UNIT 5 REVISION
Chapter 9 – Responses to Stimuli
Fill in the blanks to show your understanding of stimulus and response:
effector  response
receptor co-ordinator
Stimulus  __________
__________  __________
Write a definition for taxes:
A directional response
to a stimulus
Explain the difference between:
• positive taxes:
Movement is towards the
stimulus e.g. algae move
towards light
• negative taxes:
Movement is away from the
stimulus e.g. earthworms
move away from light
Write a definition for kineses:
Random movement in
response to a stimulus. The
more unpleasant the stimulus
the faster the organism
moves.
Give an example of kinesis:
Woodlice move more rapidly
and change direction more
often when they are in
unfavourable, dry conditions.
This is to increase their
chances of finding a more
favourable area. In moist
conditions they slow down
and change direction less
often to remain in that area.
Write a definition for tropism:
Growth movement of a
plant in response to a
directional stimulus
List the different types of tropism:
• .Phototropism
• .Geotropism
• .Hydrotropism
Explain the difference between:
• positive tropism:
Growth of the plant
towards the stimulus.
• negative tropism:
Growth of the plant
away from the
stimulus.
Chapter 9 – Nervous Control
Complete the diagram to show the
organisation of the nervous system:
Nervous system
Peripheral
Nervous System
Central Nervous
System
Brain
Label the diagram to show how the nerves are arranged in the
spinal cord:
Relay
Sensory
(Intermediate)
neurone
Contains
neurone
sensory
neurones
Spinal
Cord
Spinal
nerve
Sensory Nervous
System
Motor nervous
system
Voluntary
nervous system
Autonomic
Nervous System
Contains
Motor
neurones
Motor
neurone
Chapter 9 – Reflex Arcs
• Describe a reflex arc
• Stimulus (heat from a hot object)
•  Receptor (temperature receptors in the skin)
•  Sensory Neurone (passes nerve impulse to the spinal cord)
•  Relay Neurone (links the sensory to the motor neurone)
•  Motor Neurone (carries nerve impulse from the spinal cord to the
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muscle)
 Effector (muscle contracts)
 Response (hand moves away)
Why are reflex arcs important?
Involuntary – they do not require input from the brain, leaving it free to
carry out more complex responses
Protection – the fast response helps to protect the body from harm
Rapid – short neurone pathway, only 2 synapses
Chapter 9 – Control of Heart Rate
• State and describe the two parts to the autonomic
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nervous system
Sympathetic Nervous System – stimulates effectors and
therefore speeds up activity. Helps us cope in stressful
situations (e.g. strenuous exercise) by heightening
awareness and preparing us for activity (fight or flight)
Parasympathetic Nervous System – inhibits effectors and
therefore slows down activity. It is our ‘rest and digest’
response.
The two nervous systems are antagonistic. What does
this mean?
They normally oppose each other. If one system contracts
a muscle, the other relaxes it.
The Brain Controls Changes in the Heart Rate
Medulla oblongata (in
the brain)
Cardiac Centre
Cardiac Inhibitory
Centre
Parasympathetic
nerve linked to
SAN
Decreases heart rate
Cardiac Accelerator
Centre
Sympathetic
nerve linked to
SAN
Increases heart rate
Chapter 9 – Control of Heart Rate
• Explain the role of chemoreceptors in control of heart rate
• Found in the wall of the carotid arteries (arteries to the brain) and
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aorta
If there is increased CO2 in the blood then pH of the blood is lowered
Chemoreceptors detect this and increase the frequency of nerve
impulses to the medulla oblongata (centre that increases heart rate)
This centre sends more nerve impulses via the sympathetic nervous
system to the SAN. This increases heart rate.
The increased blood flow means that more CO2 is removed by the
lungs, so the CO2 levels in the blood return to normal and pH rises.
Once the pH is back to normal then the chemoreceptors detect this
and reduce the frequency of nerve impulses to the medulla oblongata.
The centre that increases heart rate sends less nerve impulses to the
SAN and so heart rate is lowered.
Chapter 9 – Control of Heart Rate
• Heart rate is also controlled by pressure receptors (in the
carotid arteries and aorta). Explain what these do when
the blood pressure is higher than normal:
• The pressure receptors send nerve impulses to the centre
that decreases heart rate (in the medulla oblongata). This
centre sends impulses via the parasympathetic nervous
system to the SAN, which decreases heart rate.
• Explain what these do when the blood pressure is lower
than normal
• The pressure receptors send nerve impulses to the centre
that increases heart rate (in the medulla oblongata). This
centre sends impulses via the sympathetic nervous
system to the SAN, which increases heart rate.
Chapter 9 - Receptors
Describe the structure and function of the Pacinian corpuscle:
Structure:
Pacinian corpuscles respond to pressure and are found in the skin.
They contain a single sensory neurone in the centre of layers of
tissue, each separated by a gel. The sensory neurone has a stretchmediated sodium channel in its membrane. Their permeability to
sodium will change when they change shape.
Function:
1. Normally the stretch-mediated sodium channels do not allow
sodium to pass. The neurone is at resting potential.
2. When pressure is applied then the membrane around the
sensory neurone stretches.
3. This stretching widens the sodium channels and allows sodium
ions to diffuse into the neurone.
4. Influx of sodium ions causes depolarisation (this is the generator
potential, which in turn causes an action potential to pass along
the neurone.
Chapter 9 - Receptors
Describe the structure and function of rod and cone cells in the retina:
Rod Cells:
Rod cells only produce black and white images.
Many rod cells share a single bipolar cell and
sensory neurone. This allows them to respond to
light at very low intensity (they can add together to
overcome the threshold needed to send an impulse).
A pigment (rhodopsin) must be broken down to
cause sodium channels to open in the neurone)
Rod
Cone Cells:
Cone cells produce colour images. Each
cone cell has its own bipolar cell and
sensory neurone. This means they will
only respond to light at high intensity (they
cannot add together to overcome the
threshold needed to send an impulse). A
pigment (iodopsin) must be broken down
to create a generator potential.
Shape
Rod-shaped
Cone
Cone-shaped
Number
Greater numbers
than cone cells
Fewer numbers
than rod cells
Distribution
More at the edge,
none at the fovea
Fewer at the edge,
most at the fovea
Visual acuity
Sensitivity in low
light conditions
Poor
Sensitive
Good
Not sensitive
Chapter 10 – Co-ordination
Hormonal System
Method of
communication?
