victims may be paralysed or permanently affected
in some way, if not killed.
Economic consequences:
 Persons who misuse alcohol often neglect their general health. Ill health,
effects of hangovers or withdrawal symptoms result in frequent
absenteeism from their jobs. This lowers the productivity at the workplace.
 Money is spent on purchasing alcohol, treating illnesses associated with
alcoholism, fines for traffic offences, compensation to victims of violence
or vandalism, replacement of wrecked vehicles and other alcohol-related
expenses. This could significantly reduce the amount of money available
for other living expenses.
55
Alternative responses include the higher risk of extreme criminal activity,
addiction, birth defects, etc.
6. (a)
(i)
Pathogens may enter a person. They bind to complementary
B-lymphocytes and antibodies and memory cells are produced. The
pathogens are destroyed and the person is immune to that particular type
of pathogens thereafter (since the memory cells are a larger population of
B-lymphocytes than before the infection). If the pathogens entered
naturally, as in a natural infection, the person shows natural immunity. If
the pathogens (a weakened form) were introduced deliberately, as in a
vaccine, the person shows artificial immunity.
Alternatively, the antibodies themselves may enter a person. They destroy
any pathogens, present in the body, to which they are complementary. If
the antibodies are transmitted from mother to foetus/child via the placenta
or breast milk, the foetus/child shows natural immunity. If artificially
prepared antibodies are deliberately injected in the bloodstream of a
person, the person shows artificial immunity.
(ii)
- The deliberate introduction of a pathogen, even if weakened or
genetically modified, into the body, carries a small risk of the person
becoming a patient of the very disease he/she was trying to be protected
from.
- A person may develop an unpredictable allergic reaction to the vaccine
itself. Allergic reactions can be severe to the point of fatal.
Alternative responses include the cost of vaccines and development of
resistance by the pathogen, etc.
(b)
(iii)
- Reduce the chance of entry of the pathogen into the body by frequent
hand washing, avoidance of hand contact with eyes, nose and mouth
(entry points), avoidance of contact with infected persons and use of a face
mask.
- Ingestion of food supplements which are known to strengthen the
immune system, e.g. zinc and vitamin C.
(i)
Disease: dengue fever
Control: destruction of the vector (mosquito) of the pathogen
Alternative responses are any pathogenic disease, which is not airborne,
and a control method appropriate for it.
(ii)
Methods used to control spread of viruses among both plants and animals
involve:
 Isolation and/or destruction of infected plants/animals
 Selective breeding of varieties/breeds of plants/animals which are
resistant to the viruses
56
 Destruction of the vectors of the viruses
Alternative responses include frequent sanitation of fields and farms, etc.
57
1. (a)
Height / cm Number of men
157
0
165
500
167
1000
168
1500
169
2000
170
2500
171
3000
172
3500
174
4000
178
3500
179
3000
180
2500
181
2000
182
1500
183
1000
185
500
193
0
TABLE SHOWING HEIGHT OF A SAMPLE OF MEN IN A POPULATION
(b)
Causes
 Differences in genotype due to meiosis, mutation
 Environmental influences such as diet
(c)
Blood type shows discontinuous variation (small number of phenotypes,
phenotypes are unaffected by environmental effects, trait coded for by small
number of alleles and/or genes), while height shows continuous variation (infinite
number of phenotypes, phenotypes are affected by environmental effects, trait
coded for by large number of alleles and/or genes).
(d)
(i)
- Nutrients such as glucose
- Gases such as oxygen
58
(ii)
(e)
During exercise, respiring tissues such as muscles need more energy and
so have a higher rate of respiration, so they need more oxygen.
Haemoglobin transports oxygen to these tissues. Abnormally low
haemoglobin levels will not deliver sufficient oxygen to muscles. This will
cause a lowered respiration rate and energy availability. This will decrease
athletic performance. Muscles will respire anaerobically as well,
producing lactic acid, which causes fatigue and pain, further reducing
performance.
(i)
DIAGRAM OF APPARATUS USED TO INVESTIGATE THE EFFECT OF AIR
MOVEMENT ON RATE OF WATER UPTAKE
Annotations are:
1) Equal volumes of dyed water are measured into equally sized
beakers
2) Identical plants are placed in beakers with their cut ends / roots
submerged to the same depth beneath the water
3) A fan is used to blow air at the leaves of one of the plants
4) A stopwatch is used to allow measurements to be recorded at
regular time intervals
5) A ruler is used to measure the distance moved by the dyed water
up the stems
(ii)
Moving air such as wind increases the rate of water movement through
xylem vessels OR Moving air increases the rate of transpiration.
59
(f)
Xylem vessels:
 Narrow lumen – allows water flow by adhesion to walls and cohesion to
each other (capillarity)
 Lignified – resists collapse of vessels due to high tension of water in them
 Contain no cytoplasm or cross-walls – no hindrance to water flow
 Continuous throughout stems – allow flow of water throughout stems
2. (a)
A SECTION THROUGH THE HUMAN EYE
(b)
(i)
Seedlings in A: gravity or water
Seedlings in B: light
(ii)
Auxin
(iii)
A: roots will grow downward and gain access to more water, minerals and
will be better anchored in the soil.
B: shoots will grow upward and be exposed to more light for
photosynthesis.
(c)
Similarity: Both show a directional positive response (towards the stimulus)
Difference: Soil invertebrates quickly move their entire bodies (locomotion)
towards moisture but seedlings’ shoots grow slowly towards light.
(d)
Alternative comparisons to those given in the table are acceptable.
NERVOUS STSTEM
Impulses are generated along neurones
Response to impulses is very quick
ENDOCRINE SYSTEM
Hormones are secreted into blood
Response to hormones often occurs
over a period of time (more slowly)
60
3. (a)
(b)
- Oxygen
- Tannins
(i)
A: prostate gland
B: testis
C: Vas deferens/sperm duct
(ii)
THE HUMAN MALE REPRODUCTIVE SYSTEM
(c)
(i)
The vas deferens can be cut and tied. Sperm released from the testis will
not pass the cut and will not enter the semen. The ejaculated fluid will not
contain sperm. The sperm will not meet the egg.
(ii)
The oviducts are cut and tied. The ovulated egg will not pass the cut.
Sperm in the oviducts are unable to pass the cut (from the opposing side).
The sperm will not meet the egg.
(d)
Asexual reproduction (potato)
One parent needed
Only mitosis involved (for
development of offspring)
4. (a)
Sexual reproduction (human)
Two parents needed
Both meiosis (for gamete production)
and mitosis (for development of
offspring) involved
- The skin provides a mechanical barrier against the entry of pathogens.
- At the openings or lining the tubes connected to the openings of the body, there
are cilia or secretions of mucus, acid or enzyme, e.g. mucus is secreted in the
trachea and bronchioles, which traps pathogens and prevent their entry into the
body.
61
(b)
(i)
Vector – mosquito
Eggs are laid by adult females in stagnant water. These hatch into larvae,
which remain close to the water’s surface for gaseous exchange, using
breathing tubes. They undergo moulting several times during their growth.
This is the feeding stage, in which they feed on micro-organisms and
organic matter in water. They eventually form pupae (non-feeding stage)
just beneath the water’s surface, still breathing through tubes. Within the
pupal case, complete metamorphosis is completed (larval tissue is broken
down and re-organised into the adult form) and each one develops into a
fully formed adult, or imago. These adults emerge from the pupal cases
and fly away. They will eventually reproduce sexually, with the female
laying eggs in water.
(ii)
The disease malaria is caused by the pathogen Plasmodium, for which the
Aedes mosquito is the vector.
Knowledge of the habits or requirements of the mosquito at each stage in
its life cycle allows one to use an appropriate strategy to kill it. If the
mosquito is killed, the pathogen cannot be transmitted from one person to
another, since the vector (mosquito) is the means of transmission of the
pathogen.
Removal or draining of stagnant water will destroy the larval and pupal
stage. If this is not possible, the addition of insecticides to the water’s
surface will kill them. Alternatively, a layer of oil, kerosene or lecithins on
the water’s surface will prevent their breathing and kill them. Biological
control can be done by introducing fish which feed on the larval and pupal
stages.
Removal of dense vegetation near human populated areas homes deprive
the mosquitoes of a habitat and the juices of plants, on which they feed.
Spraying of vegetation with insecticides will kill the adult mosquitoes.
(c)
(i)
The vaccine contains the attenuated pathogen for yellow fever, which is or
contains antigens.
The antigens bind to those B-lymphocytes which are complementary to
them (they recognise the antigens).
These B-lymphocytes undergo rapid cell division, producing plasma cells
(which secrete antibodies to destroy the pathogens) and memory cells.
Memory cells are a long-lived, larger population (than before the vaccine)
of B-lymphocytes complementary to the antigens.
If the pathogens for yellow fever then enter the student’s body, the larger
population of B-lymphocytes (memory cells) are able to produce the
62
antibody-secreting cells more rapidly and in greater numbers (recognition
happens faster).
Antibodies are therefore produced so quickly and in such large numbers in
the secondary immune response that the yellow fever pathogens are killed
before the student shows signs and symptoms of the disease. The student
shows active immunity.
(ii)
5. (a)
This vaccine contains artificially made antibodies against the antigens.
This has no effect on the student since he/she contains no antigens (is not
infected). If the student is subsequently infected, the primary immune
response is done, in which antibody production is slow enough for the
student to suffer signs and symptoms of the disease.
A population is the collection of all the organisms of one species (organisms
capable of interbreeding and producing fertile offspring) in a particular habitat,
e.g. all the worms in the swamp.
Physical factors are the non-living components of an ecosystem which affect the
distribution of living organisms in that ecosystem, e.g. water content and solute
concentration of the soil (it is waterlogged and brackish).
A habitat is the part of an ecosystem within which a particular organism resides,
e.g. blue herons live in the mangrove trees.
(b)
Argument for plan:
More people will have access to housing. The population is growing and there not
much land left which is suitable for housing. It is necessary to modify land such
as swamp areas to provide homes for the growing population.
Arguments against plan:
 Removal of the swamp will destroy an entire ecosystem. Mangrove
ecosystems are specialised and rare. They contain organisms which may
not be adapted to other, more common, habitats. Removal of the swamp
(habitats of these organisms) may eliminate them entirely, reducing
biodiversity. Living organisms, water and soil are stages of nutrient
cycling. Removal or modification of these would disrupt the cycling of
nutrients.
 Removal of the swamp will cause a decline in the revenue of the country.
Fish are harvested from swamps for sale. If there are no breeding grounds
for the types of fish found in swamps, the fishing industry will decline.
The revenue gained by the swamp being a tourist attraction will also be
lost. Both the entrance fees paid by tourists and the employment of tour
guides, etc. will be gone.
63
(c)
6. (a)
- Cutting trees reduce the number of trees available for photosynthesis.
Photosynthesis removes carbon dioxide from the air. Lack of trees will cause
carbon dioxide to accumulate in the air. Decomposers use the tree bodies as
substrate for their respiration, which releases carbon dioxide into the air.
- Cutting trees reduce the amount of carbon entering the animal component of the
carbon cycle. Grazing by herbivores is the means by which carbon enters the
animal stage.
- Burning trees release carbon dioxide in the air, causing it to accumulate there.
Carbon dioxide is a product of combustion.
(i)
DIAGRAM OF OUTLINE OF THE HUMAN BODY SHOWING
LOCATION OF TWO GLANDS
(b)
(ii)
Pancreas: insulin or glucagon
Pituitary gland: follicle stimulating hormone (FSH)
(iii)
Factors:
 pH
 temperature
If the blood glucose level falls below the norm, this is detected by the pancreas,
which secretes glucagon and decreases/inhibits secretion of insulin. Glucagon
causes the liver to convert glycogen to glucose, which is released into the blood,
raising its glucose level until the norm is regained. Glucagon secretion is then
inhibited by negative feedback.
If the blood glucose level rises above the norm, this is detected by the pancreas,
which secretes insulin and decreases/inhibits secretion of glucagon. Insulin
causes the liver to convert glucose to insoluble glycogen (stored in liver and
muscle), so removing it from the blood and lowering its glucose level until the
norm is regained. Insulin secretion is then inhibited by negative feedback.
64
(c)
The water content of the blood is too low (below the norm). This is detected by
osmoreceptors in the hypothalamus.
The hypothalamus causes the pituitary gland to secrete antidiuretic hormone
(ADH).
ADH travels in the blood and binds to the cells of the distal convoluted tubule and
the collecting duct.
This causes them (distal tubule and collecting duct) to reabsorb water from the
filtrate and urine, respectively, into the blood.
The additional water reabsorbed raises the water content of the blood (preventing
complete dehydration) and a small volume of concentrated urine is produced (less
water is excreted).
65
1. (a)
(b)
(i)
Jar I: soda lime/caustic soda/NaOH/KOH
Jar II: lime water/Ca(OH)2
Jar III: small organism (green plant or animal)
Jar IV: lime water/Ca(OH)2
(ii)
The lime water changes from colourless to cloudy/milky.
(i)
Place a plant in Jar III and remove any animal which may be there. Cover
Jar III with a black cloth.
(ii)
It prevents entry of light so the plant does not photosynthesise. It therefore
prevents the plant from using the carbon dioxide it produces while
respiring.
(iii)
- The jars must be sterilised, to remove microorganisms which will be
respiring and contributing to carbon dioxide production (changing the
colour of lime water).
- The apparatus must be airtight, to prevent carbon dioxide from the
atmosphere from changing the colour of the lime water.
Alternative responses include maintenance of an appropriate temperature,
ensuring that the gas tubing is below the level of the limewater, etc.
(c)
(d)
(i)
C6 H12O6 
 6CO2  6H2O  32 ATP
(ii)
It converts the energy of glucose (unavailable to cells) to the energy of
ATP (available to cells), enabling cells to perform biological work, e.g.
production of nerve impulses and growth.
Lactic acid, ATP, ethanol, carbon dioxide
66
(e)
Photosynthesis
Chloroplast
Organelles
Is energy produced or used? Produced
Water
Raw materials used
Carbon dioxide
Glucose
Final products
Oxygen
2. (a)
(b)
Respiration
Mitochondrion
Produced
Glucose
Oxygen
Water
Carbon dioxide
ATP
(i)
A: epidermis
B: blood capillary
C: hair follicle
D: sweat gland
(ii)
Malpighian layer
(i)
a)
An allele is one of the alternative forms of a gene.
b)
A genotype is the allele combination which an organism possesses
for the trait under consideration.
c)
The phenotype is the physical expression of the genotype,
sometimes modified by the influences of the environment.
(ii)
Let A represent the allele for normal pigmentation
Let a represent the allele for albinism
Parental phenotype
father-normal X
Parental genotype
Aa
Gametes
mother-normal
Aa
(A) and (a)
(A) and (a)
Random fertilisation
(A) (a)
(A) AA Aa
(a) Aa aa
Offspring genotype
AA
Offspring phenotype
Normal
Offspring phenotypic ratio
Aa
Aa
Normal
3 normal
Normal
:
aa
Albino
1 albino
67
There is a 25% probability that an albino child can be produced.
(iii)
3. (a)
AA X AA or AA X Aa or AA X aa
(i)
DIAGRAM OF TWO FLOWERS FROM DIFFERENT PLANTS
(ii)
X: animal/insect
Y: wind
(iii)
X: - large, conspicuous petals for attraction of insects/animals
- both stigmas and anthers located within flower
Alternative responses are any feature of animal pollinated flowers which
are visible in Figure 3.
Y:- anthers hang outside of the flower, to allow the pollen grains to be
easily blown away
- large feathery stigma for trapping wind-blown pollen
Alternative responses are any feature of wind pollinated flowers which are
visible in Figure 3.
(b)
After pollen is deposited on the stigma,
1) The pollen grain germinates into a pollen tube.
2) One male nucleus divides by mitosis into two male nuclei/gametes, which
enter the tube.
3) The tube grows into the style towards the ovary, digesting style tissue.
4) The tube enters the ovary and micropyle, then bursts, releasing the male
nuclei.
5) One male nucleus fuses with the female gamete/nucleus/ovum and the
other with the endosperm nucleus.
The ovary is now a fruit and the ovule is a seed.
68
(c)
4. (a)
- Animal dispersal
- Self dispersal
(i)
DRAWING OF A GENERALISED PLANT CELL
DRAWING OF A GENERALISED ANIMAL CELL
(ii)
- The chloroplast traps light and is the site of photosynthesis (conversion
of light energy to chemical energy by production of glucose). Glucose is
used for respiration or for synthesis of other necessary compounds.
- The large central vacuole is used for storage of a variety of materials,
including pigments (in petals – aiding reproduction), water (maintenance
of turgidity of cells) or waste materials.
Alternative responses include starch grain, etc.
69
(b)
(i)
The cytoplasm of both cells contains a lower water concentration than that
of the container of distilled water. Water moves down its concentration
gradient and enters both cells by osmosis. This causes a small expansion
of their protoplasm (swelling).
The expansion causes the cell membrane of the animal cell to rupture.
The tough, inelastic cell walls resists and limits the degree of expansion of
the plant cell, so it gets turgid, does not rupture and there is no further net
intake of water.
(ii)
Feature of process
Name
5. (a)
Water movement Gas movement
into cells
into and out of
cells
Osmosis
Diffusion
What particles are being
transported?
Water
Gases
Are the particles
transported across a semipermeable membrane (cell
membrane)?
Yes
No
(i)
SIMPLIFIED DIAGRAM OF THE NITROGEN CYCLE
70
(b)
(ii)
- Nitrogen is a major component of chlorophyll. A nitrogen deficiency
results in production of less or no chlorophyll. The yellow leaves do less
or no photosynthesis. Very little or no glucose is made, which is needed
by the plant as a respiratory substrate. The plant therefore lacks energy to
execute its metabolic processes such as growth.
- Nitrogen (as nitrates) is absorbed by plants and used in protein synthesis.
These proteins may be used to form structural compounds and enzymes.
A deficiency in nitrogen, and therefore, of protein, results in stunted
growth.
(i)
It causes eutrophication:
1) Nitrate (large amounts) from fertilisers leach into the river.
2) The rate of algal growth in the river increases such that the surface
of the pond is covered with algae.
3) Light is unable to penetrate the surface of the water because of the
cover of algae.
4) Aquatic plants beneath the surface of the water die.
5) Decomposers use the dead plants as substrates for their respiration,
multiplying rapidly and greatly decreasing the oxygen content of
the water.
6) Aquatic animals die because of the lack of oxygen.
(Death of plants and animals may also result if the fertilisers have
other effects such as change in pH of the water.)
(ii)
- Use of less inorganic fertiliser, coupled with use of more organic
fertiliser. Since organic fertiliser decomposes slowly, the rate of release of
soluble ions is slow and it is therefore unlikely that large amounts will be
leached from the soil.
- Frequent removal of excess algae from the river will allow light to
penetrate the water.
- Education of farmers about the concepts of leaching and eutrophication.
Alternative responses include offering incentives to farmers for
implementation of farming practices which do not threaten the stability of
ecosystems, such as planting crops downstream of rivers, etc.
71
6. (a)
DIAGRAM OF A CANINE TOOTH
(b)
- It reduces the probability of one suffering from deficiency diseases, since a
range of nutrients is likely to be present in a varied diet (it is more likely to be a
balanced diet).
- It increases the probability of one’s consumption of non-nutrients which are
necessary for health, e.g. fibre and antioxidants decrease the risk of constipation
and cancer, respectively.
- It increases the probability of better management of body weight and conditions
for which one may have a genetic predisposition, e.g. diabetes and cancer. A
varied diet is more likely to include carbohydrates which are digested slowly.
(c)
He is unable to bite tough food or cut food into smaller pieces. This limits the
range of his intake of food and possibly, of nutrients.
He is unable to mechanically digest his food by chewing. His food will therefore
not be broken into small pieces. Enzymes may not have adequate surface area to
efficiently hydrolyse food (chemical digestion), so his food may be poorly
digested.
He may still obtain nutrients by:
 Consumption of food supplements or food that require little or no
digestion, e.g. glucose and amino acids.
 Consumption of food that require little or no mechanical digestion, e.g.
soups and blended meals.
 Use of artificially made teeth/dentures to allow him to mechanically digest
food.
72
1. (a)
(i)
Feature
Fishes with feature Fishes without feature
Scales
IV, V, VI, VII
I,II,III,VIII
Spines
III
I,II,IV,V,VI,VII,VIII
CLASSIFICATION OF FISH BASED ON PRESENCE OF SCALES
AND SPINES
Other features are acceptable, provided they result in a clear distinction
between the groups, e.g. Number of fins, etc.
(b)
(c)
(ii)
The lamellae have thin walls, reducing the distance of diffusion of gases
and so increasing their rate of diffusion. There are numerous small
lamellae, which provide a large surface area for diffusion, increasing its
rate.
(iii)
It provides the organism with oxygen for aerobic respiration. It removes
carbon dioxide (produced in respiration) from the organism. It ensures
that the levels of these gases are within physiological limits (homeostasis).
(i)
Two of Sea grass, Phytoplankton or Corals with zooxanthellae
(ii)
They are photosynthetic or chemosynthetic (use energy of light or
inorganic simple molecules to synthesise complex organic compounds).
They are the means of entry of energy into the organisms of the food web.
(iii)
Mutualism, both partners benefit: corals get oxygen and nutrients and
zooxanthellae get carbon dioxide, shelter and protection.
(i)



Write a self- explanatory title. You can use the same title as that of
the table, that is: GRAPH OF ESTIMATED NUMBER OF
PARROTFISH AND PREDATORY FISH OVER THREE
YEARS
Use a key to identify each line in the graph OR write a label on
each line, e.g. Parrotfish
Use a scale on each axis such that maximum use is made of the
graph paper
73


Label each axis – Time /month on the x-axis and Number of fish
on the y-axis
Plot points accurately
GRAPH OF ESTIMATED NUMBER OF PARROT FISH AND PREDATORY
FISH OVER THREE YEARS
74
(ii)
The numbers of predatory fish were lower since they are higher up the
food chain. Over time, their numbers increased since there were no / few
of their predators present.
Other valid points are acceptable such as parrotfish has more predators
feeding on it and temperature/pH of the water is very suitable for the
predators.
(iii)
Strong jaws and Ability to swim quickly. Other valid points are
acceptable e.g. excellent eyesight and ability to blend in with its
surroundings.
(iv)
The population of parrotfish will decline to very low numbers or they may
be eliminated from that ecosystem. The predatory fish will then eat its
other food sources such as corals and jellyfish in greater numbers,
resulting in their decline or elimination. Biodiversity will be reduced if
they are eliminated.
Other valid points are acceptable e.g. decline in economy due to poor
fishing industry and less tourism due to decrease in marine life, etc.
2. (a)
(b)
(c)
(i)
A - Vena cava
B - Aorta
C - Ventricle
(ii)
Blockage will reduce blood flow (and so, its oxygen and nutrient supply)
to cardiac muscle. Lack of oxygen and glucose will decrease its rate of
respiration. It will not have sufficient energy to pump blood to arteries.
Cardiac arrest may result.
(i)
B is a vein.
(ii)
Valve
(iii)
Valve flaps open in the direction towards the heart only. This allows blood
to flow in that direction only. Backflow of blood is prevented by closure
of the valve if blood starts to flow away from the heart.
(iv)
Plaques decrease the diameter, and so the volume, of arteries. This
increases the blood pressure within the arteries. Since plaques remain in
the vessels for a long time, chronic high blood pressure (hypertension)
results.
P – xylem
Q – sieve tube element of phloem
75
(d)
Explanation – Since plants’ metabolic rate is lower, they do not need substances
delivered to cells at a rate as high as the animals’.
Mechanism – Loading of sieve tubes generates a high pressure in the tubes at the
source. The pressure gradient in the tubes from source to sink results in mass
flow of materials along the tubes in this direction.
3. (a)
(i)
A – Bowman’s capsule
Ultrafiltration: Filtration of some blood plasma (water, glucose, amino
acids, urea, salts) from glomerular capillaries into the lumen of the
nephron
B – Proximal convoluted tubule
Selective reabsorption: Transfer of certain useful substances e.g. glucose,
from the glomerular filtrate into the peritubular capillaries by osmosis,
diffusion and active transport
C – Loop of Henle
Reabsorption of water: from the loop to the blood capillaries
(b)
4. (a)
(ii)
Antidiuretic hormone
(iii)
Osmoreceptors in the hypothalamus detect the low water concentration in
the blood of the person who drank no water, causing the pituitary gland to
secrete ADH, so the distal tubule and collecting duct reabsorb more water,
producing urine containing a higher water: solute ratio (concentrated).
The other person’s blood has a lot of water, so little water is reabsorbed by
the nephron, causing the urine to have a lower water: solute ratio
(concentrated).
(i)
Since insulin is not produced or is not used, the glucose level of blood, and
so, of glomerular filtrate, is excessively high, so the proximal tubule is
unable to reabsorb all, hence, some is excreted in urine.
(ii)
Diet should be low in carbohydrate, especially disaccharides and
monosaccharides.
Blood glucose level should be measured regularly, followed by insulin
injection if necessary.
Draw a diagram of the carbon cycle. Ensure that it includes:
 The stages – Plants, Animals, Soil/Water and the Atmosphere containing
Carbon Dioxide
 At least one arrow leading to and from each stage
 Two additional arrows leading away from the Atmosphere containing
Carbon Dioxide
76

The name of the process on each arrow
DIAGRAM OF PART OF THE CARBON CYCLE
In the writing space, outline the processes:
Photosynthesis removes atmospheric carbon dioxide and incorporates it into the
organic molecules of plants.
Decomposition of dead bodies of plants returns carbon to the atmosphere (as
carbon dioxide) from the plants.
Respiration by animals returns carbon to the atmosphere (as carbon dioxide) from
the animals.
Combustion of vegetation returns carbon to the atmosphere (as carbon dioxide)
from the plants.
77
(b)
Cutting down forests reduce the number of trees available for photosynthesis.
This decreases the rate of removal of carbon dioxide from the atmosphere. It
therefore accumulates in the atmosphere.
Cutting down forests decreases the number of trees. It decreases the number of
habitats available for animals and so decreases the number of animals. Less trees
and animals are present to do respiration, reducing the rate of return of carbon
from plants and animals to the atmosphere. In the short term, though,
decomposition of the bodies of the killed plants and animals increases the rate of
return of carbon (as carbon dioxide) to the atmosphere.
Burning forests greatly increases the rate of return of carbon from the plants (and
the burnt animals as well) to the atmosphere, leading to its accumulation there.
(c)
5. (a)
(b)
People can reduce carbon emissions from vehicles by car-pooling, walking or
using bicycles whenever possible. They can reduce combustion of fossil fuels by
reducing electricity use, e.g. turning off light bulbs and air-conditioning units
when not needed. They can increase the rate of removal of atmospheric carbon
dioxide by planting trees where possible.
There are other /alternative correct responses such as using alternative forms of
energy e.g. solar systems, etc.
The genotype is the combination of alleles an organism possesses for a particular
trait.
The phenotype is the physical expression of a genotype, modified by
environmental influences.
“Recessive” is a description of an allele which is expressed in the phenotype only
in the absence of dominant alleles in the genotype of an organism.
“Dominant” is a description of an allele which is expressed in the phenotype once
present in the genotype, regardless of the presence of other (recessive) alleles.
(i)
Intelligence and Body weight
(ii)
Let A represent the allele for normal pigmentation
Let a represent the allele for albino
Parental phenotype
Parental genotype
Gametes
father - normal
X
mother- albino
Aa
aa
(A) and (a)
(a)
78
Random fertilisation
(A) (a)
(a) Aa aa
(c)
6. (a)
Offspring genotype
Aa
aa
Offspring phenotype
Normal
Albino
Offspring phenotypic ratio
1 normal
:
1 albino
Variation exists among the bacteria that cause tuberculosis (TB), with some
having a genotype which allows them to be resistant to antibiotics against TB.
The use of antibiotics (selection pressure) killed the non-resistant bacteria before
they could reproduce and transmit their non-resistant genotype to subsequent
generations (they had a selective disadvantage). The use of antibiotics did not kill
the resistant bacteria. These survived long enough to reproduce and transmit their
resistant genes to subsequent generations (they had a selective advantage).
Subsequent generations of bacteria therefore contained a progressively higher
proportion of resistant ones, until the majority of the TB bacterial population were
resistant to antibiotics.
(i)
Tissue types: Skeletal muscle, Bones and Connective tissue (tendons and
ligaments).
Reasons:
 To obtain food and other requirements, e.g. water
 To find mates for reproduction
 To avoid unfavourable environments, e.g. inappropriate moisture
level, presence of predators
(ii)
Plants do not need move around to obtain their food since they
manufacture their food by photosynthesis. The raw materials needed are
available in their immediate environment. Movement is unnecessary for
reproduction since agents of pollination (wind, animals, etc.) facilitate the
transfer of gametes. Agents of dispersal (wind, water, etc.) facilitate
dispersal of their offspring within seeds.
79
(b)
When the father first saw his son, he was far away. Light rays from the son,
entering the father’s eyes, were almost parallel and needed little refraction. To
facilitate this, his ciliary muscles were relaxed, suspensory ligaments pulled taut,
lens stretched thin and less convex.
As he watched his son running, his son was a progressively nearer object. Light
rays reflected off the son were progressively more diverging and needed
progressively more refraction. To facilitate this, the ciliary muscles in the father’s
eyes were increasing their contraction, progressively slackening the tension in the
suspensory ligaments and allowing the lens to assume a progressively more
convex shape.
Changes to the shape of the lens (accommodation) ensured that the degree of
refraction done was exactly the amount required to focus the light rays (which
entered the pupil) on the retina.
80
1. (a)
(i)
A: asexual
B: sexual
(ii)
1) The pollen grain germinates, growing a pollen tube within the style in the
direction of the ovary.
2) One male nucleus divides by mitosis, producing two male nuclei within
the tube.
3) The tube grows until it enters the micropyle, delivering the male gametes
within the ovule.
4) Double fertilisation occurs. One male gamete fuses with the female
nucleus, producing the zygote. The other fuses with the endosperm
nucleus, forming the endosperm. The ovule is now a seed, containing the
developing embryo. The ovary is now a fruit.
(b)
Feature
Mitosis
Meiosis
Number of daughter cells
2
4
produced
No of chromosomes in The same number as in Half the number of
each daughter cell
the parent cell (diploid)
chromosomes of the
parent cell (haploid)
Alternative responses may refer to the genetic variation in the daughter cells of
meiosis and the genetic stability of mitosis.
(c)
The allele for blue flower is dominant to that for white flower, since blueflowered plants are more abundant. The white-flowered plants inherited their
alleles from their blue-flowered parents. The parents therefore contained the allele
for white flowers but its expression was suppressed (it is recessive).
The parents are heterozygous. If any of them were homozygous, none of the
offspring would be white.
81
Diagram:
Let B represent the allele for blue flowers
Let b represent the allele for white flowers
Parental phenotype
blue
Parental genotype
Bb
Gametes
X
blue
Bb
(B) and (b)
(B) and (b)
Random fertilisation
(B) (b)
(B) BB Bb
(b) Bb bb
Offspring genotype
BB
Bb
Bb
bb
Offspring phenotype
Blue
Blue
Blue
White
3 blue
:
Offspring phenotypic ratio
1 white
74 blue-flowered plants and 26 white-flowered plants are in the ratio 3 : 1
(d)
(i)
Days after
Length of
germination Radicle (cm)
1
0.4
2
1.2
3
1.4
4
1.6
5
1.9
6
2.3
7
2.9
8
3.4
82
(ii)
GRAPH SHOWING LENGTH OF RADICLE AFTER GERMINATION
(iii)
- Oxygen
- A suitable temperature
(iv)
1) Heat some loam soil at 110 0C until all the water is removed (heat to a
constant weight).
2) Obtain two seed boxes and place the dry loam soil in them. Ensure that
both boxes are as identical as possible with respect to material of boxes,
dimensions of boxes, type of soil, volume of soil, location of boxes
(exposed to same temperature, oxygen level, light intensity, wind speed,
etc.) and as many other environmental factors as possible.
3) Obtain 40 healthy seeds which are as similar as possible (same age,
species, variety, from same plant, size, etc.).
4) Plant 20 seeds in each box at a depth of 1 cm below the surface of the soil.
5) Sprinkle the soil in one box with water until the soil is moist each day
(once per day) for a period of 10 days or until the seeds are germinated
from at least one box.
83
If the seeds of the box which is watered germinates but the seeds of the
dry box do not, the hypothesis is not rejected.
(e)
2. (a)
Cotyledons get smaller as their food stores are hydrolysed to provide nutrients for
the developing embryo and growth of the primary leaves (before the primary
leaves are developed enough to photosynthesise and manufacture nutrients for the
plant).
(i)
I: liver
II: stomach
III: pancreas
IV: large intestine
V: small intestine
(ii)
DIAGRAM OF THE HUMAN ALIMENTARY CANAL
(iii)
- They contain protease enzymes which are specific for hydrolysis of
protein substrates.
- Their pH levels are appropriate for optimal action of the particular
proteases they contain (stomach is acidic for action of pepsin and the
duodenum is alkaline for the action of trypsin).
(iv)
- It is used in synthesis of structures such as cell membranes and
organelles during cell division (as occurs during growth and repair of
tissue.)
- It is used in synthesis of functional compounds such as enzymes,
hormones and antibodies.
84
(b)
(i)
1) Ammonification - Urea is converted to ammonium ions by
putrefying bacteria.
2) Nitrification – Ammonium ions are oxidised to nitrite ions by
nitrifying bacteria, e.g. Nitrosomonas
3) Nitrification – Nitrite ions are oxidised to nitrate ions by nitrifying
bacteria, e.g. Nitrobacter
Nitrate ions are easily absorbed by, and are therefore available to,
plants.
(ii)
3. (a)
(b)
- Yellow or pale leaves
- Stunted growth
1 – Pathogenic
2 – Hereditary
3 – Microorganisms
4 – Damaged organs
5 – Active
- Regulation of diet such that there is high intake of fibre, water, protein, vitamins
and minerals but low intake of fat, excess salt and refined sugar
- Regular exercise (to improve blood circulation and delivery of oxygen and other
requirements to cells)
- Abstinence from drugs which have adverse effects on internal organs, such as
alcohol and nicotine
Alternative responses include the use of medication to alleviate signs and
symptoms, regular physical examinations, education of patients about the nature
of the diseases, etc.
(c)
(i)
Hereditary/genetic
(ii)
- Crops are genetically engineered to improve their nutritional value and
so reduce the incidence of deficiency diseases.
- Micro-organisms are genetically engineered to produce drugs which are
useful in the treatment of disease, e.g. insulin production.
Alternative responses include reference to genetically engineered
organisms to improve their resistance to disease and infection by
pathogens, genetic modification of pathogens for vaccine production,
genetic modification of stem stems which can later be used in organ
transplants, etc.
(d)
(i)
Vector: Aedes aegypti mosquito
Diseases: yellow fever, dengue fever
85
(ii)
- Introduce a species of fish or other animal into ponds, streams and other
waterways which will consume the eggs, larvae or pupae of the mosquito.
- Spray the habitats of the mosquitoes regularly with insecticide.
Alternative responses include removal of breeding grounds of mosquitoes,
coating the surface of stagnant water with oil or insecticide, etc.
4. (a)
P: Palisade mesophyll
Role: It contains chlorophyll within numerous chloroplasts to trap light. Carbon
dioxide is combined with hydrogen to form glucose.
Q: Xylem vessels
Role: It transports water (source of hydrogen for reduction of carbon dioxide) to
mesophyll cells.
R: Phloem sieve tubes and companion cells
Role: It transports sucrose (made from glucose) to sinks.
(b)
B has a waxy cuticle which is thicker than that of A. The waxy cuticle of B is
waterproof and prevents evaporation of water, so reducing water loss from B. It
probably is shiny and reflects light (and heat) and so reduces the need for cooling
(evaporation of water).
The stomata of B are sunken into pits, into which epidermal hairs grow. The pits
and hairs trap transpired water vapour adjacent to the stomata, reducing the water
concentration gradient (between the intercellular spaces and the stomata) and
thus, the rate of transpiration.
(c)
Plant A: adequate water supply, possibly a cool/shaded area
Plant B: inadequate water supply (hot and /or dry conditions)
Adaptations: - extensive root system for reaching water in remote/deep areas of
soil
- leaves reduced to spines (reduced surface area for transpiration)
- rolled leaves (trap layer of moisture adjacent to stomata, reducing
transpiration rate)
Alternative responses include storage of water in succulent stems or other organs.
86
5. (a)
DIAGRAM OF THE INTERNAL STRUCTURE OF THE SKIN
(b)
(i)
Term: homeostasis
During the day:
 Sweat is produced. It absorbs heat and evaporates from the surface
of his skin, taking some of the heat from his body.
 Vasodilation of capillaries close to the surface of his skin occurs.
This brings blood (containing heat) closer to the skin, for faster
heat loss (by radiation) from the body.
 Hair erector muscles relax. This causes hairs to lie flat against his
skin. When heat radiates from his body into the atmosphere, the
warmed air of the atmosphere (next to his skin) diffuses away
quickly, taking the heat with it. Flat hairs cause no resistance to the
flow of air. This response is of homeostatic significance in hairier/
furrier animals than humans.
During the night:
 Sweat is not produced. There is no absorption of heat by water.
No evaporation of water occurs. The heat of his body is conserved.
 Vasoconstriction of capillaries close to the surface of his skin
occurs. Vasodilation of capillaries further away from his skin
occurs. Less blood flows close to the skin, reducing radiation from
the body. Heat is retained in the blood flowing further from the
skin.
 Hair erector muscles contract. This causes hairs to stand upright.
When heat radiates from his body into the atmosphere, the warmed
air of the atmosphere (next to his skin) is trapped by the upright
hairs adjacent to the skin. This reduces the temperature gradient
from his body to the atmosphere, reducing the rate of heat loss.
This response is of homeostatic significance in hairier/furrier
animals than humans.
87
(ii)
Day:


Avoid exposure to the sun by staying in shaded areas, wearing a
hat and wearing light coloured clothes which will reflect light (and
heat) rather than absorb it.
Reduce activity during the day. A reduced metabolic rate is
associated with a reduced rate of respiration (less heat is produced
by the body).
Alternative responses include wearing of clothing of a material which
allows evaporated water to be lost to the atmosphere, intake of sufficient
water so profuse sweating is possible, etc.
Night:
 Wear warm clothing (heat insulating material). Heat lost from his
body will be prevented from being lost to the atmosphere by the
insulating material.
Alternative responses include sleeping inside a small tent of insulating
material, increased activity level, etc.
6. (a)
Organism
Natural habitat
Cactus
Desert/Rocky area
King fish
Ocean / Sea
Crayfish
Pond
Red mangrove Mangrove swamp
(b)
- Increased competition for light: Overlapping leaves of adjacent plants may
reduce the amount of sunlight reaching lower leaves, resulting in less
photosynthesis.
- Increased competition for water: Roots of adjacent plants compete for a given
amount of water, such that less is available for each plant to do photosynthesis.
- Increased competition for fertiliser: Roots of adjacent plants compete for a
given amount of mineral ions, such that less is available for each plant to do
synthesis of tissue.
Alternative responses include competition for space and the increased likelihood
of spread of pathogens among plants.
(c)
- Pesticides, herbicides, soap and other toxic chemicals which may affect the pH
and other properties of the water of the mangrove swamp. This may adversely
affect the oyster population or eliminate it altogether.
- Harvesting of oysters for consumption or pearl-hunting at a faster rate than they
reproduce and grow will deplete their population. Oysters may be totally
eliminated from the swamp.
88
- Removal of mangrove trees will destroy the habitats of oysters. They breed, feed
and live among the roots of the mangrove trees. Destruction of mangrove trees
will reduce or eliminate the oyster population.
89
1. (a)
(i)
Sampling methods:
 A quadrat is thrown randomly and all plants of the species of
interest which are found within it are counted / recorded. This is
done multiple times and the data is used to estimate the
density/species cover/frequency of the species.
 A line transect is pegged along the ground/soil in a straight line
and all plant species touching it are recorded/counted. This shows
how the abundance of the species of interest varies across the area.
(ii)
A pitfall trap (bottle containing some bleach or other chemical toxic to
small creatures) is placed in a hole in the soil, with the top of the bottle at
soil level. It is covered loosely enough to allow walking animals to fall in.
The animals are later removed and recorded.
There are alternative correct responses, e.g. use of photography, nets and
tracking devices, etc.
(b)
(i)
FOOD WEB 1:Food web of organisms in the field study
(ii)
PYRAMID 1: Pyramid of numbers to show the feeding relationships among the
organisms living in the area
90
(c)
(i)
To estimate the water-holding capacity:
1) Place a dry soil sample in a funnel lined with filter paper.
2) Add a known volume of water to the soil. The volume should be
large enough to saturate the entire soil sample.
3) When no more filtrate is dripping into the measuring cylinder, note
the volume of the filtrate.
4) Subtract the volume of the filtrate from the volume of the water
poured into the soil to obtain the water-holding capacity of the soil.
(ii)
Soil:




Provides a medium for anchorage of roots for plant support
Is a habitat for organisms
Contains air, providing oxygen for respiration of living organisms
Contains mineral ions and water for uptake by plants
(d)
Light bulb: Invertebrates move away from the bulb’s light and heat and fall into
the alcohol.
Alcohol: Kills (and so prevents escape) and preserves the collected specimens.
(e)
Forest soil contains dead organic matter (shed structures and dead bodies of plants
and animals.) Fungi and bacteria are decomposers – they convert the complex,
insoluble nutrients in dead organic matter to simple, soluble, inorganic ions in soil
and carbon dioxide in the air. These ions and carbon dioxide are absorbed by
plants, so they re-enter food chains. When animals eat these plants, they gain the
nutrients.
2. (a)
(i)
Light and a suitable temperature (or presence of chlorophyll)
(ii)
Gas: oxygen
Reactant: water
(iii)
Glucose is used as the respiratory substrate by plant cells to produce
energy. This energy is used to do work such as protein synthesis and
growth. Glucose is converted to cellulose and used to form cell walls.
There are other /alternative correct responses e.g. it is stored in the form
of starch and stored in storage organs, etc.
(iv)



An appropriate temperature for optimum enzyme function.
Carbon dioxide to provide carbon and oxygen for glucose
formation
Water to provide hydrogen for glucose formation
There are other /alternative correct responses e.g. humidity and
light intensity, etc.
91
(b)
3. (a)
Salivary amylase is secreted by the salivary glands into the mouth, where it
hydrolyses starch to maltose. Pancreatic amylase is secreted by the pancreas into
the duodenum, where it hydrolyses starch to maltose. Maltase is secreted by the
walls of the ileum into the lumen of the ileum, where it hydrolyses maltose to
glucose.
(i)
Label the ovaries.
(ii)
Label a point within the distal (closer to the ovary) third of the oviduct.
(iii)
Label the endometrium.
STRUCTURE OF THE HUMAN FEMALE REPRODUCTIVE SYSTEM
(b)
Sperms (male gametes) are ejaculated out of the sperm duct, then urethra, into the
vagina. Contractions of the uterus and oviducts aid the sperms to swim up the
uterus and both oviducts. An ovulated ovum (female gamete) leaves the ovary and
moves down its oviduct. Sperms reach the ovum and one enters it. The nuclei of
the ovum and sperm fuse, forming the zygote.
(c)
(i)
Meiosis
(ii)
Meiosis
Mitosis
Daughter cells contain half the Daughter cells contain the same
number of chromosomes as that of number of chromosomes as that of
their parent cell
their parent cell
Daughter cells are genetically Daughter cells are genetically
different from their parent cell, due identical to their parent cell, no
to crossing over
crossing over occurs
Alternative comparisons are acceptable.
(d)
(i)


A specialised cell does its function more efficiently than an
unspecialised cell.
Cell specialisation allows tissue and organ formation, with each
having different functions. This division of labour enables
physiological processes to occur more efficiently, as well as
relatively independently of each other.
92
(ii)
4. (a)
They have a high rate of cell division. If they are harvested, the embryo
can quickly replace them. They can differentiate into any type of
specialised cell/tissue. Physiological diseases are caused by malfunction of
tissues. These malfunctioning tissues can be replaced by functional ones
formed from stem cells which were induced to develop into the tissues.



(b)
(i)
A gene is a section/length of DNA which codes for the production of a
polypeptide. An allele is one of the alternative forms of as gene.
“Dominant” is used to describe an allele which is expressed in the
phenotype once it is present in the genotype, regardless of the presence of
other (recessive) alleles in the genotype. “Recessive” is used to describe
an allele which is expressed in the phenotype only in the absence of
dominant alleles in the genotype.
“Homozygous” is used to describe an organism which possesses two
identical alleles for a particular gene. “Heterozygous” is used to describe
an organism which possesses two different alleles for a particular gene.
Parental phenotype
father - normal
Parental genotype
Gametes
X
mother-normal
Aa
Aa
(A) and (a)
(A) and (a)
Random fertilisation
(A)
(a)
Offspring genotype
Offspring phenotype
Offspring phenotypic ratio
AA
normal
(A)
AA
Aa
(a)
Aa
aa
Aa
normal
3 normal
Aa
normal
:
aa
albino
1 albino
There is a 25% chance of producing a child with genotype “aa” (albino.)
(ii)


Wear sunglasses to protect against retinal damage.
Use sunscreen to protect against skin damage.
There are other /alternative correct responses e.g. wear white to
reflect the sunlight, etc.
93
(c)



5. (a)
Inheritance of different alleles for skin colour (which show continuous
variation), shown especially if the parents are of different races
Mutation of genes for skin colour or other type of skin disease
Different levels of exposure environmental effects which affect colour,
e.g. sunlight, bleaching chemicals
In your diagram,
 Ensure that the parts are in proportion
 Use thin clear continuous lines
 Include labels for: cornea, iris, pupil, lens, suspensory ligaments, ciliary
muscle, ciliary body, retina, choroid, sclera, optic nerve, fovea
 Write a title under the drawing
HORIZONTAL SECTION OF THE HUMAN EYE
Vision is achieved by:
1) Light rays from an object fall on the cornea, which refracts them through
the pupil and onto the lens.
2) The lens refracts the light rays such that they converge to a point (focus)
on the retina.
3) The rods and cones of the retina detect the light.
4) Impulses are generated in the sensory optic nerve and are transmitted to
the brain.
5) The brain interprets the impulses as an image of the object.
(b)
Ability to respond to stimuli:
 Alcohol is a depressant. It decreases the efficiency of information
processing in the brain by demyelination and dehydration of nerve cells in
the brain. Both of these cause eventual death of these nerve cells. In the
short term, they reduce the speed of transmission of impulses, increasing
one’s reaction time.
94




