unit 5 revision
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Unit 5 Revision. Chapter 9 – Responses to Stimuli. Fill in the blanks to show your understanding of stimulus and response: Stimulus __________ __________ __________ response. Write a definition for tropism: List the different types of tropism: . . . - PowerPoint PPT PresentationTRANSCRIPT
UNIT 5 REVISION
Chapter 9 – Responses to StimuliFill in the blanks to show your understanding of stimulus and response:
Stimulus __________ __________ __________ response
Write a definition for kineses:
Give an example of kinesis:
Write a definition for taxes:
Explain the difference between: •positive taxes:
•negative taxes:
Write a definition for tropism:
List the different types of tropism:•.•.•.
Explain the difference between: •positive tropism:
•negative tropism:
receptor co-ordinator effector
A directional response to a stimulus
Movement is towards the stimulus e.g. algae move towards light
Movement is away from the stimulus e.g. earthworms move away from light
Random movement in response to a stimulus. The more unpleasant the stimulus the faster the organism moves.
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.
Growth movement of a plant in response to a directional stimulus
PhototropismGeotropismHydrotropism
Growth of the plant towards the stimulus.
Growth of the plant away from the stimulus.
Label the diagram to show how the nerves are arranged in the spinal cord:
Chapter 9 – Nervous ControlComplete the diagram to show the organisation of the nervous system:
Nervous system
Brain
Voluntary nervous system
Motor nervous system
Peripheral Nervous System
Central Nervous System
Spinal Cord
Sensory Nervous System
Autonomic Nervous System
Sensory neurone
Relay (Intermediate) neurone
Motor neurone
Contains sensory neurones
Contains Motor neurones
Spinal nerve
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
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 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
Cardiac Accelerator Centre
Decreases heart rate Increases heart rate
Parasympathetic nerve linked to
SAN
Sympathetic nerve linked to
SAN
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
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 - ReceptorsDescribe 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 stretch-mediated 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.
Describe the structure and function of rod and cone cells in the retina:
Chapter 9 - Receptors
Rod Cone
Shape
Number
Distribution
Visual acuity
Sensitivity in low light conditions
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)
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.
Rod-shaped Cone-shapedGreater numbers than cone cells
Fewer numbers than rod cells
Poor Good
Sensitive Not sensitive
More at the edge, none at the fovea
Fewer at the edge, most at the fovea
Chapter 10 – Co-ordinationHormonal System Nervous System
Method of communication?
Method of transmission?
Speed of transmission?
Where does the communication travel to?
How widespread is the response?
Speed of the response?
Length of response?
Length and reversibility of effect?
Chemicals called hormones
Nerve impulses
Blood system Neurones
Relatively slow Very rapid
All parts of the body, only target organs respond
Specific parts of the body
Widespread Localised
Slow Rapid
Often long-lasting Short-lived
Effect may be permanent and irreversible
Effect is temporary and reversible
Chapter 10 – Co-ordination• What are chemical mediators?• Chemicals released from certain mammalian cells that affect cells in
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 - NeuronesDescribe the structure and function of the parts of a neurone:
Axon
Axon – Single, long fibre that carries nerve impulses away from the cell body.
Nucleus
Nucleolus
Cell Body
Cell Body – contains the nucleus and RER. Produces proteins and neurotransmitters.
Dendrites
Dendrites
Dendrites – branched fibres that carry nerve impulses towards the cell body.
Myelin Sheath
Myelin Sheath
Myelin Sheath – covers the axon and is made up of the lipid-rich membranes of the Schwann cells. Speeds up transmission of nerve impulses.
Schwann Cells Schwann Cells – surround the axon, protect it and provide electrical insulation.
Node of Ranvier
Node of Ranvier – gaps between adjacent Schwann cells where there is no myelin sheath.
Chapter 10 – The Nerve ImpulseExplain 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 ImpulseUsing the graph, explain what happens when an action potential passes down a neurone.