Nervous System
Chemicals called
hormones
Nerve impulses
Method of transmission?
Blood system
Neurones
Speed of transmission?
Relatively slow
Very rapid
All parts of the body, only
Where does the
communication travel to? target organs respond
How widespread is the
response?
Speed of the response?
Length of response?
Specific parts of the
body
Widespread
Localised
Slow
Rapid
Often long-lasting
Short-lived
Length and reversibility Effect may be permanent
and irreversible
of effect?
Effect is temporary and
reversible
Chapter 10 – Co-ordination
• What are chemical mediators?
• Chemicals released from certain mammalian cells that affect cells in
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the immediate vicinity
Normally released by infected or injured cells and cause arterioles
to dilate, leading to swelling.
Two types: histamine (response to an allergen or injury) and
prostaglandins (response to injury)
What are plant growth factors?
Plant hormones that affect growth
Made by cells throughout the plant, rather than organs
Produced in small quantities and affect the tissues close to them.
Example = IAA (indoleacetic acid) which causes cell elongation.
Chapter 10 - IAA
• Describe the sequence of events that causes a plant to
grow towards the light
1. Cells in the tip produce IAA, which is transported down
the shoot
2. Light causes the IAA to move to the shaded side of the
shoot
3. Concentration of IAA becomes high on the shaded side
4. IAA causes elongation of cells on the shaded side
5. The shaded side grows faster, causing the shoot to
bend towards the light
Chapter 10 - Neurones
Describe the structure and function of the parts of a neurone:
Axon – Single, long fibre that carries nerve
impulses away from the cell body.
Dendrites
Cell Body – contains the nucleus and RER.
Produces proteins and neurotransmitters.
Nucleolus
Dendrites
Axon
Nucleus
Cell Body
Dendrites – branched fibres that carry nerve
impulses towards the cell body.
Myelin Sheath – covers the axon and is
made up of the lipid-rich membranes of the
Schwann cells. Speeds up transmission of
nerve impulses.
Myelin Sheath
Schwann Cells
Node of Ranvier
Myelin Sheath
Schwann Cells – surround the axon, protect
it and provide electrical insulation.
Node of Ranvier – gaps between adjacent
Schwann cells where there is no myelin
sheath.
Chapter 10 – The Nerve Impulse
Explain what resting potential is, including the role of
potassium and sodium ions in its generation:
The inside of an axon is negatively charged
relative to the outside. The resting potential
is the usual state of the neurone. It is
-65mV.
The resting potential is established by the
sodium-potassium pump, which actively
transports 3 Na+ ions out of the axon for
every 2 K+ ions that are actively
transported in.
The membrane is much more permeable to
K+ ions than Na+ ions. So, potassium ions
diffuse back out of the axon faster than
sodium ions diffuse back in.
Chapter 10 – The Nerve Impulse
Using the graph, explain what happens when an
action potential passes down a neurone.
1.
Resting potential (-65mV), sodium voltagegated channels are closed.
2.
Sodium voltage-gated channels start to open
due to energy from a stimulus and Na+ ions
diffuse into the membrane.
3.
More sodium voltage-gated channels open and
more Na+ ions diffuse in, causing
depolarisation of the membrane.
4.
The membrane potential reaches +40mV and
the sodium voltage-gated channels close.
Potassium voltage-gated channels start to
open.
5.
More potassium voltage-gated channels open
and K+ ions diffuse out and the membrane
repolarises.
6.
Temporarily, the membrane becomes more
negative than the resting potential (due to extra
K+ ions diffusing out) – hyperpolarisation. The
potassium voltage-gated channels close and
the sodium-potassium pump takes back over
and returns the membrane to the resting
potential.
Chapter 10 – Passage of an Action Potential
Explain, using diagrams, how an action potential moves along an unmyelinated axon:
As Na+ ions flood into the membrane the charge
inside the membrane is reversed (-  +). This
establishes a localised circuit which causes
sodium voltage-gated channels a little further
down the axon to open (causing a -+ change).
The sodium voltage-gated channels behind now
close and the resting potential is re-established
(+-)
Explain, using diagrams, how an action potential moves along a myelinated axon:
In myelinated neurones the sheath acts as an electrical
insulator, preventing action potentials from forming here.
The action potential passes in a similar way to the
unmyelinated neurone, but it has to ‘jump’ from node of
Ranvier to node of Ranvier (saltatory conduction)
Chapter 10 – Speed of a Nerve Impulse
• List factors that will affect the speed of an action potential
• The myelin sheath, the diameter of the axon, temperature
• What are the reasons for the refractory period
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(hyperpolarisation)?
1) Action potential will only pass in one direction
2)Discrete impulses – Action potentials can’t be formed
immediately after the first one, so they are spread out
3) Limits the number of action potentials
Explain the ‘all-or-nothing’ principle
A stimulus has to be above a certain threshold value if it is to
cause an action potential. Below the threshold value no
action potential occurs and therefore no impulse is generated.
It doesn’t matter how much above the threshold value the
stimulus is it will still only generate one action potential.
Chapter 10 Synapses
Chapter 10 - Synapses
• State the functions of synapses
• Transmit impulses from one neurone  one neurone, one
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neurone  many neurones, many neurones  one neurone
How do synapses ensure the nerve impulse only travels in one
direction?
Neurotransmitter is only made in the presynaptic neurone, only
the post-synaptic neurone has receptors for the neurotransmitter
Explain Spatial and Temporal Summation
Spatial – a number of different presynaptic neurones together
release enough neurotransmitter to exceed the threshold and
trigger an action potential
Temporal – a single presynaptic neurone releases
neurotransmitter many times over a short time period, the total of
this can then exceed the threshold and trigger an action potential
Chapter 10 - Synapses
• Explain how inhibition may occur at a synapse
• Cl- ion channels may be made to open in the post-
synaptic membrane. Cl- ions diffuse into the membrane,
making it more negative (hyperpolarisation) and
decreasing the likelihood of an action potential being
created.
• How can drugs affect synapses?