The depressant effect of alcohol reduces the accuracy of
perception/interpretation of impulses from sensory neurones. This reduces
one’s ability to detect stimuli, e.g. blurred vision.
The depressant effect of alcohol reduces the efficiency of transmission of
impulses in motor neurones. This reduces one’s ability to respond to
stimuli, e.g. slurred speech.
Alcohol depresses the action of neurones in the cerebellum. This causes
loss of balance and muscular co-ordination, reducing one’s ability to
respond to stimuli.
Alcohol depresses the action of neurones in the higher centres of the
cerebrum (those areas involved in judgement and co-ordinated responses).
This includes the neurones involved in the inhibition of natural drives.
This may reduce the suitability of one’s response to a given stimulus.
Ability to maintain homeostasis:
 Homeostasis is the maintenance of the conditions of the internal
environment within physiological limits. This is done partly by removal of
excess materials taken in by the body. Removal of excess alcohol is done
by the liver. Chronic excessive alcohol intake puts a strain on the liver,
damaging it. Liver inflammation, fatty liver, liver cancer or eventual
cirrhosis may result. These render the liver unable to perform its other
homeostatic functions efficiently, such as regulation of levels of vitamins.
 Alcohol inhibits the secretion of antidiuretic hormone (ADH). This
prevents the reabsorption of water from the latter regions of the nephron.
This water is lost in urine, even when the solute concentration of the blood
is too high. Dehydration of tissues results by decreased production of
tissue fluid.
 Alcohol in low concentrations cause vasodilation of blood vessels,
including those of the skin, leading to excessive heat loss. At high
concentrations, it causes vasoconstriction, preventing adequate heat loss
from the skin. This disrupts the normal homeostatic control of body
temperature.
 At high concentrations, alcohol decreases breathing rate. This disrupts the
regulation of the carbon dioxide, oxygen and pH levels of the body.
6. (a)
Characteristics:
 Members are able to interbreed and produce fertile offspring
 Members have the same genes (not necessarily the same alleles for those
genes)
 Members have the same type and number of chromosomes
There are other /alternative correct responses e.g. members have similar
anatomy, ecological niches and biochemistry, etc.
95
Factors:
 Geographical isolation - physical separation of members of the species
 Behavioural isolation - development of different behavioural habits by
members of the species
 Reproductive isolation – fertile seasons / development and maturation of
gametes / courtship and mating rituals become different among members
of the species
There are other /alternative correct responses such as artificial
selection/genetic engineering, etc.
(b)
Variation of colour existed among members of the moth species – light and dark.
The presence of moth-eating birds placed a selection pressure on the population.
Increased darkening of tree trunks (soot deposits) in post-industrial Britain
camouflaged the black moths, giving them a selective advantage. They survived,
reproduced and transmitted their favourable genotype (for dark colour) to
subsequent generations, hence their increase in number.
(c)
1) Variation exists among the bacterial population, with some having a
genotype which allows them to be resistant to antibiotics. This variation
may have arisen by mutation.
2) The use of antibiotics (selection pressure) killed the non-resistant bacteria
before they could reproduce and transmit their non-resistant genotype to
subsequent generations (they had a selective disadvantage). The use of
antibiotics did not kill the resistant bacteria. These survived long enough
to reproduce and transmit their resistant genes to subsequent generations
(they had a selective advantage). These had a high reproductive rate since
there was less competition from the non-resistant ones (those were killed
by the antibiotics).
3) Subsequent generations of bacteria therefore contained a progressively
higher proportion of resistant ones, until the majority of the bacterial
population were resistant to antibiotics.
96
1. (a)
GRAPH OF THE GROWTH OF PLANTS UNDER VARYING CONDITIONS
OVER AN EXTENDED PERIOD OF TIME
97
(b)
(i)
In the light, the growth increased slowly during Days 5-15 to a height of
15 cm, then rapidly during Days 15-30 to a maximum height of 30 cm.
In the dark, the growth increased faster than that of the plant in light
during Days 5-15, reaching a height of 17 cm. The height remained at 17
cm till Day 20, then decreased to 15 cm by Day 30.
(ii)
The presence of light allows the plant to photosynthesise (convert light
energy to the chemical energy of glucose). The plant therefore has
adequate energy to do growth (protein synthesis and cell division.)
The plant in the dark does not make chlorophyll and is deprived of light.
Both these factors result in little or no photosynthesis. It therefore uses its
food stores obtain energy to do growth. Since its food stores are limited,
growth soon ceases. As organs such as leaves are shed, (cells die but are
not replaced as quickly), the height decreases.
(c)
Growth may be defined as a permanent increase in the mass and volume of an
organism when food is absorbed and converted to living matter.
(d)
The length of the plant from the tip of its shoots to its stem at soil level was
measured while the length of its growing roots were not. The length of the entire
baby was measured.
(e)
- Grow a larger number of plants in both light and dark. Measure the dry weight
of representative samples of the plants at 5-day intervals. This would measure the
increase in biomass.
- Measure the growth of the plants at smaller time intervals, e.g. every day. This
would allow one to note additional or subtler effects of light on growth rate.
Additional responses include measurement of surface area of leaves,
measurement of length of roots, replication of investigation and measurement of
fresh weight, etc.
(f)
Plant growth occurs at meristems (apical meristems at the tips of roots and shoots
and the cambium of vascular tissue of stems). Animal growth occurs at all points
or regions, resulting in an overall increase in size.
Alternative responses may refer to seasonal growth in plants or growing periods
for animals/growth throughout lifetime for plants.
98
(g)
(i)
Obtain two healthy potted plants of the same species and age, grown under
the same conditions and as similar as possible in physical features.
Enclose each plant in a container with a small opening on one side. One
container should be opaque and the other should be transparent. Expose
both arrangements to light. Record the direction of growth of each plant
over time.
(ii)
Plants grow taller until their leaves are exposed to light, which is used in
photosynthesis.
(iii)
How: invertebrates move away from light. This response involves
locomotion and is negative. A plant does a positive growth response.
Why: Dark conditions are likely to be cooler (not exposed to sunlight),
reducing the risk of desiccation of small organisms. Dark conditions allow
invertebrates to hide from their predators.
(h)
It is needed for synthesis of compounds such as melanin (protection against light)
and vitamin D (teeth and bone development).
An alternative response may state that photosynthesis (by plants) provides
humans with food for their growth.
2. (a)
(b)
(c)
A: photosynthesis
B: respiration
C: combustion
(i)
Photosynthesis
light absorbed by chlorophyll
Carbon dioxide  Water 
 Glucose  Oxygen
light absorbed by chlorophyll
6CO2  6H2O 
 C6 H12O6  6O2
(ii)
Aerobic respiration
Glucose  Oxygen 
 Carbon dioxide  Water  Energy
C6 H12O6  6O2 
 6CO2  6H2O  32 ATP
- Cutting trees reduce the number of them available for photosynthesis.
Photosynthesis produces oxygen, which is used by living organisms for aerobic
respiration. Lack of oxygen will kill organisms.
- Burning trees add a substantial amount of carbon dioxide to the air. Carbon
dioxide is a greenhouse gas. Its accumulation above the atmosphere contributes to
global warming.
Alternative responses include loss of habitat, less rainfall and soil erosion, etc.
99
(d)
- Ocean acidification: this impedes the development of calcium carbonate
exoskeletons of shellfish, oysters and corals, as well as dissolves existing
exoskeletons. Destruction of these animals will decrease the income gained from
tourism (visits to coral reefs) and harvesting and sale of shellfish.
- Rising sea levels: this increases beach erosion and destroys coastal ecosystems.
It may decrease the amount of land available for agriculture and habitation.
Alternative responses may outline any effect of increased temperature or other
drastic climate change.
(e)
3. (a)
(i)
- Large and non-biodegradable garbage, if strewn on the ground, persists
for a long time, clogging waterways and remaining as unsightly piles on
land.
- Soluble components of strewn garbage such as phosphates from soap
may enter waterways and change the pH or other chemical properties of
the water.
(ii)
- Garbage could be buried, either in compost (biodegradable materials) or
in a landfill.
- Materials could be re-used at home, when possible, or sent to a recycling plant, if available.
(i)
I: cell wall
II: cytoplasm
III: chloroplast
IV: stoma
V: nucleus
(ii)
I: it resists the outward pressure exerted by the protoplasm during water
intake and so prevents the cell from bursting.
Alternative responses include prevention of entry of some pathogens and
maintenance of cell shape.
IV: It allows water vapour to diffuse out of the leaf during transpiration.
Alternative responses may refer to inward and outward diffusion of
carbon dioxide and oxygen during photosynthesis and respiration.
100
(b)
DIAGRAM OF GUARD CELLS AFTER PLANT WAS WATERED DURING MORNING
(c)
The protoplasm of both animal and plant cells expand as water enters by osmosis.
In the plant guard cell, the outward pressure exerted by the expanding cytoplasm
is resisted and limited by the tough and inelastic cell wall. The cell wall therefore
exerts an inward pressure of equal size and the cell is firm and turgid, assuming a
semi-circular shape (due to the uneven thickening of the cell wall). The animal
cell has no cell wall to resist the outward pressure of the expanding cytoplasm and
its cell membrane is incapable of doing so. The cell bursts.
4. (a)
DIAGRAM OF THE HUMAN DIGESTIVE SYSTEM
Physical digestion:
The bread and chicken are mechanically digested (broken into smaller pieces) by
the chewing action of teeth. The chewed food is sent as a bolus to the stomach by
peristalsis of the oesophagus. Muscular contractions of the stomach walls grind
the food further. Mechanical digestion by the teeth and stomach increase the
surface area available for the action of enzymes.
101
Chemical digestion of bread (carbohydrate):
This occurs as outlined below:
Location
Enzyme
Enzyme secreted by
Mouth
Duodenum
salivary amylase
pancreatic
amylase
maltase
Salivary glands
Pancreas
Small
intestine
Wall
of
intestine
Hydrolysis reaction
catalysed by enzyme
starch 
 maltose
starch 
 maltose
small maltose 
 glucose
Chemical digestion of chicken (protein):
This occurs as outlined below:
Location
Enzyme
Stomach
pepsin
Stomach
renin
Duodenum trypsin
Small
intestine
peptidases
Enzyme
secreted by
Wall
of
stomach
Wall
of
stomach
Pancreas
Hydrolysis reaction catalysed by
enzyme
protein 
 polypeptides
soluble casein 
 insoluble casein
polypeptides 
 dipeptides
protein 
 polypeptides
Wall of small dipeptides 
 amino acids
intestine
Chemical digestion of chicken (lipid):
This occurs as outlined below:
(b)
Location
Enzyme
Duodenum
Small
intestine
lipase
lipase
(i)
Enzyme
secreted by
Pancreas
Wall of small
intestine
Hydrolysis
reaction
catalysed by enzyme
fat 
 fatty acids + glycerol
fat 
 fatty acids + glycerol
- Seeds store starch, protein, lipids, vitamins and minerals.
- Fruits store starch or reducing sugars (depending on stage of maturity),
lipids, vitamins, minerals and water.
Alternative responses include roots, leaves and stems and the nutrients
stored in them.
102
(ii)
5. (a)
- To avoid the need for continuous manufacture of food. Stored food can
be used when the plant is not photosynthesising, e.g. at night.
- To synthesise structures needed for certain processes, e.g. flowers and
seeds for reproduction and fruits for dispersal of offspring.
- To survive during periods of adverse environmental conditions, e.g. no
photosynthesis is possible if water is lacking from the soil. Buds survive
underground during dry or very cold conditions using nutrients from
storage organs.
- Erythrocytes
- White blood cells
- Platelets
- Plasma
1) Oxygen binds to haemoglobin in the alveolar capillaries of the lungs (area
of high oxygen concentration), forming oxyhaemoglobin inside red blood
cells.
2) Oxygenated blood is taken to the heart by the pulmonary vein.
3) Oxygenated blood is pumped out of the heart. It leaves through the aorta,
which divides into arteries, with one artery supplying oxygenated blood to
each organ.
4) Oxygenated blood flows close to respiring tissue (areas of low oxygen
concentration) in capillaries.
5) Oxyhaemoglobin dissociates into haemoglobin and oxygen, within the
capillaries.
6) Oxygen diffuses out of the red blood cell, across the capillary wall, into
the tissue fluid, across cell membranes and into cells, where it is available
for aerobic respiration.
(b)
(i)
- Fatigue: one is always tired and especially after physical activity
- Joint pain: especially the joints of the limbs
Alternative responses include jaundice, breathlessness and increased
vulnerability to infections.
(ii)
Parental phenotype
Parental genotype
Gametes
father: normal (carrier) X mother: normal (carrier)
AS
(A) and (S)
AS
(A) and (S)
Random fertilisation
(A) (S
(A) AA AS
(S) AS SS
103
Offspring genotype
AA
AS
AS
SS
Offspring phenotype Normal Normal (carrier) Normal (carrier) Sickle-cell anaemia
Offspring phenotypic ratio
1 normal: 2 carriers (sickle cell trait): 1 sickle-cell anaemia
There is a 25% (or 0.25 or 1 in 4) chance that a child with sickle-cell
anaemia will be born.
6. (a)
Genetic engineering is the alteration of the genome of an organism, at the
molecular level, by removal or addition of a gene or genes (not necessarily from
the same species). The steps are:
1) Cells are removed from the recipient organism and cultured.
2) The gene of interest is isolated (removed from a chromosome by
restriction enzymes or artificially synthesised).
3) The gene of interest is introduced into the cultured recipient’s cells either
directly (by means of a microprojectile) or indirectly (by means of a
vector). In the latter method, the gene is inserted into the vector and
sealed with DNA ligase. The vector is then placed in the cells of the
recipient.
4) Recipient cells are screened for successful uptake of DNA into their
chromosomes.
5) Successful cells (from step 4) are re-introduced into the recipient
organism. The organism is now transgenic or genetically modified.
Advantages:
- Desirable phenotypes are created in a shorter time than with artificial selection.
It eliminates the time taken for attainment of sexual maturity and the period of
gestation.
- The genome-altering possibilities are greater than with artificial selection.
Genes may be introduced from organisms with whom the recipient may not be
able to interbreed.
Alternative responses include the greater precision possible (variation associated
with sexual reproduction is removed).
Disadvantages:
- Variation may be lost within the species. People may rear/grow only genetically
modified organisms with superior traits. Other breeds /varieties (and their genes)
may be eliminated from the population. These eliminated genes may have been
advantageous in future environments, had they survived.
- There is a risk of “escape” of genes. Genetically modified organisms, especially
plants, may breed with wild relatives and transmit the transgene to a population
other than that for which it was intended. This may have unpredictable effects on
104
ecosystem stability, e.g. genes for herbicide resistance may be transferred to
weeds, transforming them into “superweeds.”
Alternative responses include the cost and expertise required, patent of genes and
ethical concerns/public acceptance.
(b)
(i)
Cause: The variation associated with sexual reproduction or environmental
differences are unlikely causes, since only one plant (from the field of
plants) has fruits of the new colour. It may have been caused by mutation
of the genes responsible for pigment production.
Implication: A mutation may produce associated disadvantageous traits,
since a mutation is a random change in functional DNA. The mutated
plant may therefore not be as well adapted to its environment as the others.
It may not survive long enough to reproduce multiple times and so, it may
not transmit its mutated gene to subsequent generations. The frequency of
this gene may decrease with subsequent generations and be eliminated
from the population. No evolution is therefore possible.
(ii)
Artificial selection program:
He could select before the plants reach the stage of seeds, as follows:
1) Identify plants with desirable features, e.g. colour of fruits.
2) Enclose them (or their flowers) in transparent bags to prevent
cross-pollination.
3) Collect pollen from plants with desirable features and transfer
them to the stigmas of the enclosed plants.
4) Screen the offspring of these crosses, select those with desirable
traits and repeat steps 2 and 3 using those plants.
Feature
Selection of alleles for
transmission to future
generations done by
Natural selection (occurs
in nature)
Environmental conditions
(selection pressures found
naturally
in
the
environment)
Alleles which make a
population well adapted to
its environment
Reason /basis for
selection of alleles to
be transmitted to future
generations
Length of time taken Longer
for
change
in
phenotype
of
population
Effect on size of gene Gradual increase for some
pool
genes and decrease for
Artificial selection
Humans
Alleles which produce
traits of benefit to humans
Shorter
Rapid decrease as some
organisms are deliberately
prevented from breeding
105
Effect
heterozygosity
population
Implication
evolution
population
others.
Overall size
remains constant.
on Remains high as a result of May be low as a result of
of outbreeding,
producing inbreeding,
possibly
hybrid vigour
causing
inbreeding
depression
for Is the basis of evolution
Prevents evolution
of
106
1. (a)
(i)
GRAPH SHOWING THE EFFECT OF TEMPERATURE ON THE
RATE OF REACTION OF THE ENZYME CATALASE
(ii)
40 °C
107
(iii)
0-10 °C: At 0°C, no reaction occurred because the enzyme and substrate
did not have enough energy to move and collide. As temperatures
increased to 10°C, there was a slowly increasing number of collisions
between enzyme and substrate, resulting in a slowly increasing rate of
product formation, that is, rate of reaction.
11-40 °C: The kinetic energy of the substrate and enzyme increased. This
caused them to have increased molecular motion and so, increased rates of
collision between enzyme and substrate. This resulted in a steadily
increasing rate of product formation, that is, rate of reaction. The rate is
fastest here. Catalase works optimally at 40 °C.
>40 °C: Enzymes are proteins and so catalase was denatured at these
temperatures. The shape of the active site was distorted and the substrate
could not bind to it. This caused a sharp decrease in the rate of product
formation (rate of reaction) until it reached zero (the reaction stopped).
(iv)
The rate of catalase activity increases with temperature until the optimum
temperature is reached.
Alternative conclusions may cite the optimum temperature for catalase
activity or the temperature(s) which causes(s) denaturation of catalase.
(v)
Body temperature is optimum for the action of its enzymes. Enzymes
catalyse metabolic reactions of physiological processes such as
respiration.
Alternative responses may refer to inactivation or denaturation of enzymes
at temperatures lower or higher, respectively, than body temperature.
(vi)
- Use equal volumes of enzyme in each test tube.
- Maintain the temperature of each water bath at its appropriate
temperature throughout the investigation.
Alternative responses include any example of a measure which should
have been taken to ensure, as far as possible, that there is no more than
one manipulated variable.
(vii)
- All the test tubes should be in one water bath, which should be
maintained at the optimum temperature for the enzyme (40 °C).
- The pH of the test tubes should vary. At least one should be acidic, one
neutral and at least one alkaline. Equal volumes of buffers of varying pH
should be added to each tube before adding the enzymes.
108
(b)
Pepsin catalyses hydrolysis of protein to polypeptides in the stomach.
Trypsin catalyses hydrolysis of polypeptides to smaller peptides in the ileum.
Alternative responses include any example of a protease and its action, e.g.
chymotrypsin.
2. (a)
(i)
(ii)
They are producers. They trap and convert light energy to chemical energy
(photosynthesis), which is then available to all organisms of the food web.
(iii)
a)
Small fish
b)
Tadpole
c)
Water Snake
d)
Eagle
(iv)
Algae 
 Tadpole 
 Small fish
(v)
The role of the organism to which the arrow point in a food chain is that of
a consumer.
(vi)
Energy enters the producers during photosynthesis and flows up to higher
trophic levels when animals ingest their food. Energy flows with the
materials consumed during feeding.
109
(b)
3. (a)
(b)
- It provides food in the form of living organisms (producers and prey).
- It provides other requirements (besides food) such as habitat, oxygen supply and
water to the community.
- It contains decomposers. They recycle waste, shed structures and dead bodies.
This prevents accumulation of dead organic matter and replenishes the supply of
needed materials for the community.
(i)
Stimulus: a change in the environment of an organism which is detected
by receptors and causes a response.
(ii)
Receptor: a specialised cell/organ containing specialised cells which
detects a particular stimulus.
(iii)
Effector: an organ which carries out a response to a particular stimulus.
(i)
SECTION THROUGH THE HUMAN EYE
(ii)
(c)
Light focused on that spot will be undetected (since it contains no light
sensitive cells/rods or cones) and no impulse will be sent to the brain.
There is no perception of the object from where the light came.
Light rays from the book are more divergent (so his lens is more convex) while
those from the helicopter are parallel (require less refraction).
The ciliary muscles relax, pulling the suspensory ligaments taut. They pull/exert
tension on the lens, stretching it flatter/less convex. This reduces the degree of
refraction done to the light rays and they are focussed on the retina.
(d)
Eye A
The circular muscles of the iris are relaxed and its radial muscles are contracted,
dilating the pupil and allowing as much light as possible to enter.
(e)
The response prevents or reduces the probability of damage to the light sensitive
cells of the retina by overexposure to light.
110
4. (a)
DIAGRAM OF THE HUMAN FEMALE REPRODUCTIVE SYSTEM
(b)
Barrier:
- It is temporary. They may have children later if they wish. Their reproductive
ability is not affected.
- It is free of the pain, possible health risks and post-operational recovery
associated with the surgical method.
Surgical:
- It provides no protection against the transmission of sexually transmitted
infections. STI’s may be of greater importance since permanent contraception
such as surgery may be associated with increased promiscuity.
- It is more expensive to undergo surgery than to purchase barriers.
(c)
Method: asexual reproduction, e.g. by growth of leaflets on the margins of leaves
Advantages:
- Desirable features of the parent plant are transmitted to the offspring.
- Large numbers of offspring are produced in a shorter time.
- Only one parent is needed, so an isolated plant can reproduce.
Alternative response – the method of cross-pollination with a male plant and the
associated advantages.
5. (a)
Photosynthesis is the conversion of light energy to chemical energy by plants
containing chlorophyll (traps light). Carbon dioxide and water are reactants.
Oxygen is a by-product and glucose is the product.
Equation:
Adaptations:
111
Feature
Leaf is thin
Adaptation for capture of sunlight
Sunlight can penetrate leaf and reach
all chloroplasts
Palisade mesophyll cells are columnar
A large number of them is present,
due to efficient “packing”
Epidermal cells contain no chlorophyll or Light can penetrate and reach all
other pigments
mesophyll cells
Leaf is broad
Large surface area for capture of light
Large number of chlorophyll molecules Large number of light absorbing
within each chloroplast and large number pigments
of chloroplasts
Leaf held by petiole at a fixed angle
Angle is appropriate for maximum
light absorption
Feature
Adaptation for capture of carbon
dioxide
Large intercellular spaces in spongy There is an unobstructed pathway for
mesophyll tissue which are continuous diffusion of carbon dioxide to all
with the atmosphere
photosynthesising cells
Leaf is thin
Carbon dioxide can diffuse within
leaf to all chloroplasts
Numerous stomata in lower epidermis
Large surface area for inward
diffusion of carbon dioxide
(b)
6. (a)
The structures and materials shed by plants and animals (leaves, bark, skin, horns,
scales, faeces, etc.) and their dead bodies contain complex carbon-containing
organic compounds. Decomposers use these compounds as substrate for their
respiration. They convert the complex organic compounds into simple inorganic
ones such as carbon dioxide. Carbon dioxide is released into the air, where it is
available to green plants for photosynthesis.
Causes:
- lack of iron or Vitamin B12 in one’s diet
- Inheritance of HbS HbS genotype (sickle cell anaemia)
Alternative responses include excess blood loss due to menstruation, internal
bleeding, etc.
Signs/Symptoms:
- low red blood cell count due to synthesis of limited number of haemoglobin
molecules, causing pale appearance of skin
- Fatigue/weakness/lack of energy due to a low rate of respiration (caused by lack
of oxygen)
Alternative responses include weight loss, retarded physical development, joint
pain, enlarged spleen, reduced vision, etc.
112
(b)
(i)
Homozygous: a condition in which an organism has two identical alleles
for a given gene.
Heterozygous: a condition in which an organism has different alleles for a
given gene.
(ii)
Let HbA represent the allele for normal haemoglobin
Let HbS represent the allele for sickled haemoglobin
Parental phenotype
carrier
X
carrier
Parental genotype
HbA HbS
HbA HbS
Gametes
(HbA) and (HbS)
(HbA) and (HbS)
Random fertilisation
(HbA)
(HbS)
HbA HbA HbA HbS
HbA HbS HbS HbS
(HbA)
(HbS)
Offspring genotype
HbA HbA HbA HbS
HbA HbS HbS HbS
Offspring phenotype
Normal
Normal Sickled
Offspring phenotypic ratio
Normal
3 normal
:
1 sickled
The child of genotype HbS HbS has sickle cell anaemia.
(iii)
The presence of the malarial parasite (for the severe form of malaria) is
the selective pressure. The parasite cannot survive in sickled red blood
cells. Persons of genotype HbA HbA die of malaria and persons of
genotype HbS HbS die of sickle cell anaemia.
In carriers (genotype HbA HbS), the parasite is found only in their normal
red blood cells. The parasite population in carriers is therefore too small to
cause malaria to kill them. Carriers have a selective advantage over
persons with normal red blood cells (genotype HbA HbA), since they
survive long enough to reproduce and transmit the HbS allele to future
generations. This causes the frequency of the HbS allele to be high in areas
where malaria is present, that is, there are many carriers in the population.
113
1. (a)
(i)
GRAPH SHOWING THE RATE OF REACTION OF ENZYME X
AT DIFFERENT TEMPERATURES
114
(ii)
30 °C
(iii)
5 - 30 °C: The kinetic energy of the substrate and enzyme increased. This
caused them to have increased molecular motion and so, increased rates of
collision between the enzyme’s active site and the substrate. This resulted
in a steadily increasing rate of product formation, that is, rate of reaction.
The enzyme works optimally at 30 °C.
>40 °C: The enzymes are proteins and so it was denatured at these
temperatures. The shape of the active site was distorted and the substrate
could not bind to it. This caused a sharp decrease in the rate of product
formation (rate of reaction) until it reached zero (the reaction stopped).
(iv)
Use equal volumes of enzyme in the test tubes at each temperature.
Alternative responses include repetition of experiment or any measure
taken to ensure that temperature was the only manipulated variable.
(b)
(c)
(i)
- Pepsin
- Trypsin
Alternative responses are chymotrypsin, erepsin, etc.
(ii)
- Stomach
- Small intestine/duodenum
(iii)
- Different enzymes work optimally at different pH levels.
Compartmentalisation of enzymes allows the digestive tract to maintain
the required pH in each compartment.
- Enzymes are specific. The product of one enzymatic reaction may be the
substrate for a subsequent one, e.g. the action of pepsin produces
polypeptides, which are the substrates for trypsin. Separation of these
enzymes in order of the stages of protein digestion allows greater
efficiency of digestion. This ensures that digestion is completed prior to
absorption.
(i)
Mutualism
(ii)
The peanut plant gains a supply of NO3- from the nitrogen-fixing action of
the bacteria. The bacteria gain a supply of energy and a habitat (protection
against extremes of environmental conditions, e.g. pH), from the plant.
Neither partner harms the other.
115
(iii)
1) Blend the peanut in as little water as possible, to obtain a
suspension.
2) Measure 2.0 mL of the suspension into a test tube.
3) Add 2.0 mL of NaOH(aq) to the tube and mix the contents of the
tube.
4) Add 4 drops of CuSO4 (aq) and mix the contents.
5) Observe any colour changes – if the CuSO4 (aq) changes from blue
to purple/violet, the peanuts contain protein.
(iv)
- Synthesis of enzymes: enzymes are protein molecules needed to catalyse
all their metabolic reactions.
- Growth: protein is used in the synthesis of compounds associated with
cell division and enlargement (tissue formation), e.g. synthesis of cell
membranes and nuclei.
Alternative responses may describe the role of protein in repair of tissue
or storage of nutrients.
2. (a)
(i)
- Plant cells contain a cell wall around their cell membrane but animal
cells do not possess a cell wall.
- Plant cells contain large, central, permanent vacuoles bot the vacuoles of
animal cells are small, temporary and may be located anywhere in the
cytoplasm.
- Plant cells store glucose as starch grains but animal cells contain
glycogen granules as their store of glucose.
An alternative response may refer to chloroplasts.
(ii)
- The DNA of prokaryotic cells is circular and free in the cytoplasm (not
enclosed in a nuclear membrane). The DNA of eukaryotic cells is linear
and enclosed in a nuclear membrane.
- Prokaryotic cells contain no membrane-bound organelles. Respiration
and photosynthesis (for photosynthetic ones) occur on infoldings of the
cell membrane. Eukaryotic cells contain membrane-bound organelles such
as mitochondria and chloroplasts.
Alternative responses include size of ribosomes, presence of flagella or
composition of cell wall, etc.
(b)
(i)
Cell wall: It prevents entry of some pathogens and prevents the cell from
bursting by limiting its intake of water (by exertion of an inward pressure
to resist the outward turgor pressure).
116
(c)
(ii)
Cell membrane: it separates the cytoplasm from the extracellular
environment and it regulates the entry and exit of materials into and out of
the cell.
(i)
Water moves down its concentration gradient by osmosis: from the
cytoplasm and into the sucrose solution. The loss of water causes the
cytoplasm to shrink and the cell is now crenated.
(ii)
Water moves down its concentration gradient by osmosis: from the
sucrose solution and into the cytoplasm. The intake of water causes the
cytoplasm to expand and exert a pressure against the cell membrane. The
cell membrane is not strong enough to resist the pressure and it bursts, that
is, lysis occurs.
(iii)
Concentrated sucrose solution:
Water moves down its concentration gradient by osmosis: from the
cytoplasm and into the sucrose solution. The loss of water causes the
cytoplasm to shrink and the cell membrane pulls away from the cell wall,
becoming visible, except at the plasmodesmata. The cell undergoes
plasmolysis and becomes flaccid.
Distilled water:
Water moves down its concentration gradient by osmosis: from the
sucrose solution and into the cytoplasm. The intake of water causes the
cytoplasm to expand and exert a pressure against the cell membrane and
cell wall. The tough, inelastic cell wall resists this pressure. It exerts an
inward pressure, equal to the outward turgor pressure, and so prevents
further expansion of the cytoplasm. The cell membrane does not burst.
The cell becomes turgid.
3. (a)
(i)
A: platelet
B: white blood cell or leucocyte
C: red blood cell or erythrocyte
(ii)
Platelets
(b)
D: platelets
E: thrombin
F: fibrin
(c)
(i)
Male parent’s genotype: XHY
Male parent’s gametes (xH) and (Y)
Gametes
XH
Y
XH
X XH
XHY
H
Xh
XH Xh
Xh Y
117
(ii)
1 in 4 or 0.25 or 25%
(iii)
Male
The males do not have a second allele for blood clotting. Any allele on
their only X chromosome will be expressed, without any chance of a
recessive allele being suppressed. The males produced from this cross
have a 50% chance of inheriting the recessive haemophiliac allele from
their mother.
All the females of the cross shown inherit a dominant allele (XH) from
their father. This will mask the expression any Xh allele they may have
inherited from their mother. None of these females will be haemophiliacs.
(d)
4. (a)
Pathogens are likely to enter and cause infections at a break in the skin. A blood
clot, at a break in the skin, dries to form a tough waterproof barrier which
prevents entry of pathogens while new skin is formed underneath. In
haemophiliacs, blood does not clot and no mechanical barrier is formed. They
have an increased risk of infection.
Humans:
1) Deoxygenated blood flows from the heart through the pulmonary artery to
the lungs.
2) Deoxygenated blood flows through the alveolar capillaries surrounding the
alveoli.
3) The external intercostal muscles contract, the internal intercostal muscles
relax and the diaphragm contracts.
4) The volume of the thorax (and lungs) increases.
5) The pressure within the lungs decreases until it is lower than atmospheric
pressure.
6) Air (containing oxygen) flows down a pressure gradient, from the
atmosphere and through the mouth/nose, trachea, bronchi, bronchioles and
into alveoli.
7) Oxygen dissolves in the moisture lining the alveoli.
8) Oxygen diffuses from the moisture in the alveoli, across the alveolar wall,
capillary wall, blood plasma and into the red blood cells.
9) Oxygen combines with haemoglobin, forming oxyhaemoglobin.
10) Oxygenated blood flows to the lungs in the pulmonary vein.
11) Oxygenated blood is pumped out of the heart into the aorta, which divides
into arteries (one artery supplying blood to each organ. The oxygen is
now available for delivery to cells.
118
Plants:
The oxygen content of respiring cells is low. Oxygen in intercellular spaces of the
leaves diffuses into respiring cells, lowering the concentration of oxygen in
intercellular spaces. This creates an oxygen concentration gradient from the
atmosphere to the intercellular spaces. Oxygen diffuses from the air, through
stomata, to the intercellular spaces.
Oxygen produced in photosynthesis is also used in respiration.
(b)
Cigarette smoke contains nicotine, tar, soot and carbon monoxide.
1) Nicotine constricts blood vessels, reducing the supply of oxygenated blood
to respiring cells.
2) Haemoglobin has a greater affinity for carbon monoxide than for oxygen.
Less oxyhaemoglobin and more carboxyhaemoglobin is formed, reducing
the amount of oxygen carried by blood to respiring cells.
3) Tar and soot line the alveoli and airways, irritating them and stimulating
mucus production. Nicotine inhibits the action of cilia, so mucus
accumulates in the lungs. This reduces the volume available for air (and
oxygen) within the lungs and airways.
4) Mucus reduces the surface area available for diffusion of oxygen from the
alveoli to the blood, as it forms a physical barrier between the oxygen and
the alveolar wall.
5) The immune system responds to the chemicals of the smoke by
stimulating inflammation. Elastase (an enzyme) is produced which
destroys the elastic fibres of airways. This reduces their ability to expand
and contract and the volume of air exchanged (and so, the amount of
oxygen entering) is reduced.
6) Lung cancers may form, especially at the start of the bronchioles. These
growths occupy space and reduce the volume available for the entry of air
(and oxygen.)
Alternative responses include the destruction of alveolar walls, reducing the
surface area for gaseous exchange.
5. (a)
(i)
Blood flows in the direction of the arrows below:
Lungs 
 Pulmonary veins
Pulmonary veins 
 Left atrium of heart
Left atrium 
 Left ventricle through bicuspid valve
Left ventricle 
 Aorta through semilunar valve
Aorta 
 Arteries supplying all organs/rest of body
(ii)
- Hypertension
- Coronary heart disease
Alternative responses: atherosclerosis, deep vein thrombosis, etc.
119
(b)
B-lymphocytes produce antibodies which destroy pathogens or neutralise their
toxins. Pathogens engulf and digest pathogens. T-lymphocytes destroy virally
infected cells and cancerous cells. Chemotherapy destroys all these white blood
cells, reducing their populations. The body therefore contains less of the cells
responsible for protection of the body against disease, that is, decreased natural
immunity.
(c)
Peter receives the injection because his body already contains the pathogens and
the antiserum injection would have an immediate effect against the quick-acting
pathogens.
A vaccine contains attenuated or genetically modified pathogens. It purpose is to
introduce antigens into the body. This would stimulate production of antibodies to
destroy the introduced pathogens, but more significantly, memory cells against
the introduced pathogens are produced. If those pathogens enter the body at a
later time, there are sufficient B-lymphocytes (memory cells) to quickly effect an
immune response.
When Peter’s foot was punctured, pathogens entered. His B-lymphocytes with
receptors complementary to the pathogens would stimulate antibody production.
He does not need a vaccine since his body already contains the pathogens.
The production of antibodies by his B-lymphocytes takes a longer time than it
takes for the tetanus pathogens to produce a severe, often fatal, effect on the body.
The tetanus pathogens are very quick-acting.
The antiserum injection contains artificially prepared antibodies against the
pathogens. They destroy the pathogens (or neutralise their toxins) as soon as they
enter the body. The pathogens are destroyed before they could exert harmful or
fatal effects on his body.
6. (a)
This is done by recombinant DNA technology, as follows:
1) The gene for insulin production is isolated by removing it from human
DNA/chromosomes using a restriction enzyme. Alternatively, knowledge
of the base sequence of the gene may be used to synthesise it.
2) The isolated gene is cloned (multiple copies are made).
3) Vectors (plasmids from E. coli) are isolated by removing them from the E.
coli. These plasmids contain a marker gene, that is, a gene which is easily
identified.
4) The vectors are cut open with the same restriction enzyme as was used in
isolation of the gene.
5) The genes are inserted in the vectors (one gene per vector) using DNA
ligase.
6) The recombinant vectors are re- introduced into E. coli.
7) The E. coli are screened for the presence of recombinant vector, using the
marker gene.
120
8) The E. coli which have taken up the vector are cultured. They contain the
human gene for insulin production. When they produce insulin, it
accumulates in their cells. At intervals, batches of these insulin-containing
bacteria are removed and the insulin is extracted from them and purified.
(b)
(i)
Artificial selection
Slower process, as it involves several
generations of cross-breeding in order
to breed out undesirable traits or breed
in desirable ones. The results of
crossing are unpredictable, to an
extent.
Done at the level of the organism –
sexual reproduction is involved
Genetic engineering
Faster process since humans have
more control over the transfer of
alleles. Organisms with desirable
traits can be produced in one
generation.
Done at the molecular level – genes are
removed from organisms and inserted
into cells of recipient organisms
Genes of a species may be inserted into
cells of organisms of another. New
traits are created for the recipient
species.
Involves intraspecific transfer of
genes, as they are transferred by sexual
reproduction. There is no import of
traits from another species. Traits
which are already present in the
species are selected.
Clusters of related genes are Recipients receive an isolated gene
transferred together.
without the accompanying genes
which were associated with the gene
(in the donor organism).
There
are
no
unknown
or There are unknown and unpredictable
unpredictable long term effects on the effects on both the genetically
organisms produced by this method
modified organisms and non-target
species.
More labour intensive, as it involves Less labour intensive, as it is done in a
rearing organisms, in fields or farms laboratory and involves working with
and breeding them
cells and molecules rather than whole
organisms.
Cheaper initial costs
More expensive (initially), as it
involves research, expertise and
purchase of laboratory supplies. The
price and demand of genetically
modified products may cause the
process to be cheaper in the long term.
121
(ii)
- It is cheaper since it is produced by bacteria. Culture of bacteria is
relatively cheap and simple. They reproduce very quickly.
- It is less likely to cause allergic reactions or other side effects, since the
insulin gene from humans is used. The insulin produced is therefore a
human protein. It is more effective on humans than insulin from another
animal.
- It is accepted by persons of various religions, cultures and lifestyles.
People are comfortable with the idea of introduction of a human protein in
their bodies, rather than one from another animal. Vegans and persons of
some religious beliefs may prefer to avoid use of protein from
animals/certain animals.
122
1. (a)
(i)
To investigate the effect of temperature on germination.
(ii)
Tube A
At 20 °C, the enzymes necessary for germination have more kinetic
energy, that is, they are more activated. They therefore show greater
activity.
(iii)
- Water is used to hydrolyse insoluble complex food stores such as starch
and protein to soluble simple molecules such as glucose and amino acids
(used for respiration and growth of embryo.)
- Water is used as a solvent for transport of hydrolysed food to the
growing embryo.
Alternative responses include splitting of the testa, etc.
(b)
(i)
Growth movement (phototropism).
An alternative response is part movement.
(ii)
Plants grow towards light or Plants are positively phototropic.
(iii)
- Cover the opening at the side with opaque material and cut an opening at
the top of the box.
- Adjust the handle/neck of the lamp such that the plants are illuminated
directly from above.
(iv)
In the light-dependent stage, chlorophyll in chloroplasts absorbs light. Its
energy is used for photolysis of water – it is split into hydrogen and
oxygen. In the light-independent stage, carbon dioxide from the
atmosphere diffuses through stomata, intercellular spaces and into the
mesophyll cells. It combines with hydrogen produced by photolysis of
water to form glucose. Some of it is used to form starch. The oxygen from
photolysis of water is released as a waste product. The process is
summarised as:
light absorbed by chlorophyll
6CO2  6H2O 
 C6 H12O6  6O2
123
(v)
Carbon dioxide diffuses through the stomata into the intercellular spaces
of the leaf. Some carbon dioxide produced in respiration is already
present in leaves.
Root hairs absorb water by osmosis from the soil. It moves by osmosis
across the root and enters the root xylem. It moves up the xylem largely by
the transpiration pull (tension generated in xylem by evaporation and loss
of water through stomata), and to a lesser extent, by capillarity (adhesion
and cohesion). In this way, it travels up the xylem of the root, stem and
leaves. It leaves the leaf xylem by osmosis and enters mesophyll cells.
(c)
(i)
1)
2)
3)
4)
Boil the leaves in water until they are soft.
Boil the leaves in ethanol until they are decolourised.
Submerge the leaves in hot water until they are soft.
Spread the leaves on a white tile and cover them with iodine
solution (I2/KI(aq))
5) If the iodine solution changes from yellow/brown to blue-black,
starch is present in the leaves.
(ii)
Seeds (cotyledon or endosperm)
Alternative responses include storage organs (corms, tubers, rhizomes,
etc.), fruits, roots and stem, etc.
2. (a)
(b)
(iii)
- Liver
- Muscles
(i)
Prokaryote
(ii)
Multicellular
(iii)
Amoeba (or any other example)
(iv)
Plantae
(v)
Animalia
(i)
A species is a group of individuals that can interbreed and produce fertile
offspring.
124
(c)
(ii)
Humans may differ very much in physical appearance, e.g. body shape,
skin colour and height. This variation is due, in part, to their possession of
different alleles of the SAME GENES. Organisms in possession of the
same genes are genetically similar enough to interbreed and produce
fertile offspring. Some variation in morphology is due to varying
environmental influences and not to differences in alleles. This means that
humans are genetically more similar than their appearance suggests.
(iii)
- Organisms may be very similar in morphology but may not be closely
related enough genetically to interbreed and produce fertile offspring.
- Organisms may be closely related enough genetically to interbreed and
produce fertile offspring, yet show tremendous variation in morphology,
due to mutation, sexual reproduction, environmental influences and stage
of development/life cycle.
(i)
They were no longer able to interbreed and produce fertile offspring with
the lizards of the place/land from where they came.
(ii)
- In the mountains, less atmospheric oxygen was a selection pressure,
causing lizards of smaller size to be selected for (they can survive with
less oxygen).
- In the mountains, steep terrain was a selection pressure, causing lizards
of smaller size to be selected for (they can climb vertical surfaces better).
Alternative responses include identification of environmental differences
between the highlands and lowlands, causing them to have different
selection pressures, e.g. more food and larger habitats in the lowlands.
3. (a)
(i)
X: oestrogen
Y: progesterone
Z: ovulation
(ii)
X repairs the uterus wall (endometrium) after the preceding menstruation
and inhibits the secretion of FSH.
Y maintains the repaired endometrium, causing it to become thicker and
glandular and preventing premature menstruation. It inhibits the secretion
of FSH.
(iii)
X: ovary (produced by secondary and Graafian follicles)
Y: ovary (produced by corpus luteum)
Alternative responses include the placenta and adrenal glands.
125
(iv)
Corpus luteum: It does not degenerate after a few days, but persists for
three months. It continues to secrete progesterone.
Uterus lining: It is not lost in menstruation, but gets thicker (increased
number of blood vessels). The developing embryo is implanted within it
and a placenta is formed at three months using foetal villi and the
endometrium.
(b)
Pills may contain hormone-like compounds which thicken cervical mucus,
preventing entry of sperm.
Alternative responses include prevention of ovulation, implantation or repair of
endometrium, etc.
(c)
4. (a)
- It may reduce the production of testosterone, preventing formation of sperm.
There is no male gamete available for fertilisation.
- It may decrease the motility of sperm, reducing their chance of making contact
with the ovum in the correct orientation.
- It may inhibit the development of sperm. Poorly developed sperm will be
unviable.
(i)
The iris contains both radial muscles and circular muscles. These control
the amount of light which enters the eye by varying the diameter of the
pupil (through which the light enters).
In dim light, the radial muscles contract and the circular muscles relax.
This increases the diameter of the pupil, allowing more light to enter the
eye.
In bright light, the circular muscles contract and the radial muscles relax.
This decreases the diameter of the pupil, allowing less light to enter the
eye.
(ii)
The retina contains photoreceptors called rods and cones.
Rods, unlike cones, are stimulated at low light intensities, as in the dimly
lit room. The retina contains numerous rods so a large number of them is
stimulated by the low amount of light entering Jenny’s eye. Their
stimulation causes an impulse to be sent to the brain, where a black and
white image of the furniture is perceived. Dilation of the pupil in the dimly
lit room increases the number of light rays which are focussed on the rods.
Cones, unlike rods, are stimulated at high light intensities, as occur when
the lights are switched on. The retina contains cones, which are especially
abundant in the fovea. These stimulated by the high amount of light
126
entering Jenny’s eye. Their stimulation causes an impulse to be sent to the
brain, where a coloured image of the furniture is perceived.
(b)
Let XB represent the allele for normal vision
Let Xb represent the allele for colour-blindness
Parental phenotype
father: colour-blind
Parental genotype
XbY
X
XBXb
(Xb) and (Y)
Gametes
mother: normal vision
(XB) and (Xb)
Random fertilisation
(XB) (Xb)
(Xb) XBXb XbXb
(Y) XBY XbY
Offspring genotype
XBXb
Offspring phenotype
Normal
Offspring phenotypic ratio
XbXb
Colour-blind
1 normal vision
XBY
XbY
Normal Colour-blind
:
1 colour-blind
There is a 50% (or 0.5 or 1 in 2) chance that colour-blind children can be
produced.
5. (a)
Carbohydrate:
Location
Enzyme
Mouth
Duodenum
Small intestine
Small intestine
Small intestine
Salivary amylase
Pancreatic amylase
Maltase
Sucrase
Lactase
Hydrolysis reaction catalysed by
enzyme
starch 
 maltose
starch 
 maltose
maltose 
 glucose
sucrose 
 glucose + fructose
lactose 
 glucose + galactose
Glucose, fructose and galactose are absorbed by villi.
127
Protein:
Location
Stomach
Stomach
Duodenum
Enzyme
Pepsin
Renin
Trypsin
Small intestine
peptidases
Hydrolysis reaction catalysed by enzyme
protein 
 polypeptides
soluble casein 
 insoluble casein
polypeptides 
 dipeptides
protein 
 polypeptides
dipeptides 
 amino acids
Hydrochloric acid provides a low pH for the optimal action of pepsin and renin in
the stomach. Sodium hydrogen carbonate neutralises the acidic chyme (when it
leaves the stomach and enters the duodenum) and provides a high pH for the
optimal action of trypsin.
Amino acids are absorbed by villi.
Lipid:
Location
Duodenum
Small intestine
Enzyme
Lipase
Lipase
Hydrolysis reaction catalysed by enzyme
fat 
 fatty acids + glycerol
fat 
 fatty acids + glycerol
Fatty acids and glycerol are absorbed by villi.
(b)
(i)
Glucose (absorbed by the blood capillaries of the villi) is taken to the liver
by the hepatic portal vein. It is then released in the general circulation and
is present in the arteries supplying all organs.
If its concentration is above 120 mg mL-1 blood, β cells of the pancreas
are stimulated and they secrete insulin into the blood. Insulin stimulates
cells to absorb glucose from the blood. In the cells, glucose is used as the
substrate for respiration, yielding ATP.
Insulin stimulates fat cells to absorb glucose molecules and convert them
to fat.
Muscle and liver cells absorb more glucose than other cells. In addition to
using it as a respiratory substrate, they convert some of it to glycogen,
under the influence of insulin.
If the blood glucose concentration is less than 80 mg mL-1 blood, α cells of
the pancreas are stimulated and they secrete glucagon into the blood.
Glucagon stimulates liver cells to convert the stored glycogen to glucose,
which is released into the blood, raising the blood glucose level.
128
(ii)
In a non-diabetic person, insulin is produced and this stimulates uptake of
glucose by cells. In a diabetic person, insulin is not produced (type I
diabetes) or the insulin has no effect on body cells (type II diabetes).
Glucose is poorly absorbed by cells and so, it remains in the circulatory
system. It remains at a high concentration in the blood while cells have
insufficient, for use in respiration. The ATP yield is therefore less. Less is
available for synthesis of fat and glycogen.
The glucose is filtered out of the blood in the kidneys. All of it is not
reabsorbed and some is expelled from the body in urine.
Glucose is taken up by cells and assimilated by diabetic patients only if
they use medication for this purpose.
6. (a)
Predator-prey: garden lizard and earthworm
A predator (garden lizard) is larger than its prey. It is a carnivorous (or
omnivorous) animal which belongs to a higher trophic level than its prey. It is
adapted for hunting and killing its prey.
A prey (earthworm) is smaller than its predator. It is an herbivorous (or
omnivorous) animal which belongs to a lower trophic level than its prey. It is
hunted and killed by its predator.
A garden lizard hunts, kills and eats an earthworm.
Parasitism: fleas on a dog
A host (dog) is larger than its parasites. It provides one or more benefits to the
parasites and suffers harm in the process. It gains no benefits.
A parasite (flea) is usually smaller than its host and lives in close association with
it. It frequently lives in or on its host. It gains benefits such as nutrients or shelter
from its host but gives none. It harms the host.
Fleas live on the skin of a dog, buried under its hair. They suck the blood of the
dog, gaining nutrients and shelter. The dog suffers loss of blood and nutrients.
(b)
(i)
Advantages:
 It is specific: it will not destroy a range of organisms. Use of
chemicals destroys non-target species as well.
 There is no risk of bioaccumulation of chemicals or
biomagnification along food chains: it does not add fat-soluble
chemicals to the environment which may enter organisms and
become concentrated in their tissues.
 The introduced predators will reproduce and maintain the numbers
of their population: chemicals would have to be re-applied at
intervals, since the effects of chemicals are short-lived.
129
Alternative responses include lower cost and less labour involved, etc.
Disadvantages:
 The introduced species may become invasive: the area to which the
predators are introduced may have no natural predators (or other
means of population control) for the introduced species. They may
overpopulate the area and cause unpredicted effects on the
population sizes of other organisms or on the environment.
 The introduced species may eat non-target species: the predators
may prefer to eat organisms of a species other than the one it was
intended to eat. This may also occur as the pest population
declines.
 The introduced species may not be adapted to the environment:
although the food source may be present, the predator may be
susceptible to endemic pathogens or may not survive under the
climatic conditions of its new environment.
Alternative responses include the cost of replenishing predator
populations after the pests are eliminated and the ecological effect of total
elimination of pests, etc.
(ii)
Evolution of prey:
1) The presence of the predators is the selection pressure.
2) Prey which are adapted for escape from predators (e.g. quicker
ones) survive long enough to reproduce and transmit their
advantageous genotype to their offspring. Prey which are unable to
escape are eaten and so, cannot reproduce and transmit their
disadvantageous genotypes to subsequent generations.
3) Subsequent generations of prey contain an increasing proportion of
the genotypes that enable escape from the predators. The
disadvantageous genotypes are eventually eliminated.
4) All members of the prey population will eventually be adapted to
escape from the predator. The prey population has evolved.
An alternative response may describe evolution of the predator for more
efficient capture of prey.
130
1. (a)
A BAR GRAPH OF THE NUMBER OF ORGANISMS FOUND ON A TREE
131
(b)
Pitfall trap – a bottle containing some bleach (or other chemical toxic to small
creatures) is placed in a hole in the soil under the tree, with the top of the bottle at
soil level. It is covered loosely enough to allow walking animals to fall in. The
animals are later removed and recorded.
Sheet and pooter – a sheet of fabric or plastic is spread under the tree. The
branches are vigourously shaken, causing small animals to fall onto the sheet. A
pooter may be used to collect them or photographs are taken of them. The animals
are counted and recorded.
There are alternative correct responses, e.g. capture/recapture, use of nets,
transects and tracking devices, etc.
(c)
Arthropod: spider or ant or caterpillar (immature stage of an insect)
Reptile: lizard
Bird: egret
Amphibian: tree frog
(d)
The wild pine is autotrophic – it converts the sun’s energy into chemical energy of
complex organic molecules such as glucose, using simple inorganic molecules. It
synthesises its own food.
The other organisms are heterotrophic – they digest complex organic molecules
which are synthesised or assimilated by other organisms, as their means of
obtaining nutrients and energy.
(e)
(i)
Relationship: Commensalism
The wild pine (commensal) gains a benefit from the tree (it uses the tree
for anchorage/support in a location at which it receives sunlight for
photosynthesis).
The tree (host) is unaffected (it receives neither harm nor benefit from the
wild pine).
(ii)
The relationship between the vine and the tree is parasitism.
Unlike the photosynthetic pine, the vine (parasite) receives nutrients from
the tree. The tree (host) is negatively affected by the vine (it suffers loss of
nutrients). The tree is unaffected by the presence of the pine.
(iii)
The vine absorbs nutrients from the phloem tubes of vascular bundles,
depriving the tree of its nutrients.
The presence of the vine blocks sunlight from reaching leaves, reducing
photosynthesis.
132
(f)
The ratio of green to brown lizards is the result of natural selection. The presence
of predators is the selection pressure in the lizards’ environment.
The favourable genotype is that for green colour, since it confers a selective
advantage on lizards of the green phenotype: they are better camouflaged by the
grass, they escape detection by predators long enough to reproduce multiple
times. They are better adapted to their environment.
The unfavourable genotype is that for brown colour, since it confers a selective
disadvantage on lizards of the brown phenotype: they are less camouflaged by the
grass, they are detected and eaten by predators and so do not survive long enough
to reproduce many times. They are ill adapted to their environment.
When organisms reproduce, they transmit their alleles to the members of the
subsequent generation.
Over several generations, the rates of survival, hence reproduction, hence
transmission of alleles to subsequent generations, of the green lizards are higher
than those of the brown lizards. The frequency of the allele for green colour
increases, while that for brown colour decreases, within the population. With each
generation, an increasing proportion of the lizards will be green until an
equilibrium is reached with a high ratio of green to brown. This ratio is then
maintained by continued natural selection. The number of green lizards will be
observed to be higher.
The population of lizards is therefore adapted to its environment (containing the
selection pressure of predators) by the high proportion of its members with the
favourable phenotype.
2. (a)
(b)
Carbohydrate – monosaccharides, e.g. glucose
Protein – amino acids
Fat – fatty acids and glycerol
Protein: Amino acids diffuse through the epithelium of the villi and into the blood
capillaries. The capillaries merge to form the hepatic portal vein which transports
the amino acids to the liver.
Fats: Fatty acids and glycerol diffuse through the epithelium of the villi and into
the lacteals. The lacteals merge to form a lymph vessel, which eventually makes a
junction with a blood vessel. The products enter the blood vessel and are taken to
the liver.
(c)
(i)
The pancreas secretes the enzymes trypsin, lipase and amylase. These are
secreted through the pancreatic duct into the duodenum. Here the trypsin
catalyses the hydrolysis of polypeptides to smaller peptides and
133
dipeptides. Lipase catalyses the hydrolysis of fats to fatty acids and
glycerol and amylase catalyses the hydrolysis of starch to maltose.
The pancreas also secretes hydrogencarbonate ions into the duodenum.
These neutralise the acidic chyme from the stomach and create an
environment of high pH, which is needed for optimum function of the
enzymes.
(ii)
(d)
Hormone X – insulin
Hormone Y – glucagon
When the blood glucose level decreases below the norm, glucagon stimulates the
conversion of glycogen (stored in liver) to glucose, which is released into the
blood, until the blood glucose level is once again within the norm.
When the blood glucose level increases above the norm, insulin stimulates the
conversion of glucose (from blood) to glycogen, which is stored in the liver, until
the blood glucose level is once again within the norm.
(e)
3. (a)
(i)
Person A
The blood glucose level before eating, at 7:00 a.m. is very high on the first
day. This value is much higher (300 mg dL-1) than that of a non-diabetic
person. Blood glucose levels fluctuate widely. Blood glucose increases
sharply and to very high levels after eating and decreases abruptly,
suggesting the use of insulin injections rather than natural release of
insulin.
(ii)
60 – 130 mg dL-1
(i)
I – bronchiole
II – arteriole (branch of pulmonary artery)
III – alveoli
IV – alveolar capillaries
(ii)
Part II brings blood (low in O2, high in CO2) to the alveoli and removes
blood (high in O2, low in CO2) from the alveoli.
Part IV facilitates gaseous exchange by containing air (high in O2, low in
CO2) in close proximity to the blood. It collects CO2 after gaseous
exchange as the first step in its expulsion.
134
(b)
(i)
A smoker’s lungs may contain large air spaces (due to breakdown of
alveolar walls) instead of the numerous small alveoli found in a nonsmoker.
A smoker’s lungs may be darker coloured (due to deposition of carbon
particles and tar) than those of a non-smoker.
Alternative responses include increased mucus, less cilia, presence of
lesions, tumours or fibrous tissue, constricted airways, etc.
(ii)
Non-smokers are exposed to less/no second-hand smoke and its associated
risks (bronchitis, emphysema, cardiovascular disease and encouragement
to smoke)
Smokers are compelled to decrease their smoking (since they will be at
public places at times).
Alternative responses include decreased fire hazard, pollution from
discarded ends of cigarettes, discomfort associated with smoke in one’s
environment, etc.
(iii)
Addictive component – nicotine
Negative impact – Withdrawal symptoms may be severe enough to cause
unconsciousness, physical and mental dysfunction or may be even fatal.
To avoid these extreme effects, an addict may invest the majority of
his/her money and time on the procurement of the drug, at the expense of
one’s responsibilities of family, health and job.
(c)
Numerous stomata provide a large surface area for diffusion of gases.
Numerous air spaces in the spongy mesophyll layer allow gases to be in close
proximity to mesophyll cells, allowing quick diffusion.
An extensive network of xylem vessels ensure that the surfaces of the mesophyll
cells are moist allowing dissolution of gases.
135
4. (a)
DIAGRAM OF A REFLEX ARC
Free nerve endings, called pain receptors, in the superficial area of the skin, are
stimulated by heat (contact with the hot surface). An impulse is transmitted from
the pain receptor to the CNS (most likely the spinal cord) by a sensory neurone
within a sensory nerve. Within the CNS, the impulse is transmitted to a relay
neurone. It is then transmitted to a motor neurone within a motor nerve. The
motor nerve ends at the effector. When the impulse reaches the effector (muscle),
it is stimulated to contract and the body is pulled away from the hot surface.
The arrows on the diagram indicate the path of impulse transmission during this
response.
(b)
Role:
Vasodilation of capillaries in the skin occurs, increasing blood flow to the skin.
This increases the volume of blood (containing heat) flowing close to the surface
of the skin, allowing heat loss from the blood to the environment.
Vasodilation also increases the volume of water (from blood plasma) which enters
the sweat glands, increasing sweat production. Sweat absorbs heat from the body
and evaporates, taking the heat into the atmosphere when it turns into water
vapour.
Hair erector muscles relax, allowing the hairs of the skin to lie flat against the
skin, reducing the amount of warm air trapped between the skin and the hairs.
This increases the temperature gradient between the body and the environment,
encouraging heat loss.
A reduction in body temperature is achieved by these heat loss mechanisms.
Reason:
Enzymes work optimally at specific temperatures. They are inactivated at lower
temperatures and denatured at higher temperatures. Every metabolic reaction is
catalysed by enzymes. If the body temperature is maintained at the optimum
136
temperature of the enzymes, metabolic reactions occur and so, physiological
processes such as respiration occur at an optimum rate.
(c)
Factors:
Release of greenhouse gases by human activity, e.g. combustion, emissions from
industrial plants and vehicles – a blanket of gases above the atmosphere prevent
heat from leaving the earth.
Deforestation – green plants are the only natural means of removal of carbon
dioxide (a greenhouse gas) from the atmosphere. Removal of plants results in
accumulation of carbon dioxide in the atmosphere.
Negative effects:
Flooding – this may destroy, or render uninhabitable, aquatic habitats such as
swamps and lakes. Bogs, marshes and swamps often contain uncommon
ecosystems. Destruction of such habitats increases the risk of elimination of
entire populations or ultimately, extinction of certain species.
Increase in incidence of pathogens, pests and carriers of disease – As temperate
areas get warmer, such tropical organisms will migrate to these areas. Their
presence may cause disruption of food chains / webs, loss of population control
for some species and elimination of populations of other species.
5. (a)
Physiological, e.g. hypertension
Genetic/hereditary, e.g. sickle-cell anaemia
Pathogenic, e.g. malaria
Alternative responses include deficiency diseases, etc. and a variety of examples.
(b)
Natural and artificial immunity
Actively acquired natural immunity – the body already experienced an infection,
and produces antibodies in response to a second infection; Actively acquired
artificial immunity – at a suitable time (not during an infection) the body is
vaccinated with a treated antigen, so that the body is stimulated to produce
antibodies.
Passively acquired natural immunity – antibodies pass across the placenta/breast
milk to provide the unborn/newborn with immunity; Passively acquired artificial
immunity – the vaccine contains ready-made antibodies which destroy pathogens
immediately.
137
Active immunity:
Pathogens may enter a person. They bind to complementary B-lymphocytes.
Antibodies and memory cells are produced. The pathogens are destroyed and the
person is immune to that particular type of pathogens thereafter (since the
memory cells are a larger population of B-lymphocytes than before the infection).
If the pathogens entered naturally, as in a natural infection, the person shows
actively acquired natural immunity. If the pathogens (a weakened form) were
introduced deliberately to a non-infected individual, as in a vaccine, the person
shows actively acquired artificial immunity.
Passive immunity:
Alternatively, the antibodies themselves may enter a person. They destroy any
pathogens, present in the body, to which they are complementary. If the
antibodies are transmitted from mother to foetus/child via the placenta or breast
milk, the foetus/child shows passively acquired natural immunity. If artificially
prepared antibodies are deliberately injected in the bloodstream of a person, the
person shows passively acquired artificial immunity.
Natural
Active: the patient experiences signs
and symptoms of the illness before
immunity is acquired, since the
concentration of pathogens introduced
into the body is large.
Active: the patient experiences no
effects other than those of the
introduced pathogens.
Active: there is financial expense
associated with alleviating signs and
symptoms of the infection.
Passive: immunity is provided during
the early life of a person, and is specific
against pathogens for which the mother
produces memory cells, since the
antibodies are secreted by the mother’s
B-lymphocytes.
Occurs without medical intervention. A
person may be unaware of all the
pathogens for which he/she has
immunity from.
Artificial
Active: the person may experience mild
signs and symptoms of an infection, e.g.
a fever, before immunity is acquired.
The introduced pathogens are low in
concentration or are modified, so the
person does not get full-blown
symptoms or signs associated with the
pathogens.
Active: the person may experience side
effects of a vaccine, since he/she may be
allergic to a component of the vaccine
preparation.
Active: there is financial expense
associated with purchase of the vaccine.
This may be high.
Passive: immunity is provided at the
time/stage of life the person receives the
artificially prepared antibodies. The
person chooses which antibody
preparation he/she will receive, that is,
which pathogen he/she will be protected
from.
Occurs as a result of deliberate
introduction of pathogens or antibodies.
The person is aware of all the pathogens
for which he/she has immunity from and
the duration of the immunity.
138
Any three points from the above table may be used.
Development of artificial immunity:
1. A vaccine is prepared. A vaccine contains the antigens of a particular
pathogen. The vaccine is introduced orally or intravenously into the body.
The antigens (in the vaccine) bind to the B-lymphocytes whose receptors are
complementary to them.
2. The binding of the antigens to the complementary B-lymphocytes stimulates
the B-lymphocytes to undergo multiple cell divisions involving mitosis. A
clone of the selected B-lymphocytes is produced.
3. Some of the B-lymphocytes produced in step 2 secrete antibodies, which bind
to and destroy the pathogens. Others secrete no antibodies but are long-lived.
These are called memory cells and they are a larger population of Blymphocytes than existed before the introduction of the vaccine. If the body is
subsequently infected by the same pathogen, there are enough B-lymphocytes
(memory cells) to execute the same response and destroy the pathogens so
quickly that the person experiences no signs or symptoms of the illness. The
person has acquired immunity to the pathogen.
6. (a)
1) Exchange of maternal and paternal alleles when homologous pairs of
chromosomes cross over during synapsis
2) Independent assortment of chromosomes into daughter cells due to random
alignment of bivalents on spindle fibres at metaphase I
(b)
1) Pollen sacs of the anther dehisce, exposing the pollen grains.
2) An agent of pollination transfers a pollen grain to the stigma of a flower of the
same species.
3) The pollen grain germinates, growing a pollen tube.
4) The male nucleus divides by mitosis, forming two male gametes.
5) The pollen tube grows down the style, under the influence of the tube nucleus.
6) The male gametes travel down the pollen tube.
7) When the tube reaches the micropyle, the male gametes leave the tube and
enter the embryo sac through the micropyle.
8) One male gamete fuses with the female gamete /ovum, forming the zygote.
(The other male gamete fuses with the endosperm nucleus).
139
(c)
Parental phenotype
father - normal
mother – normal
XHY
Parental genotype
Gametes
X
XHXh
(XH) and (Y)
(XH) and (Xh)
Random fertilisation
(XH)
XHXH
XHXh
H
(X )
(Xh)
Offspring genotype
XHXH
XHXh
XHY
Offspring phenotype
normal
normal
normal
Offspring phenotypic ratio
3 normal
:
(Y)
XHY
XhY
XhY
haemophiliac
1 haemophiliac
XhY is a haemophiliac male.
140
CHEMISTRY
141
1. (a)
This is the study of the speed at which reactions occur. The rate of a reaction can
be found by measuring the change in concentration of reactants or products per
unit time.
(b)
(c)
TABLE 1: DATA FOR EXPERIMENTS
Experiment
[HCl] (mol dm-3)
1
2
3
4
0.1
0.2
0.1
0.1
Form of
Zinc
granules
granules
powder
granules
Temperature
(oC)
30
30
30
20
Volume
(cm3)
89
171
187
46
(i)
Zn(s) + 2HCl(aq) 
 ZnCl2(aq) + H2(g)
(ii)
An oxidizing agent causes an element/compound to lose electrons. HCl is
the oxidizing agent. Oxidizing agents gain electrons.
2H+(aq) 
 H2(g)

