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Survival and response Explain the importance of reflex actions. 1. Automatic (adjustments to changes in environment)/ involuntary; 2. Reducing/avoiding damage to tissues / prevents injury/named injury e.g. burning; 3. Role in homeostasis/example; 4. Posture/balance; 5. Finding/obtaining food/mate/suitable conditions; 6. Escape from predators; Taxes A taxis (plural taxes) is a directional response to a directional stimulus. Taxes are common in invertebrates, and even motile bacteria and protoctists show taxes. The taxis response can be positive (towards the stimulus) or negative (away from the stimulus). Common stimuli include: • light (phototaxis), e.g. fly larvae (maggots) use negative phototaxis to move away from light to avoid exposure and desiccation. Adult flies use positive phototaxis to fly towards light to warm up. • gravity (geotaxis), e.g. earthworms show positive geotaxis to burrow downwards underground. • chemicals (chemotaxis) e.g. male moths show positive chemotaxis when flying towards a pheromone. • Movement (rheotaxis) e.g. moths use positive rheotaxis to fly into the wind and salmon use positive rheotaxis to swim upstream. Kineses A kinesis (plural kineses) is a response to a changing stimulus by changing the amount of activity. Sometimes the speed of movement varies with the intensity of the stimulus (orthokinesis) and sometimes the rate of turning depends on the intensity of the stimulus (klinokinesis). The response is not directional, so kinesis is a suitable response when the stimulus isn’t particularly localised. The end result is to keep the animal in a favourable environment. For example woodlice, who breathe using gills, use kinesis to stay in damp environments, where they won’t dry out. In dry environments woodlice move quickly and don’t turn much, which increases their chance of moving out of that area. If, by chance, they find themselves in a humid environment, they slow down and increase their rate of turning. This tends to keep them in the humid area.

Unit 5 Summary Notes A2 Biology

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Page 1: Unit 5 Summary Notes A2 Biology

Survival and responseExplain the importance of reflex actions.1. Automatic (adjustments to changes in environment)/ involuntary;2. Reducing/avoiding damage to tissues / prevents injury/named injury e.g. burning;3. Role in homeostasis/example;4. Posture/balance;5. Finding/obtaining food/mate/suitable conditions;6. Escape from predators;

TaxesA taxis (plural taxes) is a directional response to a directional stimulus. Taxes are common in invertebrates, and even motile bacteria and protoctists show taxes. The taxis response can be positive (towards the stimulus) or negative (away from the stimulus). Common stimuli include:• light (phototaxis), e.g. fly larvae (maggots) use negative phototaxis to move away from light to avoid exposure and desiccation. Adult flies use positive phototaxis to fly towards light to warm up.• gravity (geotaxis), e.g. earthworms show positive geotaxis to burrow downwards underground.• chemicals (chemotaxis) e.g. male moths show positive chemotaxis when flying towards a pheromone.• Movement (rheotaxis) e.g. moths use positive rheotaxis to fly into the wind and salmon use positive rheotaxis to swim upstream.

KinesesA kinesis (plural kineses) is a response to a changing stimulus by changing the amount of activity.Sometimes the speed of movement varies with the intensity of the stimulus (orthokinesis) and sometimes the rate of turning depends on the intensity of the stimulus (klinokinesis). The response is not directional, so kinesis is a suitable response when the stimulus isn’t particularly localised. The end result is to keep the animal in a favourable environment. For example woodlice, who breathe using gills, use kinesis to stay in damp environments, where they won’t dry out. In dry environments woodlice move quickly and don’t turn much, which increases their chance of moving out of that area. If, by chance, they find themselves in a humid environment, they slow down and increase their rate of turning. This tends to keep them in the humid area.

Page 2: Unit 5 Summary Notes A2 Biology

Plants can sense and respond to stimuli. Many of these responses are directional growth responses, called tropisms. Tropisms can be positive (growing towards the stimulus) or negative (growing away from the stimulus) and occur in response to a variety of stimuli:

From these experiments (1-3) the Darwins concluded that light is detected only at the shoot tip, and an “influence” was transmitted from the tip down the shoot to cause bending further down. They had no idea what this “influence” was.

Frits Went (4-6) showed that the “influence” was a chemical. He knew that seedlings with their tips cut off would not grow, while seedlings with an intact tip would. So he cut off the tips off growing seedlings, placed them on small blocks of agar for two hours, and then placed the agar blocks on top of cut seedlings in the dark. Agar jelly allows chemicals to diffuse through but contains no living cells, and a control experiment using agar blocks that had not been in contact with shoot tips did not promote growth. Went concluded that a chemical substance had diffused from the shoot tip into the agar, and that this substance stimulated growth further down the shoot. He called this substance auxin.

In the 1960s Winslow Briggs (7-9) used Went’s method (experiment 6) to assay the amount of auxin in plant material. He found that the greater the amount of auxin, the greater the bending. In the experiments below the numbers refer to the angle of bending and therefore to the amount of auxin.

These discoveries are summarised in the Cholodny-Went theory, which states that auxin is synthesized in the coleoptile tip; asymmetric illumination is detected by the coleoptile tip and this causes auxin to move into the darker side; auxin diffuses down the coleoptile; and the higher auxin concentration on the darker side causes the coleoptile to bend toward the light source. Although there is a lot of evidence supporting this theory, it is by no means certain and some recent studies using radioactive tracers have found no difference in IAA concentration on the dark and light sides of a shoot. An alternative mechanism is that IAA is present on both sides but is somehow inhibited on the light side, so there is little growth.

How does auxin work?In 1934 auxin was identified as a compound called indoleacetic acid, or IAA. It was the first of a group of substances controlling plant growth responses called plant growth regulators, or PGRs. PGRs are a bit like animal hormones, but the term hormones is not used for plants because PGRs are not made in glands and do not travel in blood. IAA is hydrophobic so it can diffuse through cell membranes and so move around the plant.IAA stimulates growth by:1. Binding to a receptor protein in the target cell membranes and activating a proton pump.2. This pump pumps protons (hydrogen ions) from the cytoplasm of these cells to their cell walls.3. The resulting decrease in pH activates an enzyme that breaks the bonds between cellulose microfibrils.4. This loosens the cell wall and so allows the cell to elongate under the internal turgor pressure.

Page 3: Unit 5 Summary Notes A2 Biology

Control of heart rate

Describe how the regular contraction of the atria and ventricles is initiated and coordinated by the heart itself(cardiac) muscle is myogenic;sinoatrial node/SAN;wave of depolarisation/impulses/electrical activity (across atria);initiates contraction of atriaatrioventricular node/AVN;bundle of His/purkyne tissue spreads impulse across ventricles;ventricles contract after atria/time delay enables ventricles to fill;

Describe how the rate of heartbeat is increased as muscle activity increases during exerciseIncrease in carbon dioxide / hydrogen ions;detection by / stimulation of chemoreceptors;pressure receptors detect changes in blood pressure;(receptors) in aorta / carotid arteries / medulla;(cardio) acceleratory centre (in medulla) / cardiovascular centre;impulses via sympathetic nerves/system;to SAN;change in rate of impulse production by SAN;

Explain why increased cardiac output is an advantage during exerciseIn exercise – More energy release / more respiration / activelyrespiring muscles / for aerobic respiration;Higher cardiac output – Increases O2 supply (to muscles);Increases glucose supply (to muscles);Increases CO2 removal (from muscles) /lactate removal;Increases heat removal (from muscles) /for cooling

Describe the role of the nervous system in modifying the heart rate in response to an increase in blood pressure.Pressure receptors;in aorta/carotid artery/sinus;send impulses (award once only);to medulla;send impulses (award once only);along parasympathetic / vagus pathway;slows heart rate;

Explain the effect of the parasympathetic division of the autonomic nervous system on cardiac output. (4)Impulses to SA node;Along (branch of) vagus nerve;Acetylcholine;Decreases activity of SA node/equivalent;Decreases rate of contraction/decreases heart rate/heartbeat;

Heart rate increases during exercise. Describe the part played by chemoreceptors and the medulla in increasing heart rate.Chemoreceptors(Located in) carotid/aortic bodies/medulla;detect high carbon dioxide levels/ H+/low pH;due to increased respiration;Medulla More impulses;(from medulla) to SAN;via sympathetic nervous system;(accept converse reference to parasympathetic)(send nervous) impulse to medulla / respiratory centre;

Page 4: Unit 5 Summary Notes A2 Biology

Explain how nervous control in a human can cause increased cardiac output during exercise.1. Coordination via medulla (of brain) / cardiac centre;2. (Increased) impulses along sympathetic (/ cardiac accelerator) nerve;3. To S.A. node / pacemaker;4. Release of noradrenalin;5. More impulses sent from / increased rate of discharge of S.A. node /pacemaker;Not “beats”; not “speeds up”6. Increased heart rate / increased stroke volume; Explain the effect of the parasympathetic division of the autonomic nervous system on cardiac output. (4)Impulses to SA node;Along (branch of) vagus nerve;Acetylcholine;Decreases activity of SA node/equivalent;Decreases rate of contraction/decreases heart rate/heartbeat;

Describe the role of the nervous system in modifying the heart rate in response to an increase in blood pressure.Pressure receptors;in aorta/carotid artery/sinus;send impulses (award once only);to medulla;send impulses (award once only);along parasympathetic / vagus pathway;slows heart rate;

Page 5: Unit 5 Summary Notes A2 Biology

Receptors

Pacinian corpuscles are situated deep in the skin, so are only sensitive to intense pressure, not light touch. Pressure distorts the neurone cell membrane, and opens stretch mediated sodium channels. This allows sodium ions to diffuse in, causing a local depolarisation, called the generator (or receptor) potential. The stronger the pressure, the greater the generator potential, until it reaches a threshold, when an action potential is triggered.