• Stimulate the nervous system by creating more action
potentials in postsynaptic neurones (may be by mimicking
a neurotransmitter)
• Inhibit the nervous system by creating fewer action
potentials in postsynaptic neurones (may be by inhibiting
neurotransmitter release)
Chapter 10 – Transmission Across
Synapses
1. An action potential arrives at presynaptic
membrane. Voltage gated calcium channels open,
calcium ions enter along a concentration gradient.
Afterwards these are pumped out using ATP.
2. Calcium ions cause synaptic vesicles to fuse with
the presynaptic membrane, releasing acetylcholine
into the synaptic cleft.
3. Acetylcholine diffuses cross the synaptic cleft and
binds to specific receptor sites in the post synaptic
membrane.
Chapter 10 - Synapses
4. Sodium channels open. Sodium ions diffuse rapidly
along a concentration gradient into the postsynaptic
membrane causing depolarisation.
5. The enzyme acetylcholinesterase hydrolyses
acetylcholine into choline and ethanoic acid (acetyl).
These diffuse back into the presynaptic neurone .
6. Acetylcholine is resynthesised using ATP from the
mitochondria. The break down of acetylcholine
prevents it from continuously generating a new action
potential.
Chapter 11 - Muscles
• Describe the 3 types of muscle
• Skeletal Muscle (voluntary, attached to skeleton via
tendons), Cardiac Muscle (myogenic, only in the heart),
Smooth Muscle (involuntary, in the gut, uterus and arteries)
• Explain the differences between fast and slow twitch fibres
• Slow twitch – contract more slowly, adapted for endurance
work and aerobic respiration (large store of myoglobin,
supply of glycogen, rich supply of blood vessels, numerous
mitochondria)
• Fast twitch – contract more rapidly, adapted for intense
exercise and anaerobic respiration (thicker muscle filaments,
high concentration of enzymes needed for anaerobic
respiration, store of phosphocreatine which can rapidly
generate ATP from ADP in anaerobic conditions)
Chapter 11 – Structure of Muscles
• Complete the diagram
Chapter 11 – Structure of Muscles
Describe the zones found in a
sarcomere
Z line
--I band-- ----A band----• Light bands are I-bands only
actin is found in these bands
• Dark bands are A-bands actin
and myosin overlap in these
H zone
bands
• In the middle of each A-band is
a lighter part called the H-zone
• Explain the difference between
• In the centre of each I-band is
actin and myosin
the Z-line, where the actin
• They are both protein filaments
filaments join
• Actin is thinner and has 2 strands
• The section of muscle between
twisted around each other
Z-lines is called a sarcomere
• Myosin is thicker and has long rodshaped fibres with bulbous heads • Pattern = Z I A H A I Z
Label the diagram of the
neuromuscular junction
Chapter 11 – Neuromuscular Junctions
• What is a neuromuscular junction?
• The point where a motor neurone meets a skeletal muscle fibre
• How does a neuromuscular junction work?
• Impulse reaches the neuromuscular junction
• Synaptic vesicles fuse with the pre-synaptic membrane, releasing
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acetylcholine.
Acetylcholine diffuses to the post-synaptic membrane, changing
the permeability to sodium ions
Influx of sodium ions results in depolarisation
An action potential occurs in the muscle fibre
Muscle fibre contracts
Acetylcholinesterase breaks down the acetylcholine to prevent
over-stimulation of the muscle
Chapter 11 – Contraction of Skeletal Muscle
• What is the sliding filament
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mechanism?
Actin and myosin slide past
one another when the
muscle contracts
What is the evidence for
this?
Sarcomere gets shorter
More overlap
Z-lines get closer together
I-band gets narrower
H-zone gets narrower
Chapter 11 – Contraction of Skeletal Muscle
• What are the 3 main proteins involved?
Myosin – globular, bulbous head and a long tail
2. Actin – a globular protein where the molecules are
twisted into a helix
3. Tropomyosin – long, thin threads wrapped around actin
1.
Explain more detail on the sliding filament mechanism
• Heads of myosin form cross-bridges with the actin
filaments (attach to binding sites)
• Myosin heads flex together and pull the actin along the
myosin
• They detach
• Return to original angle and re-attach (uses ATP)
• Repeats 100 times a second
Chapter 11 - Muscle Contraction – Sliding
Filament Mechanism
Chapter 11 - 3 Stages of Muscle
Contraction
Explain the stimulation, contraction and relaxation of muscles
1. Stimulation
• Neuromuscular junctions – acetylcholine diffuses across the cleft and
binds to receptors causing depolarisation
2. Contraction
• Action potential carried through t-tubules
• Ca2+ ions are released into the muscle cytoplasm from the ER and
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tropomyosin molecules move away from binding sites
Myosin (with ADP attached) binds to actin and move it along (releases
ADP)
ATP attaches to the myosin, causing it to detach from the actin
Ca2+ ions activate ATPase, which hydrolyses ATP  ADP giving the
energy for the myosin head to return to its original position
The Myosin (with ADP attached) can now bind further along the actin and
the cycle continues.
3. Relaxation
• Ca2+ ions actively transported back to the ER (energy from hydrolysis of
ATP) and tropomyosin blocks the actin again
Chapter 11 – Energy Supply
• How do muscles ensure they have enough energy for
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contraction?
Muscles need a lot of energy when they contract
Supplied by the hydrolysis of ATP
Because of the great demand for energy in certain cases
(e.g. Fight or flight responses) then it is required that ATP
be generated anaerobically as well
This is achieved by using phosphocreatine
Phosphocreatine is stored in the muscle and helps to
regenerate ATP
Chapter 12 - Homeostasis
Explain what homeostasis is:
Maintenance of a constant internal
environment. There are continuous variations
around a set point and homeostasis is the
ability to return to that set point.
Explain why homeostasis is important:
Essential to keep enzymes functioning,
prevent damage to cells from water potential
changes. It also allows organisms to be more
independent of the external environment.