 0

(iii)
Step 1: Convert mass of zinc to moles,
1
 0.015 moles of Zn.
65
Step 2: Refer to mole ratio of Zinc and H2 in balanced equation  1:1 .
From the mole ratio 0.015 moles of H2 will be produced.
Step 3: Convert number of moles of H2 to volume by multiplying 0.015 by
24 dm3  0.36 dm3 of H2(g).
(d)
Experiment 2: This experiment produces more gas than experiment #1 because
the concentration of reactant molecules is greater therefore the chances of
collision is more frequent.
Experiment 3: This produces the most because of increase surface area. More
molecules can react at the same time and this increases the rate of reaction.
Experiment 4: At lower temperature molecules have less energy to move and
collide therefore the volume of gas is lowered with decrease rate of reaction.
(e)
(i)
Magnesium granules (more reactive than zinc in the reactivity series)
142
(ii)
(f)
2. (a)
(i)
Construction of a simple electric cell using a conductivity meter, power
supply, measuring cylinder, beaker and test reagents:
1- Place an equal volume of diluted sodium chloride, hydrochloric acid
and acetic acid using a measuring cylinder into separate beakers.
2- Place each beaker one at a time into the electric cell and measure the
conductivity of the solution. Barium sulphate is a solid.
(ii)
Sodium chloride solution and hydrochloric acid are both strong conductors
of electricity and are strong electrolytes. They would give a high reading
on the conductivity meter. However, acetic is a weak electrolyte and will
not give a high reading and Barium sulphate is insoluble in solution, that is
it has no mobile electrons and will not conduct electricity.
(i)
Sublimation
(ii)
Light purple gas from the mixture of Iodine and sodium chloride will
emerge and condense on the outer cool surface of the beaker with water.
(iii)
(iv)
The intermolecular force of attraction in NaCl (ionic compound) is much
stronger than simple molecules where the force of attraction is weaker in
covalent bonds. The lattice structure in NaCl is giant sized and made up of
thousands of molecules while I2 has a smaller structure with weak
intermolecular forces of attraction therefore gently heating will vapourise
I2 molecules.
143
(b)
3. (a)
(i)
This method is not suitable for separating the Fe from NaCl.
(ii)
The force of attraction within a Fe molecule is very strong. Fe has metallic
bonding where strong intermolecular forces of attraction occur just like
NaCl. Therefore, both of them need high temperatures to break up their
structure. A magnet can be used to remove the iron fillings.
(i)
Compound A
Name: Ethene
(b)
(c)
4. (a)
Compound B
Name: Propane
(ii)
Compound A burns in air with a sooty flame.
N.B. alkenes contain fewer hydrogen atoms per molecule that results in a
smokey flame.
(iii)
Equation: C2H4(g) + 2O2(g) 
 CO2(g) + 2H2O(g)
(i)
Compound C is ethanoic acid. It is soluble in water because it contains a
hydroxyl group (OH) that has a polar H or hydrophilic group that bonds to
water easily/. Therefore making it very soluble in water.
(ii)
2CH3COOH(aq) + Ca(s) 
 (CH3COO)2Ca(aq) + H2(g)
(i)
A polymer is a compound that is made up of many smaller units forming a
long chain-like molecule. Polymers can consist of thousands of single
units joined together.
(ii)
a)
Addition polymerization
b)
Polypropene
c)
Plastics
Magnesium conducts electricity when solid because it has free (available) mobile
electrons. Metallic bonding is present in magnesium. Here the metal has a neat
arrangement of positive ions held together in a sea of free moving electrons.
However, MgI2 bonding is ionic and the electrons are locked in a crystal lattice
and are not free or mobile therefore it cannot conduct electricity. However, when
MgI2 is molten or in solution it conducts electricity as the molecule separates into
positive cations and negative anions. The cation can move towards the cathode
144
and accept electrons, the anion will move towards the anode and release electrons.
This allows the solution or molten MgI2 to conduct electricity.
(b)
(i)
(ii)
(c)
At the cathode: Mg2+(aq) + 2e- 
 Mg(s)
At the anode: 2I (aq) 
 I2(g) + 2e-
Current (I)  5A
Time  10 mins or 600 secs
Q  It
 5  600
 3000 Coulombs
From the equation for the discharge of Magnesium: Mg2+ + 2e- 
 Mg
2 moles of electrons are required for the formation of 1 mole of Magnesium.
That is, 2  96500 C 24g of Magnesium.
Convert Coulombs produced to moles then convert moles produced to mass by
multiplying by its (Mg) RAM.
3000
 24
2  96500
 0.37 g Magnesium
Moles produced to mass 
5. (a)
(i)
In process X Carbon dioxide dissolved in rain water and falls as carbonic
acid. Process X is called precipitation (rainfall). Process Y is called
photosynthesis. Here plants absorb carbon dioxide from the atmosphere
145
and combine it with water and minerals to make carbohydrates which is
stored in plants and used when needed to release energy.
(b)
6. (a)
(ii)
Process Z is combustion. An example of a fossil fuel is natural gas,
methane CH4. Methane undergoes combustion as shown by the following
equation:
CH4(g) + O2(g) 
 CO2(g) + 2H2O(l).
When fossil fuels burn or combust carbon dioxide is given off. Excess
amounts or quantities of gas pollute the air by trapping heat causing the
atmosphere to be hotter than its normal temperature. This warmer
atmosphere causes a global phenomenon called global warming. Global
warming is the direct cause of harmful effects such as melting of the ice
caps and climate change, which results in excessive flooding and drought.
(i)
a)
Since the formula of the oxide of metal R is RO, R cation is R2+
then the formula for R carbonate is RCO3(s).
RCO3(s) 
 RO + CO2
N.B. If R was more reactive than aluminium it could have been
potassium or magnesium. Potassium and sodium carbonate
undergo no reaction with heat. However, Mg will react to give
oxide and carbon dioxide.
b)
3M(s) + 2Al3+(aq) 
 3M2+(aq) + 2Al(s)
e.g. 3M(s) + 2AlCl3(aq) 
 3MCl2(aq) + 2Al(s)
Displacement reaction. Metal M is more reactive than Aluminium
therefore it will displace Aluminium from solution.
(ii)
M is more reactive than R because it can displace Aluminium from
solution therefore it will react more vigorously with dilute acid than with
R. M is higher up the reactivity series than R which means it is more
reactive.
(iii)
Both of them are metals. They have high melting and boiling points and a
chemical property is that they react with acids to produce salt and
hydrogen gas.
(i)
Source of chlorofluorocarbons - refrigerants, aerosols. Source of
phosphates – Fertilizers, detergents, weathering of rocks.
(ii)
Chlorofluorocarbons destroy the ozone layer in the stratosphere that
protect living organisms from the sun’s dangerous ultraviolet radiation
(U.V.) This harmful radiation causes cancer. The chlorofluorocarbons
break up the O3 (ozone) molecule to O2 and a free radical oxygen, O*.
Phosphate is a nutrient that algae feeds on. Excessive phosphate can cause
eutrophication or algae bloom. This excessive algae growth consumes all
146
the oxygen in the water, leaving none for other aquatic organisms. This
can result in fish kills.
(b)
One advantage of landfills is all refuse or garbage that can cause disease and
pollution can be contained in one area for neutralization or breakdown to harmless
compounds. Some disadvantages of landfills are leachate of toxic pollutants and
compounds can contaminate underground water supply and the mixing of
pollutants can cause a new potent compound (synergistic effect). Also, the release
of methane gas causes global warming and the landfill can harbor disease-causing
organisms like rodents and pathogens.
An advantage of an incinerator is that the heat produced can be used to heat water
to steam, then to steam turbines for electricity generation. Disadvantages of
incinerators are it produces large amounts of ash that has to be disposed of in
landfills and it releases atmospheric pollutants like carbon dioxide and volatile
organic compounds. Therefore, it not only causes land pollution but will also
cause air pollution.
Recycling reduces the need for raw material for the manufacture of new products.
Therefore it reduces the strain of exploration and extraction of new raw materials.
Disadvantages associated with recycling include the high start-up cost and large
amounts of recyclable material needed for the process to be economically viable
and a large area needed for collecting, storing and processing of recyclable
materials. Valuable productive forested and agricultural land can be lost for
building of recycling plants.
147
1. (a)
(i)
TABLE 1: MASS OF POTASSIUM IODIDE SOLUTION AT
VARYING TEMPERATURES
Temperature
(oC)
20
40
60
80
Mass of Beaker and
100cm3 of Water (g)
243
249
245
247
Mass of Beaker and
Salt Solution (g)
405
445
475
511
Mass of Salt
Dissolved (g)
162
196
230
264
(ii)
(iii)
247 gcm-3
(iv)
At 70oC mass of KI  247 g
At 30oC mass of KI  177 g
Mass of KI that will precipitate out from solution at 30oC  247 g  177 g
 70 g
148
(v)
Step 1: Find the number of moles of KI present at 30 oC i.e. in the mass
calculated above, divide by its RMM. (RMM of KI  166)
177 g
Number of moles of KI in 177 g 
166
 1.067 moles
Step 2: 1.067 moles of KI are present in 177g in 100 cm3. (Find the
amount of moles in 1000 cm3 to find the molarity.)
100 cm3 of KI  1.067 moles
1.067
1 cm3 of KI 
moles
100
1.067
 1000 cm3 of KI 
1000
100
 10.67 moles
(b)
Potassium Iodide is an ionic compound that is polar. Polar solvents will dissolve
in polar solutes. Water and potassium iodide is polar so the potassium iodide will
dissolve in water. Ethanol however, is polar but not as polar as water therefore the
potassium iodide is only slightly soluble in ethanol.
(c)
(i)
Barium Nitrate or silver nitrate, beaker, Bunsen burner, retort stand,
evaporating dish, filter paper, filter funnel and conical flask.
(ii)
Dissolve both salts in water as both of them are soluble. To this mixture
add a spatula of Barium Nitrate. This will react to give barium sulphate
which is insoluble. The barium sulphate can be removed by simple
filtration while the sodium chloride left in the solution can be obtained by
recrystallisation. Another method is to dissolve both soluble salts in water.
Then add silver nitrate to it. The silver nitrate will react to remove the
chloride leaving behind barium nitrate in solution that can be obtained by
recrystallisation.
(iii)
Both salts will dissolve to give a clear solution – Upon adding the reagent
(barium nitrate or silver nitrate) a white precipitate will be seen.
149
(d)
TABLE 2: TEST FOR IONS PRESENT IN Q
(i)
(ii)
2. (a)
(i)
Test
A small amount
of solid Q was
placed in a test
tube and heated
over a Bunsen
burner.
A solution of
potassium iodide
was added to a
solution of Q.



Observation
A brown gas was
produced.
Damp blue litmus
changed to red.
A bright yellow
precipitate was
formed.



Inference
Nitrogen gas was
given off. NO3- ions
present.
Acidic gas given off.
This indicates that
Pb2+(aq) + 2I-(aq) 

PbI2(s)(yellow)
X is more easily ionized than Mg or those above it in the periodic table.
The ease of ionization increases down the group II.
Explanation: As you go down group II metals, the atom is getting bigger
as more filled shells are added. The force of attraction between the
outermost electron and inner protons is getting weaker as the filled shells
shield or reduce the attractive force of the valence electron and the nucleus
therefore the outer electron is lost easily.
(ii)
Cl has a stronger oxidizing power than Y.
Explanation: Chlorine has a stronger oxidizing power than Y because it
has less filled shells than Y and the shield effect is less. The force of
attraction between the outer valence electrons and the nucleus is very
strong. Chlorine will gain electrons and it will cause other atoms to lose
electrons therefore chlorine is a powerful oxidizing agent.
(b)
(c)
(i)
Both X and Y is solids.
(ii)
X is a metal it is hard (metallic bonding), has a high melting and boiling
point while Y (non-metal) has covalent bonding and has a lower melting
and boiling point.
(i)
Covalent bonding
150
(ii)
(iii)
3. (a)
(b)
PY3 is the chemical formula. This covalent compound has a low melting
point and boiling point. It does not dissolve in polar solvents e.g. water
and it does not conduct electricity in solution. It is soluble in non-polar
organic solvents.
This is the breaking up of long chain hydrocarbons into small ones by the use of
heat or catalysts or by both. Shorter chains of lower molecular mass hydrocarbons
are produced.
(i)
OH
OH where
(ii)
OH
OH + OH
OH 
 OH
 C6H10O4
OH + OH
O
For continuous linkage and larger molecules:
(
O
O
(c)
O
OH
)n, where n  1
(iii)
Type of polymerization: Condensation polymerization
Family of polymers: Polysaccharides
(i)
Acid hydrolysis will break up the peptide bond releasing smaller units
called amino acids.
(ii)
151
H is added to the NH part to form the amide part while OH is added to the
CO part to form the carboxylic acid.
(d)
(i)
(ii)
4. (a)
(b)
a)
P and R
b)
Q or S
Polyesters
Four factors that influence the rate of reaction are: surface area, temperature,
catalyst and concentration of reactants. For some reactions: light intensity and
pressure.
(i)
The total volume of CO2 produced is 124cm3. N.B. each small block
represents 4cm3 of CO2.
(ii)
To find the mass of CaCO3 used you must convert the volume of Carbon
dioxide produced to moles then use the balanced equation to find the
equivalent number of moles of CaCO3 used. Finally, convert moles of
CaCO3 to mass by multiplying the number of moles by its RMM.
Volume of CO2 produced
Step 1: Convert volume of CO2 to Moles 
R.T.P.
124

24 000
 0.005 2 moles of CO2
R.T.P  24 dm3 but since CO2 produced in cm3 convert to cm3 by
multiplying by 1000.
Step 2: Use balanced equation to find the number of moles of CaCO3
required 1:1 mole ratio between CaCO3 and CO2. This means 0.0052
moles of CaCO3 is needed to produce 0.0052 moles of CO2.
Step 3: Convert moles of CaCO3 to mass by multiplying number of moles
of CaCO3 by RMM.
RMM of CaCO3  100 g
Mass of CaCO3  100 g  0.0052 moles
 0.52 g
(iii)
Using powdered Calcium carbonate instead of chipped will reduce the
reaction time for the experiment. This will occur because the surface area
in powdered Calcium carbonate is increased. This means that more
particles can react or are exposed for reaction at any given time when
compared to the chip.
152
(c)
(i)
Metals are often combined to make alloys because this combination
produces improved and more durable properties than the individual
metals.
(ii)
Metals high in the reactivity series will react and displace those lower
from solution. Manganese is higher than hydrogen therefore it will react
with dilute acids but copper is lower than hydrogen and it will not react
will dilute acids.
Step 1: React the mixture (alloy) with dilute acids, the manganese will
react but not copper.
Step 2: Filter the resulting mixture and collect the residue (this is copper.)
Step 3: Wash the residue with water and dry to collect sample of copper.
5. (a)
(i)
The name of the electrolyte is brine (concentrated sodium chloride.) The
ions present is Na+ (aq), Cl-(aq) and H+(aq), OH-(aq).
(ii)
At the cathode, H+ ions from the water are preferentially discharged,
because it is lower in the electrochemical series than the Na+ ions.
At the anode the Cl- ions are preferentially discharged because they are in
a high concentration. N.B. For positive ions the least reactive
metal/element will get discharged because they have the tendency to
become atoms while more reactive metals have a greater tendency to
become ions.
(b)
(iii)
At the anode: 2Cl-(aq) 
 Cl2(g) + 2eAt the cathode: 2H+(aq) + 2e- 
 H2(g)
(iv)
The ion exchange membrane is to prevent the Cl-(aq) and OH-(aq) ions from
mixing. It allows the Na+ ion to pass through to react with OH- from water
to get NaOH solution (sodium hydroxide.) It prevents the H+ from reacting
with Cl- to produce HCl (hydrochloric acid).
(i)
The anode will decrease its size, it will become thinner as the solid Cu(s)
will go into solution Cu(aq).
(ii)
The ionic equation at the cathode:
Cu(aq) + 2e- 
 Cu(s)
One mole of Cu(s) requires 2 moles of e-.
One mole of e-  Faraday’s constant
 96500 C
153
Two moles of e-  96500  2
 193000 C
1 mole of Cu(s)  193000 C
Step 1: Find the amount of coulombs produced
I  5 A, t  30  60  1800 s
Q  It
 5 1800
 9 000 C
Step 2: Convert coulombs produced to moles by dividing by 19 3000 C.
9 000 C
Number of moles of Cu(s) produced 
193000
 0.0466 moles of Cu
Step 3: Convert moles of Cu to mass by multiplying by RAM.
0.0466 moles  64 g  2.98 g
6. (a)
(i)
Ozone depletion – main pollutant is Ozone depleting substances such as
CFCs.
Global warming – main pollutant is carbon dioxide.
(ii)
Harmful effects of Ozone depletion are:
 Increase in skin cancer
 Increase in cataract of the eyes
 lower crop yield
 lower productivity of the ocean
(iii)
Harmful effects of global warming:
 climate change (more hurricanes/drought)
 melting of ice caps/loss of habitat for polar bears
 heat waves
 rise in sea level
 lower crop yield and ocean productivity
 death to plants and animals that cannot adapt to warmer
temperatures
154
(b)
(c)
Hoteliers going green can assist by:
(i)
Water use – use waste water to water plants and lawn
 washing of towel and sheets in large wash only
 have signs indicating to reuse towels and place to wash only when
dirty
 have signs in rooms to turn off taps when not in use e.g. while
brushing of teeth
 taps with timers to wash hands can be installed in bathrooms
(ii)
Garbage disposal – recycling bins around hotels and in rooms
 all biodegradable waste to be disposed of in compost heap
 use less plastic in hotel restaurants, use washable utensils
(iii)
Energy use – special keys to enter room that will power it (AC/lights) and
which is removed when locking and leaving the room. This will ensure
that the utilities are not left on after leaving rooms
 Use solar water heaters on roof for hot water
 Use energy saving light bulbs (led and fluorescent) throughout
hotel
 Motion sensor lights
I totally agree with the statement because it sets limits on how much one country
can pollute without damaging the environment. It sets a fair playing field for all
countries as excessive pollutants from one country can affect others and even
globally. If these standards are not followed, the country which violated it should
not be able to trade goods and services with countries who abide with the
standards. In order to maintain a clean, healthy environment for all forms of life to
live and flourish all countries must follow common rules and standards that will
allow life to survive.
155
1. (a)
(i)
Salts are formed when metal ions or ammonium ions take the place of the
replaceable hydrogen(s) of an acid. A normal salt is formed if all the
replaceable hydrogen of the acid is removed.
(ii)
Type of salt: An acid salt – only part of the replaceable hydrogen is
removed.
Chemical Formula: H2SO4(aq) + NaOH(aq) 
 NaHSO4(aq) + H2O(l)
(iii)
(b)
(i)
H3PO4(aq) acid salts produced are Na2PO4 – sodium hydrogen phosphate
NaH2PO4 – sodium dihydrogen phosphate
TABLE 1: TITRATION RESULTS
Final Volume (cm3)
Initial Volume (cm3)
Volume Used (cm3)
1
25.5
0.5
25.0
Titration Number
2
3
37.3
40.5
12.2
15.5
25.1
25.0
25.1  25.0
2
 25.05 cm3
(ii)
Volume of acid 
(iii)
Indicators are used to identify the end point. In part A the end point was
identified therefore in part B there was no need to add indicator as the end
point in A is the volume used in B.
(iv)
Collect the filtrate and evaporate the water. The solution of the salt must
be evaporated to dryness over a beaker of boiling water. N.B. in the
preparation of a hydrated salt, the solution of the salt must not be
evaporated to dryness, but left to crystallize.
(v)
2NaOH(aq)
Sodium hydroxide
+ H2SO4(aq) 
 NaSO4(aq) + 2H2O(l)
sulphuric acid sodium sulphate
156
(vi)
Convert gdm-3 to moles by ÷ by its RMM  98g.
gdm-3
Moles dm-3 
RMM
4.9

98
 0.05 moles dm-3
1000 cm3  0.05 moles of H2SO4
0.05
1cm3 
moles
1000
25.05 cm3 of H2SO4 was used
0.05
25.05 cm3 
 25.05
1000
 0.00125 moles of H 2SO 4 reacted
(vii)
From the balanced equation:
1 mole of H2SO4 will react to give 1 mole of Na2SO4.
0.00125 moles of H2SO4 will react to give 0.00125 moles of Na2SO4.
 0.00125 moles of Na2SO4 are produced.
Convert moles to mass by multiplying number of moles by its RMM.
RMM of Na2SO4  142g
0.00125moles of Na2SO4  0.00125 142 g
 0.1775 g
(viii) Experimental errors by the student in weighing, mixing and reading the
apparatus or inefficient method of preparation of the salt.
(ix)
ENERGY PROFILE DIAGRAM FOR AN EXOTHERMIC REACTION
157
(c)
TABLE 2: TESTS PERFORMED ON COMPOUND X
Test
A sample of X was
heated in a dry test
tube.
To one portion of a
solution of X, aqueous
sodium hydroxide is
added dropwise until
in excess and then
heated
2. (a)
(b)
Observation
A brown gas evolved
which turns damp blue
litmus red but does not
bleach it.
 No precipitate
formed.
 Upon heating, a
pungent gas evolves
which turns moist red
litmus blue
Inference
Acidic gas. The gas
could be NO2, nitrogen
dioxide
NH4+ (aq) + OH-(aq) 
NH3(g) + H2O(l)
Ammonia gas given off.
(i)
Electrolysis is the process by which the passage of an electric current
through a substance causes it to decompose.
(ii)
Active anode
(i)
Electrode 2
(ii)
Br-(l)
(iii)
Pb2+(l) + 2e- 
 Pb(s)
(iv)
Current  5 A
Time in seconds  5  60
 300 seconds
Q  It
 5  300
 1500 C
Calculate the number of moles of electrons this is equivalent to:
96500 C  1 mole of electrons
1

1C
C
96500
1
1500 mole of electrons
And 1500 C 
96500
 0.016 mole of electrons
From the reaction, 2 mol of electrons produce 1 mol of Pb.
1mol of electrons produce 0.5 mol of Pb.
0.016 mol of electrons produces 0.5  0.016  0.008 mols of Pb
158
Convert mols of Pb to mass:
 0.008  207 g  1.656 g of Pb
(c)
Experiment 1: magnesium is more reactive than zinc in the reactivity series
therefore zinc cannot displace magnesium from solution. The more reactive metal
can displace the less reactive metal from solution.
Experiment 2: Zinc is higher than copper in the reactivity series therefore it will
displace it from solution. The copper is deposited at the bottom of the beaker. In
addition to this as the copper gets displaced the blue solution slowly fades away.
3. (a)
Compound B: Alkanes
Compound C: Carboxylic acids/organic acid
(b)
C2H6O(l) + 3O2(g) 
 2CO2(g) + 3H2O(l)
(c)
More soluble compound: A
Reason: Compound A contains a polar OH that is hydrophilic i.e. a water loving
group. N.B. the small alcohols like methanol and ethanol are completely soluble
in water owing to the polar nature of the OH group. However, solubility decreases
as the number of carbon atoms in the alcohol increases.
(d)
More reactive compound: C
Reason: Compound C is a weak acid that dissociates to give H+ ions that will
readily react with sodium metal to give salt and hydrogen gas. The OH in the acid
is more reactive than the OH in alcohol.
(e)
Equation:
+ Na(s) 

Propanoic acid
+ H2(g)
Sodium propanoate
2C2H5COOH(aq) + 2Na(s) 
 2C2H5COONa + H2(g)
(f)
Use a glowing splint. The gas given off will relight it.
159
(g)
Name of catalyst: Concentrated sulphuric acid, conc.H2SO4



Conc. H2SO4
+
+ H 2 O(l)
Ethyl propanoate
N.B. when naming esters the alcohol name comes first.
4. (a)
(i)
The chloride  MCl2  159g
M  71 g  159 g
M  159  71 g