When pressure is applied to a Pacinian corpuscle, an impulse is produced in its sensory neurone. Explain how.(Pressure) deforms and opens (sodium) channelsEntry of sodium ions;Causes depolarisation (generator potential)Ions diffuse downstream and when threshold of nearby voltage gated channels is reached they open and sodium diffuses in causing depolarisation

Explain how the structure of the retina and its neuronal connections enable a person to have(i) a high degree of visual sensitivity in low light levels;Rod cells (responsible for sensitivity);Several rods connected to each bipolar cell;Additive effect of small amount of light striking several rod cells;creating a large enough depolarisation to generate an

Several rod cells to each neurone/bipolar cell;additive effect of light striking several rod cells;

(ii) a high degree of visual acuity.Cone cells (responsible for acuity);Each cone cell connected to an individual neurone;idea of light striking each individual cone cell to generate a separate action potential / impulse;very small area of retina stimulated, so very accurate vision;

Each cone is connected to a specific neurone;light striking cone cells generating separate action potentials;

G an g lio n ce lls

B ip o la r c e lls

R o d c e ll

C o n e ce ll

Page 6: Unit 5 Summary Notes A2 Biology

Nerve Impulses

Describe how the resting potential is established in an axon by the movement of ions across the membrane.

active transport / pump of Na+

out of axon;

diffusion of K+ out of axon / little diffusion of

Na+ into the axon;membrane more permeable to loss of potassium ions;limits entry of sodium ions;negatively charged proteins inside;sodium pump;

membrane relatively impermeable / less permeable to sodium ions / gated channels are closed / fewer channels;sodium ions pumped / actively transported out;by sodium ion carrier / intrinsic proteins; higher concentration of sodium ions outside the neurone;inside negative compared to outside / 3 sodium ions out for two potassium ions in;

Sodium and potassium ions can only cross the axon membrane through proteins.Explain why.cannot pass through phospholipid bilayer;because water soluble / not lipid soluble / charged / hydrophilic / hydrated

Describe the structure of the channels that allow sodium ions through membranes.Protein (molecules);transcending phospholipid bilayer / intrinsic / transmembrane

Explain what is meant by depolarisationInflow of sodium ions;So inside (more) positive (or outside more negative)/ changein membrane potential.

A change in potential difference will affect the gate on the channel protein and produce an action potential. Explain how the action potential is produced.Gate opens;Allowing entry of sodium ions;Brings about depolarisation/inside of neurone becomes positivewith respect to outside

Describe what happens to the sodium ions immediately after the passage of an action potential along the neurone.Sodium ions move out;By active transport/pump;

What is meant by the ‘all or nothing’ nature of a nerve impulse?All action potentials are the same size;threshold value for action potential to occur

Page 7: Unit 5 Summary Notes A2 Biology

Describe the events that take place in a neurone which produce an action potential (6)1 Stimulus to threshold / critical firing level;2 Sodium channels/gates open;3 Sodium ions enter;4 Down electrical/chemical gradient;5 Positive feedback;6 Depolarisation;7 Inside becomes positive / membrane potential reverses;8 Potassium channels/gates open;9 Potassium ions leave;10 Down electrical/chemical gradient [Note: only credit if not awarded earlier in point 4]11 Repolarisation;12 Sodium channels/gates close;13 Undershoot / hyperpolarisation;14 Sodium-potassium pump restores resting potential;

An action potential is produced in neurone A. Describe how this action potential passes along the neurone. (3)Any three from :(Depolarisation of axon membrane causes) local currents to be set up;

Change permeability (of adjoining region) to Na+

/open Na+

gates(in adjoining region);sodium ions enter adjoining region;adjoining region depolarises;

Neurones can respond to both strong and weak stimuli. Describe how a neurone conveys information about the strength of a stimulusFrequency of action potentials

What is meant by the refractory period?No (new) action potential/nerve impulse be produced in this time;

Explain what causes the conduction of impulses along a non-myelinated axon to be slower than along a myelinated axon.(3)non-myelinated – next section of membrane depolarised / whole membrane;myelinated – depolarisation / ion movement only at nodes;impulse jumps from node to node /saltatory conduction;

Fewer action potentials occur along a myelinated axon than along an unmyelinated axon of the same length. Explain why.Myelin insulates axon / ions can only pass through (plasma membraneof axon) at gaps in myelin sheath;(Gaps in sheath are called) nodes of Ranvier;

The rate of oxygen consumption of a neurone increases when it conducts a high frequency of impulses. Explain why.(4)Oxygen used in respiration; [Reject: Anaerobic reference] Valid reference to ATP/energy; [Reject: Production of energy] (For) sodium-potassium pump/ active transport of ions/ uptake/ synthesis of transmitter/ vesicle movement;(Higher rate of impulses means) more high / amount of sodium ion entry/potassium ion loss / transmitter uptake / release / vesicle movement;

Explain what causes transmission at a synapse to occur in only one direction.(Vesicles containing) neurotransmitter only in presynaptic membrane/neurone;receptor/proteins only in postsynaptic membrane/neurone;so neurotransmitter diffuses down concentration gradient;

Page 8: Unit 5 Summary Notes A2 Biology

Explain why the transmission of a series of nerve impulses along a myelinated neurone uses less energy than transmission along a non myelinated neurone A. (3)In a myelinated neuroneCorrect reference to saltatory conduction/description;Active transport of ions/ion pumps “only” used/less active transport of ions at nodes of Ranvier;Less respiration needed / less ATP needed;For repolarisation/restoration of ion balance;

Describe how transmission occurs across a synapse.1 Presynaptic membrane depolarises;2 Calcium channels/gates open;3 Calcium ions enter;4 Vesicles move to/fuse with presynaptic membrane;5 Release of transmitter / exocytosis;6 Diffusion across gap/cleft;7 Binds to receptors in postsynaptic membrane; [Reject: references to active site]8 Sodium channels open / sodium ions enter;

Describe the sequence of events which leads to the transmission of an impulse at a cholinergic synapse.calcium ions move into synaptic knobs / presynaptic membrane;causing synaptic vesicles to move;towards presynaptic membrane;where they release acetylcholine into gap;transmitter/acetylcholine diffuses across gap;binds onto receptor / protein molecules;on postsynaptic membrane;causing depolarisation / opening of sodium gates / action potential in postsynaptic cell membrane;

Explain what is meant by the tertiary structure of a protein and describe the importance of this in transmission across a synapse. (5)1 Polypeptide (chain) folds;2 Named bond; [Reject: peptide bond]3 Between R groups;4 Receptors/binding sites are proteins;5 Reference to neurotransmitter shape;6 Acetylcholinesterase/breakdown enzyme, is protein;7 Carrier/channel protein;8 Protein has a shape;9 Idea of complementary/fit/bind/attach to; [Note: in correct context]

Page 9: Unit 5 Summary Notes A2 Biology

MusclesDescribe the role of calcium in the contraction of a muscleMoves / detaches / changes position of / shape of / switch protein / blockingmolecule / tropomyosin / troponin;(Not just ‘switches on’)uncovering binding site (on actin) / allows cross bridges to form / eq;activates myosin ATP–ase / enables myosin head to split ATP;

Describe the role of ATP in mucle contractionATP provides energy for release / attachment / movement of myosin (head) (from binding site) / removal of calcium ions;

What is the role of phosphocreatine in muscle contraction? Allows the regeneration of ATP with out respiration. Produces ATP; rrelease Pi to join with ADP

At a neuromuscular junction there is a 0.6ms delay between maximum depolaristion in the presynaptic neurone and maximum depolarisation in the post synaptic neurone, explain why.

