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IB Biology B3.3: Muscle and motility worked solutions

Worked solutions to every practice question in B3.3: Muscle and motility of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

HL[1 mark]

In a diagram of a relaxed sarcomere, a thick strand lies in the centre and carries short projections along its length. Which protein is it? (A) Actin (B) Myosin (C) Titin (D) Z-line

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B. Thick filaments in the middle of the sarcomere, with projecting heads, are myosin. Actin (A) is the thin filament fixed to the Z-line, titin (C) is the coiled elastic strand joining the end of the myosin filament to a Z-line, and the Z-line (D) is not a protein filament.

Final answer

Question A2

HL[1 mark]

Which list puts these structures in order of size, smallest first? (A) myosin filament, sarcomere, myofibril, muscle fibre (B) sarcomere, myosin filament, myofibril, muscle fibre (C) myofibril, sarcomere, myosin filament, muscle fibre (D) myosin filament, myofibril, sarcomere, muscle fibre

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A. A myosin filament lies inside a sarcomere, sarcomeres join end to end to make a myofibril, and many myofibrils lie inside one muscle fibre. B puts the sarcomere before the filament it contains, C puts the myofibril before the sarcomere, and D puts the myofibril before the sarcomere as well.

Final answer

Question A3

HL[1 mark]

What is the role of titin in a muscle? (A) It binds calcium ions to uncover myosin sites (B) It pulls actin towards the centre (C) It helps the sarcomere recoil after stretching (D) It splits ATP at the head

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C. Titin is an elastic protein that recoils the sarcomere after stretching and resists overstretching. Titin does not bind calcium (A), the pulling of actin (B) is done by the myosin heads, and ATP is split by myosin (D).

Final answer

Question A4

HL[1 mark]

A myosin head is attached to actin. What happens when ATP binds to the head? (A) The head is released from actin (B) The head tilts and pulls the actin (C) The head grips actin more tightly (D) The Z-lines move closer

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A. ATP binding releases the head from actin. The tilt that pulls the actin (B) comes after the head has been re-set and has bound again, a firmer grip (C) is the state without ATP, and the Z-lines (D) move only when many heads have pulled.

Final answer

Question A5

HL[1 mark]

A skeletal muscle fibre changes from relaxed to contracted. Which change is seen in its sarcomeres? (A) The dark band is shorter (B) The light band is narrower (C) The myosin filaments are shorter (D) The Z-lines are further apart

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B. The light band narrows because the actin filaments slide inwards and the Z-lines move closer. The dark band (A) is the length of the myosin filaments and stays the same, the filaments (C) keep their length, and the Z-lines (D) move closer, not apart.

Final answer

Question A6

HL[1 mark]

Which structure covers the ends of the bones at a synovial joint and reduces friction? (A) Ligament (B) Tendon (C) Cartilage (D) Synovial fluid

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C. Cartilage covers the bone ends and gives a smooth, shock-absorbing surface. A ligament (A) joins bone to bone, a tendon (B) joins muscle to bone, and synovial fluid (D) lubricates the gap between the cartilage surfaces but does not cover the bone.

Final answer

Question A7

HL[1 mark]

Which statement compares the hip joint with the elbow joint? (A) The hip is a hinge and the elbow a ball and socket (B) The hip moves in three planes and the elbow in one (C) Both allow flexion only (D) The elbow allows abduction and the hip does not

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B. The hip is a ball-and-socket joint that moves in three planes; the elbow is a hinge joint that moves in one plane (flexion and extension). A reverses the joint types, C ignores extension and rotation, and D credits the elbow with abduction.

Final answer

Question A8

HL[1 mark]

Two muscles lie on opposite sides of a joint and form an antagonistic pair. What happens in the extensor when the flexor contracts? (A) It also contracts (B) It relaxes and is stretched (C) It pushes the bone (D) It contracts more strongly

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B. The extensor relaxes and is stretched, so the joint can bend. If both contracted (A and D) the joint would be held still, and a muscle cannot push (C).

Final answer

Question A9

HL[1 mark]

Which statement is true of both a human endoskeleton and an arthropod exoskeleton? (A) It contains living bone cells (B) It provides anchorage for muscles (C) It lies outside the muscles (D) It is made of chitin

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B. Both give muscles something firm to pull against. Only bone contains living bone cells (A), the endoskeleton lies inside the muscles (C), and chitin is found in the cuticle of an arthropod, not in bone (D).