 88 g
M  Strontium
(b)
(ii)
M(s) + Cl2(g) 
 MCl2(s)
Sr(s) + Cl2(g) 
 SrCl2(g)
(iii)
a)
The type of bonding in element M is metallic bonding the core of
positive cations are surrounded by a sea of mobile e- (electrons).
b)
The bonding in the chloride of element M is ionic bonding. The
element M will give up two electrons to two chloride atoms. The
element M will be positively charged (+2) while each chlorine
atom on accepting the electrons will turn to chloride ion anion
(-ve) negatively charged. The difference in charges will attach the
molecules together in a bond.
(i)
(ii)
Graphite is a made of carbon only and belongs to group IV. Graphite
bonding is unique as it has mobile e- (electrons) that can conduct an
electric current. Graphite is an excellent electrode as it conduct an electric
current and does not take part in the chemical reactions, it is inert. While
element M is in group 2 and it is a metal that is fairly reactive. If element
M is used as an electrode it can take part in the chemical reaction as it acts
as an active electrode. It can break down during electrolysis process.
BONDING IN CHLORINE
160
5. (a)
(b)
(c)
6. (a)
(b)
(i)
H2(g) hydrogen gas and NaOH(aq) sodium hydroxide
(ii)
a)
2Cl-(aq) 
 Cl2(g) + 2e-
b)
2H+(aq) + 2e- 
 H2(g)
(i)
Water enters the atmosphere by evaporation or respiration by plants and
animals. Water leaves the atmosphere by rain (precipitation), snow, fog,
mist, hail.
(ii)
Trees absorb water from the environment in liquid form and return it,
transfer or convert it to vapour form back to the environment and return
pure water vapour to the atmosphere where it will condense and fall as
rain. Building of homes by removal of trees will greatly hamper the
process and decrease the flow in the water cycle.
(i)
R(OH)2 
 stable, no decomposition
(ii)
The carbonate of R is stable and no decomposition. The carbonate of T
will yield oxide and carbon dioxide.
(iii)
R is more reactive than T it will bond faster to other elements or
substances in nature to form stable compounds.
(i)
Pollution is the contamination of land, air or water environment by
harmful or poisonous substances. These harmful substances are called
pollutants. Once something is contaminated with harmful pollutants it is
unfit for human use.
(ii)
The environment refers to the natural surroundings of living organisms
including living things (biotic component) and non-living (abiotic
components) that affect the organisms.
The main pollutants responsible for global warming are carbon dioxide and
methane. The pollutants responsible for acid rain are nitrogen oxide and sulphur
dioxide.
Carbon dioxide comes from the combustion of fossil fuel while methane comes
from the decomposition of dead organic matter. Nitrogen oxide and sulphur
dioxide comes from the combustion of fossil fuel in factories, power plants and
motor vehicles.
Fossil fuel methane, natural gas, CH4(g) + 2O2(g) 
 CO2(g) + 2H2O(g)
 HNO2(aq) + HNO3(aq)
2NO2(g) + H2O(l) 
161
Nitrous acid nitric acid
SO2(g) + H2O(l) 
 H2SO3(aq)
Sulphurous acid
Global warming can lower agriculture yield/increase pest infestation, increase in
temperature destroys eco-tourism in the ocean by destroying coral reefs
(bleaching). Increase chances of acquiring skin cancer due to hotter temperatures.
Acid rain kills crops, and microbes in soil which decreases productivity. It makes
lakes and streams acidic, killing aquatic life like fish and shell fish that is an
important food source from agriculture.
Acid rain changes the pH of soil, leaves acid droplets that may damage crops and
harm animals that eat vegetation. It can also cause corrosion of buildings and
machinery. Global warming results in melting of the polar ice caps, increase
flooding and changes in the weather (climate change). This unpredicted weather is
dangerous as hurricanes and other natural disasters can destroy agriculture,
wildlife, homes and endanger lives.
Acid rain destroys monuments and historic sites that tourists visit making the site
unattractive.
These pollutants can be reduced by burning less fossil fuels and developing more
efficient conversion methods in obtaining energy from fossil fuel and switching to
alternative, less polluting energy sources like solar and wind.
162
(i)
Figure 1. Temperature against volume
of acid
40
38
36
Temperature (oC)
1. (a)
34
32
30
28
26
24
22
20
0
5
10
15
20
25
30
35
Volume of HCl(cm3)
(ii)
25 cm3 of acid is needed to neutralize 25 cm3 of KOH.
(iii)
Temperature difference  Final temperature  Initial temperature
 38  27
 11°C
(iv)
c  42 kJ kg-1 °C-1
Assuming 1 cm3  1 g
25  25  50 cm3
50 cm3  50 g
m  50 g or 0.05 kg
H  11 0.05  4.2
 2.31 kJ
163
(b)
(c)
(i)
To investigate the rate of reaction between magnesium ribbon and iron III
chloride using different surface area.
(ii)
The reaction is a displacement reaction. The Mg ribbon is more reactive
than Fe in solution. The Mg will displace Fe from solution and Fe will be
deposited on the bottom of the beaker as a reddish solid.
(iii)
3Mg(s) + 2FeCl3(aq) 
 3MgCl2(aq) + 2Fe(s)
(iv)
The contents of beaker A will change to a lighter yellowish colour than
beaker B because the reaction is occurring at a much faster rate while
more Fe will be deposited in beaker A than B.
(v)
The greater the surface area of a Mg ribbon means more molecules of Mg
is exposed to the solution to react. Therefore the surface area of beaker A
is greater and will react faster, displacing Fe from FeCl3 solution faster
than beaker B so the yellowish solution of FeCl3 will fade faster in A and
slower in B.
(i)
Acidified Potassium dichromate or acidified potassium permanganate.
(ii)
The acidified potassium dichromate will change from orange to green
while acidified potassium permanganate will change from purple to
colourless.
Explanation: When dilute acid reacts with sulphite, sulphur dioxide gas is
produced. This gas is a reducing agent and will reduce acidified potassium
dichromate or potassium permanganate to the colour seen above.
 SO42-(aq) + 2MnSO4(aq) + 2H2SO4(aq)
2MnO4-(aq) + 5SO2(aq) + 2H2O(l) 
purple
colourless
 2Cr3+(aq) + 3H2SO4(aq) + H2O(l)
Cr2O72-(aq) + 3SO2(aq) + 8H+(aq) 
orange
green
2. (a)
(iii)
Solution Y will be Calcium hydroxide. The CO2 produced will turn it
milky white.
(i)
Element X will react with water more vigorously than K.
164
(b)
(ii)
The solution formed from this reaction will be basic.
When X reacts with water the hydroxide of X is produced. X is a metal.
Metal oxides and hydroxides are basic while non-metals form acid
solutions.
(i)
2,8,2
(ii)
QCO3
(iii)
QCO3(s) + HCl(aq) 
 QCl2(aq) +H2O(l) + CO2(g)
N.B. Q and Mg belongs to the same group therefore they will react the
same way.
(c)
When metals react with a non-metal the bonding is ionic. The metal ion will give
up one of the electron to become Na+ cation while the non-metal Z will accept
two electrons to become Z2- anion. Since Z needs two electrons, two Na atoms
will each give up one electron. The electrostatic force of attraction between the
two oppositely charged ions causes the ions to form a strong ionic bond.
2Na+ + Z2- 
 Na2Z
(d)
(i)
A reddish brown gas will be given off. The gas is NO2(g). If a glowing
splint is placed at the mouth of the test tube it will relight, indicating that
O2(g) is given off.
N.B. K, Na nitrate will decompose to give nitrate of metal and oxygen gas
while Ca to Cu nitrate will decompose to give metal oxide, nitrogen
dioxide and oxygen gas.
(ii)
Step 1: Convert given mass of Ca(NO3)2 to moles. This can be done by
dividing the given mass by its RMM.
5
 0.030 moles of Calcium nitrate reacted
64
Step 2: Using the balanced equation given in (d) calculate the number of
moles of NO2(g) produced using its proportional ratio.
2 moles of Ca(NO3)2(S) will produce 4 moles of NO2(g).
1:2
Therefore, 0.030moles of calcium nitrate will produce 0.060moles of
NO2(g).
165
Step 3: Convert moles of NO2(g) produced to volume by multiplying it by
RTP i.e.
0.060moles of NO2(g) 24000 cm3  1440 cm3 of NO2(g) produced.
3. (a)
(i)
This is where organic molecules have the same molecular formula but
differently arranged molecules or orientation of molecules in its structure.
(ii)
(b)
(i)
Isomer 1
Isomer 2
Butane
2-methylpropane
Test: You can bubble the two gases separately in a solution of acidified
potassium permanganate or bromine water.
Observation: Compound A will decolourise both reagents. Acidified
potassium permanganate will turn colourless from a purple solution while
bromine water will turn colourless from a brown liquid.
(ii)
Equation: C3H8(g) + 5O2(g) 
 3CO2(g) + 4H2O(l)
(iii)
Use for Compound A: This compound can be used in polymerization to
make plastics such as polypropene.
Use for Compound B: This can be used as a fuel, when it burns it releases
a large amount of energy.
(c)
(i)
X is water as steam.
(ii)
Name: propan-2-ol N.B. OH is on Carbon atom number 2.
Homologous series: Alcohols.
166
4. (a)
(b)
(i)
Magnesium oxide is a solid with a high melting point because the bond is
ionic due to the strong electrostatic force of attraction between the two
oppositely charged ions which causes the ions to form a strong bond. The
ions are pulled closer to each other to form a solid in a lattice structure. A
large amount of energy is needed to break this strong bond therefore the
melting point is very high. Oxygen and sulphur are both non-metals. The
bonding is much weaker than those of Magnesium and oxygen. The bond
between sulphur and oxygen is covalent, forming sulphur dioxide gas. The
molecules are held together by weak Van der waals’ forces of attraction
that are easily broken hence its low melting point.
(ii)
Oxides of sulphur will not conduct electricity as there are no mobile
electrons to share electrons in a molten or ionic aqueous solution.
However, MgO is an ionic compound and will conduct electricity in a
molten state as it will have mobile and free electrons available to conduct
electricity.
(i)
This circuit is not suitable for this experiment since there is no container
or vessel to put the test solution in with the electrodes and there is no
indicator that will detect if the test solution will conduct electricity or not
like a bulb or ammeter.
(ii)
167
(iii)
Conductors
Aqueous lead II nitrate
Aqueous ammonia
5. (a)
(b)
(c)
6. (a)
Non-Conductors
Ethanol
(i)
Zn(s) + CuSO4(aq) 
 ZnSO4(aq) + Cu(s)
(ii)
Zinc foil will react with copper sulphate because Zinc is more reactive
than copper in the reactivity series. Therefore, zinc will react with copper
sulphate by displacing the copper from the sulphate. The zinc will react
with the sulphate to form zinc sulphate while the copper metal will be
precipitated out of solution as copper metal and be deposited at the bottom
of the beaker. As the copper is being deposited the blue colour of copper
sulphate will fade. Silver foil will not displace copper from copper
sulphate because it is lower than copper in the reactivity series and it is
less reactive. Metals higher in the reactivity series will displace metals in
solution lower in the reactivity series.
(iii)
Aluminium will displace copper from copper sulphate solution.
Aluminium is higher than copper in the reactivity series and is more
reactive than copper.
(i)
The conditions for iron to rust are moisture (H2O) and oxygen.
(ii)
Painting iron is important in preventing it from rusting as the paint coats
the iron surface with an air and water repellent that forms a protective
coating around the iron.
Duralumin is an alloy of Aluminium. It is lighter and less dense than aluminium.
This alloy is also more corrosion resistant than aluminium therefore it is preferred
in the manufacture of aircraft.
Suggestion: Two nutrients that can be added is nitrogen and magnesium.
Reason: The lack of essential nutrients to plants can lower its resistance to
disease. The curling of the tips of leaves is an indication of disease caused by
mites. Both nitrogen and magnesium are essential to healthy growth of plants as it
is involved in the production of chlorophyll and many enzymatic reactions in
plants. Green healthy leaves in plants indicate a lot of chlorophyll being produced.
(b)
Advantage 1: Organic fertilizers contain a lot of microbes that is needed to break
down organic matter into simple ions that the plants can absorb. These microbes
such as bacteria and fungi are important in recycling of nutrients.
168
Advantage 2: Organic fertilizers also help maintain and build soil structure as it
contains humus and micronutrients. Better soil structure with humus and organic
matter retains moisture in the soil so that the plant does not wilt easily.
Advantage 3: Organic fertilizers do not leach its nutrients easily. It slowly
releases nutrients to the soil so that the plants can absorb it. No excessive
nutrients are added to the environment that can cause eutrophication.
(c)
Disadvantage: The release of the nutrients is slow and as a result the plant growth
can be slower than with commercial fertilizer.
(d)
Test for soil acidity: A small sample of the soil is obtained. This sample is taken
about from about 6cm below the surface. The sample is shaken with water and
allowed to settle. The settled solution is tested with universal indicator paper or
solution to determine its actual pH. The use of red or blue litmus will determine
its actual pH. Modern tests use a pH probe that is inserted into the moist soil and a
digital indicator is read.
(e)
The chemical name for lime is calcium hydroxide and its chemical formula is
Ca(OH)2. Lime should be used when the soil is too acidic. It decreases the acidity
of the soil i.e. it makes the soil less acidic.
Ca(OH)2(s) + 2H+(aq) 
 Ca2+(aq) + 2H2O(l)
(f)
When lime is added to soil at the same time as ammonia fertilizers it reacts with
each other to produce ammonia gas. This causes nitrogen to be lost from the soil
and a nitrogen deficiency in plants could result. Ammonia gas can also be toxic to
plants and microbes in the soil.
 Ca2+(aq) + 2H2O(l) + 2NH3(g)
2OH-(aq) + NH4+(aq) 
169
1. (a)
TABLE 1: DETERMINATION OF THE SOLUBILITY OF X AT VARIOUS
TEMPERATURES
Experiment
Mass of
Volume of (i)Temperature
(ii)Solubility of
Number
X(g)
water(cm3) at which crystals X (g/100g water)
reappear (oC)
1
2
4
91
50.00
2
2
8
57
25.00
3
2
12
44
16.67
4
2
16
32
12.50
5
2
20
26
10.00
(b)
(c)
Solubility of X increases as Temperature increases.
(d)
From the graph at 60 oC the solubility of X  27.5 g /100 g of water
170
Using the equation from 1 (a)(ii):
2
100  27.5 g
Mass of water
200
 27.5
Mass of water
200
Mass of water 
27.5
 7.27 cm3
(e)
In a solution a solute is completely dissolved in the solvent forming a uniform
solution while in a suspension the solute partially dissolves in the solvent and
particles can be seen suspended in the solvent. In suspension particles generally
settle to the bottom of the solvent.
(f)
Two other factors are pressure and particle size.
(g)
TABLE 2: TESTS CARRIED OUT ON SOLUTION M
(i)
(ii)
(iii)
(iv)
Test
To a portion of
M, add aqueous
sodium
hydroxide until
in excess
To a second
portion of M,
add aqueous
ammonia until
in excess
To a third
portion of M,
add aqueous
sodium iodide
To a fourth
portion of M,
add aqueous
silver nitrate
followed by
aqueous
ammonia


Observation
White precipitate
Soluble in excess






Inference
Al3+ or Pb2+ or Zn2+
or Ca2+ possibly
present
Al3+ or Pb2+ or Zn2+
possibly present
Zn2+, Pb2+, Al3+
possibly present
Zn2+present
White precipitate
formed
Precipitate
soluble in excess
aqueous
ammonia
Yellow
precipitate
formed


Pb2+(aq) + 2I-(aq)
PbI2(s)
No observable
change

NO3-(aq) or SO42-(aq)
possibly present

171
2. (a)
(i)
(ii)
(b)
They have weak intermolecular bonds holding the molecules together. Secondly,
the electronegativity of these two atoms is greatly different therefore they will not
really want to share an electron pair. Chlorine would rather accept an electron
than share.
(c)
Both I and Cl can exist as isotopes. Isotopes are the same element but different
forms have different mass number due to the different amount of neutrons.
Therefore, isotopes have same atomic number but different mass number hence
ICl sample can have different molar mass.
(d)
Cl2(g) + 2KI(aq) 
 2KCl(aq) + I2(aq)
(e)
(i)
2I-(aq) + 2e- 
 I20(s)
2I-(aq) 
 I2(s) – 2e-
(ii)
Chlorine is an oxidizing agent because it causes iodine to lose an electron
and itself to gain an electron.
-1 
0
172
3. (a)
(b)
Crude oil and natural gas (methane)
(i)
Fraction Number
1
2
3
(c)
Fraction Name
Refinery gas
Kerosene
Lubricating oils and waxes
(ii)
Lubricate mechanical parts in machinery and vehicle, polishing wax, wax
paper, petroleum jelly and candles.
(i)
Esters
(ii)
Structure A
Name: ethanoic acid
(d)
(e)
4. (a)
Structure B
Name: propanol
(i)
Starch
(ii)
Glucose: The glucose is soluble in water and will be in the liquid.
Polymer: This is insoluble in water and will settle to the bottom.
Type of polymer formed: Proteins, Nylon
Use: Muscles, rope
(i)
This is the passage of an electric current through an electrolyte resulting in
a chemical change in the electrolyte.
(ii)
During the electrolysis of molten sodium chloride, sodium metal will be
discharge and chlorine gas given off.
173
Reaction at the cathode: Na+(l) + e- 
 Na(l)
Reaction at the anode: 2Cl (l) 
 Cl2(g) + 2eThe Na+ ion will accept electrons from the cathode i.e. the sodium
undergo reduction. The Cl- ions give up their electrons to the anode i.e.
Chlorine ions undergo oxidation.
(b)
During electrolysis of aqueous sodium chloride the following ions are produced:
Na+, H+, Cl-, OH-. At the anode OH- ions are preferentially discharged because
they are in a dilute solution and they are lower in the electrochemical series than
Cl- ions. At the cathode the H+ ions are preferentially discharged because they are
in a dilute solution and they are lower in the electrochemical series than Na+.
At the anode: 4OH-(aq) 
 2H2O(l) + O2(g) + 4eAt the cathode: 2H+(aq) + 2e- 
 H2(g)
(c)
Anodising is a process where the thickness of this aluminium oxide layer is
artificially increased. Firstly- the protective outer layer on a sheet of aluminium is
removed by treating the aluminium with a sodium hydroxide solution. Secondlythe aluminium sheet is now placed in a dilute sulphuric acid solution as the anode
in the electrolysis of sulphuric acid.
The reaction at the anode: 4OH-(aq) 
 2H2O + O2(g) + 4e- . Thirdly, the
oxygen formed reacts with the aluminium anode forming a thicker oxide layer.
4Al(s) + 3O2(g) 
 2Al2O3(s).
5. (a)
(i)
The acid anhydride is sulphur trioxide. Firstly sulphur burns to form
sulphur dioxide.
S(s) + O2(g) 
 SO2(g). Secondly, when a mixture of sulphur dioxide and
oxygen is passed over a catalyst of vanadium (v) oxide, at a temperature of
about 500oC and atmospheric pressure, sulphur trioxide is produced.
Increased pressure increases yield. 2SO2(g) + O2(g)
2SO3(g). The
sulphur trioxide formed is dissolved in concentrated sulphuric acid to form
oleum.
174
(b)
6. (a)
(b)
(ii)
Anhydride. SO3(g) + H2SO4(l) 
 H2S2O7(l). The sulphur trioxide is
dissolved in sulphuric acid to form oleum or pyrosulphuric acid. The
required concentration of sulphuric acid is obtained by diluting the oleum.
H2S2O7(l) + H2O(l) 
 2H2SO4(l).
(iii)
Sulphur trioxide is not dissolved in water because sulphur trioxide reacts
with water vapour to form a mist a mist of acid.
(i)
The alloy stainless steel is preferred to pure iron in making cooking
utensils because it does not corrode or break down and it does not rust like
iron. Stainless steel also does not react with food while it is being cook.
Stainless steel is inert to corrosion and unreactive to food.
(ii)
The lack of iron in the human diet could result in anemia. This is caused
by a shortage of iron in your blood. It can result in the low production of
red blood cells. Therefore, the blood doesn’t carry enough oxygen to the
rest of your body, resulting in weariness, tiredness and weakness.
(i)
Phosphorus – a deficiency causes intense green colouration or reddening
in leaves due to lack of chlorophyll.
Nitrogen- yellow leaf, slow growth, chlorosis
Potassium – wilting, brown spotting, higher chance of damage from heat
and frost.
(ii)
This experiment can be used to test for all nutrient deficiency. Example,
for nitrogen:
i.
Take 4 identical seedlings, plant each one in a pot of identical size
and soil structure. 4 is used in case one dies.
ii.
2 seedlings will have all nutrients in the recommended dosage
while 2 other seedlings will have a very low quantity of nitrogen.
iii. All seedlings will be given the same amount of water and exposure
to light each day.
iv.
The four seedlings will be placed in a green house in order to
reduce attack by pests.
v.
Monitor and observe plant growth for the next month.
vi.
Make deductions based on observations.
(i)
Acidic soil kills microbes in soil that is needed in recycling of nutrients
and breakdown of organic matter.
(ii)
OH-(aq) + H+(aq) 
 H2O(l)
+
Ca(OH)2(s) + 2H (aq) 
 Ca2+(aq) + 2H2O(l)
175
(iii)
(c)
When calcium hydroxide is used to neutralize acidic soil, ammonium
fertilizers cannot be used at the same time. Calcium hydroxide being a
base, reacts with the ammonium ions to produce ammonia gas. This
causes nitrogen to be lost from the soil and a nitrogen deficiency in plants
could result.
2OH-(aq) + NH4+(aq) 
 NH3(g) + H2O(l).
From lime from fertilizer
Disadvantages- Expensive to set up and retain, nutrient solutions must be closely
monitored, workers need more technical knowledge.
To address these disadvantages:
1- Loan or government subsidy programme to start up.
2- Use test kits daily to monitor nutrients
3- Educate and read on hydroponic systems, go to courses that teach it.
176
1. (a)
(b)
(i)
Petroleum (crude oil), methane(natural gas), peat, coals.
(ii)
Ethanol, gasoline, diesel, methane.
(i)
TABLE 1: READINGS FROM FIGURE 2
Mass of conical flask and water (g)
Mass of conical flask (g)
Mass of water used (g)
Final temperature of water (oC)
Initial temperature of water(oC)
Temperature change(oC)
Initial mass of candle and watch glass (g)
Final mass of candle and watch glass (g)
Mass of candle burnt (g)
326
125
201
39
27
12
97.5
96
1.5
(ii)
Heat absorbed by conical flask  125 g  0.816 J g-1 °C-1 12°C
 1291.5 J
(iii)
Heat absorbed by water in conical flask  201 g  4.2 J g-1 °C-1 12°C
 10130.4 J
(iv)
Total heat absorbed by the calorimeter
 (heat absorbed by the conical flask) + (heat absorbed by water in the
conical flask)
 10130.4 J  1291.5 J
 11421.9 J
(v)
Heat of combustion of the candle wax
Total heat absorbed by calorimeter

Mass of candle burnt

11421.9 J
1.5 g
 7 614.6 Jg -1
177
(vi)
(c)
(i)
(ii)
(iii)
The can prevented heat loss to the environment and insulated the conical
flask with heat from the candle for maximum absorption.
TABLE 2: OBSERVATIONS AND INFERENCES
Test
Observation
Q is heated strongly Brown fumes are
in a boiling tube.
seen around the
mouth of the boiling
A glowing splint is
tube.
placed at the mouth
of the boiling tube.
The glowing splint is
The gases evolved
rekindled.
are bubbled into
aqueous calcium
hydroxide.
A white precipitate
if formed.
Dilute HCl is added A green precipitate
in excess to Q and
is formed which is
the mixture
insoluble in excess
warmed. To the
NaOH.
resulting solution,
aqueous NaOH is
added until in
excess.
Dilute nitric acid
 White precipitate
followed by a few
drops of silver
nitrate solution was
added to Q.



Inference
Nitrogen dioxide gas is given off.
NO2(g). Q is a nitrate. N.B. not
potassium or sodium.
Oxygen gas is given off O2(g).
Compound Q contains a carbonate,
the gas given off is CO2(g).
CO2(g) + Ca(OH)2  CaCO3(s) +
H2O(l).

Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s)
(green)
Iron II present
Cl- ions present
Cl-(aq) + AgNO3(aq)  AgCl(s) +
NO3(aq)
The test tube was

left standing in light
for 5 minutes.
Precipitate turns
grey
Ammonium
hydroxide was then
added to the
resulting mixture.
AgCl(s) + 2NH4OH(aq) +
White precipitate Ag(NH3)2Cl(aq) + 2H2O(l).
is soluble in
ammonium
hydroxide

178
2. (a)
(i)
Solid, liquid, gas
(ii)
The particles in a solid are very closely packed with very strong
intermolecular forces of attraction. These strong bonds hold the particles
together while in liquids, the force of attraction between particles are
weaker than a solid but stronger than a gas. In gas the force of attraction is
weak Van der Waals forces that are easily broken which causes the
particles to be scattered.
(iii)
Iodine changes to a gas from a solid. This process is called sublimation.
(iv)
As iodine molecules sublimes it absorbs heat energy which excite the
molecules which then vibrates and break loose of the bonding in its solid
state and goes directly to a gas.
(b)
Sodium chloride
(c)
3. (a)
(b)
Diamond
Two tests for ionic solids are:
1- dissolve compound in water and it will conduct an electric. Ionic solids are
soluble in water and are good conductors of electricity while molecular solids are
not.
2- heat test – molecular solids have low melting and boiling points while ionic has
much higher. Apply heat to the molecular solid and it will melt or vaporize easily
while ionic solids need much more heat to melt.
(i)
Compound A: Alkanes
Compound B: Alkenes
(ii)
Compound A: 2-methylbutane
Compound B: but-2-ene
(i)
This is where hydrocarbons have the same molecular formula but different
structural orientation.
179
(ii)
(c)
4. (a)
pentane
but-1-ene
2,2-dimethylpropane
Structural Isomer of Compound A
2-methylpropene
Structural Isomer of Compound B
(i)
Polymers are very long chain molecules or macromolecules made up of
small repeated units called monomers.
(ii)
Addition polymerization
(iii)
Plastic bags, plastic films, bottles, packaging containers, buckets, plant
pots
(iv)
Nylon, carbohydrates, protein, Styrofoam, Teflon, epoxy, polyester, silk,
wool, DNA, cellulose, rubber.
(i)
Calcium carbonate can be prepared by reacting 50mL of aqueous solution
of calcium chloride with 50mL of aqueous sodium carbonate.
CaCl2(aq) + (Na)2CO3(aq) 
 CaCO3(s) + 2NaCl(aq)




Put on safety goggles then proceed.
Weigh 2g of calcium chloride and 1g of sodium carbonate. Place
this into separate beakers and add 50mL of water and dissolve.
Pour the calcium chloride solution into the beaker with sodium
carbonate and stir.
Allow the mixture to react and settle. 5 minutes after filter the
mixture. The solid calcium carbonate in the filter paper can be
gently heated in an oven to obtain a dry sample.
180
(b)
5. (a)
(ii)
A fertilizer for plants as it contains essential elements for healthy plant
growth or soothe aches, remove odors, soften rough skin with a foot soak,
reduces swelling.
(iii)
Epsom salts (MgSO4) can be dissolved in water molecules because it is an
ionic compound that is very soluble in water or polar solvents. Its
anhydrous form is very hydroscopic which means it has the ability to
attract and hold water molecules.
Permanent hardness in water is hardness due to the presence of the chlorides,
nitrates and sulphates of calcium and magnesium. This cannot be removed by
boiling. The lime scale can build up on the inside of the pipe restricting the flow
of water or causing a blockage. Hard water is formed when water percolates
through deposits of limestone and chalk which are largely made up of calcium
and magnesium carbonates. These ions are dissolved in the rain water and are
carried away to streams and lakes where water is extracted and purified for
domestic and industrial use. Sodium carbonate can be used to remove permanent
and temporary hardness in water.
Na2CO3(aq) + Mg2+(aq) 
 MgCO3(s) + 2Na+(aq).
On reaction with the sodium carbonate, dissolved magnesium or calcium ions are
precipitated as the insoluble carbonate.
(i)
Process P: Clarification – the juice is neutralized by the addition of
calcium hydroxide and heat. This causes insoluble salts to be formed by
the reaction between the calcium hydroxide and sediment impurities.
Process Q: Precipitation/crystallization.
(b)
(ii)
A mixture of crystals and molasses forms massecuite. The centrifugation
process separates the molasses from the sugar crystals.
(iii)
Molasses
(iv)
Bagasse in the factory is used in the boiler room as a fuel to heat water, to
make steam to generate electricity for the factory or it can be used as a
fuel to heat up and evaporate the water from the cane juice.
(i)
Ethanol formation can be derived from the anaerobic fermentation of
sugar or molasses using a unicellular fungi, yeast, under the right
conditions of pH and temperature.
Enzymes supplied by yeast
C6H12O6(s) 
 2C2H5OH(l) + 2CO2(g)
37°C, right pH in aqueous solution
Glucose
Ethanol
181
(ii)
(iii)
Fractional distillation apparatus
This type of reaction is a neutralization reaction. The wine turned acidic
hence the sour taste. The magnesium oxide is a base which will neutralize
the acid. The ethanol is oxidized by oxygen in the air to form ethanoic
acid.
2CH3COOH(aq)
Ethanoic acid
6. (a)
(i)
+ MgO(s) 
 (CH3COO)2Mg(aq) + H2O(l)
magnesium oxide
magnesium ethanoate
The soap may be producing more scum at Ann’s house because the water
is hard. The side of the island Ann lives contains a lot of limestone. The
carbon dioxide dissolved in rain water makes it slightly acidic. This
dissolves calcium and magnesium ions present in the limestone. Calcium
and magnesium ions give rise to hardness in water and take a longer time
to lather.
CaCO3(s) + H2O(l) +
Limestone
(ii)
rain
CO2(g)

 Ca(HCO3)2(aq)
carbon dioxide
hard water
No- soapless detergents are not affected by the presence of calcium ions in
hard water.
182
Note: Hard water does not lather easily with soap (lather is a frothy white
mass of bubbles). Soap can be represented as NaS+ Sodium stearate when
the water is hard i.e. contains Ca2+ from limestone.
2NaS+(aq) + Ca2+(aq) 
 CaS2+(s) + 2Na+(aq)
When all the Ca2+ is removed by the soap then the excess can lather.
(b)
These second generation detergents have a high level of phosphates, which is a
nutrient which cause rapid growth of algae. Encatchment areas such as ponds and
lakes are quickly covered with these fast growing algae. This process is known as
eutrophication. Phosphates also reduce the quality of water for drinking purposes.
Detergents also contain additives such as perborates and enzymes that cause
anxiety.
(c)
(i)
Green chemistry is an area of chemistry that focuses on the designing of
products and processes that minimize the use of hazardous substances. It
maximizes the use of recycling, reducing and reusing in order to prevent
harm to our natural environment.
(ii)
Some of the benefits involved in utilizing the principles involved in green
chemistry are:
1- It help produces less waste in processes therefore there is less to clean
up or treat or neutralized before releasing into the environment.
2- Chemical processes are designed to minimize the production of toxic
substances that are harmful to the environment.
3- The designing of safer chemicals that are target oriented and not broad
spectrum. It is designed to affect their desired function while
minimizing their toxicity.
4- Safer solvent auxilliaries are manufactured and used only when
necessary therefore, less harm to the environment when seldom used.
5- Energy efficiency design maximizes the use of energy produced and
minimizes waste and economic impacts.
6- The use of renewable material reduces the strain and enhances
sustainability of natural resources by reducing its demand. Hence, less
natural resources have to be harvested or removed for manufacturing.
7- The use of catalysis saves time, speeds up reaction and reduces the
need for natural resources. Therefore it minimizes the need for natural
resources. Therefore, it minimizes the need for energy in the reaction
processes, less energy reduces the need for extra fossil fuel.
183
1. (a)
(b)
(i)
This is the change in concentration of reactants consumed or change in
concentration of products formed divided by the time taken for the change.
(ii)
Temperature, catalyst, surface area or particle size, pressure and light for
some reactions.
(i)
RAM KIO3  39.1  126.9  16  3
 214 g
1 mole of KIO3  214 g
 0.214 g of KIO3 
1
 0.214
214
 0.001 moles
Conc. in 100 cm3  0.001mols
0.001
100
0.001
1000 cm3 
1000
100
 0.01 moldm-3

(ii)
1cm3 
10
 Concentration of solution  0.0008 mol dm -3
50
50
Concentra of solution  0.0008 
100
 0.004 moldm-3
184
(c)
Experiment
Solution 1
(cm3)
Solution
2 (cm3)
Distilled
water
(cm3)
(i) Time
taken for
blueblack
colour to
appear (s)
(ii)
Reciprocal
time (s-1)
Concentration
of KIO3 after
mixing
(moldm3)
1
2
3
4
5
6
5
10
15
20
25
20
Solution
X
10
10
10
10
10
10
35
30
25
20
15
20
83.5
41.5
28.0
21.0
16.5
24.0
0.012
0.024
0.036
0.048
0.061
0.042
0.0010
0.0020
0.0030
0.0040
0.0050
0.0035
(iii)
(d)
As the concentration of KIO3 increases the time taken for the reaction decreases
hence, rate of reaction increases.
(e)
After mixing it is 0.0035 moldm-3.
(f)
(i)
Time
(ii)
Total volume of solution reacting. Concentration of sodium hydrogen
sulphite (NaHSO3)
185
2. (a)
(b)
To remove stains, sanitation (kill bacteria)
(i)
Oxidizing agent causes oxidation by causing an atom to lose electrons
while the oxidizing agent gain the electrons.
(ii)
Oxidation is the loss of an electron from an ion, atom or molecule to
another ion, atom or molecule. (Increase in oxidation state.)
(c)
Na+, Clx, O2-  0
1 + x + (-2)  0
Cl (x)  +1
(d)
(i)
Cl2 
 Cl-
(ii)
Reduction because it gained an electron to become -1.
(iii)
It will turn damp blue litmus paper red then bleaches it white.
Chlorine makes damp starch iodide paper turn blue-black.
(e)
3. (a)
(b)
(c)
0 
 -1
The oxidation number of Oxygen in H2O2 is -1.
The oxidation number of oxygen in H2O is -2. Oxygen is going from -1 to -2.
∴oxygen gains an electron and reduction is taking place.
N.B. oxygen’s oxidation numbers are -1 in hydrogen peroxide, -2 in water and 0
in oxygen and hydrogen stays the same at +1.
(i)
Propanoic acid
(ii)
Organic acids or carboxylic acids
(iii)
It reacts with bases to give salt and water (neutralization reaction). It
reacts with alcohols to give esters.
(i)
Butanol
(ii)
Alcohols
(iii)
CnH2n+1OH
(i)
They are both colourless gases with low melting and boiling point and are
very volatile and less dense than air.
(ii)
G
186
(iii)
1,2-dibromobutane
4. (a)
(iv)
Yes
(v)
2C4H10(g) + 13O2(g) 
 8CO2(g) + 10H2O(g)
(vi)
Easily transported and stored. Very flammable and give out a lot of heat
energy, readily available and cheap.
(i)
This is a negatively charged atom that is attracted to the positive electrode,
an anode.
(ii)
OH-(aq), SO42-(aq)
(iii)
(b)
Q  It
Firstly calculate how much coulombs are produced.
Q   I  60  60   965
 3474 000 C
Then, convert coulombs to moles of electrons. From the equation 2 moles of
electrons are needed to liberate one mole of H2(g) i.e. (2 x 96500)C  1 mole of
H2(g)
187
1
moles of H2(g)
2  96500
1
 3474 000 C 
 3474 000
2  96500
 18 moles of H2(g)
1C
Finally, convert the number of moles of H2(g) produced to Vol. by multiplying by
22.42:
Volume  18  22.42
 403.2 L
(c)
5. (a)
This process is called electro-refining of nickel. The electrode must be an active
electrode i.e. it must take part in the reaction. Impure nickel is made the anode
and a strip of pure nickel is made the cathode. The electrolyte is a mixture of
nickel sulphite or chloride and sulphuric acid. A large current is used during the
electrolysis. During the reaction the anode will get thinner and the cathode will
get thicker. This occurs because during electrolysis the nickel atoms will leave the
anode and enter the solution as nickel ions while nickel ions are discharged at and
deposited on the cathode.
Petroleum, Methane (natural gas), coals and peat.
(b)
Fraction: Kerosene, diesel, gasoline
Use: Fuels
Fraction: Bitumen
Use: to make roads
(c)
(i)
Breaking up of larger, long chain hydrocarbons into smaller chains or
units.
(ii)
Catalytic cracking uses a catalyst and relatively low temperatures while
thermal cracking uses large amounts of heat.
(i)
Pentane + sodium hydroxide 
 no reaction
(d)
+

 no reaction
Butanoic acid + sodium hydroxide 
 sodium butanoate + water
188
+
(ii)
6. (a)
(b)