Calcium ion/Ca2+

entry;vesicles fuse with preSM ( and rupture);exocytosis of/release (neuro)transmitter substance / named e.g.;diffuse across gap;attach to receptors on post SM; (not “fuse with....”)increase permeability to sodium (ions) open Na channels/ref. e.p.s.p.;

Explain the advantage of having both fast and slow twitch fibresFast fibres make immediate/fast contraction possible before the circulation/blood supply adjusts/ most energy anaerobically generated;fast fibres used in explosive/sprints locomotion;slow fibres allow sustained contraction/anaerobic energy generation;slow fibres used in maintaining posture/endurance events;(answers which combine features of both,without specifying which muscle type provides which benefit: max 1)

When a muscle contracts what happens to the A band, I band and H zone respectively. A-band: no change AND I-band: shorter;H-zone: shorter / disappears;

Describe the role of ATP calcium and CP in muscle contractionCalcium ions bind to troponin;Remove blocking action of tropomyosin / exposes actin binding sites;ATP allows myosin to join / bind to actin / form cross-bridge;‘Re-cocks’ myosin cross bridge / allows detachment from actin;Enables calcium ions to be pumped back in;Phosphocreatine allows regeneration of ATP without respiration;Phosphocreatine releases Pi to join ADP;

Describe how having a large number of slow twitch fibres helps and endurance athlete.Endurance athletes exercise for long periods of time;Respire / release energy aerobically;Or too much lactate would accumulate;Slow twitch fibres adapted to aerobic metabolism;As have many mitochondria;Site of Krebs’ cycle;And electron transport chain;Much ATP formed;Also are resistant to fatigue;

Page 10: Unit 5 Summary Notes A2 Biology

A

B

Vesic les co n ta in in gac e ty lc h o lin e

A ce ty lc ho lin ere cep to rs

A cety lch o line s te rase

A x o n o f m o to r n eu ro n e

M e m b ra n e ofm u sc le ce ll

M y a sth en icN o rm a l

Name the main protein found in structure BActin/tropomyosin

Name the structure in box AMyosin head

An action potential is generated at the cell body of the motor neurone. Explain how this action potential passes along the motor neurone to the neuromuscular junction.

Depolarisation of axon membrane/influx of Na+ establishes local currents;

Change permeability to Na+

/open Na+

gates of adjoining region;

Adjoining region depolarises / influx of Na+

;This process repeated along axon / self propagation;Correct reference to/description of saltatory conduction;

When the action potential arrives at the neuromuscular junction, it results in the secretion of acetylcholine into the synaptic cleft. Explain howDepolarisation of (presynaptic) membrane;

Ca2+ channels open / increased permeability to Ca2+ ;

Influx of Ca2+ ;Vesicles move towards presynaptic membrane;Vesicles fuse with presynaptic membrane;

Describe the role of the mitochondria in the synaptic bulbActive transport of ions/ ionic pump; (reject active transport of Ach)Synthesis of acetylcholine / neurotransmitter/ reform vacuole;Reabsorption of acetylcholine, or acetyl + choline (from cleft);Movement of vesicles (to membrane);Synthesis of relevant enzyme, e.g. acetylcholinesterase.(Reject - general uses of energy, or use in muscle fibril)

Give three differences between the muscle fibre and three epithelial cell lining the small intestinemulti-nucleate;striations / sarcomeres / banding;actin / myosin / contractile protein;no microvilli / surface not folded;more mitochondria;contain myoglobin;

Page 11: Unit 5 Summary Notes A2 Biology

Describe two ways in which a myasthenic neuromuscular junction differs from a normal one and explain how each difference would affect transmissions across the myasthenic neuromuscular junction.myasthenic has fewer folds/ fewer receptors;

so less chance of depolarisation/fewer Na+ channels open; wider gap/cleft;

so takes longer for transmitter to diffuse across;different ratio of receptors to esterase;so transmitter more likely to be destroyed before binding to receptor;

acetylcholinesterase in shallower folds/more exposed;so transmitter destroyed before binding (to receptor);

Explain why the pattern of protein filaments differs in Figure 2 and Figure 3.myofibril is contracting in Figure 3 / relaxing in Figure 2;movement of actin fibres between myosin fibres;In Fig 2, only actin / thin filaments present;In Fig 3, actin / thin filaments and myosin / thick filaments present;Actin /thin filaments have moved into myosin / thick filaments;

Explain the cause of the banding pattern in striated muscleThe banding pattern is a result of the overlap between thick and thin filaments

Describe 2 ways in which the muscle as a whole can produce contractions of varying force.By changing the frequency of stimulation so that the fibre receives impulses at a greater rate. Changing the number and sizes of the motor units (a motor unit is all the muscle fibres supplied by a single motor neurone). Large number of units gives a more intensecontraction

Name 2 things needed for cross bridge formation and where they come fromCalcium ions and ATP. Calcium from sarcoplasmic rteticulum and ATP produce from CP hydrolyses the glycolytic or the oxidative process.

Explain why the glycolytic system for producing energy in muscle contraction is limitedLactic acid is produced is a waste product and it inhibits the enzymes involved in glycogen breakdown and impedes muscle contraction

Endurance athletes, such as marathon runners, nearly always have a high proportion of slow fibres in their muscles. Explain the benefit of this.(6)Endurance athletes exercise for long periods of time;Respire / release energy aerobically;Or too much lactate would accumulate;Slow twitch fibres adapted to aerobic metabolism;As have many mitochondria;Site of Krebs’ cycle;And electron transport chain;Much ATP formed;Also are resistant to fatigue;

Fig 2 Fig 3

Page 12: Unit 5 Summary Notes A2 Biology

Explain the advantage of possessing both types of muscle fibres.(4)Fast fibres make immediate/fast contraction possible before thecirculation/blood supply adjusts/ most energy anaerobically generated;fast fibres used in explosive/sprints locomotion;

Slow fibres allow sustained contraction/anaerobic energy generation;slow fibres used in maintaining posture/endurance events;

Explain the advantage of having large amounts of glycogen in fast muscle fibres.Anaerobic respiration / glycolysis inefficient / produces little ATP;requires large amount of glucose to produce enough ATP;glycogen acting as glucose store / glycogen converted to glucose;energy store Slow muscle fibres have capillaries in close contact. Explain the advantage of this arrangement Requires oxygen / glucose;short diffusion pathway / rapid passage of oxygen;Removal of heat/CO2;

The myosin-ATPase of fast twitch muscle fibres has a faster rate of reaction than that in slow twitch 2 fibres. Use your knowledge of the mechanism of muscle contraction to explain how this will help type 1 muscle fibres to contract faster than type 2.1. overall rate of contraction limited by rate of ATP-splitting;2. ATPase splits ATP / hydrolyses ATP / converts ATP to ADP (+ phosphate);3. ATP-splitting provides energy for any TWO from myosin-actin interaction;myosin head movement / actin to move relative to myosin; to ‘cock’ myosin head;

Explain the mechanism of muscle contraction.actin filaments are moved;ratchet mechanism / description;bridges formed between myosin and actin;use of ATP in forming / breaking bridges;reference to role of calcium / tropomyosin;

Page 13: Unit 5 Summary Notes A2 Biology

HomeostasisBGLDescribe how insulin reduces the concentration of glucose in the blood. 3)Insulin binds to specific receptors (on membranes);insulin activates carrier proteins / opens channels / causes more channels to form;insulin increases the permeability of liver/muscle cells/tissues to glucose;insulin action results in glucose conversion to glycogen / glycogenesis

Describe the role of insulin in the control of blood glucose concentration. (4)increase in blood sugar leads (more) insulin secreted;binds to (specific) receptors on (liver/muscle) cells;leads to more glucoseentering cells/carrier activity/increased permeability to glucose;glucose leaves the blood;glucose entering cell converted to glycogen;

The hormones which control the concentration of glucose in the blood affect some cells in the body but not others. Use your knowledge of the structure of cell surface membranes to explain why. (3)Only target cells have appropriate receptors;These are the proteins in cell surface membrane;Receptor sites/hormones with a particular shape;Concept of fitting/binding between receptor and hormone;

Explain the changes which take place in the blood glucose concentration of the non-diabetic person the period after eating(4)Glucose absorbed into blood from gut causing glucose levels to rise;rise detected by pancreas;insulin is secreted into blood by pancreas;glucose converted to glycogen/fat in liver/muscle;increased glucose uptake by (respiring/muscle) cells;binds to specific membrane receptors;mild diabetic secretes some insulin;

Describe how blood glucose concentration is controlled by hormones in an individual who is not affected by diabetes. Process involves insulin and glucagon;For detail, up to a total of 6 marksInsulin / glucagon secreted by pancreas / islets of Langerhans;Hormone receptors in membrane (of target cells);(insulin stimulates) conversion of glucose to glycogen / glycogenesis:activates / involves enzymes;stimulates uptake by cells;conversion of glucose to lipid / protein;glucagon stimulates conversion of glycogen to glucose;/ glycogenolysis;glucagon stimulates conversion of lipid / protein to glucose /gluconeogenesis;

