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be removed, the T-cells would them an infected person can no longer fight off pathogens. Secondary
continue to make new viruses. infections result and the person is said to be suffering from AIDS.

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AIDS is the end stage of an HIV infection. It is caused by a severe failure
of the immune system as the HIV virus selectively infects helper T-cells.
Some infected individuals have no symptoms in the early stages of the
disease while others may be slightly unwell when first infected. Symptoms
of AIDS develop as the number of active helper T-cells decreases.The
symptoms occur as a result of secondary infections caused by bacteria,
fungi and viruses that the body is unable to resist due to its compromised
immune system (Figure 6.15, overleaf ).

HIV is transmitted in blood, vaginal secretions, semen, breast milk and
sometimes across the placenta. In some countries, HIV has been transmitted
in blood transfusions but in most places with medical care facilities, blood for

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transfusion is now screened for the virus.The virus is most frequently passed
from person to person in bodily fluids during sex and also when non-sterile
syringe needles are used to administer either legal or illegal drugs.

AIDS used to be thought of as a disease that affected only drug abusers
and homosexuals but it is now clear that the virus is also transmitted
by heterosexual sex.Transmission is most likely among individuals who
engage in casual sex with multiple partners.

AIDS is a worldwide pandemic but some regions are more seriously
affected than others. In 2008, the number of people living with HIV and
AIDS was estimated at 25 million in sub-Sarahan Africa and almost 8
million in South and South East Asia.AIDS is the main cause of death for
men and women aged between 16 and 50 years in these countries, Latin
America and the Caribbean.Those who die are farmers, workers and
parents, so the social and economic consequences are significant – in many
communities, the elderly are left caring for children, with little support.
Family incomes and national productivity decline as people of working
age fall ill. Food production is affected and causes more problems. Caring
for the sick and providing medical care form an expensive burden for
individual families, companies and governments.

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r Distinguish between ‘ventilation’,‘gas exchange’ and ‘cell respiration’.
r Explain the need for a ventilation system.
r Describe the features of alveoli that adapt them to gas exchange.
r Draw and label a diagram of the ventilation system, including trachea, lungs, bronchi, bronchioles and alveoli.
r Explain the mechanism of ventilation of the lungs in terms of volume and pressure changes caused by the internal and

external intercostal muscles, the diaphragm and abdominal muscles.

All living cells need energy for their activities. Energy is released from
the breakdown of glucose and other substances during the process
of cell respiration. Respiration is a chemical reaction that occurs in
mitochondria and the cytoplasm and releases energy as ATP, a form that
can be used inside cells.

Our cells use oxygen to carry out aerobic respiration and produce
carbon dioxide as a waste product. Oxygen is taken in from the air
and carbon dioxide is returned to it in a passive process known as gas
exchange. Gas exchange occurs in the alveoli of the lungs where oxygen
from the air diffuses into blood capillaries, and carbon dioxide passes in
the opposite direction. Gases are also exchanged in the tissues where
oxygen diffuses into respiring cells and is exchanged for carbon dioxide.

Whenever diffusion occurs, there must always be a concentration
gradient with a higher level of the diffusing substance in one area than in
another.Air inside the alveoli contains a higher concentration of oxygen
than the blood, so oxygen diffuses into the blood. Blood contains a higher
level of carbon dioxide than inhaled air, so carbon dioxide diffuses into
the alveoli.

For gas exchange to continue, these concentration gradients must be
maintained.As oxygen diffuses out of the alveoli, the level of oxygen
inside them gradually falls. Stale air must be expelled regularly and
replaced with a fresh supply.This is achieved by breathing in and out, a
process known as ventilation.

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Our lungs are protected inside the thorax in an air-tight cavity formed by
the ribs and diaphragm.The inside of the ribcage is lined with membranes
that secrete fluid to lubricate the lungs, making them slippery and

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reducing friction during breathing.Air is drawn in through the nose and Blowing a trumpet
passes down the trachea to the two bronchi (singular bronchus), one of
which passes to each lung (Figure 6.16). Bronchi divide into smaller and External and internal intercostal
smaller tubes, called bronchioles, which end in tiny air sacs or alveoli muscles are antagonistic muscles
(singular alveolus).The alveoli are covered with a network of capillaries because they have opposite
and together provide us with a very large surface area for the exchange of effects when they contract.
oxygen and carbon dioxide. During relaxed breathing, internal
intercostal muscles do not contract,
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the lungs when we sing or play a
Lungs have no muscles and cannot move by themselves. Breathing is wind instrument.
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abdomen (Figure 6.17).

During inhalation, contraction of the external intercostal muscles
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thorax.These movements increase the volume of the chest cavity and
lower the pressure on the lungs to below that of the air outside.As a result,
air is drawn down the trachea to fill the lungs.

Gentle exhalation occurs as the intercostal and diaphragm muscles
relax, reducing the volume of the chest cavity. Elastic fibres around the
alveoli return to their original length and pressure forces air out of the
lungs.

Long or forced exhalations involve the internal intercostal muscles,
which contract to lower the ribs. Muscles in the abdominal wall also

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Alveoli are the body’s gas exchanges surfaces. Formed in clusters at the
ends of the smallest bronchioles, more than 300 million alveoli in each
lung together provide a surface area of about 75 m2.Alveoli are roughly
spherical in shape and made of flat cells less than 5 μm thick (Figure
6.18).The capillaries that wrap around them also have thin walls of single
epithelial cells.These two thin layers make the distance for diffusion
of gases as small as possible. Oxygen diffuses through the alveolus and
capillary into the blood and carbon dioxide diffuses in the opposite
direction (Figure 6.19). So long as the diffusion gradient is maintained by
regular breathing, diffusion continues.

Table 6.5 summarises ways in which the alveoli are well adapted for
their role in gas exchange.

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r State that the nervous system consists of the central nervous system (CNS) and the peripheral nerves, and is composed

of cells called neurons that can carry rapid electrical impulses.

r Draw and label the structure of a motor neuron.
r State that nerve impulses are conducted from receptors to the CNS by sensory neurons, within the CNS by relay

neurons and from the CNS to effectors by motor neurons.

r Define ‘resting potential’ and ‘action potential’ (depolarisation and repolarisation).
r Explain how a nerve impulse passes along a non-myelinated neuron.
r Explain the principles of synaptic transmission.
r State that the endocrine system consists of glands that release hormones that are transported in the blood.
r State that homeostasis involves maintaining the internal environment between limits, including blood pH, carbon

dioxide concentration, blood glucose concentration, body temperature and water balance.

r Explain that homeostasis involves monitoring levels of variables and correcting changes in levels by negative feedback

mechanisms.

r Explain the control of body temperature, including the transfer of heat in blood and the roles of the hypothalamus,

sweat glands, skin, arterioles and shivering.

r Explain the control of blood glucose concentration, including the roles of glucagon, insulin and B and C cells in the

pancreatic islets.

r Distinguish between ‘type I’ and ‘type II diabetes’.

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layout of the human nervous system.
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;^\jgZ +#'% I]Z ]jbVc cZgkdjh hnhiZb# The peripheral nerves are the network of neurons that carry
information to and from the CNS. Peripheral nerves include sensory
neurons, which carry information to the CNS, and motor neurons, which
transmit impulses from the CNS to muscles and glands that then cause a
response.

Three types of neuron are found in the nervous system. Sensory and
motor neurons transmit information to and from the CNS, while relay
neurons within the CNS form connections between them.

The structure of a motor neuron is shown in Figure 6.21. Many small
dendrites receive information from relay neurons and transmit the
impulses to the cell body. One long axon then carries impulses away.The
cell body contains the nucleus and most of the cytoplasm of the cell.The
axon is covered by a myelin sheath formed from Schwann cells, which
wrap themselves around it. Myelin has a high lipid content and forms an
electrical insulation layer that speeds the transmission of impulses along
the axon.Along the length of the axon, junctions where two Schwann
cells meet can be seen.These junctions are known as nodes of Ranvier.

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Neurons transmit information in the form of impulses, which are short-
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potassium channels close.The resting potential is re-established by the
action of sodium–potassium pumps, which move ions back to restore
the resting potential.
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adjacent section of the axon.This occurs because local currents are set up
between adjacent regions and these cause ion channels to open, allowing
sodium ions in and potassium ions out of the axon.

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The action potential travels along the neuron rather like a ‘Mexican
wave’.The impulse can only pass in one direction because the region
behind it is still in the recovery phase of the action potential and is
temporarily unable to generate a new action potential.The recovery phase
is known as the refractory period.

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A synapse is the place where two neurons meet.Two neurons do not
touch one another and the tiny gap of about 20 nm between them is

&*'

known as the synaptic cleft.Action potentials must be transmitted across There are more than 40 different
this gap for the impulse to pass on its way and this is achieved by the neurotransmitters in the body.
presence of chemicals known as neurotransmitters. Neurotransmitters Acetylcholine and noradrenalin
are held in vesicles in the pre-synaptic cell until an action potential arrives. are found throughout the nervous
They are then released into the synaptic cleft, and diffuse across to the system, others (e.g. dopamine) are
post-synaptic membrane.There they can cause another action potential to found only in the brain.
be produced.
Many drugs affect synapses and
The synapse shown in Figure 6.25 uses the neurotransmitter influence the way nerve impulses
acetylcholine (ACh) and is a cholinergic synapse.ACh binds to receptors are transmitted. Nicotine has
and causes depolarisation of the post-synaptic membrane and the a similar molecular shape to
initiation of an action potential. Once an action potential is generated in acetylcholine and affects the
the post-synaptic membrane,ACh in the synaptic cleft is deactivated by post-synaptic membrane so that it
acetylcholinesterase enzymes and the products are reabsorbed by the pre- transmits an action potential.
synaptic membrane to be remade and repackaged in vesicles.

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The internal environment of the body remains constant, within certain
limits, despite changes that occur in the external environment.The
control process that maintains conditions within these limits is known as
homeostasis.

The factors that are controlled include water balance, blood glucose
concentration, blood pH, carbon dioxide concentration and body
temperature.

Each of these has a ‘normal’ or set point although they may vary
slightly above or below it. For example, the normal body temperature for
humans is about 37 °C.

Both the nervous system and the endocrine system are involved in
homeostasis.The endocrine system consists of ductless endocrine
glands, which release different hormones (Figure 6.26, overleaf ).
Hormones circulate in the bloodstream but each one is a chemical
messenger that only affects the metabolism of specific target cells.

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Feedback systems work by monitoring the level of a substance, or a
product, and feeding this level back to affect the rate of production
or use of the substance. Negative feedback stabilises the internal
environment by reversing the changes within it. For example, if metabolic
processes produce heat causing blood temperature to rise, sensors in the
hypothalamus in the brain respond and send messages to increase heat loss
or slow down heat production.