State some of the factors that are controlled by
homeostasis:
Temperature, pH, water potential, tissue fluid
Input
• Changes to the
system
Output
• System
returned to set
point
Effector
• Brings about
changes to the
system in order
to return it to a
set point
Receptor
• Measures level
of a factor
Control unit
• Operational
information is
stored here and
used to
coordinate
effectors
Chapter 12 – Regulation of Body
Temperature
State the main ways in which heat is gained by organisms:
Production of heat (higher metabolic rate),
gain of heat from environment (by CCR)
Complete the feedback loop for controlling
body temperature in mammals:
Normal body temp
State the main ways in which heat is lost by organisms:
Evaporation of water, loss of heat to the
environment (by CCR)
Explain how body temperature if regulated in ectotherms:
Adapting their behaviour to changes in the external
temperature e.g. exposing themselves to the sun,
sheltering, gaining warmth from the ground, generating
metabolic heat, colour variations
Explain how body temperature is regulated in endotherms:
Gain heat = vasoconstriction, shivering, raising hair,
higher metabolic rate, behaviour
Lose heat = vasodilation, sweating, lowering hair,
behaviour
Cold
receptors
in skin
Warm
receptors
in skin
Hypothalamus
Heat gain centre
Vasoconstriction,
shivering, hair
raised, higher
metabolic rate
Heat loss centre
Vasodilation,
sweating, hair
lowered, lower
metabolic rate
Chapter 12 – Regulation of Blood Glucose
• What are common characteristics of all hormones?
• Produced by glands and secreted directly into the blood stream
• Carried in the blood to target cells which have receptors on their cell•
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surface membranes which have a complementary shape to the hormone
Effective in very small quantities, but often have widespread and longlasting effects
Explain the second-messenger model of hormone action
The hormone is the first messenger and binds to receptors on target cells
(forms a hormone-receptor complex)
The hormone-receptor complex activates an enzyme in the cell that
causes the production of a chemical (the second messenger)
The second messenger causes a series of chemical changes to get to the
response needed.
Give 2 examples of hormones that work via this second messenger model
Adrenaline and Glucagon
Chapter 12 – Regulation of Blood Glucose
Explain the role of the pancreas in regulating blood
glucose:
Explain the role of glucagon in regulating blood
glucose:
The pancreas has islets of Langerhans which
contain α cells and β cells. α cells produce
glucagon and β cells produce insulin.
If the α cells detect a fall in blood glucose then
they secrete glucagon into the blood.
Glucagon combines with receptors on liver
cells and causes:
- Activation of an enzyme that converts
glycogen to glucose (glycogenolysis)
- Increase in the conversion of amino acids
and glycerol into glucose
(gluconeogenesis)
State the sources of blood glucose:
The diet - from breakdown of other
carbohydrates. From the breakdown of
glycogen (glycogenolysis) which has been
stored in the liver. From gluconeogenesis –
production of new glucose from sources other
than carbohydrates.
Explain the role of insulin in regulating blood glucose:
Explain the role of adrenaline in regulating blood
glucose:
At times of excitement or stress adrenaline is
If the β cells detect a rise in blood glucose
produced by the adrenal glands and raises the
then they secrete insulin into the blood. Insulin blood glucose level by:
combines with receptors on cells and causes:
- Activating an enzyme that causes
- Opening of glucose transport protein
breakdown of glycogen to glucose in the
channels (glucose enters cells)
liver (glycogenolysis)
- Activation of enzymes that convert glucose to - Inactivating an enzyme that synthesises
glycogen and fat
glycogen from glucose.
Chapter 12 – Regulation of Blood
Glucose
Complete the feedback loop for controlling body temperature in mammals:
To increase
To decrease
Detected by…
Detected by…
β cells, which release insulin
α cells, which release glucagon
Blood glucose
falls
Blood glucose
rises
Response…
Response…
Conversion of
glycogen to glucose,
conversion of amino
acids to glucose
Conversion of glucose to
glycogen, conversion of
glucose to fat, absorption
of glucose into cells, more
respiration
Normal blood glucose level
90mg100cm-3 blood
Blood glucose
rises, negative feedback
Blood glucose
falls, negative feedback
Chapter 12 - Diabetes
Type of Diabetes
Description
How is it
controlled?
Type I (insulin
dependent)
Normally begins in
childhood. The body can’t
produce enough insulin.
Detected by glucose in
urine. May be due to the
body attacking the β cells
of the islets of
Langerhans.
By injections of insulin
(2/4 times a day).
Injections have to match
glucose intake, so blood
glucose is monitored by
biosensors. Also
carbohydrate intake is
managed.
Type II (insulin
independent)
Normally develops in
people over 40. The
receptors on body cells
don’t respond to insulin
anymore (or due to low
supply of insulin from the
pancreas).
By regulating
carbohydrate intake and
matching this to exercise
taken. May be
supplemented by insulin
injections in some cases.
Chapter 13 - Negative Feedback
• What is negative feedback?
• Negative feedback is when the feedback causes the
corrective measures to be turned off.
• This returns the system to a normal level.
Input
Fall in
some
parameter
Receptors
Detect the
change
Control
Centre
Coordination
Effector
Have an
effect on the
system
Negative Feedback = Corrective
measures turned off
Output
Rise in
some
parameter
Chapter 13 – Positive Feedback
• What is positive feedback?
• Positive feedback is when the feedback causes
the corrective measures to stay turned on
• Examples?
• Neurones: influx of sodium ions increases the
permeability of the neurone, causing more sodium ions
to move in, which further increases the permeability etc.
This allows a very fast build-up of action potential to
respond very quickly to a stimulus
Chapter 13 – Control of the oestrous cycle
State the role of each
hormone:
FSH:
Stimulates the development
of follicles in the ovary,
which contain eggs, and
stimulates the follicles to
produce oestrogen.
Fill in the diagram to show how the hormones
interact in the menstrual cycle:
Causes a follicle to
develop
LH:
Causes ovulation to occur,
stimulates the corpus luteum
to produce progesterone
Oestrogen:
Causes the rebuilding of the
uterus lining after
menstruation and stimulates
the pituitary gland to
produce LH.
Progesterone:
Maintains the lining of the
uterus and inhibits
production of FSH from the
pituitary gland.
Causes ovulation and
corpus luteum to develop
Pituitary Gland
FSH
LH
Oestrogen
Progesterone
Ovary
Repairs the lining of the
uterus
Maintains the uterus
lining, ready for a fertilised
egg
Chapter 13 – The Human Menstrual Cycle
Explain the human menstrual cycle
1. The menstrual cycle begins with the shedding of the uterus
lining (Days 1-5)
2. The pituitary gland releases FSH to stimulate follicle growth
(Day 1 onwards)
3. The follicles release low levels of oestrogen. This causes the
build up of the uterus lining. Oestrogen also inhibits FSH and
LH release from the pituitary gland (NEGATIVE FEEDBACK).