(b) Suggest how diet and exercise can maintain low glucose concentrations in the blood of type II diabetics. (3)feed on polysaccharides / named example (not cellulose);slower digestion therefore no surge in blood sugar level;exercise - increased respiration / BMR;

Page 14: Unit 5 Summary Notes A2 Biology

The diagram shows the changes caused in a cell by the hormone glucagon when it combines with a receptor in the cell surface membrane (cascade). Using the diagram, explain how an injection of

a small amount of

glucagon into the body could cause a rapid increase in the concentration of glucose in the blood plasma. Ref to cascade / amplification effect; 1>1 molecule of cyclic AMP formed per glucagon (molecule);each cyclic AMP activates >1 enzyme(molecule) ;each enzyme causes breakdown of >1 glycogen (molecule);each glycogen gives >1 glucose / glycogen is a polymer;glucose diffuses into blood /glucose moves high to low concentration;

The graph shows changes in plasma glucose concentration that occurred in a person who went without food for some time.Use evidence from the graph to explain the role of negative feedback in the control of plasma glucose concentration. (5)

1. Deviation of a value from norm initiates corrective mechanisms;2. fluctuations in plasma glucose concentration detected by hypothalmus/islet cells in pancreas;3. initial decrease, no food given (in plasma glucose) stimulates (increased) secretion of glucagon;4. increases (in plasma glucose) stimulate (increased) secretion of insulin;5. correct ref. to role of and/or cells as secretors;6. correct ref. to interconversion of glycogen / glucose;7. increased/decreased uptake of glucose by cells (as appropriate)/correct ref to change in membrane permeability;

G lu ca g o n

R ec ep to r

C ell su rfa ce m e m b ran e

AT P C y clic A M P

G ly co g en G lu co se

A ctiv ee n zy m e

In ac tiv ee n zy m e

A d en y la tec y c lase

}

P lasm ag lu co seco n cen tra tio n

N o fo o dfro m th isp o in t o n

K ey C h an g e in th is p e rio dd u e to g lu cag o n

C h an g e in th is p e rio dd u e to in su lin

T im e

Page 15: Unit 5 Summary Notes A2 Biology

Explain the role of negative feedback in the control of plasma glucose concentration.

1. Deviation of a value from norm initiates corrective mechanisms;2. fluctuations in plasma glucose concentration detected by hypothalmus/islet cells in pancreas;3. initial decrease, no food given (in plasma glucose) stimulates (increased) secretion of glucagon;4. increases (in plasma glucose) stimulate (increased) secretion of insulin;5. correct ref. to role of and/or cells as secretors;6. correct ref. to interconversion of glycogen / glucose;7. increased/decreased uptake of glucose by cells (as appropriate)/correct ref to change in membrane permeability;

Page 16: Unit 5 Summary Notes A2 Biology

Body Temp

Explain how athletes produce heat when they run. (2)Respiration for muscular activity; (energy ‘needed/used’ for respiration’ etc, disqualifies)respiration inefficient / releases waste heat / all energy ‘ends up as ‘heat’

Explain why runners are more likely to overheat in humid conditions. (3)Humidity reduces diffusion gradient / less difference in water potential;less evaporation of sweat;less cooling due to use of heat energy for evaporation of sweat.

Describe how the body responds to a rise in core body temperature. (5)Temperature receptors stimulated in; (in skin disqualifies)hypothalamus;heat loss centre stimulated;nerve impulses to sweat glands;increase rate of / start sweat production;nerve impulses to skin arterioles;vasodilation (ref to vessels moving disqualifies)

The small mammal has ears which are usually pink, but they appear pale when the environmental temperature is low. Explain the pale appearance of the mammal’s ears when the environmental temperature is low. (3)constriction / narrowing / shunt effect;of arterioles;less blood flow to capillaries;reduces heat loss via radiation / conduction / convection;

Explain how this arrangement of the blood vessels would help the seal to maintain a constant body temperature. (4)(b) EITHER1. increased (blood) temperature results in increased blood flow through capillaries in blubber / vasodilation in blubber;2. increased skin temperature;3. increased loss of heat from skin;4. decreased temperature results in reduced blood flow through blubber capillaries/ vasoconstriction in blubber;5. correct reference to (sphincter/circular) muscles of arterioles;6. correct reference to role of shunt vessels;OR1. decreased (blood) temperature results in decreased blood flow through capillaries in blubber / vasoconstriction in blubber;2. decreased skin temperature;3. decreased loss of heat from skin;4. increased temperature results in increased blood flow through blubber capillaries/ vasodilation in blubber;5. correct reference to (sphincter/circular) muscles of arterioles;6. correct reference to role of shunt vessels;

S k in su rface

B lu b b e rC ap illa ryn e tw o rk s

Page 17: Unit 5 Summary Notes A2 Biology

Where are the receptors that detect the rise in temperature? (1)Hypothalamus;

Explain how the body of a mammal may respond to a rise in the environmental temperature.Hot receptors in skin;nervous impulse;to hypothalamus;blood temperature monitored;heat loss centre involved;vasodilation / dilation of arterioles;more blood to surface / heat lost by radiation;piloerector muscles relax;hairs flatten on skin surface;less insulation;sweating initiated / increased;panting / licking;evaporation removes latent heat;drop in metabolic rate / use less brown fat;accept long term changes such as less fat deposition;thinner fur;migration;accept one behavioural process;

b) Describe the important differences between the nervous and hormonal co–ordination systems found in a mammal.(4)Rapid / slow;direct / broadcast;short lived/ long term;mainly electrical ; chemical;delivery via nerves / blood vessels;cause depolarisation of target cell membrane /receptors in membrane of target cell;

Explain why the core temperature of a large animal like a hippo would probably rise if it stayed on land during the daytime. (3)Metabolism/respiration produces heat;Small surface area to volume ratio;Environmental/air temperature high;Heat gained/lost by radiation;Fat limits heat loss;[Note: small surface area to volume ratio=large volume to surface area ratio]

The blood vessels in the skin play an important part in allowing a mammal to conserve heat. Describe how. (2)(Vaso) constriction of arterioles / correct reference to shuntvessels or sphincters; ignore contractionReject this first mark if any reference to moving blood vessels.Less radiation / conduction / convection;Less blood to surface / more blood flows beneath fat;

Explain two advantages of endothermy over ectothermy.Any two from :Enzymes at optimum temperature;(Metabolic) reactions proceed more quickly;More independent of environment/better able to survive in differentenvironment/equivalent;

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During exercise, much heat is generated. Describe the homeostatic mechanisms that restore normal body temperature following vigorous exercise.(5)Receptors in hypothalamus detect increase in core temperature /temperature of blood;Heat loss centre stimulated;Skin arteries / arterioles dilate / vasodilation;Shunt vessels / pre-capillary sphincters constrict;More blood flows to surface (capillaries);Heat loss by radiation;Heat loss by evaporation of sweat;Reduced metabolic rate;Remove clothing / seek cooler area / cold drink;

Explain how normal core body temperature is maintained when a person moves into a cold room. (5)1.Sensors in skin/hypothalmus detect reduced temperature;2. heat gain centre activated/inhibition of heat loss centre;3. vasoconstriction/constriction of arterioles in skin surface;(R capillaried)4. dilation of shunt vessels/constriction of – capillary sphincter;5. less blood to skin surface/capillaries6. reduced heat loss by radiation;7. incresed heat gain by increased metabolic rate/respiration/movement/shivering;8. decreased heat loss by putting on clothes/huddling/reducedsweating;

How does maintaining a constant body temperature allow metabolic reactions in cells to proceed with maximum efficiency? (5)1. Body temp./37°C is optimum temp for enzymes;2. excess heat denatures enzymes/alters tertiary structure/alters shape of active site/enzyme;3. substrate cannot bind/eq,;4. reactions cease/slowed;5. too little reduces kinetic energy of molecules / molecules move more slowly;6. fewer collisions/fewer ES complexes formed’

One effect of getting into a cold shower is a reduction in the amount of blood flowing through the capillaries near the surface of the skin. Explain how the cold water causes this response. (4) (thermo)receptors in skin;(accept receptors in hypothalamus if after reference to cooled blood)impulses via nerves/neurones to or from; (once only)hypothalamus;heat gain/temperature centre (in hypothalamus);contraction /constriction of arterioles; (not capillaries, or justvasoconstriction)diversion through shunt vessels;