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amount of glucose is absorbed from the intestine, responses are initiated to
bring the levels back to normal.The graph shown in Figure 6.27 shows
how the level of glucose in the blood changes from its normal level and
how the normal level is re-established.
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Body temperature is monitored and controlled by the hypothalamus
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fluctuations in their blood glucose above and below the normal limits
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diabetes, which must be controlled by insulin injections. Symptoms Causes of diabetes
usually begin in childhood, which is why type I diabetes is sometimes
known as ‘early onset’ diabetes. Evidence that type II diabetes is
influenced by genes has come
Type II diabetes accounts for 90% of all cases. It is caused by body from studies of ethnic groups
cells failing to respond to the insulin that is produced.Again, the result worldwide. Native Americans,
is that blood glucose level remains too high.This type of diabetes can be Native Australian Aborigines and
controlled by a change to a low carbohydrate diet. It is often associated the Polynesian Maori people all
with obesity, age, lack of exercise and genetic factors.The pancreas does have a higher incidence of diabetes
produce insulin although levels may fall as the disease progresses.Type II than would occur by chance.
diabetes is sometimes known as ‘late onset’ diabetes.
In the last decade, there has been
Diabetics must monitor their blood glucose level carefully so that they a large increase in the number of
can control it, since the body’s internal control mechanism is not working people in industrialised countries
properly. affected by type II diabetes. Can
you suggest a reason for this?
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r Draw and label diagrams of the adult male and female reproductive systems.
r Outline the role of hormones in the menstrual cycle, including FSH, LH, estrogen and progesterone.
r Annotate a graph showing hormone levels in the menstrual cycle, illustrating the relationship between changes in

hormone levels and ovulation, menstruation and thickening of the endometrium.

r List three roles of testosterone in males.
r Outline the process of in vitro fertilisation (IVF).
r Discuss the ethical issues associated with IVF.

Reproduction is one of the important characteristics of living things.
Human male and female reproductive systems produce the gametes (the
sperm cell and egg cell) that must come together to begin a new life.
The two reproductive systems enable the gametes to meet and the female
reproductive system provides a suitable place for fertilisation to occur and
an embryo to develop.The ovaries and testes also produce hormones that
regulate sexual development and reproduction.

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The testes are held just outside the male body in the scrotum (Figure
6.29, overleaf ). Each testis produces both sperm cells and the hormone

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The temperature in the scrotum testosterone. Sperm cells are carried from the testes in the sperm ducts,
is about 35 °C, which is ideal for which transport them past the seminal vesicles and prostate glands.These
sperm production. glands together produce the seminal fluid in which the sperm cells
travel. Seminal fluid containing sperm cells is known as semen.This fluid
&*- travels down the penis in the urethra and is transferred to the vagina

during intercourse.The penis contains spongy tissue that can fill with
blood during sexual arousal to cause an erection. Sperm production is a
continuous process, which begins at puberty and continues throughout a
man’s life.

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The hormone testosterone, which is produced by the testes, has important
roles in the sexual development and reproductive behaviour in males.

r During fetal development, testosterone causes the development of the

male genitalia.

r At puberty, levels of testosterone rise and cause the development of male

secondary sexual characteristics including growth of muscle, deepening
of the voice, enlargement of the penis and growth of body hair.

r Testosterone stimulates the continuous production of sperm, and

behaviour associated with the sex drive.

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The female reproductive system includes the ovaries, which produce the
female gametes, the oviducts, where fertilisation usually takes place, and
the uterus, where the baby grows and develops (Figure 6.30, overleaf ).The
vagina is the passageway for sperm cells to enter the body, and the way out
for the baby when it is born.

Production of female gametes is a cyclical process, which lasts
approximately 28 days. During the first half of this menstrual cycle the
egg cell is produced, and in the second half the uterus lining thickens to
prepare for implantation of a fertilised egg.The cycle involves hormones
that are released by the ovaries and the pituitary gland.

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Ovaries produce two hormones, estrogen and progesterone.These
hormones stimulate the development of female characteristics at puberty
and also influence the changes in the uterus lining during the menstrual
cycle and pregnancy.The pituitary gland in the brain produces two
further hormones, luteinising hormone (LH) and follicle-stimulating
hormone (FSH). FSH stimulates the development of immature follicles
in the ovary, one of which will come to contain a mature egg cell. LH
stimulates the follicle to release the egg and subsequently to form the
corpus luteum.

The sequence of events begins at the start of menstruation, which
is often called a period (Figure 6.31, page 161). During the first four or
five days of the cycle, the endometrium (lining) of the uterus is shed and
leaves the body through the vagina.This indicates that fertilisation has not
occurred during the previous month.

In this early part of the cycle, the pituitary gland secretes FSH, which
stimulates the development of an immature follicle in the ovary.The
follicle then secretes estrogen, which enhances the follicle’s response to
FSH.As the level of estrogen rises, it also stimulates the repair of the
uterus lining.

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As the follicle grows, estrogen levels rise to a peak at around day 12, when Figure 11.22 on page 276 shows
they stimulate the release of LH from the pituitary gland.As LH levels the stages in the development of
reach their highest point, ovulation – the release of the egg cell from the a follicle in the ovary, right up to
follicle – takes place. Ovulation usually occurs at around the day 14 of the ovulation and then the formation
cycle. Immediately afterwards, LH stimulates the empty follicle to form of the corpus luteum.
the corpus luteum. Levels of estrogen begin to rise and as a result FSH
and LH levels fall.

The corpus luteum secretes progesterone, which stimulates the
thickening of the endometrium and prepares the uterus to receive an
embryo. It also inhibits the production of FSH and LH.

If the egg cell is not fertilised, the LH level falls, the corpus luteum
degenerates and progesterone and estrogen levels fall.The fall in
progesterone stimulates the breakdown of the uterus lining. FSH is no
longer inhibited, so a new follicle is stimulated and the cycle begins again.

?d l_jhe [Zgi^a^hVi^dc The first successful IVF baby
was Louise Brown, born in
In vitro fertilisation (IVF) is a technique used to help couples who have 1978 following her mother’s
been unable to conceive naturally.There are many reasons for infertility. IVF treatment. Professor Robert
Males may have a low sperm count, blocked or damaged sperm ducts Edwards, who pioneered the
or be unable to achieve an erection. Females may fail to ovulate or have technique, was awarded the Nobel
blocked or damaged oviducts, or produce antibodies in cervical mucus prize in Physiology or Medicine
that destroy sperm. in 2010.

The first step in IVF treatment is an assessment of whether the couple
are suitable for treatment. If so, the woman is injected with FSH for about
10 days.This hormone causes a number of egg cells to mature at the
same time in her ovaries. Just before the egg cells are released from the

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& Are the increased risks of When IVF treatment is carried out, a woman is given drug
birth defects acceptable? Is it treatments to interrupt her own natural reproductive cycle and induce
even possible to assess what an her body to release more eggs than is normal. Healthy sperm are
acceptable level of risk is? selected and, in some treatments, injected into the eggs.There is some
evidence that babies born after IVF treatment may be at a greater risk
' What would you define as a ‘birth of birth defects than children born following natural conception.The
defect’? largest study into the number of congenital abnormalities in babies
born following fertility treatment found that 4.2% were affected by
( The causes of the birth defects major malformations, compared with a rate of 2–3% for babies not
are unknown.They could be due conceived through IVF.
to the treatment or to factors
associated with the infertility of Although this rate is still low, ever more babies are born each year as a
the parents.Would people’s view result of fertility treatment. In Great Britain, about 200 000 such babies
of the risk be changed if we had have been born since 1991, suggesting that up to 8000 children may be
enough evidence to identify the affected.These figures include not only IVF, but also other treatments
causes? such as using drugs to stimulate ovulation.

) Is it important who carries out In France, a survey of 15 000 births in 33 hospitals was carried out
reviews and studies of these between 2003 and 2007.These results also showed increased numbers of
risks? Can scientific evidence be malformations. Geraldine Viot from the Maternité Port-Royal hospital
evaluated differently by people in Paris, speaking at an annual meeting of the European Society of
who have strong views about Human Genetics, said ‘Given that our study is the largest to date, we
the acceptability of assisted think that our data are more likely to be statistically representative
reproduction? of the true picture… At a time when infertility is increasing and
more couples need to use ART [assisted reproductive technology] to
conceive, it is vitally important that we find out as much as we can
about what is causing malformations in these children.’

follicles, they are collected using a laparoscope.The egg cells are ‘matured’
in culture medium for up to 24 hours before sperm cells are added to
fertilise them. Fertilised egg cells are incubated for about three days until
they have divided to form a ball of cells.These embryos are checked to
make sure they are healthy and developing normally. Either two or three
will be selected and placed into the woman’s uterus for implantation.The
pregnancy is then allowed to continue in the normal way.Any remaining
embryos can be frozen and stored for use later. Figure 6.32 summarises
the stages in IVF treatment.

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IVF has enabled men and women who would naturally be infertile to
have children but it has also produced some serious ethical issues. Some
of these are outlined in Table 6.8. Each society needs to think about these
issues and decide what should be done about them.

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l]^X] bVn WZ eVhhZY dc id X]^aYgZc Wdgc Vh V gZhjai d[ >K;#

IVWaZ +#- 6g\jbZcih [dg VcY V\V^chi >K; igZVibZci#

+ =JB6C =:6AI= 6C9 E=NH>DAD<N & &+(

FjZhi^dch id Xdch^YZg 7dgc id \^kZ

& Should IVF conceptions such as My Sister’s Keeper is a novel by Jodi Picoult, published in
Anna’s be permitted? 2004.The book, which was made into a film in 2009, follows the life
of 13-year-old Anna, who enlists the help of an attorney to sue her
' Who should have rights over a parents for rights to her own body.Anna’s older sister Kate suffers from
child’s genetic material in cases leukemia and Anna was conceived through IVF to be a genetic match
like this? and donor for her sister.Anna donates bone marrow and blood for her
sister throughout her life.As Kate’s condition worsens, their parents
( Is it right that a child should be want Anna to donate a kidney to Kate after she goes into renal failure,
created simply to keep another but Anna files a lawsuit against her parents for medical emancipation
person alive? from them despite the consequences for her sister’s health.

This novel is a fictional story but already there have been real examples
of children being conceived by IVF and born to provide stem cells for
siblings who would otherwise die.

:cY"d["X]VeiZg fjZhi^dch (1)
(1)
1 Which of the following statements is correct? (1)

A Large food molecules need to be digested so that they can be absorbed through
the gut wall.

B The liver and pancreas both release their digestive enzymes into the small intestine.
C The approximate pH of the stomach is 8.
D Assimilation is the transfer of the end products of digestion from the gut lumen

to the blood.

2 Which of the following statements is correct?

A Blood passes from the atria to the ventricles through the semilunar valves.
B The coronary and pulmonary arteries supply oxygenated blood to the heart muscle.
C The pacemaker, brain and adrenal gland are all involved in the regulation of the

heart beat.
D A red blood cell travels from the lungs to the stomach via the right atrium, right

ventricle and aorta.

3 Which of the following statements is correct?

A Antibiotics are effective against viruses but not bacteria.
B Antigens are made by lymphocytes.
C Phagocytes are white blood cells that can be found in the tissue fluid around cells.
D HIV is a bacterium that reduces the number of active lymphocytes in the blood.

&+)

4 Which of the following statements is correct? (1)

A During breathing in, the rib cage moves up and out and the pressure inside the thorax
increases.