4. More oestrogen is released from growing follicles. It reaches a
peak where it now stimulates the pituitary gland to release
more FSH and LH (POSITIVE FEEDBACK) (Day 10).
5. This causes a surge in LH and FSH. The surge in LH causes a
follicle to release its egg – ovulation. (Day 14).
Chapter 13 - The Human Menstrual Cycle
6.
7.
8.
9.
10.
LH now causes the empty follicle to develop into the
corpus luteum. This secretes progesterone and a small
amount of oestrogen.
Progesterone maintains the thick uterus lining and inhibits
FSH and LH release (NEGATIVE FEEDBACK).
If fertilisation does not occur, the corpus luteum
degenerates and stops producing progesterone.
The drop in progesterone stops maintaining the uterus
lining, so it breaks down - menstruation. FSH is also no
longer inhibited.
FSH can be released again and the cycle continues.
Chapter 14 - RNA
• Give a brief overview of how the sequence of bases in DNA
determines the structure of a protein
• The sequence of DNA bases in a gene codes for the
sequence of amino acids in a protein
• Sections of the DNA code are transcribed onto a single
stranded ribonucleic acid (RNA) molecule in the nucleus
(TRANSCRIPTION)
• In eukaryotic cells messenger RNA (mRNA) carries the
genetic information from the nucleus to the cytoplasm
• Proteins are synthesised in the cytoplasm
(TRANSLATION)
• The sequence of bases in mRNA (GENETIC CODE)
determines the sequence of amino acids in a protein
Chapter 14 - RNA
• Explain the main features of the genetic code
Label the structures of RNA:
• Each amino acid is coded for by a sequence of
Phosphate
group
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3 bases on mRNA (codon)
A few amino acids have a single codon
Most amino acids have more than 1 codon (the
code is degenerate)
3 codons are stop codons to mark the end of
the polypeptide chain
The code is non-overlapping (each base is read
only once)
It is a universal code (same codon = same
amino acid in almost all organisms)
Ribose
BASE:
Adenine
BASE:
Cytosine
BASE:
Guanine
BASE:
Uracil
Chapter 14 - RNA
Describe the structure and role of
messenger RNA:
Describe the structure and role of transfer RNA:
Long strand, single helix. Forms a
mirror copy of DNA. Once formed
mRNA leaves the nucleus via nuclear
pores, enters the cytoplasm and
associates with ribosomes.
Small molecule, single stranded chain folded into a
clover-leaf shape. Has a point where the aa
attaches and opposite to that is a sequence of 3
bases that make up the anticodon. This will pair
with the codon on mRNA during protein synthesis.
DNA
mRNA
tRNA
_______
Double polynucleotide
chain
Single polynucleotide
_______
chain
Single polynucleotide
_______
chain
Largest
Size: ________
DNA and tRNA
Size:Between
________________
Smallest
Size: ________
Shape:_____________
Double-helix
Single-helix
Shape:_____________
Clover-shape
Shape:_____________
Pentose sugar:
deoxyribose
___________________
Pentose sugar:
ribose
___________________
Pentose sugar:
ribose
___________________
A, T, C, G
Bases: _________
A, U, C, G
Bases: _________
A, U, C, G
Bases: _________
Chapter 14 - Transcription
• Explain the process of transcription (the process of making pre-
mRNA)
1. The DNA is unwound at a certain point (the start of a gene)
2. DNA Helicase (an enzyme) breaks the hydrogen bonds
between the DNA bases
3. RNA polymerase then moves along the coding (template)
strand of the DNA attaching RNA nucleotides to it by
complementary base pairing (G-C, C-G, T-A, A-U)
4. A new strand is formed - pre-mRNA. DNA reforms its normal
double helix structure behind RNA polymerase as
transcription progresses
5. When a stop codon is reached RNA polymerase detaches
and the completed pre-mRNA is released
Chapter 14 - Splicing
• Explain what splicing is and why it is important
• Both the DNA gene sequence and the pre-mRNA consist
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of exons (code for proteins) and introns (non-coding)
The pre-mRNA is therefore spliced to remove introns
(non-protein coding sequences) and join exons together
The exons can be rejoined in a variety of ways – so a
single section of DNA can code for up to a dozen
polypeptides.
This forms mRNA.
Mutations can affect the splicing of the pre-mRNA and
thus form non-functional proteins
Chapter 14 – Translation (polypeptide synthesis)
Describe the process of translation
1. mRNA passes out of the nuclear pore, into the cytoplasm
2. A ribosome attaches to the starting codon on the mRNA
3. The tRNA with the complementary anticodon moves to the
ribosome and pairs up with the codon on the mRNA. This tRNA
carries an amino acid.
4. A tRNA with the complementary anticodon pairs with the next
codon on the mRNA. This tRNA carries another amino acid.
5. The ribosome moves along the mRNA, bringing together 2 tRNA
molecules at a time.
6. The 2 amino acids on the tRNA join by a peptide bond (requires
an enzyme and ATP)
7. The ribosome moves onto a third codon and the first tRNA
molecule is released and is free to collect another amino acid
from the pool in the cell.
8. The process continues (with more ribosomes passing behind the
first, allowing many identical proteins to be assembled
simultaneously).
9. The synthesis of a polypeptide continues until the ribosome
reaches a stop codon. Then, the ribosome, tRNA and mRNA
molecules all separate and the polypeptide chain is complete.
Chapter 14 – Gene Mutation
Explain what gene mutation is and give some of the
causes of gene mutation:
A gene mutation is any change to one or
more nucleotide bases, or rearrangement
of the bases in DNA.
Causes – random/spontaneous change
during DNA replication, high energy
radiation, chemicals that alter DNA
structure or interfere with transcription.
Explain the process of gene mutation through
deletion of bases:
A nucleotide is lost from the normal DNA
sequence. It makes a completely different
polypeptide (as whole codon sequence
shifted down one).
E.g. TGC AGC TAC G
TCA GCT ACG
Lose the 2nd base:
Explain the process of gene mutation through
substitution of bases:
A nonsense mutation:
Change in base forms a stop codon.