1. Resting potential (-65mV), sodium voltage-gated 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.
Explain, using diagrams, how an action potential moves along an unmyelinated axon:
Chapter 10 – Passage of an Action Potential
Explain, using diagrams, how an action potential moves along a myelinated 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 (+-)
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
(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
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
• Explain the difference between actin and myosin
• They are both protein filaments• Actin is thinner and has 2 strands twisted
around each other• Myosin is thicker and has long rod-shaped
fibres with bulbous heads
Describe the zones found in a sarcomere• Light bands are I-bands only
actin is found in these bands• Dark bands are A-bands actin
and myosin overlap in these bands
• In the middle of each A-band is a lighter part called the H-zone
• In the centre of each I-band is the Z-line, where the actin filaments join
• The section of muscle between Z-lines is called a sarcomere
• Pattern = Z I A H A I Z
--I band-- ----A band-----
H zone
Z line
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
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
• What is the sliding filament 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
Chapter 11 – Contraction of Skeletal Muscle
• What are the 3 main proteins involved?
1. 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
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 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
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 - HomeostasisExplain what homeostasis is: Input
• Changes to the system
Receptor• Mea
sures level of a factor
Control unit• Oper
ational information is stored here and used to coordinate effectors
Effector• Bring
s about changes to the system in order to return it to a set point
Output• Syst
em returned to set point
Explain why homeostasis is important:
State some of the factors that are controlled by homeostasis:
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.
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.
Temperature, pH, water potential, tissue fluid
Chapter 12 – Regulation of Body Temperature
State the main ways in which heat is gained by organisms:
State the main ways in which heat is lost by organisms:
Explain how body temperature if regulated in ectotherms:
Explain how body temperature is regulated in endotherms:
Complete the feedback loop for controlling body temperature in mammals:
HypothalamusHeat gain centre Heat loss centre
Normal body temp
Cold receptors in skin
Warm receptors in skin
Vasodilation, sweating, hair lowered, lower metabolic rate
Vasoconstriction, shivering, hair raised, higher metabolic rate
Production of heat (higher metabolic rate), gain of heat from environment (by CCR)
Evaporation of water, loss of heat to the environment (by CCR)
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
Gain heat = vasoconstriction, shivering, raising hair, higher metabolic rate, behaviourLose heat = vasodilation, sweating, lowering hair, behaviour
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-
surface membranes which have a complementary shape to the hormone• Effective in very small quantities, but often have widespread and long-
lasting 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 GlucoseExplain the role of the pancreas in regulating blood glucose:
State the sources of blood glucose:
Explain the role of insulin 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.
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.
If the β cells detect a rise in blood glucose then they secrete insulin into the blood. Insulin combines with receptors on cells and causes: - Opening of glucose transport protein channels (glucose enters cells)- Activation of enzymes that convert glucose to glycogen and fat
Explain the role of glucagon in regulating blood glucose:
Explain the role of adrenaline in regulating blood glucose:
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)
At times of excitement or stress adrenaline is produced by the adrenal glands and raises the blood glucose level by:- Activating an enzyme that causes
breakdown of glycogen to glucose in the liver (glycogenolysis)
- Inactivating an enzyme that synthesises glycogen from glucose.
Chapter 12 – Regulation of Blood GlucoseComplete the feedback loop for controlling body temperature in mammals:
Normal blood glucose level90mg100cm-3 blood
To increase To decrease
Blood glucose Blood glucose
Detected by… Detected by…
Blood glucose Blood glucose
Response…Response…
falls
α cells, which release glucagon
Conversion of glycogen to glucose, conversion of amino acids to glucose
rises, negative feedback
rises
β cells, which release insulin
Conversion of glucose to glycogen, conversion of glucose to fat, absorption of glucose into cells, more respiration
falls, negative feedback
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 12 - Diabetes
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.
ReceptorsDetect the change
Control CentreCoordination
EffectorHave an effect on the system
Output Rise in some parameter
Input Fall in some parameter
Negative Feedback = Corrective measures turned off
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
Fill in the diagram to show how the hormones interact in the menstrual cycle:
State the role of each hormone:FSH:
LH:
Oestrogen:
Progesterone:
Stimulates the development of follicles in the ovary, which contain eggs, and stimulates the follicles to produce oestrogen.
Causes ovulation to occur, stimulates the corpus luteum to produce progesterone
Causes the rebuilding of the uterus lining after menstruation and stimulates the pituitary gland to produce LH.
Maintains the lining of the uterus and inhibits production of FSH from the pituitary gland.