B Alveoli are small sac-like structures that have walls only two cells thick.
C To assist with breathing out, the abdominal and external intercostal muscles contract.
D The purpose of ventilation is to increase the rate of gas exchange in the alveoli.

5 Which of the following statements is correct? (1)

A In a motor neuron, the impulse travels from the axon to the cell body.
B A nerve impulse travels from a receptor to the CNS via a sensory neuron and from

the CNS to an effector via a motor neuron.
C The endocrine system produces hormones, which pass from the endocrine glands

to the blood through ducts.
D The resting potential is divided into two parts, depolarisation and repolarisation.

6 Which of the following statements is correct? (1)

A In the male reproductive system, the sperm ducts are connected to the ureters.
B During the process of in vitro fertilisation, a woman is injected with FSH to stimulate

egg cells to mature.
C FSH and oestrogen are hormones released from the ovaries whereas LH and

progesterone are hormones released from the pituitary gland.
D In the menstrual cycle, the levels of progesterone begin to rise around day 5.

7 Outline the functions of the stomach, small intestine and large intestine in a table like the one below. (4)

HidbVX] HbVaa ^ciZhi^cZ AVg\Z ^ciZhi^cZ
egdk^YZh VX^Y^X Zck^gdcbZci
egdk^YZh lZV`an Va`Va^cZ
Zck^gdcbZci

8 Explain the relationship between the structure and function of arteries, veins and capillaries. (9)

9 Explain antibody production. (3)

10 Explain how the skin and mucous membranes prevent entry of pathogens into the body. (3)

11 Explain how blood glucose concentration is controlled in humans. (8)

12 Discuss the ethical issues of in vitro fertilisation (IVF) in humans. (8)

+ =JB6C =:6AI= 6C9 E=NH>DAD<N & &+*

13 The body mass index (BMI) takes into account the LZ^\]i $ `\ 7dYn BVhh >cYZm =Z^\]i $ Xb
weight and height of a person and can be used to
determine if a person is overweight or obese.The &*% *% KZgn dWZhZ &'*
nomogram shows the body mass index for a range &)% )% &(%
of weights and heights. &(% &(*
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68CE ! [gdb < 7gVn! EX[i_jo" <Wj _djWa[ WdZ 9^hed_Y :_i[Wi[! '%%%

a State the weight description of a 75 kg man who is 145 cm tall. (1)

b A woman of height 150 cm has a BMI of 40. Calculate the minimum weight she must lose to be (1)
considered ‘normal’.

c Outline the relationship between height and BMI for a fixed weight. (1)

(total 3 marks)
© IB Organization 2009

&++

14 Blind mole rats (Spalax ehrenberghi) are adapted to live in underground burrows with very low oxygen
conditions. Scientists compared blind mole rats and white rats in order to determine whether these
adaptations are due to changes in their ventilation system.

Both types of rat were placed on a treadmill and the amount of oxygen consumed was measured at
different speeds.This study was done under normal oxygen conditions and under low oxygen conditions.
The results are shown in the scatter graphs below.

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) BVgX] &.., ! kda# .)! ^hhjZ )! ee# '%+'Ä'%+,! ;^\ &!  '%%( CVi^dcVa 6XVYZbn d[ HX^ZcXZh! JH6 cdgbVa dmn\Zc

a Compare the oxygen consumption of blind mole rats and white rats when the treadmill is not moving. (1)

b Compare the effect of increasing the treadmill speed on the oxygen consumption in both types of rats (3)
under normal oxygen conditions. (2)

c Evaluate the effect of reducing the amount of oxygen available on both types of rat.

The lungs of both types of rats were studied and the features important to oxygen uptake were compared.
The results are shown in the bar chart below.

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d Using your knowledge of gas exchange in lungs, explain how these adaptations would help the (3)
blind mole rats to survive in underground burrows.

(total 9 marks)
© IB Organization 2009

+ =JB6C =:6AI= 6C9 E=NH>DAD<N & &+,

15 Near the middle of pregnancy in humans, the placenta begins to secrete a hormone called corticotrophin-
releasing hormone (CRH). CRH influences the production of hormones that stimulate the development
of the fetus.

A study was carried out to determine if levels of CRH were correlated with the timing of the baby’s
delivery. Blood samples were taken from 500 women during their pregnancies and the concentration of
CRH was measured.The women were then divided into three groups according to whether their baby
was delivered prematurely, at full-term or late.

The graph below shows how the concentration of CRH varied in the mothers’ blood (maternal blood)
in each of the three groups during pregnancy.

BViZgcVa 8G= XdcXZcigVi^dc $ &%% @Zn
ebda YbÄ( eaVhbV .% LdbZc l]d
-% YZa^kZgZY
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+% I^bZ h^cXZ hiVgi d[ egZ\cVcXn $ lZZ`h LdbZc l]d
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hdjgXZ/ G Hb^i]! IY_[dj_\_Y 7c[h_YWd BVgX] &... ee# +-Ä,*

a Outline how the concentration of CRH in the blood changes during the pregnancies of women who (2)
delivered prematurely. (2)
(1)
b Compare how the concentration of CRH changes during the pregnancies of women who delivered (2)
late with those that delivered at full-term.
(3)
c Measure the difference in CRH concentration at 30 weeks between the women who delivered
prematurely and those that delivered at full-term.

d Suggest how knowledge of the CRH concentration in maternal blood might be used by doctors
monitoring pregnancies.

The Western spadefoot toad (Scaphiopus hammondii) lives in desert areas in California and lays its eggs in
pools formed by rain.When the egg first hatches, its body form is referred to as the tadpole stage.At some
point, it undergoes metamorphosis (a change in body form) to develop into the adult toad.

If the pools where the eggs have been laid shrink due to a lack of rain, the tadpoles quickly develop into
small adult toads. If there is sufficient rain and the pools persist, the tadpoles develop more slowly and grow
large before developing into adult toads.

e Suggest how undergoing metamorphosis at different times in response to water levels helps the survival
of the toad.

It has been suggested that CRH control of development might have evolved in amphibians long before
mammals appeared. In toads, increase in CRH concentration leads directly to an increase in the level of
the hormone thyroxine and indirectly to an increase in corticosterone levels.

&+-

An experiment was carried out to determine what hormones might be involved in triggering

development in response to pond drying.Tadpoles were raised in a constant high-water-level

environment.They were then divided into two groups. One group was transferred to a tank containing
10 dm3 of water – a high-water environment.The other group was transferred to a tank of the same
size containing only 1 dm3 of water – a low-water environment.The concentrations of thyroxine and

corticosterone were measured in each group.The results are shown below.

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f Compare the concentrations of thyroxine and corticosterone in the two groups. (1)
g Predict how the concentration of CRH would be different in the two groups.
(total 13 marks)
© IB Organization 2009

+ =JB6C =:6AI= 6C9 E=NH>DAD<N & &+.

, CjXaZ^X VX^Yh VcY egdiZ^ch

>cigdYjXi^dc

Nucleic acids are very large macromolecules composed of a backbone of sugar and phosphate
molecules each with a nitrogenous base attached to the sugar. In Chapter 3, the role of DNA and
RNA is outlined. In this chapter, the detailed structure of different nucleic acids is considered, as
well as the vital role of nucleic acids in converting the genetic code contained in the chromosomes
into the protein molecules that are needed to make all cells function.

,#& 9C6 higjXijgZ

6hhZhhbZci hiViZbZcih

r Describe the structure of DNA, including the antiparallel strands, 3'–5' linkages and hydrogen bonding between the

purines and the pyrimidines.

r Outline the structure of nucleosomes.
r State that nucleosomes help to supercoil chromosomes and help to regulate transcription.
r Distinguish between ‘unique or single-copy genes’ and ‘highly repetitive sequences’ in nuclear DNA.
r State that eukaryotic genes can contain exons and introns.

8* D I]Z ( Ä* a^c`V\Z
8)
8& A nucleotide consists of the sugar deoxyribose to which are attached a
phosphate group and a nitrogenous base.The carbons in the sugar are
8( 8' numbered from 1 to 5 in a clockwise direction starting after the oxygen at
;^\jgZ ,#& I]Z higjXijgZ d[ V cjXaZdi^YZ# the apex (Figure 7.1).

r The base is attached to carbon 1.
r Carbon 2 has just a hydrogen attached instead of an OH group – this is

the reason the sugar is called deoxyribose.

r Carbon 3 is where the next nucleotide attaches in one direction.
r Carbon 5 has a phosphate group attached to it, which is where the next

nucleotide attaches in the other direction.
This means that each nucleotide is linked to those on either side of it
through carbons 3 and 5.The linkages are called 3'–5' linkages.

6ci^eVgVaaZa higVcYh

Look at Figure 3.10 on page 48.This shows part of a DNA molecule,
in which two polynucleotide strands, running in opposite directions,
are held together by hydrogen bonds between the bases. Notice that the
deoxyribose molecules are orientated in opposite directions. One strand
runs in a 5' n 3' direction whereas the other runs in a 3' n 5' direction.
Figure 7.2 also shows this.The strands are described as being antiparallel.

&,%

I]Z WVhZh VcY ]nYgd\Zc WdcY^c\ An easy way to remember the two
groups of DNA bases is to look at
The four DNA bases are cytosine, thymine, adenine and guanine, and they the words.Thymine and cytosine,
fall into two chemical groups called pyrimidines and purines. Cytosine which both contain a letter y, are in
and thymine are pyrimidines, and adenine and guanine are purines. the group pyrimidines, which also
contains a letter y.
Cytosine pairs with guanine and thymine pairs with adenine – that is,
a pyrimidine always pairs with a purine.This is because they are different hj\Vg"e]dhe]ViZ
sizes: pyrimidines are smaller than purines.The pairing of a pyrimidine ÈWVX`WdcZÉ
with a purine ensures that the strands are always the same distance apart.
]nYgd\Zc
It is important to notice that cytosine, a pyrimidine, cannot pair with WdcYh
adenine, even though it is a purine. Cytosine and guanine have three
hydrogen bonds whereas thymine and adenine only have two hydrogen
bonds (Figure 7.2).

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CjXaZdhdbZh

A eukaryotic chromosome is composed of a double strand of DNA
combined with proteins. Some of these proteins, called histones, combine
together in groups of eight to form a bead-like structure (Figure 7.3).
The strand of DNA takes two turns around this bead before continuing
on to the next bead. It is held in place on the bead by a ninth histone.
The group of nine histones with the DNA is called a nucleosome.The
function of nucleosomes is to help supercoil the chromosomes during
mitosis and meiosis and also to help regulate transcription.