Cross-channel swimmers experience a large decrease in external temperature when they enter the water. Describe the processes involved in thermoregulation in response to this large decrease in external temperature.(7)1. hypothalamus (contains the thermoregulatory centre);2.has receptors which detect temperature changes of blood;3.receives impulses from receptors in skin;4.nerve impulses transmitted (from hypothalamus / brain);5.results in vasoconstriction / constriction of arterioles / dilationof shunt vessels;6.diversion of blood to core / specified organ / less blood to skin;7.muscular contraction /shivering generates heat via respiration;8.release of thyroxine / adrenaline;9.increase in metabolic rate / respiration;10.correct reference to negative feedback mechanisms;

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The ears of a rabbit play an important part in helping the animal to keep its body temperature constant. After a period of exercise, the insides of a rabbit’s ears become redder in colour as the blood flow to the skin surface increases. Explain how the different components of nervous communication are involved in the process leading to the response shown by the rabbit’s ears. (6)Stimulus is increased blood temperature;Increase in temperature results from exercise/respiration/metabolism;Detected by receptors in hypothalamus;Hypothalamus is coordinator;In this case, the heat loss centre;Effectors are muscles;Of arteriole;Response involves vasodilation;Increased blood flow to capillaries;Allowing heat loss by radiation/convection;Correct reference to action potential/nerve impulse;

Oestrous cycle

The relationship between oestrogen and LH is an example of positive feedback. Explain how.Answer showing understanding of positive feedback i.e. more produces more / differs further; Answer showing understanding of positive feedback correctly linked to oestrogen and LH i.e. more oestrogen produces more LH;;

Explain how changing concentrations of oestrogen and progesterone regulate the oestrous cycle.Any six from :1 Progesterone inhibits (release of) FSH/LH;2 Once progesterone falls (on day 16) FSH increases;3 FSH increase causes follicles to develop;4 Developing follicles produce oestrogen;5 Oestrogen inhibits FSH (release);6 High oestrogen/approx. day 18 stimulates FSH (release);7 High oestrogen stimulates LH (release);8 LH causes ovulation/causes progesterone (release)/formation of corpus luteum

The oestrous cycle in a female mammal is controlled by hormones. Describe the part played by FSH and LH in the control of the oestrous cycle. (5)

FSH stimulates growth of a follicle;Developing follicle produces oestrogen;(FSH) and LH bring about ovulation / oestrus;LH stimulates formation of corpus luteum;LH stimulates production of progesterone;Fall in LH / FSH means oestrogen production no longer stimulated;

During the oestrous cycle in a mammal, one or more follicles mature. Ovulation then takes place. Describe the part played by hormones in controlling these events. (6)FSH secreted by pituitary gland;Stimulates growth of follicle;Ovary/follicle cells produce oestrogen;Negative feedback/inhibits secretion of FSH;Oestrogen stimulates secretion of LH/LH from pituitary;LH stimulating ovulation;Second increase in FSH also associated with ovulation;

Explain how oral contraceptives containing progesterone and oestrogen work. (5)Oestrogen inhibits FSH;prevents follicle developing;progesterone inhibits LH;also inhibits FSH;inhibits ovulation;FSH and LH bring about ovulation

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Genetic code

Describe the molecular structure of DNA and explain how a sequence of DNA is replicated in the bacteria. (9)nucleotides;composition of a nucleotide,4 bases named;sugar-phosphate ‘backbone’;two (polynucleotide) strands;specific base-pairing;example e.g. A–T / C–G;hydrogen bonding;‘uncoiling’ / ‘unzipping’;semi-conservative replication;DNA polymerase;new complementary strands form / identical DNA molecule produced;DNA inserted into plasmids;which are self-replicating;

Describe how the structure of DNA allows it to carry out its functionSugar – phosphate backbone gives strength;Coiling gives compact shape;Sequence of bases allows information to be stored;Long molecule / coiling stores large amount of information;Complementary base pairing enables information to be replicated /transcribed;Double helix protects weak hydrogen bonds / double helix makesmolecule stable;Many hydrogen bonds together give molecule stability;Prevents code being corrupted;Hydrogen bonding allows chains to split for replication / transcription OR molecule unzips easily for replication / transcription.

Explain why the genetic code is described as non-overlapping and degenerateEach base is part of only one codon 1Some amino acids are coded for by more than one codon/ base sequence;

Compare tRNA vs mRNA1) tRNAClover shapedStandard lengthHas an amino acid binding siteanticodontRNA has H bonds between complementary base pairsLimited number of types (64)2) mRNALinearVariable length (depends on the length of gene)Many different types (depends on the gene)No H-bondingNo base pairs

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Compare DNA Vs RNASimilarities:Contain phosphateMade up of nucleotidesContains organic bases (A, C and G) (not T as it is replaced by U in RNA)Pentose sugarDifferencesRNA single strandedRNA has non-coding strands (introns) removed Ribose sugar in RNA deoxyribose in DNAU in riobose replaces the T3 types of RNA, only one DNASmaller than DNA

Compare the structures of RNA and DNA;Alikeboth have phosphate/phosphoric acid/PO4;bases/named bases/accept letters;nucleotides;pentose sugar;DifferentDNA deoxyribose;DNA thymine;DNA double stranded;DNA larger/longer;DNA one form RNA 3 types;

Describe translation and transcriptionTranscriptionSection of DNA unwinds / uncoils;DNA separates, h-bonds breakRNA nucleotides align;complementary base pairing / example of pairing;U replaces TmRNA polymerase (joins nucleotides);mRNA is modified, introns are removedTranslationmRNA moves into cytoplasm / through nuclear pore / to ribosome;tRNA carries specific amino acid;mRNA read in codons / triplets;anticodon of tRNA matches codon of mRNA;ATP used in activation / joining amino acids;amino acids join by peptide bonds;tRNA used repeatedly;sequence of bases / codons determines sequence of amino acids;

Describe what happens during translation.codons on mRNA;anticodons on tRNA;20 types tRNA moleculespecific amino acid attached to tRNA;peptide bonds formed

Describe the role of tRNA in the process of translation.anticodon complementary to codon/reads message on mRNA;specific amino acid;carried/transferred (to ribosome);correct sequence of amino acids along polypeptide;

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Comparison of replication and transcriptionSimilaritiesH bonds break and the DNA unzipsDNA acts as a template for complimentary basePolymerase enzymes are involvedDifferencesU replaces T in RNAIn replication all the DNA is copied, in transcription on sections are copiedOnly one strand is used as a template in transcription (antisense strand), both strands are used in replicationRNA polymerase in transcription whereas DNA polymerase is sued in replicationmRNA is produced in transcription, DNA is produced in replication

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MutationsA common question is to explain how mutations result in non-functioning or different proteins. Look at the examples below and lo0k for common marking points within.

What is a gene mutation?Change in base/nucleotide;

Describe one way in which the structure of the DNA of a gene may be changed as a result of a mutation.Addition / deletion / substitution;Of nucleotide / base

Substitutions: mis-sense (codon changed so codes for new amino acid), non-sense (codon changed to a stop codon), silent (change in 3rd base of a codon and still results in the same amino acid due to degenerate genetic code)Addition/deletion: frame shift is caused, alters the base sequence from the point of mutation and can change many amino acids

Name mutagens/factors that increase frequency of mutationsHigh energy ionized particles/X-rays/ultraviolet light/high energyradiation/uranium/plutonium/gamma rays/tobacco tar/caffeine/pesticides/mustard gas/base analogues/free radicals;(reject radiation)High energy radiation / X rays / ultraviolet light / gamma rays;high energy particles / alpha particles / beta particles;named chemical mutagens e.g. benzene / caffeine / pesticide / mustard gas /tobacco tar / free radicals;(two named examples of any of the above = 2 marks)length of time of exposure (to a mutagen);dosage (of mutagen);

Explain how exposure to a mutagenic agent may result in an inactive enzyme being produced by a cell.Change in the sequence of nucleotides/bases/addition/deletion/substitution;changed order of amino acids/different protein/different tertiary;structure;inactive enzyme if shape of active site is changed/enzyme-substratecomplex does not form;

Explain how mutation of a gene can result in an organism lacking a particular enzyme.mutation results in incorrect sequence of bases/nucleotides in DNA/frame shift of nucleotides;incorrect codons/base triplets on mRNA;so incorrect amino acids brought to ribosome/incorrect tRNA bring amino acids;wrong sequence of amino acids changes tertiary structure or active site(of enzyme)/no longer functions as enzyme/no or different enzyme formed/protein non-functional;

Explain how the gene mutation results in failure to produce the enzyme phenylalanine hydroxylase.change in base sequence in mRNA / different mRNA codons;different tRNA molecules pair with mRNA;with different amino acids / change in primary structure;(reject produces different amino acids)change in tertiary structure of protein;change in shape of active site;

Explain why mutation of a mitochondrial gene might result in no functional cytochrome oxidase being produced.change in base/nucleotide (in DNA);change in base sequence of mRNA/change in codons/idea of frame shift following deletion or addition;incorrect tRNA/anticodon;incorrect amino acids/ different primary structure/formation of newstop codon;different tertiary structure/different 3D structure/different polypeptide/shortened polypeptide;different shape of active site/no active site present;Explain why a mutation involving the deletion of a base may have a greater effect than one involving substitution of one base for another.