+ H2O(l)
C5H12(g) + 8O2(g) 
 5CO2(g) + 6H2O(g)
Calcium hydroxide Ca(OH)2 and ammonium sulphate (NH4)2SO4.
Ammonia gas is a base. If sulphuric acid is used a neutralization reaction will
occur. That is the sulphuric acid will react with the ammonia gas and you will not
get dry ammonia.
2NH3(g) + H2SO4(aq) 
 (NH4)2SO4(aq) ammonium sulphate
N.B. anhydrous calcium chloride will also react with ammonia gas.
(c)
(i)
Its density and solubility in water.
(ii)
The method that is best suited is upward delivery. Downward delivery is
not suited as ammonia is less dense than air and it will rise to the top and
escape into the atmosphere. Displacement of water is not suited because
ammonia gas is very soluble in water and will dissolve in it causing the pH
to increase.
(d)
Diagram showing the laboratory preparation of dry ammonia
189
1. (a)
(i)
TABLE 1: RESULTS OF EXPERIMENT
Time (min)
Temperature(oC)
0
80
1
73
2
68
3
65
4
65
5
65
6
64
7
60
8
55
9
50
10
48
(ii)
(iii)
Solid, liquid, gas
(iv)
The constant temperature at which a solid changes from a solid to a liquid.
(v)
65 oC
190
(vi)
3-5 minutes
(vii)
Hold boiling tube away with wooden forceps and point away from body to
avoid injury. Add anti-bumping granules into water bath for uniformed
boiling.
(viii)
Particles of A at 0 minute
Particles of A at 10 minutes
191
(b)
(i)
(ii)
(iii)
(iv)
(v)
2. (a)
(b)
Test
Place a small portion
of solid, B in a test
tube and heat strongly
using a Bunsen
burner
Dissolve the
remainder of B in
about 10cm3 of
distilled water, stir,
then filter. Collect the
filtrate and divide it
into three equal
portions for use in the
following tests
To the first portion of
the filtrate from
(b)(ii), add aqueous
NaOH solution
slowly until in excess
To the second portion
of the filtrate from
(b)(ii) add aqueous
NH3 slowly until in
excess
To the third portion of
the filtrate from
(b)(ii), add aqueous
KI






Observation
A reddish brown
gas evolved
White
precipitate
formed
Soluble in
excess
White
precipitate
formed
Insoluble in
excess
Yellow
precipitate
formed

Inference
NO2, nitrogen
dioxide gas


Al3+, Ca2+, Pb2+, Zn2+
Al3+, Pb2+, Zn2+


Al3+, Pb2+, Zn2+
Al3+, Pb2+


Pb2+
Pb2+(aq) + 2I-(aq) 
PbI2(s)
An oxidizing agent causes a substance to lose electrons and itself gains electrons
(reduced).
A reducing agent causes the substance to gain electrons and itself lose electrons
(oxidized).
(i)
Acidified potassium dichromate (VI)/ acidified potassium manganate
(VII)
(ii)
Test 1: C is a reducing agent. N.B. manganate is an oxidizing agent.
Test 2: C is an oxidising agent. N.B. KI is a reducing agent.
(iii)
Test 1: the colour change is from purple owing to the presence of the
mangante (VII) ion (MnO4-) to colourless owing to the formation of the
manganese (II) ion (Mn2+). The mangante (VII) ion gains electrons from
the reducing agent.
192
Test 2: the colour change is from colourless owing to the presence of the
iodide ion (I-) to brown owing to the formation of Iodine in solution.
3. (a)
(b)
(iv)
Fe(s) + Cu2+(aq) 
 Fe2+(aq) + Cu(s)
(v)
In the electrochemical series the element higher in the series will displace
the element lower. In test 3, iron is higher than copper in the
electrochemical series, therefore iron will displace the copper in the
solution. In test 4, no reaction took place because silver is lower than
copper in the electrochemical series and therefore cannot displace the
copper from the solution.
Compounds that have the same molecular formula but their structure are oriented
differently.
(i)
Pentene
(ii)
It is a gas. It has a low boiling point. It is less dense than air. It is volatile.
(iii)
Decolourised acidified potassium mangante from purple to colourless.
Decolourised bromine solution from red-brown to colourless.
(iv)
Pentene
Compound P
(c)
(i)
(ii)
(d)
2-methyl-but-2-ene
Isomer of compound P
The properties of liquid. Volume and density (low).
Pent-2-ol
193
(e)
4. (a)
170oC, concentrated sulphuric acid
An allotrope is different forms of the same element in the same physical state.
Diamond
(b)
Graphite
(i)
Atoms in metals are tightly packed together and as a result the valence
electrons become delocalized and cations are formed. This sea of cations
and electrons are held together by strong forces of electrons.
(ii)
In graphite the 4th electron of carbon atom becomes delocalized allowing
electricity to move through it.
(c)
Bonds in Ca2+ is stronger therefore it requires more energy to breakdown than Na+
which only has one bond which requires less energy to breakdown. Since Ca2+
and O2- have relatively higher charges than Na+ and Cl-, there is a stronger force
of attraction.
5. (a)
(b)
Covalent bonding because they are both non-metals by which carbon has 4
valence electrons and would rather share than gain or give up electrons.
(i)
 CCl4 + 4HCl
CH4 + 4Cl2 
194
(ii)
Substitution reaction, UV light, sunlight.
(c)
Alkene: C4H8
(d)
(i)
Acid: C4H8O2
Lighting splint: alkanes burn with a clean, blue flame while alkenes burn
with with a smoky orange-red flame.
An alkene will decolourize KMnO4 (VII) rapidly while an alkane will not.
6. (a)
(b)
(ii)
Carboxylic acids react with alcohols to produce esters in a process called
esterification whereas alkanes do not.
(i)
Extraction of the mined bauxite ore in Jamaica occurs through the Bayer
process which involves treating the crushed bauxite with moderately
concentrated sodium hydroxide solution. This results in a white product
called alumina or aluminium oxide. This is then taken to another
processing plant to be converted into aluminium via electrolysis, which is
conducted in a large tank lined with carbon which acts as the negative
electrode. Huge blocks of carbon are hung above the middle of the tank
acting as the positive electrode. In order to reduce the melting point of the
pure alumina it is dissolved in molten cryolite (Na3AlF6) which has a
much lower melting point. Once dissolved, its ions are free to move. At
the cathode Al3+ ions undergo reduction and gain electrons.
 Al(l)
Al3+(l) + 3e- 
(ii)
Since the majority of the Caribbean islands are relatively small, it is
difficult to extract aluminium mainly because there is not enough space on
the island. The waste size of aluminium is large and its extraction must
take place somewhere where it is able to be absorbed like forests with a lot
of trees.
Duralumin is better to use because it is strong, lightweight and able to withstand
greater stress than pure aluminium. Magnalium is better because it is stronger,
harder and more corrosion resistant than pure aluminium. It has a lower density
thus it is also lighter than aluminium.
195
(c)
Though metals such as mercury are useful in the manufacturing of many products
(thermometers, car batteries etc.) it poses a threat to the environment. It can
damage the CNS of the body causing degenerative neurological disorders and
death.
196
1. (a)
This is a solution of known concentration. It is made by dissolving a known
weight of solute in a specific volume.
(b)
17.11
13.95
3.16
Mass of beaker and sample (g)
Mass of beaker (g)
Mass of sample (g)
(c)
(i)
(ii)
(iii)
Burette Readings(cm3)
Final volume
Initial volume
Volume used
Titration 1
25.6
1.00
24.6
Titration 2
26.3
2.05
24.25
Average volume of KMnO4 used in the titrations 
Titration 3
24.5
0.15
24.35
 24.5  24.35  cm3
2
 24.3 cm
3
(iv)
0.01 moles of KMnO4 in 1000 cm3
0.01
No. of moles in 1cm3 
1000
 0.00001 moles
No. of moles in 24.3 cm3 KMnO4  0.00001 24.3
 0.000243 moles
(d)
(i)
According to the balanced equation 1:5 mole ratio
25 cm3 of Fe2+ will contain 5  0.000243  0.001215 moles of Fe2+
(ii)
In 25c m3 of Fe2+  0.001215 moles
0.001215
In 1cm3 of Fe2+ 
25
 0.0000486 moles
3
2+
 250 cm of Fe  0.0000486  250
 0.01215 moles
197
(e)
Mass of anhydrous FeSO4  152  0.01215
 1.8468 g
(f)
Mass of water in the hydrated FeSO4  3.16 g 1.8468 g
 1.3132 g
(g)
No. of moles of water in the hydrated sample 
1.3132
18
 0.0730 moles
0.073
0.01215
6
(h)
n
(i)
The colour change at the endpoint is from purple to colourless/light pink.
(j)
One of the reagent or reactant is coloured. KMnO4 is purple. On addition to
FeSO4 the colour of the solution changes to pink.
(k)
(i)

(ii)


(iii) 

2. (a)
Test
Aqueous sodium
hydroxide was added
dropwise
And then in excess
The resulting mixture
from (i) was left to
stand in air
Aqueous barium nitrate
was added
Followed by dilute
nitric acid





Observation
Dirty green
precipitate
Insoluble in
excess
Turns brown on
exposure to air
Inference
Fe ions present
White precipitate
insoluble in acid
SO42- ions present
2+
Fe2+ ions oxidized
to Fe3+
(i)
This is the passage of an electric current through an electrolyte causing a
chemical change such as decomposition.
(ii)
This is the coating of or depositing a thin layer of one metal on top of
another.
(iii)
1- Purification of metals eg. Copper
2- Anodising of metals eg. Aluminium
3- Extraction of metals from ores eg. Aluminium
198
(b)
(i)
Both electrodes are made of carbon.
3. (a)
(ii)
Equation at anode: 2Br- 
 2e- + Br2 (g)
 Pb (s)
Equation at Cathode: Pb2+(l) + 2e- 
(i)
Alkane
(ii)
Fuel, Fuel in cigarette lighters, propellant in aerosols, a refrigerant
(iii)
Sunlight or UV rays
(iv)
+ Cl2(g) 
 HCl(g) +
(b)
(c)
(v)
It will react with ammonia gas to form ammonium chloride which is seen
as white fumes.
(i)
A polymer is a long chain of combined smaller units called monomers.
The monomers are attached to each other via their functional group to
form larger molecules. Polymers can be made up of millions of smaller
units.
(ii)
Type of polymerization: Addition polymerization
Name of polymer: polypropane
Use of polymer: plastics
(i)
Ester linkage
199
4. (a)
(ii)
Type of polymerization: condensation
Use of polymer: cosmetics, perfumes, synthetic flavor
(iii)
Water
Isotopes are elements that have the same atomic number but different mass
number due to the different number of neutrons. Both carbons have the same
atomic number, 6, i.e. they have six electrons and six protons 126 C has 12  6  6
neutrons. 136 C has 13  6  7 neutrons. Both
12
6
C and
13
6
C has the same atomic
13
6
mass but different mass number. C has one extra neutron.
(b)
Cobalt 60 is used to arrest the development of cancer.
Iodine 131 is used to treat hyperthyroidism.
Carbon 14 is used in breath test to detect the ulcer causing bacteria and in carbon
dating eg. age of fossil.
Plutonium 238 is used to provide energy for heart pacemakers.
Thallium 99 is used to monitor heart disease. Sodium 24 is used to trace the flow
of blood and to locate obstructions in the circulatory system.
(c)
(i)
W belongs to group 7 (halogen) period 2.
X belongs to group 2 (alkaline earth metals) period 3.
The type of bonding is ionic bonding. X2+, W-. XW2, metal first followed
by non-metal.
(ii)
The compound above will dissolve in water. Ionic compounds are of high
solubility in polar or ionic solvents. When the oppositely charged ions in
the solid ionic lattice are surrounded by the opposite pole of a polar
molecule, the solid ion is pulled out of the lattice and into the liquid.
5. (a)
(b)
Colourless gas, less dense than air, boiling point is -33.3 oC, freezing point is –
77.7 °C, alkaline gas, pungent smell.
(i)
200
(ii)
Ca(OH)2(s) + 2NH4Cl(s) 
 CaCl2(s) + 2H2O(l) + 2NH3(g)
(iii)
Concentrated sulphuric acid will react with the ammonia gas in a
neutralization reaction as the ammonia gas is alkaline to give ammonium
sulphate.
H2SO4 + NH3 
 NH4SO4
Calcium oxide and ammonia are both basic compounds so they do not
react with each other.
(c)
Ammonia gas will turn red litmus paper blue and it will react with hydrogen
chloride vapour to form a white dense gas, ammonium chloride which has a
pungent smell.
(d)
Eutrophication
Poison water which becomes hazardous to health and can lead to
metheamoglobinemia or blue baby syndrome
Potential cancer risk (nitrates can react with amines or amides in the body to form
nitrosamines which is known to cause cancer.
Anoxia- this is a lack of oxygen due to high levels of nutrients eg. Nitrates. It
causes fish kills.
6. (a)
High specific heat capacity - helps in the absorption of heat from the body and
cools it via sweating and evaporation.
Lower density of ice causes lakes and rivers to freeze from the top which insulates
the water below preventing it from freezing which allows life to persist below.
It is attracted to polar molecules. Cohesion causes surface tension that allows
insects to walk on water (habitat). Adhesion – capillary action helps plants absorb
water and nutrients from soil.
Water has a high polarity which helps to dissolve a wide range of solutes. It is an
ideal solvent eg. Blood.
(b)
(c)
(i)
Ca(HCO3)2(aq) + (Na)2CO3(aq) 
 CaCO3(s) + 2NaHCO3(aq)
(ii)
Water can be softened by ion exchange resins. Water is slowly passed
through a column of ion exchange resins. The dissolved calcium and
magnesium ions displace the sodium ions in the resin.
Ca2+(aq) + Na2Z(s) 
 CaZ(s) + 2Na+(aq)
Hard water does not lather with soap.
1- Place 100 cm3 of water sample in a boiling tube with a small piece of soap.
2- Cork the boiling tube and shake for 1 minute.
3- Observe the content of the tube. If it lathers it is soft water. If it does not lather
it is hard water. Lather is a frothy white mass of bubbles.
OR
201
1-Place 10 cm3 of pure water in a test tube A and 10 cm3 of hard water in another
in B
2-Insert equal amounts of soap into each test tube and shake vigorously
3-Observe the test tubes. Test tube A should have lather while test tube B
shouldn’t.
202
1. (a)
TABLE 1: RATE OF REACTION BY MASS OF MAGANESE (IV) OXIDE
Experiment
1
2
3
4
5
6
Mass of Maganese (IV)
oxide (g)
0.0
0.5
1.5
2.2
3.6
4.0
Rate of reaction (mLs-1)
0.0
0.6
1.8
2.7
4.5
5.0
(b)
The rate of a reaction is given by either the change in concentration of reactants or
products with time at a stated temperature.
(c)
2H2O2(l ) 
 2H2O(l) + O2(g)
(d)
203
(e)
(i)
As the mass of the catalyst increases the rate of the reaction increase.
More alternate pathway and orientation of successive collision for reactant
to react occurs therefore the rate of the reaction increases. This implies
that the mass of catalyst is directly proportional to the rate of reaction.
(ii)
From the graph, 3.0 g of the catalyst will give a reaction rate of 3.7 mLs-1.
(iii)
Rate  3.7 mLs-1
After 10 s the volume of oxygen produced is 3.7 mL 10  37 mL
(f)
(g)
Glass tube
(h)
Temperature, surface area, pressure, light
204
2. (a)
(i)
Mole: this is the amount of a substance which contains as many
elementary entities (atoms, molecules, ions, electrons or protons) as there
are carbon atoms in 12 g (0.012 kg) of Carbon-12.
Molar mass: this is the sum of all masses of each element in a molecule or
compound and is expresses as g/mol or kg/mol OR the mass of 1 mole of
any substance expressed in grams is called the molar mass and has units of
gmol-1.
(b)
(ii)
1 mole of aspirin  180 g
0.2 moles of aspirin  180  0.2
 36 g
(iii)
180 g of aspirin  1mol
1
1g
moles
180
1
 18 g of aspirin 
18
180
 0.1 moles
From the given equation, when one mole of aspirin is hydrolysed, one mole of
ethanoic acid is produced i.e. 1:1 or 0.1 moles : 0.1 moles.
0.1 moles of ethanoic acid is produced.
1 mole of ethanoic acid  60 g
60
0.1 moles of ethanoic acid   0.1
1
 6 g of ethanoic acid
(c)
(i)
This is a substance that dissociates into mobile ions in a solution. It has
mobile ions and can conduct an electric current.
(ii)
CH3COO- (aq) and H+ (aq)
(iii)
This is the negatively charged electrode in an electrolytic cell. At this
point e- are discharged into the electrolyte solution. N.B. At the cathode
positive ions called cations are attracted to it. They accept electrons from
the cathode.
(iv)
H+(aq) cation
(v)
This is a weak electrolyte because it partially dissociated in water.
CH3COOH(aq)
CH3COO-(aq) + H+(aq)
205
3. (a)
(i)
Name of Homologous series: Alkanes
General formula: CnH2n+2, n ≥ 1.
(ii)
This is a gas. It is used as a primary blowing agent in the production of
polystyrene foam and other foams. It is an ingredient for petrol fuel and is
used as solvents.
(iii)
This is when molecules have the same molecular formulae but different
orientation of atoms in the structure (or different structural formulae.)
(iv)
Pentane
2- methylbutane
(b)
(c)
2,2-dimethylpropane
(i)
Compound Q is soluble in water because it has a hydrophilic group – OH.
It is a polar compound with a water loving end. The polar OH group forms
hydrogen bonds with H2O.
(ii)
This compound Q is acidic because it belongs to carboxylic acids. The
functional group is acidic COOH and H+ from the functional group will
dissociate into solution which will make it acidic.
(i)
This will change from red-brown to colourless.
(ii)
2,3-dibromobutane
206
4. (a)
A salt is an ionic compound which is formed from a neutralization reaction
between an acid and a base. Calcium carbonate is used to manufacture dry walls,
chalk for writing, agricultural lime to reduce acidity in soil and tablet in medicine.
Sodium nitrate is used for the manufacture of fertilizers, pyrotechnics, smoke
bombs, pottery enamels and food preservation.
(b)
Calcium chloride + sodium carbonate 
 calcium carbonate + sodium chloride
CaCl2(aq) + Na2CO3(aq) 
 CaCO3(s) + 2NaCl(aq)
Put on safety goggles. Get two 250 mL beakers. To each one add 50 mL of water.
To one beaker add 10 g of sodium carbonate. To the other add 10 g of CaCl2. Stir
to dissolve each component. Add the contents of one beaker into the other and
mix for 5 minutes. Allow this mixture to settle for 5 minutes then filter using
simple filtration method. The calcium carbonate salt will remain in the filter paper
as it is an insoluble salt. Rinse with distilled water and allow to air dry.
(c)
Na2CO3(s) + 2HNO3(aq) 
 2NaNO3(aq) + H2O(l) + CO2(g)
Sodium carbonate + nitric acid (V) 
 sodium nitrate + water + carbon
Dioxide
Add Na2CO3 in excess to neutralize all acid, then filter out excess and collect
filtrate as NaNO3(aq).
5. (a)
Fermentation is a process of making ethanol from food groups full of sugars and
carbohydrates by yeast. The starting material can be any food full of natural
glucose and carbohydrates. Yeast and water is added to make a mixture. Yeasts
are unicellular fungi and are the source of the enzyme which convert the
carbohydrates to ethanol. Yeast grows exponentially and multiply rapidly when
water, nitrogen, vitamins and mineral salts are present.
207
Enzymes (zymase), 37°C
C6 H12O6 
 2C2 H5OH 2CO2
Right pH in aqueous solutions

 ethanol + carbon dioxide
Sugar, starchy material 
Enzymes supplied by yeast
(b)
Distillation is necessary because the ethanol and remaining water are miscible. In
order to separate the two from the filtered mixture distillation is used.
208
(c)
The reagent used is acidified orange potassium dichromate (vi) crystals. A colour
change with reaction of ethanol goes from orange to green. This is the oxidation
of ethanol.

 CH3COOH
CH3CH2OH + (O) 
K Cr O /H +
2
Ethanol
6. (a)
(i)
2
7
Ethanoic acid
Least reactive to most: B, Fe, A.
A(s) + FeSO4(aq) 
 A2SO4(aq) + Fe(s)
2+
A(s) + Fe (aq) 
 2A+(aq) + Fe(s)
(ii)
A + CuSO4(aq) 
 A2SO4(Aq) + Cu(s)
Cu(s) is lower in the reactivity series than Fe(s). Since A is more reactive
than Fe(s) it will displace Cu(s) from solution. The blue solution of copper
II sulphate will slowly fade away as the copper is displaced and deposited
as shiny brown solid.
(iii)
(b)
A is very reactive and since vast amounts of energy is required,
electrolysis is used to extract it from its ore while B is less reactive and
reduction with coke (C) is used instead.
An alloy of iron is steel (Fe 50%, chromium 10-30%, small amounts of carbon,
nickel, manganese). This is used in the construction industry because of its
strength and cost. Used in buildings and bridges.
209
1. (a)
(i)
TABLE 1: TITRATION OF SOLUTION M (HCl) WITH SOLUTION N
(NaOH)
Burette Readings
Titration 1
Titration 2
Titration 3
(cm3)
(Rough)
Final Volume
25.70
38.30
40.50
Initial Volume
0.25
13.30
15.50
Volume of solution
25.45
25.00
25.00
M used
(ii)
25  25
 25 cm3
2
(iii)
Mass concentration  3.6 gdm-3
gdm-3
Molar concentration 
RMM
3.6 gdm-3

36.5 gmol-1
 0.1 moldm-3
(iv)
1000 cm3  0.1 moles
0.1
1 cm3 
1000
0.1
 25 cm3 
 25
1000
 0.0025 moles
(v)
HCl(aq) + NaOH(aq) 
 NaCl(aq) +H2O(l)
(vi)
From the balanced equation , 1 mol of HCl will react with 1 mol of NaOH
1:1 mole ratio
0.0025 moles of HCl will react with 0.0025 moles of NaOH.
210
(vii)
In 25cm3 of NaOH  0.0025moles
0.0025
1cm3 of NaOH 
25
0.0025
 1000cm3 of NaOH 
1000
25
 0.1 moldm-3
gdm-3
RMM
Mass concentration (gdm-3)  Molar concentration  RMM
 0.1 moldm-3  40 g
Molar concentration 
 4 gdm-3
(b)
(i)
Van der Waals forces
(ii)
NaCl’s melting point is higher because it forms an ionic crystal lattice
structure when it forms bonds. The bonds between the Na+ and Cl- are
ionic which makes it a very strong structure and requires a lot of energy to
break it. NaCl has very strong intermolecular forces of attraction that bind
the structure together.
(iii)
2. (a)
(i)
Acid - A substance that dissolves in aqueous solution to liberate H+ ions.
Alkali - A substance that liberates OH- ions when dissolved in aqueous
solution.
211
(ii)
Acid salts - A salt which is able to liberate H+ ions when dissolved in
solution because the salt is formed by incomplete replacement of the
hydrogen of an acid.
Normal salts - A salt which all of the acid hydrogen atoms have been
replaced by a metal or the hydroxide radical of a base are replaced by the
acid radical.
(iii)
Na3PO4(aq)
(iv)
3NaOH(aq) + H3PO4 (aq) 
 Na3PO4(aq) + H2O(l)
(v)
Na2HPO4(aq) or NaH2PO4(aq)
(vi)
2NaOH(aq) + H3PO4(aq) 
 Na2HPO4(aq) + 2H2O(l) or NaOH(aq) +
H3PO4(aq) 
 NaH2PO4(aq) + H2(g)
(b)
0.05M H2SO4(aq). Note: lower the pH, stronger the acid.
(c)
(i)
Ascorbic acid, citric acid. Note: all citrus has citric acid, all Vitamin C is
ascorbic acid.
(ii)
Neutralization
(iii)
NaHCO3, sodium hydrogen carbonate. This is the main ingredient in
baking soda.
3. (a)
(i)
(ii)
(iii)
(b)
(i)
Addition reaction
UV light
CH4(g) + Cl2(g) 

 CH3Cl(g) + HCl(g)
UV light
CH3Cl(g) + Cl2(g) 
 CH2Cl2(g) + HCl(g)
212
(c)
4. (a)
(b)
(ii)
Substitution reaction
(i)
A polymer is made up of many smaller units joined together to form a
macromolecule. These smaller units are called monomers.
(ii)
This is when a long chain molecule is formed by combining unsaturated
monomers. This is done by rearrangement of bonds without the loss of any
atom or molecule under specific conditions of heat, pressure and catalysts.
(i)
Endothermic is when heat is absorbed from surroundings for a reaction or
process to occur while exothermic is when heat is released to the
environment when a reaction occurs.
(ii)
Bond making – Exothermic
Bond breaking – Endothermic
(i)
RMM of KNO3  101 g
1 mol of KNO3  101 g
OR
101g of KNO3  1 mol
1
1g of KNO3 
moles
101
1
12 g of KNO3 
12
101
 0.12 mols
N.B.: put what is given off on the L.H.S. and what you want to find on the
R.H.S.
(ii)
Using temperature change as Initial – final, you do not have to change the
sign at the end,
T  4.20 C
100 cm3  100 g , assuming the density of water  1gcm-3.
H  mcT
 100 g  4.2 J  4.2°C
 1764 J
Sign is positive, therefore reaction is endothermic.
(iii)
0.12 mols  1.764 J
1.764
1 mol 
0.12
 14.700 J
213
(iv)
Styrofoam cup - used as an insulating container for the reactants.
Thermometer - used to measure the temperature change.
(v)
H   ve
=1764 J
5. (a)
(i)
Esters
(ii)
(iii)
C4H9COOH – acid
C2H5OH – alcohol
(iv)
Concentrated H2SO4(aq)
(v)
An alcohol is very soluble in water due to the presence of polar OH
(hydroxyl) functional group in the alcohol which is able to form hydrogen
bonds with water while the ester has little or no solubility since it has a
non-polar structure which is unable to form H bonds with water.
(b)
Condition and reagent: An excess NaOH(aq) and heat under reflux.
Balanced chemical equation:
CH3COOC2H5(l) + NaOH(aq) 
 CH3COONa(aq) + C2H5OH(aq)
(c)
Process: Saponification
Source: Animal fat or vegetable oil
214
6. (a)
- It has a high specific heat capacity
-It has a high latent heat of vaporization
-It can dissolve many substances (universal solvent)
-It acts as a shock absorber
(b)
Place blue CoCl2 (Cobalt chloride) paper in the vapor. If water vapor is present
the paper will change from blue to pink OR anhydrous copper II sulphate crystals
from white to blue.
(c)
Ca(HCO3)2(aq) + Na2CO3(aq) 
 CaCO3(s) + 2NaHCO3(aq)
(d)
Ions to anode: OH-(aq) and SO42-(aq) but OH-1 will be preferentially discharged.
Ions to cathode: H+(aq)
Balanced equation at the cathode: 2H+(aq) + 2e- 
 H2(g)
Balanced equation at the anode: 4OH (aq) 
 2H2O(l) + O2(g) + 4eOverall: 4H+(aq) + 4OH-(aq) 
 2H2(g) + O2(g) + 2H2O(l)
Note: Hydrogen gas is given off at a 2:1 ratio to oxygen i.e. for every 2 moles of
hydrogen gas given off. 1 mole of oxygen gas is given off.
215
PHYSICS
216
1. (a)
Length of air column L/mm
152.0
158.0
163.0
170.0
179.0
182.0
Temperature  /ºC
Temperature T / K
14.0
287.0
29.0
302.0
40.0
313.0
57.5
330.5
78.0
351.0
85.0
358.0
(b)
Graph of L/mm vs T/K
217
y2  y1
x2  x1
175  153

342  290
22

52
 0.42 mm K -1
(c)
Slopes, S 
(d)
The slope tells how the length of air column and effectively the volume of air
changes with absolute temperature.
(e)
Value of the length of the air column at 273 K  146 mm
(f)
Charles’ law states that the volume of a fixed mass of gas is directly proportional
to the absolute temperature provided the pressure remains constant.
(g)
Using
V1 V2

T1 T2
V2
2

 273  35  273  75

V
2
 2
308 348
2  348
V2 
308
 2.26 L
2. (a)
(i)
Quantity
Force
Formula
F  ma
Potential energy
E  mgh
Momentum
(b)
p  mv
Unit
N (kg ms-2 )
J  Nm or kg m 2 s -2 
kg ms-1
(ii)
For bodies undergoing collision, the total momentum before collision is
equal to the total momentum after collision, provided no external force is
acting.
(i)
The linear momentum is conserved in the crash. The total momentum
before collision is zero since the momentum of each truck is equal and
opposite (mv  mv) . After collision, the momentum is zero since the
velocity of each truck is zero.
218
(ii)
By the principle of conservation of momentum
Total momentum before impact  total momentum after impact
mb vb  mt vt   mb  mt  v
 0.1 vb    5.0  0    0.1  5.0  6.0
0.1vb  30.6
vb  306 ms -1
3. (a)
(b)
A longitudinal wave is one in which the vibration of the parties is parallel to the
direction of travel of the wave.
(i)
A wavelength – A to E (or (B to F, C to G etc.)
(ii)
Amplitude – B to W (or D to X etc.)
(i)
Using  
(c)
(d)

v
f
340
0.350 1000
  0.97 m
(ii)
The frequency remains unchanged in travelling from one medium to
another i.e. 0.350 KHz.
(iii)
Refractive index of water, nw 
water
air
1.29
0.97
nw  1.33
The features of the vacuum flask reduce the heat loss due to conduction,

4. (a)
219
convection and radiation.
 The vacuum between the double walls prevents the heat loss due to
conduction and convection.
 The silvered glass wall reduces heat loss due to radiation.
 The cork support and cork stopper reduces heat loss due to conduction and
convection.
(b)
(i)
Energy per day collected  Intensity  Area  Efficiency
 5  5  0.95
 23.75 kWh
(ii)
Energy per day to heat water  0.5  23.75
(iii)
Energy per day available 
 11.88 kWh
80 92

 23.75
100 100
 17.48 kWh
(iv)
Using E  mc
E

m
c
17.48  3.6 106

4 200   55  25 
 499 kg
5. (a)
The current and voltage are recorded each time the rheostat is varied. A graph of I
vs V is plotted.
220
(b)
(i)
R1 R2
R1  R2
1000 1000
 1000 
1000  1000
Total resistance, RT  R1 
RT  1500
(ii)
V
RT
110

1500
Current, I 
 0.073 A
(iii)
Using
P
V
1100

110
 10 A
I
 Fuse rating  10A  13A
6. (a)




The GM tube is first placed next to the source without any shielding and the count
rate observed.
A sheet of paper is then placed between the source and the GM tube. A significant
reduction in the count rate will confirm the presence of alpha particles.
The GM tube is then placed behind the aluminum sheet and a further reduction in
count rate indicates the presence of beta particles from the source.
When the GM tube is placed behind the lead sheet, a further reduction in count
rate confirms the presence of gamma rays.
221
C 
 147 N  1e
(b)
14
16
(c)
Mass of  21 H  21 H   2.014 0  2
 4.0280 u
Mass of 21 He  4.0026 u
Mass defect,
m  0.0254 u
 0.0255 1.66 10 27 kg
 4.216 4 10 29 kg
Energy released, E  mc2
 4.216 4 10 29   3 108 
2
 3.795 1012 J
222
1. (a)
Angle of incidence, iˆ /°
Angle of reflection, r̂ /°
sin iˆ
sin rˆ
30.0
23.5
0.500
0.398
40.0
30.5
0.643
0.508
50.0
38.0
0.766
0.616
60.0
43.7
0.866
0.691
70.0
48.5
0.940
0.749
(b)
Graph of sin iˆ vs sin rˆ
223
(c)
Gradient 
y2  y1
x2  x1
0.850  0.550
0.680  0.440
0.300

0.240
 1.25

(d)
(i)
The incident ray, refracted ray and normal at the point of incident are all
on the same plane.
(ii)
For a wave travelling from one medium to another the value of
sin iˆ
is a
sin rˆ
constant called refractive index, n.
(e)
Refractive index, n  gradient  1.25
sin iˆ
Using n 
sin rˆ
sin 90
1.25 
sin rˆ
sin 90
 sin rˆ 
1.25
1.00

1.25
 0.800
rˆ  sin 1 0.800
 53.1
(f)
Refractive index of coating, n  nL

nL  n2
 1.252
 1.56
2. (a)
(i)
Velocity is the rate of change of displacement.
(ii)
Acceleration is the rate of change of velocity.
(iii)
Linear momentum is the product of mass and linear velocity.
224
(b)
Total distance
Total time
60

6.5
 9.23 ms -1
(i)
Average speed 
(ii)
Using distance  Average speed  Time
 vu 
s
t
 2 
v0
60  
t
 2 
2  60
v
t
2  60

6.5
 18.46 ms-1
v u
t
18.46  0

6.5
 2.84 ms -1
(iii)
Acceleration, a 
(iv)
a)
He possessed kinetic energy.
b)
Kinetic energy, k.e. 
1 2
mv
2
1
  86  18.462
2
 14 650 J
 14.65 kJ
3. (a)
(i)
225
(ii)
(iii)
(b)
(iv)
Both (ii) and (iii) represent direct current (dc).
(i)
The truth table represents a NAND gate.
(ii)
A
0
0
1
1
4. (a)
B
0
1
0
1
C
0
1
1
1
D
1
0
1
0
E
0
0
1
0
(i)
As the switch is closed the current flows through the brushes and through
the coil.
From Fleming’s left hand rule, the magnetic acting on AB causes a
downward force and on CD an upward force.
The momentum of rotation of the coil allows it to cross the vertical
position after which the current in the coil is reversed to produce
continuous rotation.
(ii)
The purpose of the commutator is to reverse the current in the coil every
half turn thus allowing continuous flow of current and continuous rotation.
226
(b)
Work done
Time
mgh

t
25 10  30

5
 1500 W
(i)
Power provided by the motor, P 
(ii)
Using P  IV
1500  I  24

I
1500
24
 62.5 A
(iii)
5. (a)