CjXaZVg 9C6

The nucleus of a human cell contains nearly 2 m of DNA, but genes make
up only a proportion of this.The DNA in a eukaryotic cell can be divided
up into two types:

r unique or single-copy genes, which make up 55–95% of the total

, CJ8A:>8 68>9H 6C9 EGDI:>CH &,&

cjXaZdhdbZ XdgZ Xdch^hi^c\ d[ 9C6 a^c`Zg r highly repetitive sequences, or satellite DNA, which account for
Z^\]i ]^hidcZ egdiZ^c bdaZXjaZh
5–45%.
9C6 YdjWaZ ]Za^m Vcdi]Zg ]^hidcZ The repetitive sequences are typically between 5 and 300 base pairs long
ldjcY il^XZ VgdjcY egdiZ^c ]daY^c\ and may be duplicated as many as 100 000 times in a genome.
i]Z cjXaZdhdbZ XdgZ i]Z cjXaZdhdbZ
9C6 ^h ldjcY id\Zi]Zg Many genes include sections called introns that are transcribed but not
VgdjcY V hZg^Zh 9C6 a^c`Zg Xdci^cj^c\ translated. Only sections of genes known as exons are both transcribed
d[ cjXaZdhdbZh! idlVgYh i]Z cZmi and translated.You will learn more about introns and exons in Section 7.3
a^`Z V hig^c\ d[ cjXaZdhdbZ (page 175).
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& Djia^cZ l]Vi ^h bZVci Wn i]Z iZgb ÈVci^eVgVaaZaÉ#
cjXaZdhdbZh
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cjXaZdhdbZh
XVc WZ iV\\ZY L]Vi ^h _jc`4
l^i] egdiZ^ch
id egdbdiZ dg When the highly repetitive sequences of DNA were first
gZegZhh discovered, they appeared to have no function. Scientists at the
igVchXg^ei^dc time called them ‘junk DNA’ and thought they were simply excess
baggage. Before scientists began mapping several animal genomes,
;^\jgZ ,#( I]Z higjXijgZ d[ V cjXaZdhdbZ# they had a rather restricted view about which parts of the genome
were important.According to the traditional viewpoint, the really
crucial things were genes and a few other sections that regulate gene
function were also considered useful.

But new findings suggest that this interpretation was not correct. In
2004, David Haussler’s team at the University of California, Santa
Cruz, USA, compared human, mouse and rat genome sequences.
They found – to their astonishment – that several long sequences
of repeated DNA were identical across the three species.As David
Haussler exclaimed:

‘It absolutely knocked me off my chair.’

When the Human Genome Project was planned, there were calls
from some people to map only the bits of genome that coded
for protein – mapping the rest was thought to be a waste of time.
Luckily, entire genomes were mapped and have proved vital to the
study of so-called ‘junk DNA’. It is now thought that the most
likely scenario is that junk DNA controls the activity of vital genes
and possibly embryo development.

FjZhi^dch id Xdch^YZg
& Do you think it is appropriate to label something as ‘junk’ simply

because it is thought to have no function?
' Does such labelling hinder scientific progress?
( Should experiments be carried out to answer fundamental

questions even if they have no obvious application?
) Who should decide which research is most likely to be valuable?

&,'

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6hhZhhbZci hiViZbZcih

r State that DNA replication occurs in a 5' n 3' direction.
r Explain the process of DNA replication in prokaryotes, including the role of enzymes (helicase, DNA polymerase,

RNA primase, and DNA ligase), Okazaki fragments and deoxynucleoside triphosphates.

r State that DNA replication is initiated at many points in eukaryotic chromosomes.

DNA replication ensures that exact copies of existing molecules are dg^\^cVa
produced before a cell divides.The process is said to be semi-conservative 9C6
and each strand of an existing DNA molecule acts as a template for the
production of a new strand. cZl
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As Figure 7.2 shows, at one end of a DNA strand there is a free 3' carbon daY 9C6 higVcYh
and at the other there is a free 5' carbon. (Ignore the fact that there is ;^\jgZ ,#) 9C6 gZea^XVi^dc ^h hZb^"
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therefore be described as running in a 5' n 3' direction or in a 3' n 5' higVcY WZXdbZh eV^gZY l^i] dcZ cZl higVcY#
direction. DcZ dg^\^cVa higVcY ^h XdchZgkZY ^c ZVX] cZl
9C6 bdaZXjaZ! ]ZcXZ hZb^"XdchZgkVi^kZ#
The process of replication is semi-conservative – that is, each original
strand acts as a template to build up a new strand (Figure 7.4).The DNA Replication fork the point
double helix is unwound to expose the two strands for replication by the where the DNA double helix is
enzyme helicase, at a region known as a replication fork. Eukaryotic being separated to expose the two
chromosomes are very long compared to prokaryotic chromosomes and strands as templates for replication
so, to speed up the process, replication starts at many replication forks Leading strand the new strand
along a chromosome at the same time. that is synthesised continuously and
follows the replication fork
Replication must occur in the 5' n 3' direction (just as in transcription Lagging strand the new
and translation, described in Sections 7.3 and 7.4), because the enzymes strand that is synthesised in short
involved only work in a 5' n 3' direction.As the two strands are fragments in the opposite direction
antiparallel, replication therefore has to proceed in opposite directions on to the movement of the replication
the two strands. However, the replication fork where the double helix fork
unwinds moves along in one direction only.This means that on one of Okazaki fragments short
the strands replication can proceed in a continuous way, following the fragments of a DNA strand formed
replication fork along, but on the other strand the process has to happen on the lagging strand
in short sections, each moving away from the replication fork (Figure
7.5, overleaf ).The strand undergoing continuous synthesis is called the
leading strand.The other strand, in which the new DNA is built up in
short sections, is known as the lagging strand.

AZVY^c\ higVcY

Replication to produce the leading strand is straightforward. First RNA
primase adds a short length of RNA, attached by complementary base
pairing, to the template DNA strand.This acts as a primer, allowing the
enzyme DNA polymerase III to bind. DNA polymerase III then adds
free building units called deoxynucleoside triphosphates (dNTPs) to

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) 9C6 a^\VhZ hZVah je gZeaVXZh GC6 eg^bZg gZea^XVi^dc cZmi id eg^bZg!
WgZV`h WZilZZc 9C6 l^i] 9C6# VcY VYYh YCIEh ^c V
[gV\bZcih# *É (É Y^gZXi^dc#

;^\jgZ ,#* 9C6 gZea^XVi^dc#

the existing strand of DNA in a 5' n 3' direction, following the progress
of helicase as it moves the replication fork along the DNA double helix.
The RNA primer will later be removed by DNA polymerase I.

The dNTPs have two extra phosphate groups attached, and are said
to be ‘activated’.They pair up with their complementary bases on the
exposed DNA strand and DNA polymerase III then links together the
sugar and the innermost phosphate groups of adjacent nucleotides.The
two extra phosphate groups are broken off and released. In this way, a
continuous new DNA strand is built up on the leading strand.

AV\\^c\ higVcY

Synthesis of the lagging strand is a little more complicated, as it has to
occur in discontinuous sections, which are then joined together.
1 As for the leading strand, RNA primase first synthesises a short RNA

primer, complementary to the exposed DNA.This happens close to the
replication fork.

&,)

2 DNA polymerase III starts replication by attaching at the 3' end of ( HiViZ i]Z Y^gZXi^dc ^c l]^X] 9C6
the RNA primer and adding dNTPs in a 5' n 3' direction. It moves gZea^XVi^dc dXXjgh#
away from the replication fork on this strand.
) :meaV^c l]n D`VoV`^ [gV\bZcih bjhi
3 DNA polymerase I now removes the RNA primer and replaces it WZ egdYjXZY dc dcZ 9C6 higVcY#
with DNA using dNTPs. Short lengths of new DNA called Okazaki
fragments are formed from each primer.The new fragment grows
away from the replication fork until it reaches the next fragment.

4 Finally, DNA ligase seals up each break between the Okazaki fragments
by making sugar–phosphate bonds so that a continuous strand of new
DNA is created.

There are several chemical names involved in replication, so it is helpful to list them.
Nucleoside triphosphate (NTP) a building unit for RNA – a ribose nucleotide with two additional
phosphates, which are chopped off during the synthesis process
Deoxynucleoside triphosphate (dNTP) the same as NTP but with deoxyribose instead of ribose, so it is used
to build DNA rather than RNA through base-pairing to the parent DNA strand
Helicase an enzyme found at the replication fork, with two functions – to unwind the two DNA strands, and to
separate them by breaking the hydrogen bonds
RNA primase this enzyme adds NTPs to the single-stranded DNA that has been unzipped by helicase, in a
5' n 3' direction, to make a short length of RNA base-paired to the parent DNA strand
DNA polymerase III this enzyme adds dNTPs in a 5' n 3' direction where RNA primase has added a short
length of complementary RNA as a primer; it is unable to add dNTPs directly to the single-stranded parent DNA
that has been unzipped by helicase
DNA polymerase I this enzyme removes the RNA nucleotides of the primers on the lagging strand in a
5' n 3' direction and replaces them with DNA nucleotides using dNTPs
DNA ligase this enzyme joins adjacent Okazaki fragments by forming a covalent bond between the deoxyribose
and the phosphate of adjacent nucleotides

,#( IgVchXg^ei^dc

6hhZhhbZci hiViZbZcih

r State that transcription is carried out in a 5' n 3' direction.
r Distinguish between the ‘sense’ and ‘antisense’ strands of DNA.
r Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase,

nucleoside triphosphates and the terminator.

r State that eukaryotic RNA needs the removal of introns to form mature mRNA.

The process by which the DNA code is used to build polypeptides occurs
in two stages.The first is transcription, which transfers sections of the
genetic code from DNA to an mRNA molecule.Transcription happens
in the nucleus.The second stage, known as translation, occurs in the
cytoplasm and uses the mRNA, together with ribosomes, to construct the
polypeptide.Translation is discussed in the next section.

, CJ8A:>8 68>9H 6C9 EGDI:>CH &,*

A single chromosome contains IgVchXg^W^c\ 9C6 ^cid bGC6 ^c egd`VgndiZh
DNA that codes for many proteins.
During transcription, genes (short At the start of transcription, the DNA molecule is separated into two
lengths of DNA that code for strands by the enzyme RNA polymerase, which binds to the DNA near
single polypeptides) are used to the beginning of a gene. Hydrogen bonds between the bases are broken
produce mRNA. Most genes are and the double helix unwinds (Figure 7.6).
about 1000 nucleotides long, a few
are longer and a very small number Transcription begins at a specific point on the DNA molecule
are less than 100 nucleotides.The called the promoter region. Only one of the two strands is used as a
size of the gene corresponds to the template for transcription and this is called the antisense strand.The
size of the polypeptide it codes for. other DNA strand is called the sense strand. RNA polymerase uses
free nucleoside triphosphates (NTPs) to build the RNA molecule, using
Nucleoside triphosphates (NTPs) complementary base pairing to the DNA and condensation reactions
are the molecules used by RNA between the nucleotides.This produces a primary mRNA molecule that
polymerase to build mRNA is complementary to the antisense strand being transcribed, and has the
molecules during transcription. same base sequence as the sense strand (except that it contains the base U
As they move into place, two in place of T). RNA polymerase moves along the antisense DNA strand
phosphates are removed from them in a 3' n 5' direction.As it does so, the 5' end of a nucleotide is added to
so they become converted into the 3' end of the mRNA molecule so that the construction of the mRNA
nucleotides. proceeds in a 5' n 3' direction.