Change in base sequenceChanges the mRNA (codons)Binds different tRNAChanges amino acid sequence (primary structure)Change tertiary structureChange the active site/receptor site no longer bind…

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Deletion causes frame shift / alters base sequence (from point of mutation);changes many amino acids / sequence of amino acids (from this point);substitution alters one codon / triplet;one amino acid altered / code degenerate / same amino acid coded for;

A gene mutation may cause no change in the structure of the protein coded for. Explain why.Degenerate code / clear description;(New triplet) codes for same amino acid;

Mutations and cancerSometimes errors occur during the copying of a sequence of bases. Explain why some errors have less severe consequences than others.changes in base sequence will not necessarily affect amino acid coded for/most amino acids have more than one code;these codes differ only in the third base;so changes in the third base are likely to cause no change in the amino acid sequence/ protein;changes in first/second base result in an incorrect amino acid in the sequence/ formation of the incorrect protein/example of mutationcausing this type of change;change in amino acid present may have no effect on functioning of protein/some amino acids more important in tertiary structure than others;

Describe how altered DNA may lead to cancer. (6)1 (DNA altered by) mutation;2 (mutation) changes base sequence;3 of gene controlling cell growth / oncogene / that monitors cell division;4 of tumour suppressor gene;5 change protein structure / non-functional protein / protein not formed;6 (tumour suppressor genes) produce proteins that inhibit cell division;7 mitosis;8 uncontrolled / rapid / abnormal (cell division);9 malignant tumour;

Cells contain suppressor genes, which code for proteins that control cell division and growth. Describe what is meant by a mutation, and explain how a mutation in suppressor genes might lead to the development of a malignant tumour. (6)Mutation of suppressor gene – up to 4 marks1. Mutation is a change in the DNA / sense strand;2. Base sequence altered / e.g.;3. Suppressor gene produces wrong instructions / has different code;4. (Therefore) different amino acid sequence;5. Different protein structure / non-functional protein;

Malignant tumour – up to 2 marks6. Cell division by mitosis;7. Tumour cells growth abnormal / continuous / uncontrolled / rapid;8. Tumour cells spread / invade other tissues / form secondary tumours / metastasis;9. Via blood / lymph system;

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Describe and explain how gene expression is controlled (pre and post transcription)Pre transcriptional regulationTranscription depends on the activation of transcription factorsUsually by hormones (like oestrogen)The transcription factors attach to the promoter region of the DNA forming a transcription factor initiation complexThis complex is recognised by RNA polymerase which binds and begins transcription

Post transcriptional regulationRNA interference (RNAi) prevents translation of the mRNAsiRNA is double stranded and shorter than RNA It is split by an endonuclease into a passenger and guide strand The guide strand is loaded into the RISC complex which then attaches to the complementary section of bases on the mRNA and cleaves the mRNA

Uses of siRNAIdentifying the function of genes in cells or simple organisms by silencing the gene and observing the functional effects.Genetically-modifying crop plants by silencing undesirable genes e.g. those that make toxins or allergens. In the Flavr Savr tomato a gene causing ripening was silenced, causing the tomatoes to stay fresh for longer.Treating human patients to silence essential genes in cancer cells, or in antiviral herapies e.g. silencing the gene for the receptor protein used by HIV.

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Mutations (insertion deletion = fame shifts) (substituions = mis-sense (new codon=new amino acid), non-sense (stop codon) and silent (no effect)What is a gene mutation?Change in base/nucleotide;

Describe one way in which the structure of the DNA of a gene may be changed as a result of a mutation.Addition / deletion / substitution;Of nucleotide / base

Name mutagens/factors that increase frequency of mutationsHigh energy ionized particles/X-rays/ultraviolet light/high energy radiation/uranium/plutonium/gamma rays/tobacco tar/ caffeine/pesticides/mustard gas/base analogues/free radicals;(reject radiation)High energy radiation / X rays / ultraviolet light / gamma rays;high energy particles / alpha particles / beta particles;named chemical mutagens e.g. benzene / caffeine / pesticide / mustard gas / tobacco tar / free radicals;

Length of time of exposure (to a mutagen);dosage (of mutagen)

Explain how exposure to a mutagenic agent may result in an inactive enzyme being produced by a cell.Change in the sequence of nucleotides/bases/addition/deletion/substitution;changed order of amino acids/different protein/different tertiary;structure;inactive enzyme if shape of active site is changed/enzyme-substratecomplex does not form;

Explain how mutation of a gene can result in an organism lacking a particular enzyme.mutation results in incorrect sequence of bases/nucleotides in DNA/frame shift of nucleotides;incorrect codons/base triplets on mRNA;so incorrect amino acids brought to ribosome/incorrect tRNA bring amino acids;wrong sequence of amino acids changes tertiary structure or active site(of enzyme)/no longer functions as enzyme/no or different enzyme formed/protein non-functional;

Explain how the gene mutation results in failure to produce the enzyme phenylalanine hydroxylase.change in base sequence in mRNA / different mRNA codons;different tRNA molecules pair with mRNA;with different amino acids / change in primary structure;(reject produces different amino acids)change in tertiary structure of protein;change in shape of active site;

Explain why mutation of a mitochondrial gene might result in no functional cytochrome oxidase being produced.change in base/nucleotide (in DNA);change in base sequence of mRNA/change in codons/idea of frame shift following deletion or addition;incorrect tRNA/anticodon;incorrect amino acids/ different primary structure/formation of newstop codon;different tertiary structure/different 3D structure/different polypeptide/shortened polypeptide;different shape of active site/no active site present;

Explain why a mutation involving the deletion of a base may have a greater effect than one involving substitution of one base for another.Deletion causes frame shift / alters base sequence (from point of mutation);changes many amino acids / sequence of amino acids (from this point);substitution alters one codon / triplet;one amino acid altered / code degenerate / same amino acid coded for;

A gene mutation may cause no change in the structure of the protein coded for. Explain why.Degenerate code / clear description;(New triplet) codes for same amino acid;

Change in base sequenceChanges the mRNA (codons)Binds different tRNAChanges amino acid sequence (primary structure)Change tertiary structureChange the active site/receptor site no longer bind…

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In Vivo gene cloningA gene is obtained from mRNA (using reverse transcriptase), rather than DNA. Suggest why.Idea that mRNA is present in large amounts in cell making the protein /mRNA has been edited / does not contain introns / mRNA codes for single protein;Difficulty of finding one gene among all the genes in the nucleus / large amounts of mRNA coding for insulin will be present in insulin producingcells / idea that mRNA will be ‘edited’

Describe a plasmid.Circular DNA; separate from main bacteria] DNA; contains only a few genes

Explain what is meant by a vectorCarrier; DNA/gene; (context of foreign DNA); Into cell/other organism/host;

Explain the function of a vectorTransfers / carries (foreign) gene / DNA;Into bacteria / carrot cell / host cell;

Transformation is the process of trying to introduce the recombinant plasmid into the bacteria. Explain how you could identify bacteria containing the plasmidreplica plating;use of pad/velvet surface to transfer bacteria;use of agar plate containing ampicillin/no tetracycline and agarplate containing tetracycline;in bacteria with human DNA tetracycline gene no longer functional/not resistant to tetracycline;bacteria with human DNA grow on plate with ampicillin/notetracycline but are killed by tetracycline;bacteria with no extra DNA in plasmid not killed,

Explain how modified plasmids are made by genetic engineering and how the use of markers enables bacteria containing these plasmids to be detected.isolate wanted gene/DNA from another organism/mRNA from cell/organism;using restriction endonuclease/restriction enzyme/reverse transcriptase to get DNA;produce sticky ends;use ligase to join wanted gene to plasmid;also include marker gene;example of marker e.g. antibiotic resistance;add plasmid to bacteria to grow (colonies);(replica) plate onto medium where the marker gene is expressed;bacteria/colonies not killed have antibiotic resistance gene and (probably) the wanted gene;bacteria/colonies expressing the marker gene have the wanted gene as well;