Production of the polypeptide chain is
stopped early. The final protein would be
very different and would not perform its
function.
A mis-sense mutation:
Change in base changes the amino acid
coded for. The polypeptide will differ by
one amino acid. The significance of this
mutation will depend on the precise role of
that amino acid.
A silent mutation:
Change in base still codes for the same
amino acid. This is due to the degenerate
nature of the code.
Chapter 4 – Gene Mutation
• What are the 2 types of genes that control cell division?
• Proto-oncogenes and tumour-suppressor genes
• Explain the role of proto-oncogenes
• To stimulate cell division. Growth factors attach to a receptor
on the cell-surface membrane and via relay proteins in the
cytoplasm ‘switch on’ genes needed for DNA replication
• What can a mutation cause to happen to proto-oncogenes?
• They mutate into oncogenes. This can cause the receptor to
be permanently activated, so cell division is switched on
even without growth factors. Or they can code for a growth
factor to be produced in excessive amounts, stimulating cell
division (tumour forms).
Chapter 14 – Gene Mutation
• Explain the role of tumour-suppressor genes
• To inhibit cell division. They maintain normal rates
of cell division and prevent tumours
• What can a mutation cause to happen to tumoursuppressor genes?
• They may mutate to become inactivated. Cell
division increases. The mutated cells formed from
this are usually different to normal cells. Most
mutated cells die, but those that survive can
clone themselves and form tumours.
Chapter 15 – Totipotency and Cell
Specialisation
• All cells contain the same genes, so how do certain cells
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do certain jobs?
Although all cells contain all genes only certain cells are
expressed (‘switched on’) in any one cell at any one time.
Some genes are permanently switched on/off and others
are turned on and off as needed.
Explain totipotency and cell specialisation
Cells which can mature into any body cell are called
totipotent cells. These later differentiate and become
specialised for their function.
During cell specialisation only some of the genes are
expressed. It would be wasteful to make proteins that are
not needed, so genes coding for these are switched off.
Chapter 15 - Totipotency
Explain totipotency in animal cells
•Mature animals possess very few totipotent cells. These
are called adult stem cells
• Adult stem cells are found in the lining of the small
intestine, skin and bone marrow
• During the very early stages of embryonic development
animals also possess embryonic stem cells
Explain totipotency in plant cells
•In contrast to animals, most mature plants have many
totipotent cells, which retain the ability to differentiate
• Many plant cells can potentially differentiate into all other
types of plant cell
• Hence plants can be cloned by taking cuttings or using
tissue culture techniques
Chapter 15 – Regulation of Transcription and
Translation
Explain how transcription is usually started:
Explain the effect of oestrogen on gene
transcription:
•
Oestrogen can switch ON genes by
starting transcription. Oestrogen is a
lipid-soluble hormone and so passes into
cells by diffusing through the
phospholipid bilayer of the membrane.
• In the cytoplasm oestrogen combines
with a receptor on the transcriptional
factor (complementary shapes)
• By binding here, oestrogen changes
the shape of the receptor, which then
causes the inhibitor molecule to be
released.
• The transcriptional factor can now
enter the nucleus through a nuclear
pore and bind to the DNA, stimulating
transcription.
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The gene needs to be stimulated by
molecules called transcriptional factors.
These move from the cytoplasm into the
nucleus.
Each transcriptional factor has a site that
binds to a specific region of the DNA.
When it binds, it stimulates this region of
DNA to begin the process of transcription.
mRNA is produced and then translated
into a polypeptide.
When a gene is not being expressed, the
site on the transcriptional factor that binds
to the DNA is blocked by an inhibitor
molecule.
This prevents the transcriptional factor
from binding to DNA and so prevents
transcription.
Chapter 15 – Regulation of Transcription and
Translation
Explain the effect of siRNA on gene expression:
siRNA inhibits translation of mRNA and
turns genes OFF.
siRNA (small interfering RNA) is a
small double-stranded section of RNA.
How it works:
• Enzyme 1 breaks up RNA making
siRNA molecules
• Enzyme 2 combines with one of the
two strands of siRNA
• siRNA guides the enzyme to mRNA
by pairing up its bases with
complementary ones on the mRNA
• The enzyme cuts the mRNA into
smaller sections and it can’t be
translated
• The gene has not been expressed.
RNA
Enzyme 1
siRNA
siRNA +
enzyme
siRNA +
siRNA +
enzyme
enzyme
siRNA +
enzyme
joined to
mRNA
Smaller
sections of
mRNA
Chapter 16 – DNA Technology
• What are the 5 stages in the process of making a protein
using the DNA technology of gene transfer?
1. Isolation of the DNA fragments that have the gene for
the protein
2. Insertion of the DNA fragment into a vector
3. Transformation – transfer of the DNA into suitable host
cells
4. Identification of host cells that have successfully taken
up the gene
5. Growth/Cloning of the population of host cells
Chapter 16 – Isolation (Producing DNA Fragments)
Explain how DNA fragments can be produced using
reverse transcriptase:
This method uses an enzyme that
‘works backwards’. It can produce
DNA from mRNA.
1. A cell that readily produces the
protein is selected
2. These cells will contain large
quantities of the relevant mRNA
for that protein, which is extracted
3. If reverse transcriptase is added, it
can make DNA from this RNA.
4. This DNA is known
complementary DNA (cDNA).
5. DNA Polymerase can then
synthesise the other strand of
DNA from free nucleotides.
Explain how DNA fragments can be produced using
restriction endonuclease:
Restriction endonucleases are enzymes
that cut DNA at a specific sequence of
bases (a ‘recognition sequence’)
Some restriction endonucleases cut a
straight line through the DNA leaving
‘blunt ends’
Some restriction endonucleases cut a
staggered fashion through the DNA
leaving ‘sticky ends’
Chapter 16 – In vivo gene cloning: the use
of vectors
• Explain the importance of ‘sticky ends’
• They leave exposed bases – this is due to the staggered
nature of the cut.
• Due to the complementary base-pairing rules of DNA,
sticky ends on one gene, will pair up with sticky ends of
another bit of DNA that has also been cut by the same
restriction endonuclease.
• DNA ligase joins the sugar-phosphate backbones
together
• If two ‘cut’ pieces of DNA are mixed (from two different
organisms), recombinant DNA has been produced.