RNA polymerase checks the mRNA molecule as it forms to ensure
that bases are paired correctly.As the mRNA molecule is extended,
the DNA is rewound into a helix once a section has been transcribed.
Eventually RNA polymerase reaches another specific sequence on the
DNA called the terminator region, which indicates the end of the gene.
The RNA polymerase releases the completed RNA strand and finishes
rewinding the DNA before breaking free.

6 8 edda d[ [gZZ
< cjXaZdh^YZ
6 8 ig^e]dhe]ViZh
J 8
GC6 edanbZgVhZ gZbdkZh ild e]dhe]ViZh [gdb cjXaZdh^YZ 8 < <
ig^e]dhe]ViZ VcY a^c`h i]Z *É ZcY d[ i]Z cjXaZdi^YZ id i]Z 8 < <
(É ZcY d[ i]Z \gdl^c\ GC6 bdaZXjaZ [gZZ cjXaZdh^YZ
ig^e]dhe]ViZh Va^\c Wn
hZchZ higVcY 8 866<< 8< I68 6 I I 8 8 8< WVhZ eV^g^c\ VcY VgZ ]ZaY
Vci^hZchZ higVcY Wn ]nYgd\Zc WdcYh

(É *É (É *É
<< 8 8 < J 6 8 6 J J 8 8 8 < (É

]Za^m gZ"[dgbh << I8 8< 8 6 I< I 66<<< 8 GC6
edanbZgVhZ
Vh i]Z GC6 jcW^cYh VcY
edanbZgVhZ i]Z GC6
bdkZh Vadc\ 8 Vci^hZchZ higVcY gZVY ^c bdaZXjaZ ^h
i]Z (É *É Y^gZXi^dc gZaZVhZY Vi i]Z
8 iZgb^cVidg
< Y^gZXi^dc d[ bGC6 hnci]Zh^h gZ\^dc
bGC6 higVcY 6
GC6 *É 8 < 8 6 J GC6 edanbZgVhZ
edanbZgVhZ
W^cYh Vi V ZconbZ
egdbdiZg
gZ\^dc

;^\jgZ ,#+ IgVchXg^ei^dc#

&,+

>cigdch VcY Zmdch ^cigdc Zmdc edhi"igVchaVi^dcVa
bdY^[^XVi^dc
In eukaryotes, many genes contain sequences of DNA that are transcribed
but not translated.These sequences, which appear in mRNA, are known bGC6
as introns.After transcription of a gene, the introns are removed in a
process known as post-transcriptional modification.The sequences of ;^\jgZ ,#, >cigdch VcY Zmdch ^c bGC6#
bases that remain are known as exons and these are linked together to
form the mature mRNA that is translated (Figure 7.7). Mature mRNA
leaves the nucleus via the nuclear pores and moves to the cytoplasm.

,#) IgVchaVi^dc

6hhZhhbZci hiViZbZcih

r Explain that each tRNA molecule is recognised by a tRNA-activating enzyme that binds a specific amino acid to the

tRNA, using ATP for energy.

r Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA

binding sites and mRNA binding sites.

r State that translation consists of initiation, elongation, translocation and termination.
r State that translation occurs in a 5' n 3' direction.
r Draw and label a diagram showing the structure of a peptide bond between two amino acids.
r Explain the process of translation, including ribosomes, polysomes, start codons and stop codons.
r State that free ribosomes synthesise proteins for use primarily within the cell, and that bound ribosomes synthesize

proteins primarily for secretion or for lysosomes.

Translation is the process by which the information carried by mRNA ed^ci d[ ViiVX]bZci
is decoded and used to build the sequence of amino acids that eventually d[ Vb^cd VX^Y
forms a protein molecule. During translation, amino acids are joined 6
together in the order dictated by the sequence of codons on the mRNA 8
to form a polypeptide.This polypeptide eventually becomes the protein 8
coded for by the original gene.
cjXaZdi^YZh
IgVch[Zg GC6 iGC6
]nYgd\Zc
The process of translation requires another type of nucleic acid known as WdcYh
transfer RNA or tRNA. tRNA is made of a single strand of nucleotides
that is folded and held in place by base pairing and hydrogen bonds Vci^XdYdc
(Figure 7.8).There are many different tRNA molecules but they all have a ;^\jgZ ,#- IgVch[Zg GC6 Ä V ÈXadkZg aZV[É
characteristic ‘clover leaf ’ appearance with some small differences between h]VeZ#
them.

At one position on the molecule is a triplet of bases called the
anticodon, which pairs by complementary base pairing with a codon
on the mRNA strand.At the 3' end of the tRNA molecule is a base
sequence CCA which is the attachment site for an amino acid.

An amino acid is attached to the specific tRNA molecule that has its
corresponding anticodon, by an activating enzyme.As there are 20
different amino acids, so there are also 20 different activating enzymes in

, CJ8A:>8 68>9H 6C9 EGDI:>CH &,,

the cytoplasm.The tRNA activating enzymes recognise the correct tRNA
molecules by their shapes. Energy for the attachment of an amino acid
comes from ATP.A tRNA molecule with its attached amino acid is called
a charged tRNA.

G^WdhdbZh

Ribosomes are the site of protein synthesis. Some ribosomes occur free
in the cytoplasm and these synthesise proteins that will be used within
the cell. Others are bound to the endoplasmic reticulum, forming rough
endoplasmic reticulum, and these synthesise proteins for secretion from
the cell or for use within lysosomes.

Ribosomes are composed of two subunits, one large and one small.The
subunits are built of protein and ribosomal RNA (rRNA). On the surface
of the ribosome are three tRNA binding sites (site 1, site 2 and the exit
site), and one mRNA binding site (Figure 7.9).Two charged tRNA
molecules can bind to a ribosome at one time. Polypeptide chains are
built up in the groove between the two subunits.

7j^aY^c\ V edaneZei^YZ

Translation is the process that decodes the information of mRNA into

the sequence of amino acids that eventually form a protein.Translation

consists of four stages:

1 initiation 3 translocation

2 elongation 4 termination

bGC6 ha^YZh i]gdj\] \Ve Vh i]Z
XdYZ ^h ÈgZVYÉ ^c i]Z g^WdhdbZ

hbVaa hjWjc^i

aVg\Z hjWjc^i g^WdhdbZ bdkZh h^iZ & Ä l]ZgZ i]Z eZei^YZ WdcY
a^cZ d[ hZXi^dc i]gdj\] Vadc\ i]Z bGC6 ^c ^h [dgbZY! Wn l]^X] i]Z cZl Vb^cd
g^WdhdbZ id h]dl hZfjZcXZ V *É id (É Y^gZXi^dc! VX^Y ^h ViiVX]ZY id i]Z \gdl^c\
d[ ^ciZgVXi^dc h^iZh ÈgZVY^c\É i]Z XdYdch edaneZei^YZ

;^\jgZ ,#. I]Z higjXijgZ d[ V g^WdhdbZ# *É bGC6
&,- (É
ÈZm^i h^iZÉ Ä l]ZgZ
i]Z [gZZ iGC6 aZVkZh h^iZ ' Ä l]ZgZ i]Z
i]Z g^WdhdbZ Vci^XdYdc d[ Vc
^cXdb^c\ X]Vg\ZY
iGC6 WVhZ eV^gh
l^i] V XdYdc dc
i]Z bGC6 higVcY

i]Z \ZcZi^X XdYZ ^c i]^h bGC6 ^h gZVY ^c i]^h Y^gZXi^dc

* (
6J<<6JJ8 8J<8J66

i]^h XdYdc gZegZhZcih i]Z Vb^cd i]^h XdYdc gZegZhZcih ÈhideÉ!
VX^Y bZi]^dc^cZ! l]^X] h^\cVah VcY h^\cVah i]Z ZcY d[ igVchaVi^dc

i]Z ÈhiVgiÉ d[ igVchaVi^dc , CJ8A:>8 68>9H 6C9 EGDI:>CH &,.

;^\jgZ ,#&% bGC6 ÈhiVgiÉ VcY ÈhideÉ XdYdch#

>c^i^Vi^dc

Translation begins at a start codon (AUG) near the 5' end of the mRNA
strand.This codon codes for the amino acid methionine and is a signal
to begin the process of translation (Figure 7.10).This is called initiation.
The mRNA binds to the small subunit of a ribosome.Then an activated
tRNA molecule, carrying the amino acid methionine, moves into
position at site 1 of the ribosome. Its anticodon binds with the AUG
codon using complementary base pairing. Hydrogen bonds form between
the complementary bases of the mRNA and tRNA and, once this has
happened, a large ribosomal subunit moves into place and combines with
the small subunit.

:adc\Vi^dc

Initiation is followed by elongation and the formation of peptide bonds
(Figure 7.11). tRNA molecules bring amino acids to the mRNA strand in
the order specified by the codons.To add the second amino acid, a second

dcZ Vb^cd VX^Y Vcdi]Zg Vb^cd VX^Y

G G

='C 8 8DD= ='C 8 8DD=
=
= ='D
G eZei^YZ WdcY
Ild Vidbh d[ ]nYgd\Zc
VcY dcZ d[ dmn\Zc VgZ G
adhi Ä i]Zn [dgb V
bdaZXjaZ d[ lViZg#

='C 8 8D =C 8 8DD=

==

I]Z ild Vb^cd VX^Yh VgZ _d^cZY Vi i]^h ed^ci Wn V eZei^YZ WdcY#
;^\jgZ ,#&& I]Z [dgbVi^dc d[ V eZei^YZ WdcY Wn V XdcYZchVi^dc gZVXi^dc#

Look back at Table 4.1 on page 68 charged tRNA with the anticodon corresponding to the next codon
to see all the mRNA codons and enters site 2 of the ribosome and binds to its codon by complementary
what they represent. base pairing.The ribosome catalyses the formation of a peptide bond
between the two adjacent amino acids (Figure 7.12).The ribosome and

>c^i^Vi^dc Zm^i bGC6
bGC6 ViiVX]ZY id h^iZ h^iZ & h^iZ '
hbVaa hjWjc^i d[
g^WdhdbZ 6J<6668 88 (É
J68
*É
XdYdc

BZi

:adc\Vi^dc Zm^i bGC6
h^iZ h^iZ & h^iZ '
*É
XdYdc 6J<6668 88 (É
J6 8 JJJ

iGC6 Wg^c\h
hZXdcY Vb^cd
VX^Y ^cid
edh^i^dc

aVg\Z hjWjc^i BZi Anh
d[ g^WdhdbZ

eZei^YZ WdcY
[dgbZY

IgVchadXVi^dc Zm^i g^WdhdbZ bdkZh ^c
h^iZ h^iZ & h^iZ ' V *É (É Y^gZXi^dc Vadc\
*É i]Z bGC6! id i]Z cZmi
XdYdc 6J<6668 88 XdYdc
J6 8 JJJ
(É
<
< iGC6
< bdk^c\ ^c
Egd
iGC6
aZVk^c\

Anh

\gdl^c\ BZi
edaneZei^YZ eZei^YZ WdcY

;^\jgZ ,#&' I]Z hiV\Zh d[ igVchaVi^dc#

&-%

tRNA molecules now hold two amino acids.The methionine becomes
detached from its tRNA. Now the ribosome moves along the mRNA and
the first tRNA is released to collect another methionine molecule.