Describe how bacteria may be produced which have the foreign genes in their plasmids.1 cut desired gene (from DNA);2 using restriction endonuclease/restriction enzyme;OR1 use mRNA from oat which will code for resistance;2 and use reverse transcriptase to form desired DNA;OR1 make artificial DNA with correct sequence of bases;2 using DNA polymerase;3 cut plasmid open;4 with (same) restriction endonuclease/restriction enzyme;5 ref. sticky ends/unpaired bases attached;6 use (DNA) ligase to join / ref. ligation;7 return plasmid to (bacterial) cells;

8 use of Ca2+/calcium salts/electric shock;

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What is recombinant DNA?contains genes/nucleotides/sections of DNA/artificial DNA from two species/2 types of organisms;

Explain the importance of marker genes.Allows transformed bacteria to be separated from non-transformed;Further detail e.g. transformed bacteria survivewhen antibiotic applied to medium;

Restriction enzymes /restriction endonucleasesThese are enzymes that cut DNA at specific sites, recognition sequences (palindromic sequences)GAATTCCTTAAG

Some restriction enzymes cut straight across both chains, forming blunt ends, Most enzymes make a staggered cut in the two strands, forming sticky ends.The products are called restriction fragments

Explain the meaning of the term ‘sticky ends’Cut ends of DNA;one strand longer than the other / staggered cut / unpaired bases;Can attach to complementary DNA / bases;

DNA ligase completes the DNA backbone by forming covalent phosphodiester bonds. Restriction enzymes andDNA ligase can therefore be used together to join lengths of DNA from different sources. Useful when joining restriction fragments.

The enzyme reverse transcriptase does the reverse of transcription: it synthesizes DNA from an RNA template (so it is an RNA-dependent DNA polymerase enzyme). Reverse transcriptase is produced naturally by a group of viruses called the retroviruses (which include HIV), and it helps them to invade cells. In biotechnology reverse transcriptase is used to make an “artificial gene”, called complementary DNA (cDNA), from an mRNA template as shown in this diagram:

Reverse transcriptase has several uses in biotechnology:• It makes genes without introns. Eukaryotic genes with many introns are often too big to beincorporated into a bacterial plasmid, and bacteria are unable to splice out the introns anyway. Theartificial cDNA gene is made from mRNA that already has the introns spliced out of it, so it can beexpressed in bacteria.• It makes a stable copy of a gene, since DNA is less readily broken down by enzymes than RNA.• It makes genes easier to find. There are some 20 000 genes in the human genome, and finding the DNAfragment containing one gene out of this many is a very difficult task. However a given cell onlyexpresses a few genes, so only makes a few different kinds of mRNA molecule. For example the b cellsof the pancreas make insulin, so make lots of mRNA molecules coding for insulin. This mRNA can beisolated from these cells and used to make cDNA of the insulin gene.

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In vitro gene cloning (PCR)The polymerase chain reaction (PCR) can be used to produce large quantities of DNA. Describe how the PCR is carried out. (6)

1. DNA heated to 90 to 95°C;2. Strands separate;3. Cooled / to temperature below 70°C4. Primers bind;5. Nucleotides attach;6. By complementary base pairing;7. Temperature 70 - 75°C;8. DNA polymerase joins nucleotides together;9. Cycle repeated;

What are primers?(short) length of DNA;single stranded; (reject reference to RNA)with specific base sequence/complementary base sequence;indicates where replication starts/stops annealing;

Why are primers neededTo mark beginning and/or ends of the part of DNA needed / forattachment of enzymes or nucleotides / initiator / keeps strands apart;Attaches to / complementary to start of the gene / end of fragment;Replication of base sequence from here;

GMOs (advantages and disadvantages)See notes and past paper question examples

DNA sequencing/Dideoxy sequencing/chain termination

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Evaluating BiotechnologyThe whole point of creating genetically-modified organisms is to benefit humans, and the benefits areusually fairly obvious, but nevertheless there has been some vocal opposition to GMOs. Opposition isoften based on ethical, moral or social grounds, such as harm to animals or the environment, though therecan also be more practical issues, such as distrust of large corporations.

Benefits• Medicines and drugs can be produced safely in large quantities from microbes rather than fromslaughtered animals. These medicines benefit humans and can spare animal suffering as well.

• Agricultural productivity can be improved while using less pesticides or fertilisers, so helping theenvironment. GM crops can grow on previously unsuitable soil or in previously unsuitable climates.

• GM crops can improve the nutrition and health of millions of people by improving the nutritional qualityof their staple crops.

Risks• Risks to the modified organism. Genetic modification of an organism may have unforeseen geneticeffects on that organism and its offspring. These genetic effects could include metabolic diseases orcancer, and would be particularly important in vertebrate animals, which have a nervous system and soare capable of suffering. The research process may also harm animals.

• Transfer to other organisms. Genes transferred into GMOs could be transferred again into otherorganisms, by natural accidents. These natural accidents could include horizontal gene transmission inbacteria, cross-species pollination in plants, and viral transfer. This could result in a weed being resistantto a herbicide, or a pathogenic bacterium being resistant to an antibiotic. To avoid transfer via crosspollination,

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genes can now be inserted into chloroplast DNA, which is not found in pollen.

• Risks to the ecosystems. A GMO may have an unforeseen effect on its food web, affecting otherorganisms. Many ecosystems are often delicately balanced, and a GMO could change that balance.• Risk to biodiversity. GMOs may continue to reduce the genetic biodiversity already occurring due toselective breeding.

• Risks to human societies. There could be unexpected and complicated social and economicconsequences from using GMOs. For example if GM bananas could be grown in temperate countries,that would be disastrous for the economies of those Caribbean countries who rely on banana exports.

• Risks to local farmers. Developing GMOs is expensive, and the ownership of the technology remainswith the large multi-national corporations. This means the benefits may not be available to farmers inthird world countries who need it most.

Process usesNucleotides (normal A, T, C and G)DNA to be sequenced (amplified by PCR)Primers (to serve as the initiation point for chain synthesis)Polymerase (joins nucleotides)

Then special dideoxynucleotides are used that can only bond to one nucleotide. These are the terminators of DNA synthesis. These can be labelled radioactively or fluorescently for identification after electrophoresis.

In the reaction vessels chains of different lengths are formed and can be separated by electrophoresis

The above diagram shows the results of a particular sequencing process and the order of bases that made up the DNA is as follows

ACGCCCGAGTAGCCCAGATT

Gene/DNA probesShort single stranded pieces of DNA (20 bases long)

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Complementary to bases in known sequences of DNAProbes are often labelled (radioactively/fluorescently) to allow identificationProbes will hybridise with complementary base pairsUsed to identify certain alleles in a person’s genome (cystic fibrosis)Used to look for the presence of particular genes in different species

What is a DNA probe?Piece of DNA;Single stranded;Complementary to/binds to known base sequence/gene;

Strand of DNA;Short strand / up to 20 bases long;With base sequence that is complementary to part of target gene;Radioactive labelling / fluorescent labelling;

Explain how the use of a gene probe could enable the presence of a mutant allele of the cystic fibrosis gene to be detected.Probe will attach (to mutant allele);attaches to one DNA strand;as a result of complementary base pairing;radioactivity detected on film/X-ray / by autoradiography(if mutant allele present);

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ElectrophoresisSeparates DNA fragments of different sizes.DNA samples are placed into wells cut in an aragose gel, covered in a buffered ionic solutionCurrent is passed through the gel and the negative charged DNA moves toward the positive electrode. DNA fragments diffuse through the pores in the gel, and the larger pieces face a greater resistance to movement than smaller pieces and so will travel a shorter distance.

Rate of movement depends upon: strength of the field, temperature of the buffered solution, strength of the ionic solution, pore sizeDNA fragments cannot be seenDNA can be stained with chemicals, azure A (makes it blue) or ethidium bromide which makes it fluoresces under UV light.DNA could be radioactively labelled (using P32 isotope), and a photographic film is placed over the gel in a dark room and the radioactivity exposes the film (autoradiography)

Marker genes can be added to the gel. These have a known base length and can thus be used to determine the size of fragments

The fragments will continue to diffuse after the current is off, and the DNA bands will start to blur, so the DNA is usually transferred to a nylon sheet in southern blotting (see below)

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Genetic fingerprinting Genome contains the same repetitive sequences (satellite DNA) of non-coding DNA. The number of bases in the repeat sequence varies (ATG or AGCTTCA) and the length of the repeat sequences (the number of times it repeats), and it is the length of the repeats that are unique to each person.This can happen in non-coding DNA as mutations will arise and be retained as this has no functional impactThese repeating core sequences are called VNTRs (variable number tandem repeats) if the sequence is between 10-80 bases or STRs (short tandem repeats) if the sequence is between 2-9 bases.