Final answer

Question A10

HL[1 mark]

What is a motor unit? (A) A fibre and its myofibrils (B) A motor neuron and the fibres it supplies (C) A sarcomere and its Z-lines (D) A muscle and its tendons

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B. A motor unit is one motor neuron and all the muscle fibres it supplies. A single fibre with its myofibrils (A) is only part of a unit, a sarcomere and its Z-lines (C) is part of a myofibril, and a muscle with its tendons (D) is much larger.

Final answer

Question A11

HL[1 mark]

Reindeer travel hundreds of kilometres each year between summer and winter ranges. Which reason for locomotion is this? (A) Foraging (B) Escape from danger (C) Searching for a mate (D) Migration

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D. A long, regular, seasonal trip between two regions is migration. Foraging (A) is searching for food within a range, escape (B) is a response to danger and mate search (C) is a journey to find a partner.

Final answer

Question A12

HL[1 mark]

During exhalation, which muscles contract and where does titin store energy? (A) Internal contract; titin in the stretched external layer stores energy (B) External contract; titin in the stretched internal layer stores energy (C) Internal contract; titin in the contracting layer releases ATP (D) Both contract; titin joins the layers

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A. In exhalation the internal intercostals contract and stretch the external layer, and the titin in the stretched external layer stores elastic energy. B has the layers the wrong way round, C has titin releasing ATP, which it cannot do, and D wrongly has both layers contracting.

Final answer

Tutorial practice

Question B1

HL[5 marks]

Draw a labelled diagram of one relaxed sarcomere.

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Award one mark each for: two Z-lines drawn equal in length and parallel, one at each end of the sarcomere and labelled (1); thin filaments labelled actin, fixed to the Z-line and reaching in towards the centre (1); thick filaments labelled myosin, central, with projecting heads drawn (1); titin drawn as a coiled strand from the end of each myosin filament to the nearest Z-line and labelled (1); a dark band spanning the myosin and a light band across the Z-line, correctly bracketed (1). Do not let actin touch myosin at the ends and do not join the thin filaments to the thick ones.

Final answer

Two vertical Z-lines labelled; thin actin filaments fixed to each Z-line and reaching inwards; thick myosin filaments in the middle, with heads; titin from each end of the myosin to the nearest Z-line; the dark band (myosin length) and a light band labelled.

Question B2

HL[6 marks]

A rower's thigh muscle shortens as she pulls on the oars. Explain how a sarcomere in this muscle shortens.

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Any six of: actin filaments are anchored to the Z-lines (1); myosin filaments lie in the centre of the sarcomere (1); the myosin heads bind to actin, forming cross-bridges (1); the heads tilt and pull the actin filaments towards the centre (1); ATP binds to the head, which releases it from actin (1); ATP is split and the energy re-sets the head so that it binds again further along (1); the cycle is repeated by many heads (1); the filaments slide, so the overlap increases and the Z-lines come closer (1); the filaments do not shorten (1).

Final answer

Actin is fixed to the Z-lines and myosin lies in the centre. Myosin heads bind actin and tilt, pulling the actin towards the centre. ATP binding releases the heads, and splitting ATP re-sets them for another pull. The overlap increases, the Z-lines move closer and the sarcomere shortens; the filaments keep their length.

Question B3

HL[4 marks]

A bottlenose dolphin spends its whole life in the sea but breathes air. Describe two adaptations of a dolphin for swimming, each with its function.

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Each adaptation scores one mark and its function one mark: a streamlined body (1) reduces drag so that less energy is used (1); a tail fluke beaten up and down (1) pushes the animal forward by thrust (1); limbs shaped as flippers (1) steer and brake (1); airways that allow periodic breathing (1) give a large, quick exchange of air at the surface between dives (1). Maximum 4. An adaptation without its function scores one mark only.

Final answer

Any two linked pairs, for example: a streamlined body, which reduces drag; a tail fluke moved up and down, which gives thrust; flippers, which steer; airways that allow rapid breathing at the surface between dives, which supports long dives.

Question B4

HL[7 marks]

Eight volunteers did stretching exercises for six weeks. The maximum angle of hip abduction was measured with a goniometer before and after. Mean maximum hip abduction angle in degrees: before 36 with SD 3, after 45 with SD 4. (a) Calculate the percentage change in the mean angle. [2] (b) Describe how a goniometer is used to measure the angle. [2] (c) Use the mean \pm SD values to decide whether the increase is likely to be real. [2] (d) State one limitation of the investigation. [1]

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(a) Change =\dfrac{45-36}{36}\times100=25\,\% (2; one mark for 9, one for dividing by the original value). (b) The centre of the goniometer is placed over the hip joint and its arms along the two body segments (1); the leg is moved to its limit, the angle is read in degrees, and the reading is repeated and averaged (1). (c) Before: 36\pm3 gives 33 to 39; after: 45\pm4 gives 41 to 49; the ranges do not overlap, so the increase is probably real (2). (d) There was no control group of volunteers who did not stretch, so the increase might have another cause (1).