IgVchadXVi^dc

Translocation is the movement of the ribosome along the mRNA strand
one codon at a time.As the ribosome moves, the unattached tRNA moves
into the exit site and is then released into the cytoplasm, where it will
pick up another amino acid molecule.The growing peptide chain is now
positioned in site 1, leaving site 2 empty and ready to receive another
charged tRNA molecule to enter and continue the elongation process.
Figure 7.12 shows how initiation, elongation and translocation occur as
mRNA is translated.

IZgb^cVi^dc

Translocation and elongation are repeated until one of the three ‘stop’
codons aligns with site 2, which acts as a signal to end translocation.
There are no tRNA molecules with anticodons corresponding to these
stop codons.The polypeptide chain and the mRNA are released from the
ribosome and the ribosome separates into its two subunits.This final stage
of translation is called termination.

EdanhdbZh

Translation occurs at many places along an mRNA molecule at the same
time.A polysome like that in Figure 7.13 is a group of ribosomes along
one mRNA strand.

g^WdhdbZh bdk^c\ bGC6 ^c
Vadc\ bGC6 XnideaVhb

hbVaa VcY
aVg\Z hjWjc^ih
d[ g^WdhdbZ
edaneZei^YZ

\gdl^c\
edaneZei^YZ
X]V^ch
;^\jgZ ,#&( 6 edanhdbZ#

* HiViZ l]Vi ^h bZVci Wn Vc ÈZmdcÉ#
+ 9^hi^c\j^h] WZilZZc ÈigVchXg^ei^dcÉ VcY ÈigVchaVi^dcÉ#
, 9^hi^c\j^h] WZilZZc i]Z ÈhZchZÉ VcY ÈVci^hZchZÉ higVcYh d[ 9C6#
- Djia^cZ i]Z gdaZ d[ i]Z egdbdiZg gZ\^dc ^c igVchXg^ei^dc#
. 9gVl V eZei^YZ WdcY WZilZZc ild Vb^cd VX^Yh#
&% HiViZ l]ZgZ i]Z egdiZ^c i]Vi ^h hnci]Zh^hZY Wn [gZZ g^WdhdbZh ^h jhZY#

, CJ8A:>8 68>9H 6C9 EGDI:>CH &-&

,#* EgdiZ^ch

6hhZhhbZci hiViZbZcih

r Explain the four levels of protein structure, indicating the significance of each level.
r Outline the difference between fibrous and globular proteins, with reference to two examples of each protein type.
r Explain the significance of polar and non-polar amino acids.
r State four functions of proteins, giving a named example of each.

Eg^bVgn E]Z eZei^YZ EgdiZ^c higjXijgZ
higjXijgZ <aj WdcYh
Vb^cd VX^Y Ing Vb^cd Proteins are large, complex molecules, usually made up of hundreds of
hZfjZcXZ HZg VX^Yh amino acid subunits.The way these subunits fit together is highly specific
>hd to each type of protein, and is vital to its function. Figure 7.14 illustrates
BZi the structure of the protein hemoglobin.

HZXdcYVgn ]nYgd\Zc The first stage of protein production is the assembly of a sequence
higjXijgZ WdcYh of amino acid molecules that are linked by peptide bonds formed by
ild condensation reactions.This sequence forms the primary structure of
i]Z ]Za^m edaneZei^YZ a protein.There are many different proteins and each one has different
h]VeZ ^h X]V^ch numbers and types of amino acids arranged in a different order, coded for
bV^ciV^cZY B eaZViZY by a cell’s DNA.
l^i] h]ZZi
]nYgd\Zc The secondary structure of a protein is formed when the
WdcYh polypeptide chain takes up a permanent folded or twisted shape. Some
polypeptides coil to produce an B helix, others fold to form C pleated
A ]Za^m sheets.The shapes are held in place by many weak hydrogen bonds.
IZgi^Vgn Depending on the sequence of amino acids, one section of a polypeptide
higjXijgZ may become an B helix while another takes up a C pleated form.This is
the case in hemoglobin.
Y^hja[^YZ WZiV X]V^c
Wg^Y\Z &)+ Vb^cd The tertiary structure of a protein forms as the molecule folds still
VX^Yh further due to interactions between the R groups of the amino acids
edaneZei^YZ (Figure 7.16, page 184) and within the polypeptide chain.The protein
X]V^c takes up a three-dimensional shape, which is held together by ionic bonds
FjViZgcVgn between particular R groups, disulfide bridges (covalent bonds) between
higjXijgZ sulfur atoms of some R groups, and by weaker interactions between
hydrophilic and hydrophobic side chains. Figure 7.15 shows the different
]Zbd\adW^c Vae]V X]V^c types of bond involved in maintaining the tertiary structure of proteins.
bdaZXjaZ &)& Vb^cd Tertiary structure is very important in enzymes because the shape of an
VX^Yh enzyme molecule gives it its unique properties and determines which
egdhi]Zi^X ]ZbZ substrates can fit into its active site.
\gdje
The final level of protein structure is quaternary structure, which
;^\jgZ ,#&) I]Z higjXijgZ d[ ]Zbd\adW^c# links two or more polypeptide chains to form a single, large, complex
protein.The structure is held together by all the bonds that are important
in the previous levels of structure. Examples are collagen (which has three
polypeptide chains), hemoglobin (which has four), antibodies (which also
have four) and myosin (which has six).

In addition, many proteins contain prosthetic groups and are called
conjugated proteins. Prosthetic groups are not polypeptides but they

&-'

=nYgd\Zc WdcYh [dgb WZilZZc higdc\an edaVg \gdjeh# 9^hja[^YZ WdcYh [dgb WZilZZc XnhiZ^cZ bdaZXjaZh#
I]Zn XVc WZ Wgd`Zc Wn ]^\] iZbeZgVijgZ dg Wn e= X]Vc\Zh# I]Z WdcYh XVc WZ Wgd`Zc Wn gZYjX^c\ V\Zcih#
XnhiZ^cZ
WdcY id gZhi EÄ E E WdcY id gZhi
d[ bdaZXjaZ d[ bdaZXjaZ 8=' 8='
C= 8
D H= H

EÄ Y^hja[^YZ WdcY

h]VgZY ZaZXigdch h]VgZY ZaZXigdch H= H
heZcY bdgZ i^bZ heZcY bdgZ i^bZ
VgdjcY C VgdjcY D

]nYgd\Zc WdcY

>dc^X WdcYh [dgb WZilZZc VheVgV\^cZ 8=' 8='
^dc^hZY Vb^cZ VcY XVgWdmna^X
VX^Y \gdjeh# I]Zn XVc WZ 8=' XnhiZ^cZ
Wgd`Zc Wn e= X]Vc\Zh#

8 =nYgde]dW^X ^ciZgVXi^dch dXXjg WZilZZc cdc"edaVg
h^YZ X]V^ch#
D C='
ingdh^cZ
^dc^X WdcY
8=' D=
D DÄ

8

8=' =8 8=(
8=(
8=' kVa^cZ

\ajiVb^X VX^Y
;^\jgZ ,#&* IneZh d[ WdcY i]Vi VgZ ^bedgiVci ^c egdiZ^c higjXijgZ#

are able to bind to different proteins or parts of them. For example,
hemoglobin is a conjugated protein, with four polypeptide chains, each
containing a prosthetic heme group.

;^Wgdjh VcY \adWjaVg egdiZ^ch
Protein molecules are categorised into two major types by their shape.
Fibrous proteins are long and narrow and include collagen, keratin
(which is found in hair and nails) and silk. Fibrous proteins are usually
insoluble in water and in general have secondary structure. Globular
proteins have a more rounded, three-dimensional shape and have either
tertiary or quaternary structure. Most globular proteins are soluble in
water. Globular proteins include enzymes, such as pepsin, and antibodies.
Myoglobin and hemoglobin are also globular proteins.

EdaVg VcY cdc"edaVg Vb^cd VX^Yh

Amino acids are divided into two groups according to the chemical
properties of their side chains or R groups (Figure 7.16, overleaf ). Polar

, CJ8A:>8 68>9H 6C9 EGDI:>CH &-(

<ZcZgVa higjXijgZ d[ Vc Vb^cd VX^Y I]^h \gdje kVg^Zh ^c
Y^[[ZgZci Vb^cd VX^Yh#
G >i ^h `cdlc Vh i]Z G \gdje
=D dg h^YZ X]V^c#

C8 8
= D=

=

HigjXijgZ d[ i]Z h^beaZhi Vb^cd VX^Y! \anX^cZ

G \gdje ^h ]nYgd\Zc

=
=D

C8 8
= D=

=

Vb^cZ \gdje XVgWdmna^X VX^Y \gdje

;^\jgZ ,#&+ I]Z \ZcZgVa higjXijgZ d[ Vc Vb^cd VX^Y VcY i]Z higjXijgZ d[ \anX^cZ#

C=' and non-polar amino acids have different properties and their positions in
=8 8='D= a molecule affect the behaviour and function of the whole protein.

8DD= Amino acids with non-polar side chains are hydrophobic.Those with
polar side chains are hydrophilic. Non-polar amino acids are found in
hZg^cZ Ä V ]nYgde]^a^X Vb^cd VX^Y parts of proteins that are in hydrophobic areas, while polar amino acids are
in areas that are exposed to an aqueous environment such as cytoplasm or
C=' blood plasma.
=8 8=(
For membrane proteins, the polar hydrophilic amino acids are found
8DD= on the outer and inner surfaces in contact with the aqueous environment,
while the non-polar hydrophobic amino acids are embedded in the
VaVc^cZ Ä V ]nYgde]dW^X Vb^cd VX^Y core of the membrane in contact with the hydrophobic tails of the
phospholipid bilayer (Figure 7.17).This helps to hold the protein in place
in the membrane. Some integral proteins act as channels, and the pore
is lined with hydrophilic amino acids to enable polar substances to pass
through.

egdiZ^c dc dcZ h^YZ egdiZ^c i]Vi X]VccZa egdiZ^c
d[ bZbWgVcZ igVkZghZh bZbWgVcZ l^i] edgZ

Djih^YZ d[ XZaa

e]dhe]da^e^Y W^aVnZg

>ch^YZ d[ XZaa
@Zn

edh^i^dc d[ cdc"edaVg Vb^cd VX^Y h^YZ X]V^ch Ä ^ch^YZ i]Z egdiZ^c
edh^i^dc d[ edaVg Vb^cd VX^Y h^YZ X]V^ch Ä dc i]Z ZmedhZY [VXZh d[ i]Z egdiZ^c

;^\jgZ ,#&, >c bZbWgVcZ egdiZ^ch! edaVg ]nYgde]^a^X Vb^cd VX^Yh VgZ [djcY dc i]Z hjg[VXZh ^c XdciVXi l^i] i]Z VfjZdjh Zck^gdcbZci!
l]^aZ cdc"edaVg ]nYgde]dW^X Vb^cd VX^Yh VgZ ZbWZYYZY ^ch^YZ i]Z e]dhe]da^e^Y W^aVnZg#

&-)

Polar and non-polar amino acids are also important in enzymes, where
they assist in the binding of substrates.An enzyme that acts on a polar
substance (for example, amylase) has polar amino acids in its active site,
whereas lipases have non-polar amino acids in the active site.