The VNTRs can be cut using restriction enzymes; these will cut at recognition sequences either side of the VNTR, so it will cut the same section for all people, but the number of repeats and thus the distance between the recognition sites will vary and should be unique to each person, this will produce different sized fragments (restriction fragment length polymorphism)

Below person A (on top) has the same repeat sequence as person B, ATG, but in A this repeats only 4 times, in B it repeats 13 times. When the fragments are separated on the electrophoresis gel, person A’s DNA fragment will move further as it is smaller..

Genetic fingerprint process:Extract the DNA: using a solvent like chloroform or phenolAmplify the DNA: using PCR and primers designed to copy the desired VNTR sequenceDigestion of DNA: using restriction enzymes Separation: using gel electrophoresisSouthern blotting: transfer of the genetic material from the gel to a more stable and durable materialThe gel is placed in alkali to separate the DNA strands and make them single stranded (so gene pobe can adhere)Nylon membrane or nitrocellulose membrane is laid down on the gel, then absorbent paper towels are placed onto of the sheet and it is weighted to ensure a good contactThe paper absorbs the alkali and draws the DNA with itThe negatively charged DNA will adhere to the positive nylon membraneThe sheet is separated and the DNA fixed with UV light

Hybridisation: the nylon sheet is covered with gene probes (complementary to the VNTR) which will anneal with any complementary base sequences. The gene probe can be radioactively/fluorescently labelledDevelopment: to visualise the DNA using UV light or autoradiographyInterpretation

Used in paternity tests, CSI investigations.In paternity tests a child will inherit VNTRs in some combination form mother and father, so bands should match both parents.

Describe how the technique of genetic fingerprinting is carried (6)1. DNA is cut;2. Using restriction enzyme;3. Electrophoresis;4. Separates according to length/mass/size;5. DNA made single-stranded;6. Transfer to membrane/ Southern blotting;7. Apply probe;8. Radioactive/ single stranded/ detected on film/ fluorescent;9. Reference to tandem repeats/VNTRs/minisatellites;10. Pattern unique to every individual;

ATGATGATGATG

ATGATGATGATGATGATGATGATGATGATGATGATGATG

12 bases long

39 bases long

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Describe how genetic fingerprinting is carried out.1 DNA extracted from sample;2 DNA cut/hydrolysed into segments using restriction endonucleases;3 must leave minisatellites/required core sequences intact;4 DNA fragments separated using electrophoresis;5 detail of process e.g. mixture put into wells on gel and electric current passed through;6 immerse gel in alkaline solution / two strands of DNA separated;7 Southern blotting / cover with nylon/absorbent paper (to absorb DNA);8 DNA fixed to nylon/membrane using uv light9 radioactive marker/probe added (which is picked up by required fragments)/complementary to minisatellites;10 (areas with probe) identified using X-ray film/autoradiography;

Gene therapyTransfer of genetic material into cells or tissue to cure diseases, aims to replace defective genes with healthy genes.

Most likely to treat single gene disorders (cystic fibrosis, SCID, Huntington’s, muscular dystrophy)

Types of Gene TherapyIt is important to appreciate the different between somatic cell therapy and germ-line therapy.• Somatic cell therapy means genetically altering specific body (or somatic) cells in order to treat the disease. This therapy may treat the disease in the patient, but any genetic changes will not be passed on the offspring of the patient.• Germ-line therapy means genetically altering those cells (sperm cells, sperm precursor cells, ova, ova precursor cells, zygotes or early embryos) that will pass their genes down the “germ-line” to future generations. Alterations to any of these cells will affect every cell in the resulting human, and in all his or her descendants.

Treatment using gene therapyGene replacement: a healthy gene replaces a defective geneGene supplementation: copies of the healthy gene are added alongside the defective gene. The added dominant genes mask the effect of the recessive allele.If the condition is caused by a dominant allele then try to added base sequences within the gene sequence to prevent functional protein being made

Delivery of the geneRequires a vector to carry the gene, many types of vectors exist, no perfect vector of all jobs so each disease and target area needs to be considered.Liposomes: DNA enclosed in a lipid vesicle which fuses with the cell membrane to deliver DNA. Difficult to create a fine enough spray to reach the tissues.

Viruses: adapted to inject DNA into cells and are often specific to certain regions. Modified so that it cannot reproduce itself (non-harmful)

Stem cells: remove stem cells from the person and modify these then replace the cells in person. Not all organs have stem cells to target.

DisadvantagesSuccess of gene uptakeSuccess of gene expressionShort lived (if cells die and are replaced and it is not copied)Multiple treatments and thus cost./inconvenience)Vectors can cause primary immune response and then repeated treatment causes a secondary immune response dangerous.If allele is up taken in wrong place, within a gene, like a tumour suppressor gene and causes cancer)Multiple gene disorders difficult to treatCosteugenics

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Restriction mapping

Process of identifying the types of restriction sites present in a piece of DNA, the number of these sites and the location of the sites relative to each other.

Different restriction enzymes and combinations of enzymes are used to cut DNAThe fragments are separated by gel electrophoresisThe size of the fragments is used to determine the location of the restriction sites (recognition sequences)The fragment size can be determined by using marker fragments of known sizes on the gel and comparing the unknown fragments. Remember the smaller a fragment the further it will travel.A partial digest is possible if the enzymes are not left long enough to make their cuts, thus producing fragments of different lengths; there would be large fragments present that would add up to more than the total lengthThere is also the risk that fragments of the same size are cut and would overlap on the gel, so the total fragment length sum for the complete combined digest must be the same as for each individual digest.

1 kilobase (kb) = 1000 bases/nucleotides)

Examples

1) Linear DNA was cut using two enzymes A and B, the resulting fragment lengths are shown belowA= 8, 6 and 3 kbB = 10 and 7 kbA + B (double digest on one piece of DNA) = 8, 6, 2 and 1 kb

We can see that enzyme A cuts twice to produce three fragments, B cuts once to produce two fragments.We can see that the total length of DNA is 17KB long as all fragments add up to that number

Q1. Sketch what the electrophoresis gel would look like; show the restriction fragments for each combination of cuts.

Q2. Sketch the location of the fragments on the linear piece of DNA shown below (may be best to sketch the possible locations for A and B individually, then for A and B together

2) Linear DNA is cut with the following enzymes and the fragments produced are also notedA = 0.8 and 6.2B = 1.2 and 5.8A + B = 5.8, 0.8 and 0.4

Q1. Sketch the gelQ2. Show the position of the restriction sites on a stretch of linear DNA

3) Linear DNA is cut with 2 enzymes and the results are listed belowA = 2, 4 and 6B = 4, 3 and 5A + B = 2, 2, 2, 1 and 5

Q1. Sketch the gelQ2. Show the position of the restriction sites on a stretch of linear DNA

A B A + B

17kb

7 kb

12 kb

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4) A circular plasmid was cut with a combination of enzymesA = 20KB (made 1 cut to produce a linear fragment)B = 12, 2 and 6A + B = 8, 4, 2 and 6

Q1. Sketch the gelQ2. Show the position of the restriction sites on the circular plasmid

5) A circular plasmid was cut with a combination of enzymesA = 10, 30 and 60B = 100A + B = 10, 20, 30 and 40

Q1. Sketch the gelQ2. Show the position of the restriction sites on the circular plasmid

6) A circular plasmid was cut with a combination of enzymesA = 70 and 30B = 40 and 60A + B = 40, 30, 20 and 10

Q1. Sketch the gelQ2. Show the position of the restriction sites on the circular plasmid

7) A circular plasmid was cut with a combination of enzymesA + B = 45 and 15A + C = 35 and 25B + C = 50 and 10

Q1. Sketch the gelQ2. Show the position of the restriction sites on the circular plasmid

7) A circular plasmid was cut with a combination of enzymesA + B = 90 and 10A + C = 60 and 40B + C = 70 and 30

Q1. Sketch the gelQ2. Show the position of the restriction sites on the circular plasmid

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Answers

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7.

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Genetic screening1. A few cells are taken from a patient by biopsy.2. DNA is extracted from the biopsy and if necessary amplified by PCR.3. The DNA is labelled with a fluorescent chemical.4. The DNA is denatured (i.e. separated into single strands) by heating or strong alkali.5. The single-stranded DNA is added to the microarray and mixed for a few hours. DNA that has a complementary sequence to any of the probes fixed to the microarray will hybridise to the probes by complementary base pairing.6. The microarray is washed with a buffer, washing away any loose fluorescent DNA that is not hybridised to the probes on the chip.7. If the microarray is illuminated with ultra-violet light, any spots where the subject’s DNA has hybridised to a probe will fluoresce and can be seen, while the spots where the probes haven’t hybridised will remain dark. In practice the microarray is “read” with a laser, which illuminates each spot in turn, recording the amount of fluorescence in each case on a computer.8. The locations of the fluorescent, “positive”, spots are matched with the name of that probe’s allele, giving a detailed genetic profile of the patient.