Since P  I , greater power will increase the current.
Determine the mass of the metal block using a balance after the holes were
made for the heater and thermometer.
The initial temperature of the block was measured before heating.
Turn on the switch and allow the block to be heated for a measured period
of time.
Ensure that the current and voltage are constant during the heating. Adjust
the thermostat if necessary. Record the readings of the ammeter, voltmeter
and final temperature.
Calculation:
Provided no heat loss,
Heat supplied by heater  Heat gained by metal block
IVt  mc
IVt
 Specific heat capacity, c 
m
(b)
(i)
Heat supplied  Heat gained by liquid
E  mc
13.6 1000  0.1 c   50  25 
13600

c
0.1 25
 5440 J kg-1 °C -1
(ii)
There will be no change.
227
(iii)
The specific heat capacity is fixed for a particular substance and its value
remains constant.
6. (a)
(b)
Radiation
Range in air
Alpha (α)
About 5cm of air
Gamma (γ)
Travels much further
Behaviour in an
electric field
Deflected opposite to
direction of electric
field
No deflection
(i)
Number of neutrons in Xenon (Xe)  143  54
 89
(ii)
Mass defect, m  1.00867  235.04393 
Type of track in a
cloud chamber
Bold and straight
Short faint tracks
 142.93489  89.907 30  3 1.00867  u
 236.052 60  235.868 20
 0.184 40 u
m , in kg  0.18440 1.66 10 27
 3.06104 10 28 kg
Energy released, E  mc2
 3.06104 10 28   3 108 
2
 2.75 1011 J
(iii)
The preferred method would be the artificial decay.
(iv)
This is because it produces much more energy  2.75 1011 J  , than
natural decay  9.98 1013 J  .
228
1. (a)
Velocity is the rate of change of displacement.
Acceleration is the rate of change of velocity.
(b)
Graph of velocity vs time
229
y2  y1
x2  x1
7.5  2.5

6.0  2.0
 1.25 ms -2
(c)
Slope 
(d)
Slope of a velocity vs time graph is equal to the acceleration.
(e)
Resultant force, F  ma
 60  1.25
 75.0 N
(f)
Distance covered after 10 seconds  Area under graph
 Area of trapezium
1
 sum of parallel sides   height
2
1
  2.0  10.0  10.0
2
 60.0 m

2. (a)
(i)
a)
Emr with wave length longer than visible light –
infrared/radiowave
b)
Emr with wavelength shorter than visible light – Gamma rays/
X-rays / ultraviolet rays
(ii)
Name of wave
X-ray
Gamma ray
Radio wave
(b)
(i)
Source
X-ray tube
Radioactive Nuclei
Radio / TV transmitters
Use
To take X-ray pictures
Radiotherapy
Communication
v f
 f 
v

3.0 108

3.0 1012
 1.0 1020 Hz
230
(ii)
sin iˆ (air)
sin rˆ (glass)
sin 
1.5 
sin 35
sin   1.5sin 35
 1.5  0.5736
 0.860 4
  59.4
Refractive index, n 


3. (a)
Circuit symbol
Name of components
Cell
Variable resistor
or
Filament lamp/bulb
a.c. supply
Semi-conductor diode
Fuse
or
(b)
(i)
V  IR
(ii)
Total resistance in circuit, RT 
V
I
24

2
 12 
R2  RT  R1
 12  4
 8
(iii)
Voltmeter reading, V  IR2
 28
 16 V
(iv)
With switch S closed, new total resistance, RT  4 
88
88
 44
RT  8 
231
V
RT
24

8
 3.0 A
New current, I 
4. (a)




(b)
The boiling tube is heated in a water bath until all the naphthalene is completely
melted.
The boiling tube is removed from the water bath and allowed to cool.
The temperature is recorded at fixed intervals (e.g. every minute using a stop
watch) while stirring continuously to ensure equilibrium temperature.
A graph of temperature vs time is plotted to produce the cooling curve.
(i)
Mass of melted ice  110  100
(ii)
Heat lost by water  mc
 100  4.2   30  20 
 10 g
 4 200 J
Heat required to Heat to warm melted

melt ice
ice to 20°C
 mi l f  mi  cw  
(iii)
Total heat gained by ice 
(iv)
Heat lost by water  Total heat gained by ice
4 200  mi l f  mi  cw  
4 200  10l f  10  4.2   20  0 
4 200  10l f  840
4 200  840
10
 336 J g -1
lf 
5. (a)
232



(b)
When the magnet is pushed into the coil, the change in magnetic flux
causes an induced e.m.f.
If the magnet is pushed in at a faster speed, the rate of change of magnetic
flux is increased which in turn causes an increased induced e.m.f.
If a more powerful magnet is used there will also be an increased induced
e.m.f.
(i)
The changing speed of the car will cause a changing magnetic flux and
therefore an induced e.m.f. (current).
(ii)
Increasing speed of a car will increase the rate of magnetic flux cutting
which in turn will increase the charging current to the batteries. Reducing
the car speed reduces the charging current.
(iii)
Conversion efficiency 
Power output
100%
Power input
600 100
100%
200 000
 30%

6. (a)
J.J. Thomson postulated the ‘Plum Pudding’ model. In this model the atom was
looked at as a mass of positive charge dotted with negative electrons that
neutralized the positive charge. The structure was similar to a pudding (+ve)
dotted with currants.
Ernest Rutherford’s nuclear model of the atom from the α – particle scattering
experiment conducted by his students, Geiger and Marsden. From the result of the
experiment, he concluded that the atom consists of a tiny positively charge mass
or nucleus. The rest of the atom was mainly empty space that carried a negative
charge.
(b)
(c)
(i)
Nuclides with identical mass number -
(ii)
The heaviest nuclide is
238
92
Number of neutrons in
238
92
(iii)
Isotopes –
(i)
First half life, t 1 :
40
19
P and
40
19
P and
40
18
R
Xe .
Xe  238  92
 146
39
19
Y (same atomic number, different mass number)
2
t1
t1
2
2
4000 
 2000 
1000
233
2t 1  55.0 s
2
 t1 
2
55.0
2
 27.5 s
Second half life, t 1 :
2
t1
2
1000 
 500
t 1  80.0  55.0
2
 25.0 s
(ii)
Average t 1 
2
27.5  25
2
 26.3 s
234
1. (a)
Graph of Velocity, v/ms-1 vs Time, t/s
235
y2  y1
x2  x1
57.0  11.0

50.0  10.0
 1.15 ms -2
(b)
Slope 
(c)
The slope of a velocity vs time graph is equal to the acceleration.
(d)
(i)
Total distance travelled  Area under graph
 Area under trapezium
1
 sum of parallel sides   height
2
1
  60  180   69
2
 8 280 m

Total distance travelled
Total time taken
8 280

180
 46.0 ms -1
(ii)
Average velocity of taxi 
(iii)
Momentum of taxi  Mass  Velocity
 1500  69
 103500 kg ms -1
(e)
(i)
Displacement is the distance moved in a specific direction.
(ii)
Quantity
Displacement
Acceleration
2. (a)
Scalar
Vector


(i)
Quantity
Specific heat capacity
Symbol
c
S.I. Unit
J kg-1 K-1
Specific latent heat of
vapourisation
lv
J kg-1
236
(b)
(ii)
The heat capacity of substance is the heat energy required to raise the
temperature of the substance by 1 degree Celsius or Kelvin.
(iii)
Heat capacity  Mass  Specific heat capacity
C  mc
(i)
Assuming no heat loss:
Total energy required  Energy to change ice at 0°C to water at 0°C
+Energy to heat water from 0°C to 3°C
 ml f  mc
  0.025  340 000    0.025  4 200  3
 8500  315
 8815 J
3. (a)
Energy
Time
8815

300
 29.38 J s -1 or W
(ii)
Rate of heat energy received 
(i)
In a simple cell the current is due to the flow of positive and negative ions
(i.e. cations and anions).
In the copper wire the current is due to the flow of free electrons.
(ii)
Positive terminal – Carbon rod
(iii)
Negative terminal – Zinc (casing)
(iv)
Using Q  It
Q  0.1 60
 6.0 C
(v)
Since the same charge flows through the circuit, the charge that flows
through the cell  6.0 C.
237
(b)
(i)
Waveform
Wave A
Wave B
Wave C
(ii)
Type of current
d.c.
d.c.
a.c.
For waveform C:
Period,
T  20 s
1
Frequency, f 
T
1

0.20
 5.0 Hz
4. (a)
The double slit is placed directly in front of the ray box and the white screen
placed approximately 1 m from the double slit as show in the diagram above. As
the light passes through each double slit, diffraction occurs i.e. the light spreads
out. The diffracted light from each slit interferes to produce bright and dark bands
due to constructive and destructive interference respectively.
(b)
(i)
Refractive index, n 
sin iˆ (air)
sin rˆ (prism)
sin 30
sin 20
0.500

0.342
 1.46

238
(ii)
n
1.46 
Speed in air, v1
Speed in prism, v2
3 108
v2
 v2  2.05 108 ms-1
5. (a)
v
(iii)
Using f 
(i)
Advantages of using a.c. to transmit electrical power:
 There is less transmission loss with low current a.c.
 Ability to step-up or step-down voltages using transformers

3.0 108
f 
430 10 9
 6.98 1014 Hz
(ii)
Features which enhance efficiency:
 Secondary coil is wound on top of primary to maximize flux
linkage
 Thick copper wire for coils (less resistance) to reduce energy loss
due to heating.
 Soft iron core to reduce energy loss due to magnetization and
demagnetization.
 Laminated core to reduce energy loss due to Eddy currents.
239
(b)
(i)
Using
Vs N s

Vp N p
110 000 900

11000
Np

N p  900 
11000
110 000
 90 turns
(ii)
Using
Vs I p

Vp I s
110 000 8000

11000
Is

I s  8000 
11000
110 000
 800A
(iii)
Efficiency,  

Power output
Power input
Transmission power
Vp I p
 Transmission power   Vp I p
 0.7  11000  8000
 61.6  106 W
 61.6 MW
6. (a)
(i)
Three uses of radioisotopes in medicine:
 Radio therapy - treatment of cancer
 Diagnostics - tracer studies
 Sterilization of medical instruments and bandages.
(ii)
Safety precaution when using radioactive substances:
 Avoid eating or drinking near radioactive substances.
 Use gloves, tongs, body suite, masks or robotics when handling
radioactive substances.
 Use radioactive shielding e.g. lead embedded glass.
 Keep exposure time as short as possible.
 Keep large distance away from radioactive source.
240

(b)
(i)
Use photographic plate badges if working in radioactive
environment.
Number of half lives t 1 in going from 16 dis/min to 1 dis/min
2
t1
t1
t1
t1
2
2
2
2
16 
 8 
 4 
 2 
1
Number of half lives  4
Probable age of plant  4  5600
 22400 years
(ii)
Energy lost by the sun in 1 second, E  mc2
 2.0 109   3 108 
2
 1.8 1026 J
Power output  1.8 1026 W

1.8 1026
1000
 1.8 1023 kW
241
1. (a)
p/cm
10.0
20.0
30.0
35.0
40.0
45.0
q/cm
86.0
76.2
66.0
62.8
57.9
53.6
x/cm
40.0
30.0
20.0
15.0
10.0
5.0
y/cm
36.0
26.2
16.0
12.8
7.9
3.6
(b)
Graph of y/cm vs x/cm
242
y2  y1
x2  x1
33.5  2.5

38.0  4.0
31.0

34.0
 0.91
(c)
Slope, z 
(d)
When plasticine is 27.5 cm from pivot, R:
 x  27.5 cm
From the graph, y  24.0 cm
(e)
The principle of moments states that when a body is in equilibrium, the sum of the
clockwise moments about a point is equal to the sum of the anti-clockwise
moments about the same point.
(f)
(i)
Wp  z  Wm
 z  mg
 0.9  0.05 10
 0.46 N
(ii)
2. (a)
mp 
Wp
g
0.46

10
 0.046 kg
(i)
Name
Mass
Time
Current
Temperature
Length
Symbol
m
t
I
T
l
Base (S.I.) unit
kg
s
A
K
m
(ii)
A linear scale is one which has equal spacing between intervals.
A non-linear scale is one which has unequal spacing between intervals.
(iii)
Non-linear scale – Conical flask
243
(b)
3. (a)
Mass
Volume
Mass

l bh
15000

2  1.1 2.5
 2 727.3 kg m -3
(i)
Density 
(ii)
Pressure 
(i)
E  mc2
E  Change in energy
m  Change in mass
c  Speed of light
(ii)
For:
1) Large amount of nuclear energy produced from small
quantity of raw material
2) There are no emission of smoke or greenhouse gases.
Force
Area of base
15000 10

1.1 2
 68181.8 Nm -2
Against:
1) Accidents can be catastrophic.
2) Hazardous waste produced.
3) Weaponizing the energy threatens the existence of living things.
(b)
Conservation of mass number, A
1  235  A  90  3
 A  143
Conservation of atomic number, Z
0  92  56  Z
 Z  36
(c)
2  3  P 1
P4
11  Q
Q2
244
Hence, X is He (Helium).
4. (a)
(b)
When the switch S is closed, the current flows through the brushes and through
the coil. From Fleming’s left hand rule, the magnetic field acting on the coil
causes one side to move up and the other side to move down.The momentum of
rotation of the coil allows it to move it cross the vertical position after which the
commutator reverses the current in the coil to produce continuous rotation.
(i)
Period, T  30 ms
1
Frequency, f 
T
1

30 10 3
 33.3 Hz
(ii)
Peak current, I o  4 A
Vo  I o R
 4  45
 180 V
Peak-to-peak voltage 180 
180  360
5. (a)
(iii)
Maximum power  IoVo
 4  180
 720 W
(i)
Heat flows from Block B (higher temperature) to Block A (lower
temperature).
(ii)
T / K   / °C  273
T / K  52  273
 325 K
(b)
(i)
Total heat energy transferred to water 
40
 300 103
100
 120000 J
Using E  mc
120000  2  4.2 103  
120 000
  
8.4 103
 14.3 °C
245
Final temperature of water  27 14.3
 41.3 °C
(ii)
6. (a)
(b)
a)
The black surface makes it a good absorber of heat.
b)
In the collector reservoir, the hot water is at the top because of
convection.
c)
The glass cover of the solar collector allows short wavelength
radiation to enter and prevents the long wavelength radiation from
leaving.
d)
The copper tubing is a good conductor of heat and allows the
transfer of heat from the black surface absorber plate to the water
in the tubing.
e)
The insulation reduces the transfer of heat by conduction from the
collector to the surroundings.
(i)
In Longitudinal waves the vibration of the parties are parallel to the
direction of wave travel.
In transverse waves the vibration of the particles are perpendicular to the
direction of wave travel.
(ii)
Using v  f 
v
 f 

3 108

700 10 9
 4.3 1014 Hz
(i)
A series of sound pulses was transmitted into the water from the ship.
The time taken, t for the reflected sound pulses to return to the ship was
measured. The depth of the oil plume was determined by multiplying the
t
speed of the pulse by .
2
(ii)
Depth of oil plume  Speed in water 
t
2
0.3
2
 217.5 m
 1450 
246
1. (a)
Graph of Potential difference (V) vs Current (I)
247
(b)
Points used (0.56, 42.0) and (0.1, 7.5).
y y
Gradient  2 1
x2  x1
42.0  7.5

0.56  0.1
34.5

0.46
 75 V/A
(c)
The gradient represents the resistance.
(d)
The potential difference is directly proportional to the current. (Graph is a straight
line through the origin.)
(e)
(f)
The readings from the voltmeter and ammeter are taken.
The rheostat is adjusted and new readings are recorded.
This is repeated at least five times. Then a graph of V vs I is plotted.
(g)
For parallel arrangement of resistors:
1
1
1
1
 

RT R1 R2 R3
1 1 1 1
  
RT 8 12 15
1 15  10  8

RT
120
33
120
11

40

248
 RT 
40
11
 3.6 
2. (a)
(i)
Physical Quantity
Area
Volume
Density
Derived S.I. unit
m2
m3
kg m-3
Physical Quantity
Mass
Time
Derived S.I. unit
kg
s
(ii)
Or
Current
Temperature
Length
(b)
A
K
m
Mass
Volume
102

150
 0.68 g cm-3 / 680 kg m-3
(i)
Density 
(ii)
Volume 
(iii)
Relative density of gasoline 
Mass
Density
325

13.6
 23.9 cm3
Density of gasoline
Density of mercury
0.68

13.6
 0.05
249
3. (a)
(b)
Weight of yacht  mg
 8300 10
 83000 N
83000
7 000
 11.9
Minimum number  12 types
Number of tyres required 

(c)
1. The resultant force in any direction is zero.
2. The sum of the moments about any point is zero.
(d)
(i)
For a body totally or partially immersed in a fluid, the weight of the fluid
displaced is equal to the upthrust.
(ii)
The entire yacht is not solid steel but constitutes other materials including
air. So its overall density is less than sea water. This means that it will
displace its own weight in sea water while it is still afloat.
(iii)
Weight of sea water displaced  Weight of yacht

 83000 N
Mass of sea water  8300 kg
Mass
Density
8300

1025
Volume of sea water displaced 
 8.1 m3
4. (a)
(b)
A narrow beam of α particles was directed towards a very thin sheet of gold foil.
The scattering was monitored using a rotating scintillation microscope which can
detect α particles.
This was enclosed in an evacuated chamber to avoid interference from air
molecules. The results showed that:
1) Most of the α particles went straight through deflected which indicates that
most of the atom was empty space.
2) Few α particles were deflected between 0 - 90º which indicates that the atom
has a small positively charged nucleus.
3) Even fewer α particles were rebounded which indicates that the nucleus has a
mass.
(i)
235
92
144
1
U  01 n 
 90
36 Kr  56 Ba  2 0 n  Energy
250
(ii)
Mass defect   235.118  1.009    89.947  143.881  2.018 
 0.281 u
E  mc2
Using
E   2811.66 10 27    3.0 108 
2
 4.189 101 J
5. (a)
(b)
p1V1 p2V2
pV

or
 constant
T1
T2
T
(i)
General Gas Law:
(ii)
According to the kinetic theory, the air molecules inside the balloon are in
continuous random motion. These molecules strike the walls of the
balloon causing a change of momentum. The rate of change of momentum
is equivalent to the force exerted by the molecules on the balloon. This
force acting on the surface of the balloon gives the pressure.
(i)
Energy used, Q  Energy required to (change ice at 0°C to water at 0°C
 Heat water 0°C to water at 100 °C + change water at
100°C to steam at 100 °C)
Q  ml f  mc  mlv
 2 000 g  2 kg 
  2  330 000    2  4 200 100    2  2 250 000 
 660 000  840 000  4500 000
 6 000 000 J
(ii)
Energy
Time
6 000 000

6 000
Power 
 1000 W
6. (a)
(i)
Three differences between ‘light waves’ and ‘sound waves’:
 Light waves do not require a medium whereas sound waves require
a medium for travel.
 Light waves are transverse waves whereas sound waves are
longitudinal waves.
 Light waves can be polarized whereas sound waves cannot be
polarized.
 Light waves belong to the e.m.r. spectrum whereas sound waves
do not.
251
(ii)
(b)
Electromagnetic waves mare transverse waves which means that they can
be polarized. They travel at a speed of 3.0 108 ms-1 in a vacuum and do
not require a medium for travel. The progressive electromagnetic waves
can transfer energy from one place to another and undergo reflection,
refraction and diffraction.
Time for 50 claps
 30.3 s
Time for each echo 
30.3
50
 0.606 s
Distance sound travelled  100  2
 200 m
Distance
Time
200

0.606
 330 ms -1
Speed 
(c)
Using v  f 
v
 
f
3.0 108

100 106
 3.0 m
 300 cm
252
1. (a)
Graph of Vs/V vs Vp/V
253
(b)
Points selected for gradient  (7.2, 65) and (1.6, 15)
y y
Gradient, S  2 1
x2  x1
65  15

7.2  1.6
50

5.6
 8.9
(c)
(i)
Vs
Vp
Vs N s

Using
Vp N p
750
8.9 
Np
750

Np 
8.9
Gradient, S 
 84.3
(ii)
Using

(d)
(i)
Vs I s

Vp I p
1.6
8.9 
Is
1.6
Is 
8.9
 0.18 A
Power in secondary
 100%
Power in primary
IV
 s s 100%
I pV p
Efficiency,  
0.15
 8.9 100%
1.6
 83.4%

(ii)
Ideal transformer has an efficiency of 100% i.e. no power loss.
254
(e)
For efficient function of transformer:
1) Laminated core (eliminate Eddy current)
2) Soft iron core (easy magnetization and demagnetization)
3) Thick copper wire windings (low resistance)
4) Secondary coil wound on top primary (better magnetic linkage)
2. (a)
Quantity
Diameter of wire
Volume of liquid
Temperature
Weight/Force
Time
(b)
(c)
(i)
Force acting on stone – weight due to gravity
(ii)
The velocity is increased from zero to a maximum as it hits the ground i.e.
it is accelerated.
(i)
Weight  mg
 0.06  10
 0.6 N
(ii)
a)
b)
3. (a)
Instrument
Micrometer screw gauge or Vernier
caliper
Measuring cylinder/Burette
Thermometer
Spring balance
Stop watch/clock
Resultant, R  11.8 cm  23.6 ms-1
Direction from OA  9.6
(i)
255
(ii)
Type of Thermometer
Clinical thermometer
Laboratory thermometer
Thermocouple
(b)
(i)
Operating Temperature
Range/°C
20 to 110
250 to 800
35 to 43
Pressure at 20 m below surface  Atmospheric pressure + Pressure due to
Depth
 PA   gh
 1.01105  1025 10  20 
 3.06 105 Pa
(ii)
Using

p1V1

T1
V2

V1

p2V2
T2
p1T2
p2T1
3.06 105   25  273
1.01105  10  273
 3.19
4. (a)
(i)
The normal is an imaginary line draw perpendicular to the reflecting
surface at the point of incidence.
(ii)
Angle of incidence is the angle between the incident ray and the normal at
the point of incidence.
(iii)
Angle of reflection is the angle between the reflected ray and the normal at
the point of incidence.
(b)
Features of the image produced by plane mirror:
 Image is laterally inverted
 Image is erect
 Image is same size as object
 Image is virtual
 Image distance is the same as object distance
(c)
To reverse the laterally inverted word when viewed on the rear-view mirror. This
will allow the correct wording ‘AMBULANCE’ to be seen from the reflection of
the mirror.
256
(d)
(i)
Angle of incidence, iˆ  90  30
 60
sin iˆ
sin rˆ
sin 60
1.5 
sin rˆ
sin 60
 sin rˆ 
1.5
0.866

1.5
 0.577

rˆ  35
Using n 
(ii)
5. (a)
(i)
ˆ  60 
Angle of refraction on QR boundary  CBD
The semi-conductor diode is defective if:
 It conducts in both forward and reverse bias orientations
 It does not conduct in both forward and reverse bias orientations.
(ii)
(iii)
Not-gate
Input
0
1
(b)
(i)
Output
1
0
Equivalent resistance (series), RT  R1  R2  R3
 2  6  12
 20 
257
(ii)
Equivalent resistance (parallel),
1
1
1
1
 

RT R1 R2 R3
1 1 1 1
  
RT 2 6 12
6 2 1
 
12 12 12
9

12
3

4
4
RT  
3


(c)
6. (a)
The decision for series circuit was not wise because:
In series circuit, if one bulb blows all the others would not light
The p.d. across each bulb is reduced in a series circuit.
Alternative sources of Energy in the Caribbean:
 Solar – Water heating, generate electricity
 Wind – Generate electricity
 Geothermal – Generate electricity
 Hydro – Generate electricity
Importance:
 Less reliance on depleting fossil fuel
 Less emission of CO2 and other pollutants
 Renewable sources of energy
 Cheaper production of energy
(b)
(i)
When the ball is kicked, it is given kinetic energy. As it moves towards the
goalkeeper the kinetic energy is converted into potential energy as it rises.
At the highest point the kinetic energy is minimum and the potential
energy is maximum. The potential energy decreases to zero and kinetic
energy to a maximum on striking the ground.
(ii)
Difference in energy  P.E.  K.E.
1
 mgh  mv 2
2
(iii)
Momentum of ball  Mass  Velocity
 0.43  7
 3.01 kg ms-1
258
1. (a)
Graph of Length/m vs Load/N
259
(b)
Points considered for gradient: (9.6, 0.50), (2.0, 0.29)
y y
Gradient  2 1
x2  x1
0.50  0.29

9.6  2.0
0.21

7.6
 0.028 m N -1
(c)
The gradient can be used to determine the spring constant, k, where k 
(d)
(i)
The original length of this spring is the length when F  0 N .
Form graph when F  0 , Length  0.24 m
(ii)
Using F  kx (Hooke's Law)
(where x is extension)
0.7 10  1 x

x  0.20 m
1
.
Gradient
 New length  0.24  0.2
 0.44 m
(e)
(iii)
The graph would change from straight line to curve.
(i)
The region of proportionality exist in the straight line section of the graph.
(ii)
Quantity
Load
Extension
2. (a)
(i)
Scalar
Vector


The specific heat capacity of a substance is the heat energy required to
change the temperature of 1 kg of the substance by 1 degree Kelvin or
Celsius.
(ii)
Quantity
Heat capacity
Specific latent heat of fusion
(iii)
General Gas Law:
Symbol
C
S.I. Unit
J K -1
lf
J kg
p1V1 p2V2

T1
T2
260
where p1  intial pressure
V1  initial volume
T1  initial temperature
(b)
(i)
p2  final pressure
V2  final volume
T2  final temperature
Energy, Q1 , required to heat water to 100°C:
Q1  mc
 8  4 200  100  33
 2 251200 J
 22.5 105 J
(ii)
Energy, Q2 , required to change water at 100°C into steam at 100°C:
Q2  mlv
 8  2300 000
 18 400 000 J
(iii)
Total energy, QT , required to heat 8 kg of water at 33º C to steam at
100ºC:
QT  Q1  Q2
 2 251200  18 400 000
 20 651200 J
 20.7 MJ
3. (a)
(i)
Snell’s law states that for light rays passing from one transparent medium
to another, the ratio of the sine of the angle of incidence to the sine of the
sin iˆ
angle of refraction i.e.
is a constant called the refractive index, n.
sin rˆ
(ii)
261
(b)
4. (a)
(b)
(i)
The ray of light will enter the prism undeviated since it enters at 90 º. The
ray is then incident on AC at an angle of 45º. Since this angle is greater
that the critical angle for glass (42º), total internal reflection occurs at AC.
This reflected ray strikes BC at an angle of incidence of 45º, again causing
total internal reflection.
The reflected ray from BC is incident on AB at 90º and therefore passes
through undeviated.
(ii)
After emerging, the ray had turned through 180º.
(i)
When two or more bodies collide, the total momentum of the bodies
before collision is equal to the total momentum of the bodies after
collision provided no external forces are acting on the bodies.
(ii)
For a launching rocket, the momentum of the rocket in the upward
direction is equal to the momentum of the extruded burnt fuel in a
downward direction.
(i)
Initial momentum of track  mv
 1250  25
 31250 kg ms -1
(ii)
Initial momentum of car  625  30
 18750 kg ms-1
(iii)
By the principle of conservation of momentum:
Total momentum before collision  Total momentum after collision
31250  18750  1250  625   v
12500  1875v

v  6.7 ms-1 in direction due North
5. (a)
262




(b)
The circuit is set up as shown in diagram with the ammeter in series with
the test resistor R and the voltmeter in parallel to R.
The variable resistor Q is adjusted and the values of I and V are recorded
from the ammeter and voltmeter respectively.
This is repeated to obtain at least 5 pairs of values for I and V.
A graph V vs I is plotted and the gradient will give the resistance of the
metallic conductor R.
R2 R3
 R4
R2  R3
3 3
RT  3 
3
33
 3  1.5  3
Total resistance, RT  R1 
(i)

 7.5 
V
RT
12

7.5
 1.6 A
(ii)
Using I 
(iii)
Using V  IR
 V  I   Combined resistance of R2  R3 
 1.6  1.5
 2.4 V
6. (a)



The background count is measured without any radioactive source in
place.
With the β source in place the count rate is obtained for increasing number
of Aluminium sheets until it drops to the background count.
The β – source is replaced by the γ source, the process is repeated and the
results compared.
263
(b)
(i)
1
0
n  23290Th 
 23390Th
Th 
 01 e  233
91 Pa
233
90
233
91
(ii)
Pa 
 01 e  233
92 U
Mass defect  L.H.S.  R.H.S.
132.91525  97.91033  
  233.03964  1.00867   

 3  1.00867

 234.04831  233.85159
 0.19672 u
Energy released, E  mc2
 0.19672 1.66 10 27   3.0 108 
2
 2.94 1011 J
264
1. (a)
Graph of Image size, I vs Object size, O
265
(b)
Points selected for gradient: (2.8, 1.40), (0.8, 4.0)
y y
Gradient, G  2 1
x2  x1
14.0  4.0

2.8  0.8
10.0

2.0
 5.0
(c)
The gradient G represents magnification.
(d)
The focal length in the distance between the principal focus and the optical centre
of the lens.
(e)
Focal length is associated with lens.
(f)
(i)
For a plane mirror:
(ii)
Magnification 
2. (a)
Image size  Object size  10.0 cm
Image size
Image distance
or
Object size
Object distance
10.0

10.0
1
(i)
266
(ii)
1)
2)
3)
(b)
Force
Gravitational force (weight)
Drag force (air resistance)
Upthrust
Centripetal, magnetic,
nuclear, tension, etc.
Situation
A javelin falling in the air
An object immersed in a liquid
Object in circular motion, between
magnets or current carrying
conductors, nucleus of atoms,
stretched springs, etc.
(i)
(ii)
By the principle of moments:
Clockwise moments  Anti-clockwise moments
W  0.5   500  1   400  0.5 

0.5W  700
700

W
0.5
 1400 N
(iii)
Taking moments about X:
500  y  1400  0.5

y
700
500
 1.4 m (on the same side of the fulcrum)
3. (a)
(i)
A laboratory thermometer – volume of a liquid
(ii)
A thermocouple – E.m.f.
(b)
Use of thermometer
To measure body temperature
To measure temperature lower than
40C
Rapidly changing temperature
Design feature
Small temperature range, constriction
in the bore
Alcohol thermometer
Junction of small mass
267
(c)
(i)
Charles’ law states that the volume of a fixed mass of gas is directly
proportional to the absolute temperature provided the pressure remains
constant.
(ii)
From Charles’ Law:
V1 V2

T 1 T2
V2
40


 273  30   273  70 
40  343

V2 
303
 45.28 cm3
Increase in volume  45.28  40
 5.28 cm3
5.28
100%
40
 13.2%
Percentage increase 
4. (a)
(b)
(i)
Properties of electromagnetic waves:
 They are all transverse waves
 They travel at the same speed in a vacuum i.e. 3  108 ms-1
 Can travel in a vacuum
 Can be polarized
(ii)
Other types of electromagnetic waves:
 Gamma rays - lower
 X-rays
- lower
 UV
- lower
 IR
- higher
 Radio
- higher
(i)
a)
Using f 

c

3 108
f1 
2 10 7
 1.5 1015 Hz
3 108
6.5 10 5
 4.6 1012 Hz
f2 
268
b)
Decrease in frequency, f  1.5 1015  4.6 1012
 1.495 1015 Hz
(ii)
E1  k 1.495 1015  J
(iii)
New, frequency f 
…Equation 
3 108
6 10 7
 0.5 1015 Hz
f  0.5 1015  4.6 1012
 4.954 1014
E  k  4.954 1014  J
…Equation 
Equation   Equation :
14
E k  4.954 10 

E1 k 1.495 1015 
 0.33
 E  0.33E1
5. (a)




The circuit is set up as shown in diagram with the ammeter in series with
the filament lamp, L and the voltmeter in parallel to L.
The variable resistor Q is adjusted and the value of I and V are recorded
from the ammeter and voltmeter respectively.
This is repeated to obtain at least 5 pairs of wide range values of I and V.
A graph of I vs V is plotted.
269
(b)
(i)
Total resistance, RT  2 
3 5
35
 3.785 kΩ
Using V  IR
 V  1103  3.875 103
 3.875 V
(ii)
 3 
Current in 5 k   
1 mA
 35
 0.375 mA
6. (a)
(iii)
If the 2 kΩ resistor ,burns out the circuit is now open and no current will
flow.
(i)
226
86
(ii)
Number of neutrons in
43
Ra 
 222
84 Rn  2 He  
226
86
Ra  A  Z
 226  86
 140
(b)
An atom is normally neutral because it has equal number of protons and
electrons i.e. equal number of oppositely charged particles. The atoms are stable
because the nuclear forces holding the protons and neutrons together in the
nucleus is greater than the repulsive electrostatic forces. Larger atoms require a
greater neutron : proton ratio for stability.
(c)
Isotopes are atoms of the same element with the same atomic number but
different mass number i.e. same number of protons but different number of
neutrons.
270
(d)
For element X:
(i)
Mass number  5  6
 11
Atomic number  5
Charge on nucleus  Positive (protons)
(ii)
Isotope of X  125 X
271
1. (a)
Graph of Temperature,  /°C vs Time, t/min
(b)
Melting point of substance  70.0°C
(c)
(i)
The substance is changing between liquid and solid state between B and
C.
272
(d)
(ii)
As the substance is changing state, latent heat of fusion is removed from it
resulting in no temperature change.
(i)
At C, the substance is in solid phase.
(ii)
Between C and D the substance is cooling.
Heat loss, Q  Heat loss from A to B + Heat loss from B to C + Heat loss
from C to D
Q  mc  ml f  mc
(e)
  0.015 1763  90.0  70.0   0.015  215000 
  0.015 1760   70.0  57.5
 528.9  3225  330.56
 4 084.46 J
 4.08 kJ
(f)
Physical quantity
Heat capacity
Symbol
C
S.I. Unit
J K-1
lv
J kg-1
Specific latent heat of
vapourisation
2. (a)
(i)
Equivalent derived unit for Joule:
 kg m2 s-2
 Nm
(ii)
Application of solar energy:
 Solar water heater
 Watches
 Calculators
 Satellites
(iii)
Advantage of using solar energy in the Caribbean:
 plenty of sunlight in the Caribbean
 renewable source
(iv)
Alternative energy
Geothermal
Wind
Hydro-electricity
Source
Hot rocks deep in the earth
Wind, especially along coastline
Flowing rivers, waterfalls
273
(b)
(i)
Gravitational potential energy, E p  mgh