Polar amino acids on the surface of a protein increase its solubility
while non-polar amino acids help a protein maintain its structure. Lipases
are proteins that have polar groups on the outside so they are soluble in
the gut, but non-polar groups in the their active site so that lipids can bind
to them.

EgdiZ^c [jcXi^dch

Some proteins build the structure of an organism’s body, some act as
chemical receptors and others act as enzymes or hormones.Table 7.1
summarises some important protein functions.

EgdiZ^c ;jcXi^dc 8dbbZcih

]Zbd\adW^c igVchedgi ]Zbd\adW^c W^cYh dmn\Zc ^c VgZVh d[ ]^\] XdcXZcigVi^dc VcY gZaZVhZh ^i ^c gZhe^g^c\ i^hhjZh
VXi^c VcY bndh^c bdkZbZci i]ZhZ egdiZ^ch ^ciZgVXi id XVjhZ i]Z XdcigVXi^dc d[ bjhXaZh
eZeh^c Y^\Zhi^dc eZeh^c ^h Vc ZconbZ i]Vi Y^\Zhih egdiZ^c ^c i]Z hidbVX]
XdaaV\Zc higjXijgVa XdaaV\Zc higZc\i]Zch WdcZ VcY ^h V bV_dg XdbedcZci d[ iZcYdch! a^\VbZcih VcY i]Z h`^c
^bbjcd\adWja^c YZ[ZcXZ ^bbjcd\adWja^ch VXi Vh Vci^WdY^Zh

IVWaZ ,#& HdbZ [jcXi^dch d[ egdiZ^ch#

,#+ :conbZh

6hhZhhbZci hiViZbZcih

r State that metabolic pathways consist of chains and cycles of catalysed reactions.
r Describe the induced-fit model.
r Explain that enzymes lower the activation energy of the chemical reactions that they catalyse.
r Explain the difference between competitive and non-competitive inhibition, with reference to one example of each.
r Explain the control of metabolic pathways by end-product inhibition, including the role of allosteric sites.

The basics of enzyme action were introduced in Chapter 3. It would be
useful to re-read that section before beginning this Higher Level topic.

BZiVWda^X eVi]lVnh

Metabolic pathways consist of chains or cycles of reactions that are
catalysed by enzymes. Metabolism includes all the chemical activities that
keep organisms alive. Metabolic pathways may be very complex, but most
consist of a series of steps, each controlled by an enzyme. Simple pathways
involve the conversion of substrates to a final product:

enzyme 1 enzyme 2 enzyme 3

substrate X ~~n substrate Y ~~n substrate Z ~~n end product

, CJ8A:>8 68>9H 6C9 EGDI:>CH &-*

ZconbZ ZconbZÄhjWhigViZ
VXi^kZ h^iZ XdbeaZm

egdYjXih

hjWhigViZ

6 hjWhigViZ bdaZXjaZ ^h YgVlc I]Z ZconbZ X]Vc\Zh h]VeZ l]Zc i]Z I]Z egdYjXih VgZ gZaZVhZY! VcY i]Z ZconbZ
^cid i]Z VXi^kZ h^iZ d[ i]Z ZconbZ# hjWhigViZ [^ih ^cid i]Z VXi^kZ h^iZ! [dgX^c\ gZijgch id ^ih cdgbVa h]VeZ! gZVYn id gZXZ^kZ
i]Z hjWhigViZ id gZVXi# Vcdi]Zg hjWhigViZ bdaZXjaZ#

;^\jgZ ,#&- I]Z ^cYjXZY"Ò i bdYZa d[ ZconbZ VXi^dc#

@Zn Each arrow represents the specific enzyme needed to catalyse the
& 2 VXi^kVi^dc ZcZg\n l^i]dji XViVanhi conversion of one substrate to the next.
' 2 VXi^kVi^dc ZcZg\n l^i] XViVanhi
Other metabolic pathways, such as photosynthesis or respiration,
:cZg\n aZkZa jcXViVanhZY involve chains of reactions and cycles of reactions – as you will learn in
& gZVXi^dc Chapter 8.
'
>cYjXZY"Ò i bdYZa d[ ZconbZ VXi^dc
XViVanhZY ZcZg\n
gZVXi^dc X]Vc\Z The lock-and-key hypothesis discussed on page 54 explains enzyme
action by suggesting that there is a perfect match between the shape of
8djghZ d[ gZVXi^dc the active site of an enzyme and the shape of its substrate.This theory was
proposed in 1890 by Emil Fischer.
ZcZg\n 2 dkZgVaa ZcZg\n X]Vc\Z V[iZg gZVXi^dc Ä
X]Vc\Z bdhi W^dad\^XVa gZVXi^dch gZaZVhZ bdgZ In the last century, research published by Daniel Koshland (1958)
suggested that the process is not quite this straightforward.The lock-and-
ZcZg\n i]Vc i]Zn iV`Z ^c Vh VXi^kVi^dc key hypothesis cannot account for the binding and simultaneous change
ZcZg\n VcY VgZ hV^Y id WZ Zmdi]Zgb^X that is seen in many enzyme reactions, nor the fact that some enzymes can
bind to more than one similarly shaped substrate.
;^\jgZ ,#&. <gVe] id h]dl VXi^kVi^dc
ZcZg\n [dg Vc Zmdi]Zgb^X gZVXi^dc l^i] VcY A more likely explanation of enzyme action is that the shape of an
l^i]dji V XViVanhi# enzyme is changed slightly as a substrate binds to its active site (Figure
7.18).The substrate causes or induces a slight change in the shape of the
Enzymes do not change the active site so it can fit perfectly.As the enzyme changes shape, the substrate
quantity of product that is formed, molecule is activated so that it can react and the resulting product or
only the rate at which the product products are released.The enzyme is left to return to its normal shape,
is formed. ready to receive another substrate molecule.

This hypothesis is known as the induced-fit model of enzyme action.

6Xi^kVi^dc ZcZg\n

Enzymes work by lowering the activation energy of the substrate or
substrates. In order for a metabolic reaction to occur, the substrate has to
reach an unstable, high-energy ‘transition state’ where the chemical bonds
are destabilised, and this requires an input of energy, which is called the
activation energy. When the substrate reaches this transition stage, it
can then immediately form the product. Enzymes can make reactions
occur more quickly because they reduce the activation energy of reactions
they catalyse to bring about a chemical change (Figure 7.19).

Metabolic reactions that occur in living organisms have to occur at the
body temperature of the organism, which is never high enough to bring
substrates to their transition state.The active site of an enzyme is very

&-+

important because it can lower the amount of energy needed to reach a hjWhigViZ XdbeZi^i^kZ
transition state, so the reaction can occur at the temperature of the organism. VXi^kZ h^iZ ^c]^W^idg

8dbeZi^i^kZ VcY cdc"XdbeZi^i^kZ ^c]^W^i^dc ;^\jgZ ,#'% 8dbeZi^i^kZ ^c]^W^i^dc#

Enzyme inhibitors are substances that reduce or prevent an enzyme’s
activity. Some inhibitors are competitive and others non-competitive.

Competitive inhibitors have molecules whose structure is similar to
that of the substrate molecule that normally binds to the active site.They
compete with the substrate to occupy the active site of the enzyme, and
prevent the substrate molecules from binding (Figure 7.20).The inhibitors
are not affected by the enzyme and do not form products, so they tend
to remain in the active site.This means that the rate of reaction is lower
because the substrate cannot enter an active site that is blocked by an
inhibitor.At low concentrations of substrate, competitive inhibitors have a
more significant effect than at higher concentrations, when the substrate
can out-compete the inhibitor (Figure 7.21).

8dbeZi^i^kZ ^c]^W^idg Cdc"XdbeZi^i^kZ ^c]^W^idg

l^i] ^c]^W^idg cd ^c]^W^idg
l^i] ^c]^W^idg
cd ^c]^W^idg l^i] bdgZ ^c]^W^idg

I]Z ^c]^W^i^dc egdYjXZY
Wn V XdbeZi^i^kZ ^c]^W^idg
^h cZ\ViZY Wn ]^\]
XdcXZcigVi^dch d[
hjWhigViZ#

l^i] bdgZ ^c]^W^idg
>c^i^Va gViZ d[ gZVXi^dc
>c^i^Va gViZ d[ gZVXi^dc

HjWhigViZ XdcXZcigVi^dc HjWhigViZ XdcXZcigVi^dc

;^\jgZ ,#'& <gVe]h id h]dl i]Z Z[[ZXih d[ XdbeZi^i^kZ VcY cdc"XdbeZi^i^kZ ^c]^W^idgh dc gZVXi^dc gViZ! Vh hjWhigViZ XdcXZcigVi^dc

^cXgZVhZh#

Non-competitive inhibitors also combine with enzymes but not at hjWhigViZ
the active site.They bind at another part of the enzyme where they either VXi^kZ h^iZ
partly block access of the substrate to the active site or cause a change in
the shape of the enzyme so that the substrate cannot enter the active site
(Figure 7.22). Increasing the concentration of substrate in the presence of
a non-competitive inhibitor does not overcome inhibition (Figure 7.21).

Table 7.2 (overleaf ) compares the nature and effects of competitive and
non-competitive inhibitors.

8dcigdaa^c\ bZiVWda^X eVi]lVnh Wn ZcY"egdYjXi ^c]^W^idg h^iZ cdc"XdbeZi^i^kZ
^c]^W^i^dc ^c]^W^idg

End-product inhibition means that an enzyme in a pathway is inhibited ;^\jgZ ,#'' Cdc"XdbeZi^i^kZ ^c]^W^i^dc#
by the product of that pathway.This prevents a cell over-producing a

, CJ8A:>8 68>9H 6C9 EGDI:>CH &-,

8dbeZi^i^kZ ^c]^W^idgh Cdc"XdbeZi^i^kZ ^c]^W^idgh

™ higjXijgVaan h^b^aVg id i]Z hjWhigViZ bdaZXjaZ ™ higjXijgVaan jca^`Z i]Z hjWhigViZ bdaZXjaZ
™ dXXje^Zh VcY WadX`h i]Z VXi^kZ h^iZ ™ W^cYh Vi V h^iZ VlVn [gdb i]Z VXi^kZ h^iZ! gZYjXZh
™ ^[ XdcXZcigVi^dc d[ ^c]^W^idg ^h adl! ^cXgZVh^c\ i]Z XdcXZcigVi^dc
VXXZhh id ^i
d[ hjWhigViZ l^aa gZYjXZ i]Z ^c]^W^i^dc ™ ^[ XdcXZcigVi^dc d[ hjWhigViZ ^h adl! ^cXgZVh^c\ i]Z
™ ZmVbeaZh ^cXajYZ/
XdcXZcigVi^dc d[ hjWhigViZ ]Vh cd Z[[ZXi dc W^cY^c\ d[
Ä dmn\Zc! l]^X] XdbeZiZh l^i] XVgWdc Y^dm^YZ [dg i]Z VXi^kZ i]Z ^c]^W^idg hd ^c]^W^i^dc hiVnh ]^\]
h^iZ d[ g^WjadhZ W^he]dhe]ViZ XVgWdmnaVhZ ^c e]didhnci]Zh^h ™ ZmVbeaZh ^cXajYZ/
Ä XnVc^YZ VcY XVgWdc bdcdm^YZ! l]^X] WadX`
Ä Y^hjaÒ gVb! l]^X] XdbeZiZh l^i] VXZiVaYZ]nYZ [dg i]Z VXi^kZ
h^iZ d[ VaYZ]nYZ YZ]nYgd\ZcVhZ XnidX]gdbZ dm^YVhZ ^c VZgdW^X gZhe^gVi^dc! aZVY^c\
id YZVi]

IVWaZ ,#' 8dbeVg^c\ XdbeZi^i^kZ VcY cdc"XdbeZi^i^kZ ^c]^W^idgh#

substance it does not need at the time. Many products may be needed by
the cell at a specific time or in specific amounts and over-production not
only wastes energy but may also become toxic if the product accumulates.