Chapter 16 – Insertion into a vector
(inserting genes into plasmids)
• Explain how DNA fragments can be inserted into a vector:
• DNA from the cell that makes the
protein is removed
• The DNA and the bacterial plasmid
are cut with the same restriction
endonuclease
• The DNA fragments and open
plasmid are mixed together with
DNA ligase (the complementary
sticky ends pair up)
• The gene is incorporated into the
plasmid – they now have
recombinant DNA.
Chapter 16 – Transformation (introducing
DNA into host cells)
• How are plasmids introduced into bacterial cells?
• The plasmids and bacteria are mixed together in a
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medium containing calcium ions. The Ca2+ ions, and
changes in temperature, make the bacteria permeable
and allow the plasmids to pass through the membrane
into the cytoplasm.
Why won’t all the bacteria possess the DNA fragments?
Only a few bacterial cells will take up the plasmids
Some plasmids will have closed up without taking up the
DNA fragment.
What do we need to identify which bacteria possess the
DNA fragments?
Gene markers
Chapter 16 – Identification of host cells that
have successfully taken up the gene
• How can we use antibiotic resistance to identify those
bacteria that have taken up the plasmid?
• The plasmids contain genes for resistance to antibiotics.
Grow the bacteria on a medium containing the antibiotic.
Those that have not taken up the plasmid will die.
• Why do we also need to use gene markers?
• Because some plasmids will have closed before taking up
the DNA fragment and therefore will be resistant to the
antibiotic but not produce the protein.
Chapter 16 – Identification of host cells
that have successfully taken up the gene
Gene markers can identify if a bacteria has taken up the plasmid. They all involve a
second, separate gene on the plasmid that is easily identifiable.
Explain what Antibiotic-resistance markers are and
how they work:
This is an old technology and has now been
taken over by new methods. A second
antibiotic-resistance gene was cut to
incorporate the DNA fragment. So, if the
bacteria contain a plasmid that has the DNA
fragment, they will no longer be resistant to
the antibiotic. Test this by growing colonies
on agar containing the antibiotic.
Explain what Fluorescent markers are and how
they work:
Transference of a gene from a jellyfish that
produces a green fluorescent protein. Using
restriction enzymes, the gene of interest is
then inserted into the middle of the
fluorescence gene, so the latter cannot be
expressed. Bacteria that have taken up the
plasmid alone will fluoresce under a
microscope BUT those containing the
recombinant plasmid will not.
Explain what enzyme markers are and how they work:
This method involves inserting your gene of interest (e.g. Insulin), into a gene that codes for an
enzyme such as lactase. There is a substrate that is usually colourless, but turns blue when
lactase acts upon it. If you insert your chosen gene into the gene that makes lactase, you will
inactivate the lactase gene.
If you now grow bacterial cells on an agar medium containing the colourless substrate, any
bacteria that have taken up the recombinant plasmid, will form white colonies not blue ones.
Chapter 16 – In vitro gene cloning (PCR)
Explain what the polymerase chain reaction is and what is
required for it:
A method of copying fragments of DNA. It is rapid and
efficient. It requires: the DNA fragment to be copied,
DNA polymerase (enzyme that joins nucleotides
together), primers (short sequences of nucleotides
complementary to those at the end of the DNA),
nucleotides, thermocycler (computer-controlled machine
that varies temperature precisely over a period of time).
Describe the advantages of in vitro gene cloning
It is extremely rapid.
It does not require living cells.
Describe the advantages of in vivo gene cloning
Useful when we want to introduce a gene into
another organism.
Almost no risk of contamination.
Very accurate.
Cuts out specific genes.
Produces bacteria that can produce large quantities
of gene products.
Describe the 3 stages of PCR:
1. Separation of the DNA strand –
DNA fragments, primers and DNA
polymerase are placed in a vessel
at 95°C and the 2 DNA strands
separate.
2. Addition (annealing) of the primers –
The mixture is cooled to 55°C and
the primers join to their
complementary bases on the DNA
fragment. These provide the starting
point for DNA polymerase and
prevent the two strands from
rejoining.
3. Synthesis of DNA – The
temperature is increased to 72°C
(the optimum for DNA polymerase)
and DNA polymerase adds
complementary nucleotides along
each of the separated DNA strands.
This copies the two strands
simultaneously.
Chapter 16 – Use of recombinant DNA
technology
Explain what genetic modification is:
Altering the genetic make-up of organisms
e.g. increasing yield, improving nutrient
content of food, introducing resistance or
tolerance to pesticides/herbicides, making
vaccines, producing medicines.
Give some examples of how microorganisms have
been genetically modified:
Antibiotics – produced bacteria that make
more antibiotics and faster
Hormones – bacterial cells can produce
insulin, human growth hormone, cortisone
and the sex hormones.
Enzymes – bacteria produce enzymes for the
food industry e.g. amylase for beer
production, lipase to improve cheese flavour
Give some examples of how plants have been
genetically modified:
GM tomatoes that prevent a softening
enzyme being produced, herbicide-resistant
crops, disease-resistant crops, pest-resistant
crops.
Give some examples of how animals have been
genetically modified:
Transfer of genes from an animal with natural
resistance to a disease into a different animal.
Introduction of growth-hormone into sheep and
fish.
Production of proteins for human medicine, such
as anti-thrombin
• insert gene for the protein alongside the gene
that codes for protein in goats milk in a
fertilised egg
• Implant into a female goat and then crossbreed kids with the gene
Chapter 16 – Gene Therapy
Explain what cystic fibrosis is:
A genetic disorder caused by a mutant recessive allele where 3 bases (AAA) are missing
(deletion mutation). This removes a single amino acid from the protein coded for and makes the
protein unable to perform its function. Usually the protein transports Cl- ions across epithelial
membranes, with water following by osmosis, keeping epithelial membranes moist. With CF the
epithelial membranes are dry and produce sticky, viscous mucus. This causes mucus
congestion in the lungs, accumulation of thick mucus in pancreatic ducts and sperm ducts.
Explain how gene therapy is used to treat CF:
Two ways:
1. Gene replacement: the defective gene is replaced with a healthy gene
2. Gene supplementation: the healthy gene (CFTR) is added alongside the defective gene.
Two techniques:
1. Germ-line gene therapy: replacing or supplementing the defective gene in the fertilised egg
so that all cells of the organism will develop normally, as will their offspring. Process is
currently prohibited due to moral and ethical issues.