E p  0.44  9.8 12
 51.74 J
(ii)
Assuming no energy loss:
Kinetic energy, Ek  Initial potential energy, E p
1
Ek  mv 2
2
 51.74
2  51.74
v
m

2  51.74
0.44
 15.3 ms -1

(iii)
Momentum, p  mv
 0.44 15.3
 6.7 kg ms -1
3. (a)
(b)
(i)
p1V1 p2V2

T1
T2
(ii)
p1  initial pressure
T1  initial temperature
V1  initial volume
(i)
Using
p2  final pressure
T2  final temperature
V2  final volume
p1 p2

T1 T2
p2
2  105

 273  23  273  34 

p
2  105
 2
296
307
2  105  307
p2 
296
 2.07 105 Nm -2
274
(ii)
An increase in pressure causes an increase in the kinetic energy of the air
molecules in the tyre. This causes the air molecules to strike the inner
walls of the tyre with greater frequency and momentum, thus increasing
the pressure.
(iii)
Using

4. (a)
(b)
V1 T1

V2 T2
V2 273  34

V1 273  23
307

296
 1.04
Laws of reflection:
 The incident ray, refracted ray and normal at the point of incidence are all
on the same plane.
 For light rays passing from one transparent medium to another, the ratio of
the sine of the angle of incidence to the sine of the angle of refraction
sin iˆ
is a constant called refractive index.
sin rˆ
(i)
Assuming Nemo is along the straight line with BA:
ˆ
Angle C  NBA
 90  42
 48
(ii)
Using n 
1
sin cˆ
1
n
sin 48
1

0.7431
for critical angle
 1.35
(iii)
For Nemo to see Bruce it means that is there is total internal reflection i.e.
angle of incidence on the water-air boundary is greater than the critical
angle c. Therefore the horizontal distance from Bruce’s eye to B is 5m.
So that the distance from Bruce’s eye from Nemo’s eye is 10m.
275
5. (a)
(b)
Ways of conserving existing energy sources:
 Replace incandescent bulbs with CFL bulbs to conserve electrical energy.
 Switch off lights and appliances when not in use to conserve electrical
energy.
 Use car-pooling and more walking to save fuel or chemical energy.
 Use energy efficient appliance to conserve electrical energy.
(i)
Number of hours in two weeks  2  7  24
 336 hours
Power of bulbs in kWh  0.06  336
 20.16 kWh
6. (a)
(ii)
Electrical charges for two weeks  20.16  $0.26
 $5.24
(iii)
Efficiency 
(i)
‘Half-life’ is the time taken for half the number of radioactive atoms to
disintegrate.
Useful energy output
 100%
Energy input
20.16  15.5

100%
20.16
4.66

100%
20.16
 23%
t1
(ii)
N t 12 N t 12 N t 12 N
N 
 
 
 

2
4
8
16
 5t 1  20 days
2
2
t1 
2
20
5
 4 days
(b)
Uses of radioactive isotopes:
 Tracers e.g. detect leaking gas lines
 Dating i.e. estimating age of artifacts
 Nuclear energy
 Radio therapy
276
Precautions when handling radioisotopes:
 Uses thick gloves or remote control mechanical arms.
 Store radioactive materials in thick lead containers.
 Use protective clothing that must not be removed from lab.
 Use proper labelling.
(c)
Using E  mc2
E

m  2
c
6.7 1010

2
 3 108 
 7.4  10 7 kg
 New mass  1  7.4 10 7
 0.9999926 kg
277
1. (a)
Graph of Activity (A) vs Time (t)
278
(b)
t  1.5s
t  9.5s  t 1  8.0 s
At 36s-1 ,
At 18s-1 ,
2
t  7.0 s
t  14.0s  t 1  7.0 s
-1
At 24s ,
At 12s-1 ,
2
t  10.5s
t  18.0s  t 1  7.5s
-1
At 16s ,
At 8s-1 ,
2
 t 1  average  
2
8.0  7.0  7.5
3
 7.5 s
(c)
After 25 days, activity  4.5 s-1
(d)
(i)
Radioactive emissions:
 Alpha particles
 Beta particles
 Gamma Rays
(ii)
Most dangerous radioactive emission is alpha particles.
(i)
C  mc
(ii)
C  Heat capacity
c  Specific heat capacity
2. (a)
(iii)
Specific heat capacity, c
The quantity of heat energy
required to change the temperature
of 1kg of a substance by 1 degree
C or K.
Unit: J kg-1 K-1
E
c H
m
Each specific substance has a
constant value
Heat capacity, C
The quantity of heat energy
required to change the temperature
of the total mass of a substance by
1 degree C or K.
Unit: J K-1
E
C H

Varies according to the mass of the
substance
279
(b)
(c)
(i)
Using E  mc
 2  4 200  100  37 
 529 200 J
(ii)
Using E  mlv
 2  2.3 106
 4.6 106 J
3. (a)
(iii)
Total heat energy  529200  4.6 106
 5129200 J
(i)
‘Electrical resistance’ is the opposition to the flow of an electrical current.
V
It is calculated from R  .
I
(ii)
Meter
Ammeter
How connected in a
circuit (series or
parallel)
Series
Resistance
(high or
low)
Low
Voltmeter
Parallel
High
Reason for size of
resistance
So as not to affect the
current in the circuit
So as to draw as little
current as possible
from the circuit
280
(b)
(i)
2 6
26
12

8
 1.5 
RT 
I  A1  
V
RT
12
1.5
8 A

(ii)
4. (a)
(b)
Since power supply is connected directly across the 2  resistor, then
V
Current I  A2  
R
12

2
6 A
Newton’s three laws of motion:
 A body will continue in a state of rest or of uniform motion in a straight
line unless compelled by an external force to act differently.
 The rate of change of momentum is directly proportional to the applied
force and takes place in the direction in which the force acts.
 If an object A exerts a force on object B then object B will exert an equal
but opposite force on object A i.e. to every action there is an equal and
opposite reaction.
(i)
Initial momentum  mv
 70  26
 1820 kg ms-1 (in a direction towards the wall)
(ii)
Change in momentum, p
Time
mv  mu
F
t
m v  u 

t
70  0  26 

0.1
 18 200 N (in a direction against the dummy)
Average force, F 
281
(iii)


5. (a)
(b)
p
t
p
t
F
1820

45000
t  0.040 s
Using F 
(i)
The graph of volume versus temperature in degrees Celsius will produce
A straight line intercepting the volume axis (i.e. not through the origin).
When this graph is extrapolated to the temperature axis, it cuts the axis at
273°C which represents the absolute zero on the Kelvin scale.
(ii)
T / K   / C  273
(i)
Using pV
1 1  p2V2
5  50  1V2

V2  5  50
 250 ml
(ii)
Using
p1 p2

T1 T2
p2
5

 273  25  273  60 
p
5
 2
298 333
333  5
p2 
298
 5.6 atm
282
6. (a)
(b)
(i)
(ii)
Magnification, m 
(i)
Using



Image distance, v
height of image
or
Object distance, u
height of object
1 1 1
 
f u v
1
1 1


10 20 v
1 1 1
 
v 10 20
2 1

20
1

20
v  20 cm
Image is on opposite side of lens.
(ii)

(iii)
v
u
20
m
20
1
Using m 
The image formed is real.
283
1. (a)
Length of
Pendulum
Ɩ (m)
0.20
0.30
0.40
0.50
0.60
0.70
Time for 20
Oscillations,
t (s)
18.00
21.91
25.40
28.28
31.10
33.80
Time for 1
Oscillation
(period) T (s)
0.90
1.10
1.27
1.41
1.56
1.69
Period
squared
T2 (s2)
0.81
1.21
1.61
1.99
2.43
2.86
Graph of Period squared ( T 2 ) vs Length (L)
284
(b)
Points selected for gradient: (0.10, 0.40), (0.62, 2.50)
y y
Gradient  2 1
x2  x1
2.50  0.40
0.62  0.10
2.1

0.52
 4.0 s 2 m -1

(c)
1
2
g  4 2  T
l
1
gradient
1
 4  3.142 
4.0
-2
 9.86 ms
 4 2 
(d)
2. (a)
(i)
Forms of energy
Mechanical
Chemical
Thermal
(ii)
Example
A moving car/object
A car battery
A pot of boiling water
Unit of energy : Joule (J)
285
(b)
(iii)
A Joule is the work done by a force of one Newton when its point of
application moves one metre in the direction of action of the force.
(i)
Assuming no friction:
Amount of energy  Work done
 Force  Distance
 F s
 mgs
 60 10 150
 90 000 J
(ii)
Power 
Work done
Time taken
90 000
25
 3600 W
 3.6 kW

(iii)
3. (a)
More power would be needed since energy would be lost due to friction.
(i)
Temperature-Fixed point
Upper fixed point
Lower fixed point
(ii)
Value from Dr. T’s Bag
100°C
0°C
The lower fixed point is the temperature of pure melting ice at normal
atmospheric pressure.
(iii)
Type of thermometer
Liquid-in-glass thermometer
Platinum resistance thermometer
OR
Constant volume gas thermometer
Thermocouple
(b)
Physical quality
Volume of a liquid
Resistance of platinum
Pressure of gas
E.m.f. between the junctions
Initial temperature, T1  27  273  300 K
Initial pressure,
p1  220  100  320 kN m-2
Final pressure,
p2  250  100  250 kN m-2
286
p1 p2

T1 T2
320 350

300 T2
350
T2 
 300
320
Using


4. (a)
 328 K
T2  382  273
 55°C
(i)
The laws of reflection :


(ii)
The angle of incident, iˆ is equal to the angle of reflection, r̂ .
The incident ray, the reflected ray and the normal at the point of
incidence are all on the same plane.
The raindrops on the windscreen act like prisms which cause dispersion of
light to produce the ‘glare’.
287
(b)
(i)
Using refractive index, nw 
sin bˆ (air)
sin aˆ (water)
sin 45
sin 32
0.7071

0.5229
 1.33
nw 
(ii)
Using n 
sin bˆ
sin aˆ
 sin bˆ  n sin aˆ
 1.36  sin 32
 1.36  0.5299
 0.720 7
bˆ  46

5. (a)
(b)
(iii)
The pencil ‘bends’ more in ethanol than in water.
Ethanol has a higher refractive index which implies a decrease in angle of
refraction i.e. more ‘bending’.
(i)
V  IR , where R  resistance of conductor
(ii)
For resistors in parallel:
1
1
RT  
R1 R2
(i)
Using V  IR
V

R
I
3.0

0.30
 10.0 
288
(ii)
Total resistance, RT 
R1 R2
R1  R2
10 100
10  100
1000

110
 9.1

V
R
3.0

9.1
 0.33 A
Using I 
(iii)
If the rheostat is reduced too much the overall resistance in the circuit will
decrease causing the current to increase. If this reaches above the current
rating of the bulb it can cause it to blow as well as the heating up of the
connecting wires.
(iv)
For small currents the resistance is constant, hence the straight line. As the
current increases, heat is produced causing the resistance to increase hence
the curved line in the graph.
289
6. (a)
(i)



(ii)
(b)
(c)
The alpha particles are deflected in the same direction as the
electric field i.e. away from positive and towards negative.
The beta particles are deflected opposite to the direction of the
electric field with greater deflection since they are lighter than
alpha particles.
Gamma rays carry no charge and are therefore undeflected by the
electric field.
For a uniform magnetic field into the page:
 Alpha particles would be deflected towards the left (Fleming’s left
hand rule) in a circular arc.
 Beta particles would undergo greater deflection towards the left.
 Gamma rays would not be deflected.
210
82
0
Pb 
 210
83 Bi  1 e (beta)
210
83
0
Bi 
 210
84 Po  1 e
210
84
Po 
 82 Pb  24 He (alpha)
206
Mass defect, m  0.214 1.66 10 27
 3.3432 10 28 kg
Energy released, E  mc2
 3.3432 10 28   3 108 
2
 3.0 1011 J
290
1. (a)
Graph of Velocity, V vs Time, t
291
(b)
(i)
Acceleration during AB  Slope of AB
Vertical displacement

Horizontal displacement
30

20
 1.5 ms -2
(ii)
Velocity after 70 seconds  15 ms-1
(iii)
Total distance travelled  Area under graph (trapezium)
1
  sum of parallel sides   height
2
1
  40  80   30
2
1
 120  30
2
 1800 m
(c)
Over the period BC the car is travelling with a constant velocity of 30 ms-1. There
is no acceleration.
(d)
Velocity is defined as the rate of change of displacement or the rate of change of
distance in a particular direction.
2. (a)
Physical quantity
(b)
Unit
(i)
Kinetic energy is the energy of a body due to its motion.
(ii)
Potential energy is the energy of a body due to its state or position.
(iii)
E p  mg h
292
(c)
(d)
(iv)
At the top of the waterfall the water has potential energy and as it fall
down the waterfall the potential energy is converted into kinetic energy.
(i)
A - potential energy
(ii)
B – kinetic energy
(iii)
C – potential energy
(iv)
At, D, the type of energy increasing is potential energy.
Reasons: The pendulum bob is gaining height as it approaches A
1 2
mv
2
1
  0.4  52
2
5 J
Ek 
3. (a)
Type of thermometer
Design feature
Mercury-in-glass
Narrow bore
laboratory
thermometer
Clinical thermometer
Constriction in bore
Thermocouple
Small junctions with
a low heat capacity
Reason for design feature
To detect small changes in
temperature i.e. sensitive
Retaining a measured
temperature
Ability to measure rapidly
changing temperatures
(b)
The upper fixed point is the temperature of steam at standard atmospheric
pressure and is 100 ºC.
The lower fixed point is the temperature of pure melting ice and is 0º C.
(c)
(i)
Pressure (Pa)
1.1105
1.2 105
1.3 105
(ii)
Temperature(°C)
35
63
91
Temperature(K)
308
336
364
According to the Pressure Law p  T (absolute temperature) provided V
p
is constant   constant, k
T
Testing data:
293
p1 1.1105

 357
T1
308
p2 1.2 105

 357
T2
336
p3 1.3 105

 357
T3
364
 Pressure Law is supported by the set of data.
4. (a)
(b)
(c)
(i)
The incident ray, reflected ray and normal at the point of incidence are all
on the same plane.
(ii)
Properties of the image formed by a plane mirror:
 Same size as object
 Same distance from mirror as object
 Virtual
 Laterally inverted
 Upright
(i)
Image distance  Object distance
 85
 13 m
(ii)
Since light travels in straight lines the truck driver can only view to car
from the side mirror as the truck is opaque. Car drives on the other hand
can use the rear view mirror that can view vehicles through the back
windscreen.
(i)
Using n 
(ii)
An increase in refractive index implies a greater angle of refraction and
therefore increased lateral displacement.
sin iˆ
sin rˆ
sin iˆ
 sin rˆ 
n
sin 30

1.3
0.5

1.3
 0.384 6

rˆ  22.6
294
5. (a)
(i)


(ii)
A real image is formed when the object distance is greater than the focal
length.
A virtual image is formed when the object distance is less than the focal
length.
Real images are inverted and virtual images are upright.
(iii)
(b)
Image size
Object size
3.6

2.4
 1.5
(i)
Magnification, m 
(ii)
Using m 

(iii)
Using


v
u
v  mu
 15  2.0
 30 cm
1 1 1
 
f u v
1 1
1


f 20 30
3 2

60
5

60
1

12
f  12 cm
295
6. (a)
(b)
The circuit consists of a 3 V battery power supply which is connected in series
with a switch, a rheostat, a fixed resistor and an ammeter. A voltmeter is
connected in parallel to the fixed resistor.
(i)
R1 R2
R1  R2
1 2
R
1 2
2
 
3
Resistance across BC: R 

(ii)
 1 2 
Total resistance in circuit, RT  
3
 1 2 
2

RT   3
3
2
3 
3
 3.67 
V
R
12
I
3.67
 3.27 A
Using I 

(iii)
Using P  I 2 R
 3.27 2  3
 32.1 W
296
1. (a)
Graph of Activity (A) vs Time (t)
297
(b)
(i)
For activity
A0  80, t0  0 hours
For half the activity, A1  40, ti  1.6 hours
t 1  1.6 hours
 Half life,
2
(ii)
When
A2  50, t2  1.0 hour
For half the activity, A3  25, t3  2.6 hours
t 1  1.6 hours
 Half life,
2
When
A4  20, t4  3.0 hours
For half the activity, A5  10, t5  4.5 hours
t 1  1.5 hours
 Half life,
2
Average, t 1 
2
1.6  1.6  1.5
3
 1.6 hours
(c)
For A  10 diss-1 , t  4.5 hours
(d)
The line is not perfectly smooth because of the random nature of decay.
(e)
Atomic number is the number of protons in the atom of the element.
Mass number is the number of protons and neutrons in the atom of the element.
(f)
The number ‘123’ represents a specific isotope of Iodine having a mass number
of 123.
2. (a)
Physical quantity
(b)
(i)
S.I. unit
If a body A , exerts a force on body B, then body B will exert an equal and
opposite force on body A. i.e to every action there is an equal and opposite
reaction.
298
(ii)
As the plane engine expels the air with a force towards the tail end of the
plane, an equal and opposite force is exerted on the plane that propels it
forward. The weight of the plane is offset by the lift force caused by the
wing which keeps it flying horizontally.
(c)
(i)
Linear momentum is the product of mass and velocity of a moving body.
(ii)
By principle of conservation of momentum:
Total momentum before collision  Total momentum after collision
8 10    2  5  8  2  v
80  10  10v
70
v
10
 7 ms-1 due East

3. (a)
EH :
Heat energy
Unit: Joule (J)
c:
 :
Specific heat capacity
Change in temperature
Unit: J kg-1 K-1
Unit: K or °C
(b)
The symbol ‘Ɩ’ represent the specific latent heat.
(c)
(i)
Heat lost by water:
Using E  mc
 100  4.2   30  20 
 100  4.2  10
 4 200 J
(ii)
Energy required to change melted ice at 0º C to water at 20º C:
Using E  mc
 10  4.2  20
 840 J
299
(iii)
Heat lost by water  Heat gained by ice
4 200  840  mi l f
4 200  840  10l f
l f  336 J g -1
4. (a)
(b)
Electromagnetic waves and use: (any three)
 Gamma rays radiotherapy (cancer treatment)
 X-rays
radiography (imaging)
 UV
medical treatment of skin disorder
 Visible
analytical equipment to detect elements
 IR
radiator heating
 Radio
satellite communication
Using v  f 
v


f
For rattlesnakes:
3 108

3.5 1014
 8.6 10 7 m
For honeybee:
3 108

11015
 3 10 7 m
Distance
Time
750

2.3
 326 ms -1
Speed 
5. (a)
(i)
The electric current in a metal is due to the free electrons as the only
charge carrier. The direction of flow of these electrons is opposite to the
conventional current direction. In electrolyte the electric current is due to
the flow of negative and positive ions. The conventional current direction
is the same as the direction of flow of positive ions and opposite to the
flow of negative ions.
(ii)
The current flow in semiconductor is similar to that of electrolyte since the
semiconductor has both free electrons and positive holes a charge carriers.
300
(b)
Using Q  It
Q

t
I
For 4320 C battery:
4320
t1 
0.6
 7 200 s
(i)
For 9000 C battery:
9 000
t2 
0.6
 15000 s
Time difference, t  t2  t1
 15000  7 200
 7800 s
Charging time for old model  7 200 s
Q
Current rating,
I
t
9 000

7 200
 1.25 A
(ii)
6. (a)
(b)
The candle wax in the test tube was first melted to liquid state by placing it in the
hot water bath using a test tube holder. The tube was then placed on a rack and a
thermometer was inserted. While in liquid state it was stirred for even distribution
of heat. Starting from t = 0 s, the temperature was measured at regular time
intervals. A graph of temperature vs time was then plotted.
(i)
(ii)
As the water drops from the top of a waterfall, its potential energy is
converted to kinetic energy. This energy is transferred to the rotating
blades of the turbine which is converted to electrical energy by the
generator.



The presence of waterfall in Dominica ensures a renewable source of
energy.
Hydroelectricity produces no waste or greenhouse gases and therefore
does not affect global warming.
The long term cost of hydroelectricity production is cheaper than energy
from fossil fuel which makes it a viable alternative.
301
1. (a)
Graph of Electrical energy, E vs Temperature rise, T
302
(b)
Points selected for slope: (2.0, 2.4), (13.4, 16.0)
y y
Slope, S  2 1
x2  x1
16.0  2.4

13.4  2.0
13.6

11.4
 1.2 kJ K -1
(c)
The slope represents heat capacity, C.
(d)
Specific heat capacity, c 
(e)
(i)
The procedure will minimize the error since at 10º C below room
temperature heat is transferred from the environment to the liquid. But at
10º C above room temperature, heat is transferred from the liquid to the
environment, resulting in negligible net heat transfer.
(ii)
The liquid should be stirred while heating to ensure equilibrium measured
temperature.
C
m
1.2 1000

250
 4.8 J g -1 K -1
Energy, E
Time, t
E
Time, t 
P
18000

40
 450 s
Using power, P 
(f)

 7.5 minutes
2. (a)
(b)
1) Carbon rod (+ve electrode)
2) Electrolyte (paste of Ammonium Chloride)


A primary cell cannot be recharged whereas a secondary cell can be recharged.
A primary cell has high internal resistance whereas a secondary cell has low
internal resistance.
303

(c)
3. (a)
The chemical reaction in a primary cell is irreversible whereas in a secondary cell
the chemical reaction is reversible.
(i)
Charge added, Q  It
 1 4  60  60
 14 400 C
(ii)
Energy added, E  QV
 14 400 12
 172800 J
 172.8 kJ
(iii)
The charging voltage for a solar module can be 12, 24 or 48 VDC in order
to handle the current (A) from the solar module.
(i)
The principle of moments states that when a body is in equilibrium, the
sum of the clockwise moments about any point (pivot) is equal to the sum
of the anticlockwise moments about the same point.
(ii)
(b)
(i)




A force applied to a spanner to rotate a nut.
A force applied to swing-open a door.
A force applied to push a swing.
A driver turning a steering wheel.
By the principle of moments:
Clockwise moments (Kyle)  Anti-clockwise moments (Keion)
500 1  300  x
500

x
300
 1.67 m
(ii)
304
At equilibrium:
Total upward force  Total downward forces
R  500  300
 800 N
(iii)
4. (a)
(i)


Moment of reaction force about the pivot  0 Nm
A swimming pool seemed shallow when viewed from above
A straight stick seemed to bend when dipped in water.
(ii)
(b)
(iii)
The speed of light will decrease in travelling from air to water since water
is a denser transparent medium than air.
(i)
The prism causes dispersion of white light.
(ii)
305
(c)
(i)
Angle of
incidence, iˆ
30°
50°
60°
(ii)
Using n 
 sin rˆ 

5. (a)
(b)
Angle of
refraction, r̂
sin iˆ
sin rˆ
20°
31°
35°
0.50
0.77
0.87
0.34
0.52
0.57
sin iˆ
sin rˆ
sin iˆ
n
sin 70

1.52
0.94

1.52
 0.62
ˆr  38
(i)
NOT gate
(ii)
AND gate
(iii)
NOR gate
(i)
NOT gate
Input
0
1
(ii)
(i)
Output
1
0
NOR gate
Input
0
0
1
1
(c)
sin iˆ
sin rˆ
1.47
1.48
1.53
0
1
0
1
Output
1
0
0
0
A 1
B0
C 1
306
(d)
6. (a)
(ii)
A 1
B 1
C 1
(i)
The use of technology has improved transportation in land, air, sea and
space travel.
(ii)
Advances in electronic technology have improved radio, television,
internet and satellite communication.
(iii)
Improved operating efficiency in equipment and household appliances.
(i)
A – Thin sheet of gold foil
B – Rotating scintillation microscope
C – Alpha source (in lead container)
(b)
(c)
(ii)
Gold foil was used because gold is very malleable and therefore very thin
sheets of gold foil can be produced.
(iii)
1) Most of the atom was empty space
2) The atom has a positively charged center that contains most of its mass.
(i)
Number of neutrons  28 14
 14
(ii)
29
14
(i)
The decay process represents alpha decay.
(ii)
p  222  4
Si or
30
14
Si or
31
14
Si
 218
q  76  2
 74
0
Y 
 218
75 Z  1 e
(iii)
218
74
(iv)
The particle’s mass is reduced in an alpha decay and almost unchanged in
a beta decay.
307
1. (a)
Graph of Induced E.m.f./V vs Time/ms
(b)
(i)
Induced e.m.f. after 12.5 ms  1.50 V
(ii)
Another time when induced e.m.f. was 0V  27.75 ms
308
(c)
(i)
(ii)
A changing magnetic flux due to the approaching magnet causes an
induced e.m.f. in the coil which in turn produces an induced current.
(iii)
The sensitive galvanometer can detect and measure the current flowing in
both directions.
(iv)
(d)
(v)
To increase the induced e.m.f:
 Move the magnet faster
 Use a stronger magnet
 Use a coil with more turns
(vi)
With the magnet stationary in the coil there would be no change in
magnetic flux and therefore no induced current.
(i)
309
(ii)
1
T
1
f 
0.02
 50 Hz
Using f 
(iii)
2. (a)
(b)
(i)
A transverse wave is one in which the displacement of the particles is at
right angles to the direction of travel of the wave. A longitudinal wave is
one in which the displacement of the particles is parallel to the direction of
travel of the wave.
(ii)
Transverse wave: surface water waves, electromagnetic waves, waves in
strings.
Longitudinal wave: Sound waves, Pressure waves, Seismic P-waves.
(iii)
v f
(iv)
v  10  250
 2500 ms-1
(i)
(ii)
Amplitude, A  0.5 103 m
(iii)
Period, T  4.0 103 s
310
(iv)
Frequency, f 
1
T
1
4.0 10 3
 250 Hz

(c)
3. (a)
(b)
(c)
Property of electromagnetic waves:
 Transverse waves
 Travel at a speed of 3 x 108 ms-1 in a vacuum
 Carry no charge
 Do not need a vacuum to travel
(i)
Nuclear fission is the splitting of large unstable nucleus into smaller, more
stable nuclei with the release of energy.
(ii)
Two advantages of utilizing nuclear energy:
 No greenhouse gases or smoke produced
 Small quantities of raw material produce large quantities of energy
(i)
Precautions taken by workers in a nuclear power plant:
 Use protective clothing and shielding
 Handle radioactive material with robotic arms
 Use film badge to indicate exposure to radiation
(ii)
Disadvantages of using nuclear reactor to generate energy:
 Nuclear waste is difficult and expensive to process.
 Nuclear accidents can be devastating
 Nuclear plants have limited life and therefore has to be dismantled
at a cost.
 Nuclear power plants generate external dependence since not many
countries have Uranium.
(i)
P  92  36
 56
(ii)
‘c’ represents the speed of light.
(iii)
Total mass of elements  Total mass of product side of equation
  232.560  152.620  3 1.670   10 27
 390.190 10 27 kg
(iv)
Mass defect, m   398.350  1.670   10 27  390.190  10 27
 9.8310 27 kg
311
Energy released, E  mc2
 9.83 10 27   3 108 
2
 8.85 1010 J
4. (a)
(b)
(v)
This energy can be used to boil water to make steam which drives turbine
to generate electricity.
(i)
For carriage moving horizontally at a constant speed in a straight line:
Newton’s first law of motion applies: Everybody continues in a state of
rest or of uniform motion in a straight line unless compelled by an external
force to act differently.
(ii)
Carriage is in free fall:
Newton’s second law of motion applies: The rate of change of momentum
is proportional to the applied force and takes place in the direction in
which the force acts.
(i)
Using
v  u  at
64.8 1000 m
64.8 kmh -1
60  60 s
64800

3600
 18.0 ms-1


(ii)
5. (a)
(i)
18.0  0  10t
a  g 
t  1.8 s
1
Using s  ut  at 2
2
1

s  gt 2
u  0
2
1

s  10 1.82
2
 16.2 m
A car radiator: needs to lose heat efficiently
 Painted black since a black surface is a better radiator of heat than
a white surface.
 Has a large surface area with the use of ‘fins’ to ensure greater
emission.
312
(b)
(ii)
The roof of a Caribbean home: needs to reflect radiation
 Usually a shiny surface since it reflects radiation better than a dull
surface.
 It is painted white since this colour is a poor absorber of heat
radiation.
(i)
Assuming negligible heat loss:
Heat supplied by immersion heater  Latent heat required to evaporate
Water
Pt  mlv


(ii)
6. (a)
150  5  60   0.28  0.26   lv
45000  0.20  lv
45000
lv 
0.02
 2250000 J kg-1
If the coil of the immersion heater is not completely submerged it will
take a larger time to evaporate the same mass of water. If the same power
rating value (150 W) is used in calculation, the value of the specific latent
heat of vaporization will increase.
(i)
The principal focus of a converging lens is a point on the principal axis
which all rays initially parallel to the principal axis will converge on after
refraction by the lens.
313
(ii)
The principal focus of a diverging lens is a point on the principal axis
which all rays initially parallel to the principal axis appear to diverge from
after refraction by the lens.
(b)
(i)
Using


1 1 1
 
f v u
1 1 1
 
v f u
1
1


12.0 18.0
3 2

36
1

36
v  36.0 cm
v
u
36.0

18.0
2
(ii)
Magnification, m 
(iii)
Height of image  m  7.2
 2  7.2
 14.4 cm
(iv)
The image formed is real.
314
1. (a)
(i)
Angle of reflection, r̂
10.0
30.0
50.0
70.0
90.0
(ii)
Angle of incidence, iˆ
6.5
19.0
30.0
38.0
41.0
sin rˆ
0.174
0.500
0.766
0.940
1.000
sin iˆ
0.113
0.326
0.500
0.616
0.656
Graph of sin rˆ vs sin iˆ
315
(iii)
Points selected for gradient
(0.60, 0.91), (0.10, 0.15)
y y
Gradient  2 1
x2  x1
0.91  0.15

0.60  0.10
 1.52
(b)
The physical property represented by gradient is refractive index.
(c)
The critical angle is the angle of incidence that produces an angle of refraction of
90º for light travelling from one medium to an optically less dense medium.
(d)
From Table 1, critical angle for glass  41.0º
(e)
Using


n(cladding) sin 90

n(core)
sin c
1
1.03 
sin c
1
sin c 
1.03
 0.9709
c  76.1
2. (a)
Solid
Liquid
Gas
(b)
(i)
Shape
Volume
Movement of
molecules
Definite/Fixed
Takes shape of
container
No fixed shape
Definite/Fixed
Definite/Fixed
Vibrate
Move amongst
one another
Move freely
Full space
Intramolecular
forces
Strong
Very weak
Negligible
weak
Time of heating   5  60   45
 345 s
(ii)
Mass of water evaporated  375  360
 15 g
 0.015 kg
316
(c)
3. (a)
(b)
Assuming no heat loss:
Energy supplied by electric heater  heat energy required to evaporate water
Pt  mlv
Pt
lv 
m
100  345

0.015
 2.3106 J kg-1
Energy is the stored ability to do work.
S.I. unit of energy  Joule (J)
(i)
Forms of energy
Nuclear Energy
Electromagnetic Radiation Energy
Kinetic Energy
(ii)
(c)
(i)
Example
Radioactive decay
Radio waves, X-rays
Objects in motion
Chemical energy stored in the battery is converted to light energy (and
some heat energy).
1
Maximum kinetic energy, Ek  mv 2
2
1
  0.5 1.82
2
 0.81 J
(ii)
Assuming no energy is lost in moving from Y to X
Maximum gravitational potential energy  Maximum kinetic energy
 0.81 J
(iii)
Maximum E p  mgh

h
Ep
mg
0.81

0.5  9.8
 0.165 m
 16.5 cm
317
4. (a)
Arguments for nuclear fission reactors:
1) There is a reduction in environmental pollution as a result of little or no
carbon dioxide emission.
2) Produce high amounts of nuclear fission energy using relatively small
quantities of raw materials.
Arguments against nuclear fission reactors:
1) Accidents can be devastating, resulting in radiation exposure.
2) They create harmful nuclear waste.
(b)
Advantages of nuclear fusion over nuclear fission:
1) Nuclear fusion produces much more energy than nuclear fission.
2) Nuclear fusion does not produce the level of toxic radioactive waste, like
nuclear fission.
(c)
2
1
(d)
H  21 H 
 31 H  11 H
Mass of 21 H  31 H  2.01410178  3.01604927
 5.03015105 u
Mass of 42 He  01 n  4.00260325  1.00866492
 5.01126817 u
Mass defect   5.03015105  5.01126817  1.66 10 27 kg
 3.135 10 29 kg
Energy released  mc2
 3.13455808 10 29   3.0 108 
2
 2.82 1012 J
5. (a)
Newton supported the particle theory of light based on the evidence that light
travels in straight lines and can travel through a vacuum. It is also supported by
the phenomena of photo electric effect.
Huygens supported the wave nature of light based on the evidence that light can
undergo reflection, refraction, diffraction and interference.
(b)
(i)
Objects placed at a distance between f and 2f would produce a real
magnified image. The phone should therefore be placed at a distance
greater than 15.0 cm but less than 30.0 cm.
318
(ii)
Using


1 1 1
 
f u v
1 1 1
 
v f u
1 1
 
15 20
1

60
v  60 cm
v
u
60

20
3
(iii)
Magnification of image, m 
(iv)
Dimension of image of phone screen  11.0  6.0  3
 33.0 cm 18.0 cm
6. (a)
(v)
Image formed in (b)(ii) is inverted.
(vi)
The image can be made larger by moving the phone closer to the lens or
moving the screen further from the lens.
(i)
Magnetic field due to a current carrying conductor
319
(b)
(ii)
Resulting magnetic field when a current carrying conductor is placed
between the poles of the magnet.
(i)
The part labeled X is the split ring or commutator.
(ii)
When the switch is closed the current flows through the carbon brushes
and commutator into the coil. The current flows from D to C and B to A.
From Fleming’s Left Hand Rule, a downward force is exerted on DC
while an upward force is exerted on AB. This is due to the magnetic fields
produced by the coil and the magnet. The momentum of the coil allows it
to cross the vertical position. The commutator reverses the direction of the
current in the loop as the contact changes from one brush to the other. At
this point AB then moves down while DC moves up resulting in
continuous rotation in one direction.
320