In an assembly-line reaction, such as those described in Figure 7.23,
each step is controlled by a different enzyme. If the end-product begins
to accumulate because it is not being used, it inhibits an enzyme earlier in
the pathway to switch off the assembly line. In most cases, the inhibiting
effect is on the first enzyme in a process, but in other cases it can act at a
branch point to divert the reaction along another pathway.

When the end-product starts to be used up, its inhibiting effect reduces,
the inhibited enzyme is reactivated and production begins again.This is an
example of negative feedback (see page 154).

End-product inhibition may be competitive or non-competitive.
Competitive inhibition will only work if the product is a similar shape
to the normal substrate and there can be an induced fit of the product or
inhibitor onto the enzyme. In most cases, the product will be a different

^c^i^Va hjWhigViZ ^c^i^Va hjWhigViZ
ZconbZ &
^ciZgbZY^ViZ 6 ^ciZgbZY^ViZ 8
^ciZgbZY^ViZ 6
ZconbZ ' ^ciZgbZY^ViZ 7

^ciZgbZY^ViZ 7 ZcY"egdYjXi & ZcY"egdYjXi '
ZconbZ (
I]Z ZcY"egdYjXi ^c]^W^ih Vc ZconbZ ^c i]Z eVi]lVn! l]^X] XVjhZh V
ZcY"egdYjXi Y^[[ZgZci ZconbZ id XdbZ ^cid eaVn VcY i]Z eVi]lVn ^h Y^kZgiZY Ydlc
I]Z ZcY"egdYjXi ^c]^W^ih i]Z ZconbZ V Y^[[ZgZci gdjiZ#
XViVanh^c\ i]Z [^ghi gZVXi^dc ^c i]Z hZg^Zh!
hd Vaa i]Z hjWhZfjZci gZVXi^dch hide#
;^\jgZ ,#'( :cY"egdYjXi ^c]^W^i^dc#

&--

shape and therefore this has to be non-competitive inhibition. In this case, hjWhigViZ egdYjXih
the enzyme is known as an allosteric enzyme, the product is called an VXi^kZ h^iZ
allosteric inhibitor and the place where it binds to the enzyme is called
the allosteric site (Figure 7.24). VaadhiZg^X h^iZ

&& Djia^cZ i]Z [djg aZkZah d[ egdiZ^c higjXijgZ#

&' A^hi i]gZZ Y^[[ZgZcXZh WZilZZc Ò Wgdjh VcY \adWjaVg egdiZ^ch#
&( A^hi [djg [jcXi^dch d[ egdiZ^ch#

&) Djia^cZ l]Vi ^h bZVci Wn ÈVXi^kVi^dc ZcZg\nÉ#

&* :meaV^c ]dl i]Z ^cYjXZY"Ò i bdYZa Y^[[Zgh [gdb i]Z adX`"VcY"`Zn
]nedi]Zh^h [dg ZconbZ VXi^dc#

VaadhiZg^X cdc"XdbeZi^i^kZ ^c]^W^idg

;^\jgZ ,#') 6aadhiZg^X Xdcigda# 6aadhiZg^X
^c]^W^idgh egZkZci i]Z VXi^kZ h^iZ [jcXi^dc^c\#

:cY"d["X]VeiZg fjZhi^dch

1 The components of a nucleosome are: (1)

A ribosomal RNA and DNA
B 8 histone proteins and DNA
C 8 histones proteins in a ball + 1 further histone
D 9 histone proteins and DNA

2 Which of the following statements is correct about the structure of DNA? (1)

A The purine base cytosine is linked to the pyrimidine base guanine through three
hydrogen bonds.

B The sugar–phosphate strands are antiparallel and linked by complementary base
pairing.

C The bases are linked to each other through a 3'–5' linkage.
D Complementary base pairing of guanine with cytosine and adenine with uracil

means that the two sugar–phosphate strands lie parallel.

, CJ8A:>8 68>9H 6C9 EGDI:>CH &-.

3 Which of the following statements is correct about DNA replication? (1)

A The enzymes DNA ligase and RNA primase can be found on the lagging strand.
B Okazaki fragments are produced by DNA polymerase I and DNA polymerase III

on the leading strand.
C On the lagging strand, the RNA primer is synthesised by RNA primase and then

converted into a DNA strand with the enzyme DNA polymerase III.
D The enzyme DNA polymerase III uses deoxynucleoside triphosphates to build a

new DNA strand only on the leading strand.

4 Which of the following statements is correct about transcription? (1)

A The enzyme RNA polymerase moves along the antisense strand in a 3' n 5' direction.
B The sequence of bases in the strand of RNA being synthesised is the same as the

sequence of bases in the sense strand of DNA.
C In eukaryotic cells, exons are removed from the primary RNA in the nucleus to make

mature RNA.
D Messenger RNA is synthesised by RNA polymerase in a 3' n 5' direction.

5 Which of the following statements is correct about translation? (1)

A Ribosomes that are free in the cytoplasm synthesise proteins that are primarily for
lysosome manufacture and exocytosis.

B Ribosomes are made of two subunits and mRNA binds to the larger one.
C On the larger ribosome subunit there are three binding sites that can be occupied by

tRNA molecules.
D During polypeptide synthesis, the ribosome moves along the mRNA strand in a

3' n 5' direction until it reaches a stop codon.

6 Which of the following statements is correct about protein structure? (1)

A There are four levels of protein structure.The primary level is held together by
covalent and hydrogen bonding.

B Enzymes have an active site that is a three-dimensional structure produced by
secondary level folding of the protein.

C The B helix and C pleated sheet are both types of tertiary level folding.
D Both tertiary and quaternary level proteins can form conjugated proteins.

7 Which of the following statements is correct about enzymes? (1)

A Most enzymes use the lock-and-key model of substrate interaction as it is the most
stable.

B In allosteric control of metabolic pathways, a product within the pathway can act as
a non-competitive inhibitor of an enzyme earlier in the pathway.

C Increasing the concentration of substrate has no effect on the rate of a reaction being
inhibited by a competitive inhibitor.

D Competitive inhibitors bind to an allosteric site and alter the shape of the active site.

&.%

8 Explain the process of translation. (9)

9 Explain how control of metabolic pathways can be brought about by end-product inhibition. (6)

10 It had always been assumed that eukaryotic genes were similar in organisation to prokaryotic genes.
However, modern techniques of molecular analysis indicate that there are additional DNA sequences lying
within the coding region of genes. Exons are the DNA sequences that code for proteins while introns are
the intervening sequences that have to be removed.The graph shows the number of exons found in genes
for three different groups of eukaryotes.

EZgXZciV\Z d[ \ZcZh IWYY^WhecoY[i Y[h[l_i_W[ V nZVhi
&%%

-%
+%
)%
'%

%
)%

(% :heief^_bW c[bWde]Wij[h [gj^i [an
'%

&%

%
'%

&* BVbbVah
&%

*

%
& ' ( ) * + , - . &% && &' &( &) &* &+ &, &- &. '% 1(%1)%1+%3+%
CjbWZg d[ Zmdch

hdjgXZ/ 7Zc_Vb^c AZl^c &... =[d[i Lbb! DJE! e# **

a Calculate the percentage of genes that have five or fewer exons in mammals. (1)
(2)
b Describe the distribution of the number of exons and the percentage of genes in D. melanogaster. (2)

c i Compare the distributions of the number of exons found in genes of S. cerevisiae and in mammals. (1)

ii Suggest one reason for the differences in the numbers of exons found in genes of S. cerevisiae and
in mammals.

, CJ8A:>8 68>9H 6C9 EGDI:>CH &.&

Human DNA has been analysed and details of certain genes are shown in the table below.

<ZcZ <ZcZ h^oZ $ `^adWVhZ eV^gh bGC6 h^oZ $ `^adWVhZ eV^gh CjbWZg d[ ^cigdch

^chja^c &#, %#) '
XdaaV\Zc (-#% *#% *%
VaWjb^c '*#% '#& &)
e]ZcnaVaVc^cZ ]nYgdmnaVhZ .%#% '#) &'
Ynhigde]^c '%%%#% &,#% *%

h djgXZ/ L^aa^Vb H @aj\ VcY B^X]VZa G 8jbb^c\h! '%%' ! 9edY[fji e\ =[d[j_Yi! ,i] ZY^i^dc! EgZci^XZ =Vaa! e# (&)

d Calculate the average size of the introns for the albumin gene. (2)
(2)
e Analyse the relationship between gene size and the number of introns. (1)

f Determine the maximum number of amino acids that could be produced by translating the (1)
phenylalanine hydroxylase mRNA.

Hemoglobin is a protein composed of two pairs of globin molecules. During the process of
development from conception to adulthood, human hemoglobin changes in composition.Adult
hemoglobin consists of two alpha-globin and two beta-globin molecules.Two globin genes occur on
chromosome 16: alpha-globin and zeta-globin. Four other globin genes are found on chromosome
11: beta, delta, epsilon and gamma.The graph below illustrates the changes in expression of the globin
genes over time.

g State which globin genes are the first to be expressed after fertilisation.

EZgXZciV\Z d[ ]Zbd\adW^c *%

)%

(%

'%

&%

% &% '% (% )% ' ) +
@Zn lZZ`h d[ \ZhiVi^dc W^gi] bdci] d[ V\Z

Vae]V"\adW^c WZiV"\adW^c Zeh^adc"\adW^c
\VbbV"\adW^c YZaiV"\adW^c oZiV"\adW^c

hdjgXZ/ VYVeiZY [gdb B 8jbb^c\h! >kcWd >[h[Z_jo! )i] ZY^i^dc!
LZhi$LVYhldgi] EjWa^h]^c\ 8dbeVcn

h Compare the expression of the gamma-globin gene with the beta-globin gene. (3)

i Deduce the composition of the hemoglobin molecules at 10 weeks of gestation and at 2 months (2)
after birth.

(total 17 marks)
© IB Organization 2009

&.'