2. Somatic-cell gene therapy: targets the affected tissues and is not passed on to future
generations. As the cells of the lungs are continually dying and being replaced the treatment
needs to be repeated periodically (as often as every few days). It currently has limited
success. Long-term the aim is to target undifferentiated stem cells.
Chapter 16 – Gene Therapy
Somatic-cell gene therapy
• Describe the two ways that cloned normal genes can be introduced into the
cells of the lungs
• Using a harmless virus
• Adenoviruses are made harmless (stop them replicating)
• Adenoviruses are grown in epithelial cells with plasmids that have incorporated the
CFTR gene
• The CFTR gene becomes incorporated into the adenoviruses
• Adenoviruses isolated and purified and introduced into the nostrils of the patient
• The adenoviruses inject their DNA into the epithelial cells of the lungs.
• Wrapping the gene in lipid molecules
• CFTR genes are isolated and inserted into bacterial plasmid vectors (and those taken
it up are identified and clone)
• Plasmids are extracted from the bacteria and wrapped in lipid molecules to form a
liposome
• The liposomes are sprayed into the nostrils of the patient and enter the epithelial cells
via the phospholipid bilayer
• Why are these methods not always effective?
• Adenoviruses may cause infections
• Patients may develop immunity to adenoviruses
• Liposome may not be fine enough to pass through the tiny bronchioles
• Even if the CFTR gene gets into epithelial cells, very few are expressed
Chapter 16 – Gene Therapy
• Explain how gene therapy can be used to treat severe
combined immunodeficiency
• SCID is a rare inherited disorder where people do not
produce antibodies or show a cell-mediated response
• The gene that codes for the enzyme ADA is defective
• Gene therapy:
• The normal ADA gene is isolated and inserted into retroviruses
which are then grown
• The retroviruses are mixed with the patients T cells and they inject
a copy of the normal ADA gene into T cells
• The T cells are reintroduced into the patient’s blood
• T cells live for only 6-12 months, so the treatment has to be
repeated
Chapter 16 – Locating and Sequencing
Genes
Explain how DNA probes work:
Explain what DNA sequencing is:
A DNA probe is a short, singlestranded section of DNA that has
a label attached to make it easily
identifiable (radioactively labelled
probes or fluorescently labelled
probes).
DNA probes are used to identify
genes:
- A DNA probe is made that
has bases that are
complementary to the DNA
sequence that makes up part
of the gene we want to find
- The test DNA is separated
into its two strands
- The DNA strands mix with the
probe, which binds to the
complementary bases (DNA
hybridisation)
- The site at which the probe
binds can be identified
The Sanger method uses modified nucleotides that can’t attach to the
next base in the sequence when joined. Set up 4 test tubes, each with:
1. Many single-stranded DNA fragments of the test DNA
2. A mixture of nucelotides
3. A small quantity of 1 of 4 terminator nucleotides (test tube 1 = A
terminator etc)
4. A primer to start the process of DNA synthesis (labelled)
5. DNA polymerase
Depending upon where the terminator nucleotide binds DNA synthesis
may be terminated early or late. So, the DNA fragments in each tube
will vary in length. They will all end in the same base (test tube 1 = A
etc). Can be identified by the primer. Now separate out these fragments
by gel electrophoresis.
Gel Electrophoresis
Place the DNA fragments onto an agar gel and apply a voltage across
it. The larger fragments move more slowly. This separates DNA
fragments of different lengths. Can only be done with DNA fragments
up to 500 bases. Larger genes need to be cut using restriction
endonucleases and each fragment sequenced. They then need to be
pieced back together to recreate the whole gene. Use restriction
mapping.
Chapter 16 – Locating and Sequencing
Genes
• Explain what restriction mapping is:
Cutting DNA with a series of different restriction endonucleases. The
fragments are separated by gel electrophoresis.
You can use more than one restriction endonuclease to cut in different places
and get different lengths of fragments.
It is now automated and computers analyse the data. Electrophoresis is
carried out in a single narrow gel.
Chapter 16 – Screening for clinically
important genes
Explain how genetic screening works:
1. The order of nucleotides on the mutated gene is
determined by DNA sequencing
2. A fragment of DNA with complementary bases
to the mutated sequence is produced
3. DNA probe is formed by radioactively labelling
the DNA fragment
4. PCR is used to produce lots of copies of the
probe
5. Probe is added to single stranded DNA
fragments from the test person
6. If the person has mutated DNA the probe will
bind to the complementary bases on the DNA
7. These DNA fragements are now labelled and
can be distinguished by the use of X-ray film
8. If the DNA probe has bound then the X-ray film
will be exposed
9. If the DNA probe has not bound then the X-ray
film will not be exposed
Explain the role genetic counselling plays in
genetic screening:
Genetic screening goes hand in hand with
genetic counselling.
Expert advice provided by a counsellor
helps individuals to understand the results
and implications of the screening and so
make appropriate decisions.
The family history of an inherited disease
is researched and couples are advised on
the likelihood of it arising in their children.
A counsellor can inform a couple of the
effects of a disorder and its emotional,
psychological, medical and economic
consequences.
It can make couples aware of any further
medical tests that give a more accurate
prediction of whether the children will have
the condition.
Chapter 16 – Genetic fingerprinting
Explain the process of DNA fingerprinting:
Extraction:
Explain how you would interpret the results
of DNA fingerprinting:
Extract the DNA from the rest of the cell and
increase its quantity by PCR
Compare DNA fingerprints e.g. from a
crime scene and a suspect visually and
then through an automated scanner
machine. The closer the match between
the 2 patterns, the more likely the DNA
is from the same person.
Digestion:
Cut the DNA into fragments by restriction
endonucleases
Separation:
Use gel electrophoresis then immerse the gel
in alkali (separate double into single strands)
and transfer the strands on to a nylon
membrane (Southern Blotting).
Hybridisation:
Radioactive probes bind to key sequences.
Development:
Put an X-ray film over the nylon membrane
and the film is exposed by the radioactivity.
Explain some of the uses of DNA
fingerprinting:
Forensic science – connect individuals
to crimes
Paternity tests
Determining genetic variability within a
population