Ovary
Pituitary GlandFSH
Causes a follicle to develop
LH
Causes ovulation and corpus luteum to develop
Oestrogen
Repairs the lining of the uterus
Progesterone
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 Cycle6. LH now causes the empty follicle to develop into the
corpus luteum. This secretes progesterone and a small amount of oestrogen.
7. Progesterone maintains the thick uterus lining and inhibits FSH and LH release (NEGATIVE FEEDBACK).
8. If fertilisation does not occur, the corpus luteum degenerates and stops producing progesterone.
9. The drop in progesterone stops maintaining the uterus lining, so it breaks down - menstruation. FSH is also no longer inhibited.
10. 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• Each amino acid is coded for by a sequence of
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)
Label the structures of RNA:
Ribose
Phosphate group BASE:
Adenine
BASE: Cytosine
BASE: Guanine
BASE: Uracil
Chapter 14 - RNADescribe 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
_______ polynucleotide chain
_______ polynucleotide chain
_______ polynucleotide chain
Size: ________ Size: ________________ Size: ________
Shape:_____________ Shape:_____________ Shape:_____________
Pentose sugar: ___________________
Pentose sugar: ___________________
Pentose sugar: ___________________
Bases: _________ Bases: _________ Bases: _________
Double Single Single
Largest Between DNA and tRNA Smallest
Double-helix Single-helix Clover-shape
deoxyribose ribose ribose
A, T, C, G A, U, C, G A, U, C, G
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
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 MutationExplain what gene mutation is and give some of the causes of gene mutation:
Explain the process of gene mutation through substitution of bases:
A nonsense mutation:
A mis-sense mutation:
A silent mutation:
Explain the process of gene mutation through deletion of bases:
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.
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 Lose the 2nd base: TCA GCT ACG
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.
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.
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 tumour-suppressor 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 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.
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 - Totipotency
Chapter 15 – Regulation of Transcription and Translation
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.
Explain how transcription is usually started:
• 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.
RNAEnzyme 1
siRNA + enzyme
siRNA + enzymesiRNA +
enzyme
siRNA + enzyme joined to mRNA
Smaller sections of
mRNA
siRNA
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:
Explain how DNA fragments can be produced using restriction endonuclease:
This method uses an enzyme that ‘works backwards’. It can produce DNA from mRNA. 1. A cell that readily produces the
protein is selected2. 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.
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 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 geneGene 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 Fluorescent markers are and how they work:
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.
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 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.
Describe the advantages of in vitro gene cloning
Describe the advantages of in vivo gene cloning
It is extremely rapid.It does not require living cells.
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.
Chapter 16 – Use of recombinant DNA technology
Explain what genetic modification is:
Give some examples of how microorganisms have been genetically modified:
Give some examples of how animals have been genetically modified:
Give some examples of how plants have been genetically modified:
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.
Antibiotics – produced bacteria that make more antibiotics and fasterHormones – 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
GM tomatoes that prevent a softening enzyme being produced, herbicide-resistant crops, disease-resistant crops, pest-resistant crops.
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 cross-breed kids with the gene
Chapter 16 – Gene TherapyExplain what cystic fibrosis is:
Explain how gene therapy is used to treat CF:
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.
Two ways:1. Gene replacement: the defective gene is replaced with a healthy gene2. Gene supplementation: the healthy gene (CFTR) is added alongside the defective gene.Two techniques:3. 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.
4. 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 TherapySomatic-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, single-stranded 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 DNA2. A mixture of nucelotides3. 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 polymeraseDepending 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 ElectrophoresisPlace 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: Explain the role genetic counselling plays in genetic screening:
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
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 fingerprintingExplain the process of DNA fingerprinting:Extraction:
Digestion:
Separation:
Hybridisation:
Development:
Explain how you would interpret the results of DNA fingerprinting:
Explain some of the uses of DNA fingerprinting:
Extract the DNA from the rest of the cell and increase its quantity by PCR
Cut the DNA into fragments by restriction endonucleases
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).
Radioactive probes bind to key sequences.
Put an X-ray film over the nylon membrane and the film is exposed by the radioactivity.
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.
Forensic science – connect individuals to crimesPaternity testsDetermining genetic variability within a population