Final answer

(a) (45-36)/36\times100=25\,\%. (b) Place the centre of the goniometer on the hip joint, lay the arms along the trunk and the leg, move the leg to its limit and read the angle in degrees. Repeat and take the mean. (c) The ranges do not overlap, so the difference is probably real. (d) There was no control group.

Question B5

HL[4 marks]

Adult barnacles are cemented to rocks, but their larvae swim in the plankton. (a) State what is meant by sessile. [1] (b) State one movement a sessile adult animal can make. [1] (c) Explain one advantage of a swimming larva for a sessile species. [2]

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(a) A sessile organism is fixed in one place (1). (b) One movement of a sessile adult, such as extending its legs to filter food, or a sponge opening and closing its exit opening (1). (c) The swimming larva can settle in a new place, which spreads the species into new habitats (1) and reduces competition between young and adults (1).

Final answer

(a) Fixed in place, not able to travel. (b) The barnacle extends its legs to sweep food into its mouth. (c) The larva can reach new rocks, so the species spreads and young barnacles do not compete with the adults.

Question B6

HL[5 marks]

Micrograph X of a myofibril shows a sarcomere 2.4\ \mum long with a dark band of 1.6\ \mum. Micrograph Y, of the same myofibril in another state, shows a sarcomere 1.8\ \mum long with a dark band of 1.6\ \mum. (a) Identify the contracted micrograph, giving a reason. [1] (b) Calculate the length of the light band in each micrograph. [2] (c) Calculate the percentage decrease in sarcomere length. [1] (d) Explain why the dark band is the same length in both. [1]

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(a) Y is contracted because its sarcomere is shorter (1). (b) The light-band length is the sarcomere length minus the dark-band length: X 2.4-1.6=0.8\,\mum (1); Y 1.8-1.6=0.2\,\mum (1). (c) \dfrac{2.4-1.8}{2.4}\times100=25\,\% decrease (1). (d) The dark band is as long as the myosin filaments, and the myosin filaments do not change in length when actin slides over them (1). Maximum 5.

Final answer

(a) Micrograph Y. (b) Light band: X 2.4-1.6=0.8\,\mum; Y 1.8-1.6=0.2\,\mum. (c) (2.4-1.8)/2.4\times100=25\,\%. (d) The myosin filaments keep their length; actin slides over them.

Question B7

HL[4 marks]

Give the reason for locomotion shown in each case. (a) A lion runs after a zebra. [1] (b) The zebra gallops away. [1] (c) A male newt crosses a field to a pond where females gather. [1] (d) Arctic terns fly to Antarctic waters each year. [1]

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(a) A lion running after a zebra is foraging for food (1). (b) The zebra gallops away to escape a predator (1). (c) A male newt crossing a field to a pond where females are found is searching for a mate (1). (d) Arctic terns that fly to southern waters each year are migrating (1). Name the gain, not the action: running and flying are not reasons.

Final answer

(a) Foraging. (b) Escape from danger. (c) Searching for a mate. (d) Migration.

Question B8

HL[5 marks]

A fast motor unit F and a slow motor unit S in a limb muscle were compared. Unit F: 200 fibres, force 1.00 N, contraction time 20 ms. Unit S: 50 fibres, force 0.05 N, contraction time 90 ms. (a) Define a motor unit. [1] (b) Calculate the force per fibre for each unit. [2] (c) Suggest which unit would be used to hold a posture, with a reason. [2]

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(a) A motor unit is one motor neuron and all the muscle fibres it supplies (1). (b) F: 1000\ \text{mN}/200=5 mN per fibre (1); S: 50\ \text{mN}/50=1 mN per fibre (1). (c) Unit S, because posture needs a small force held for a long time (1), and the slow unit can contract slowly without tiring (1).

Final answer

(a) A motor neuron with all the muscle fibres it supplies. (b) F: 1.0/200=5 mN per fibre; S: 50/50=1 mN per fibre. (c) Unit S: it gives a smaller force that can be kept up, so it holds the posture.

Question B9

HL[6 marks]

A section through a human hip joint shows: P the rounded end of the femur; Q a thin smooth layer covering P; R a tough band joining the femur to the pelvis; S a tough cord joining a muscle to the femur; T slippery liquid between the layers covering the bones. (a) Identify structures P, Q, R, S and T. [5] (b) State the name of the movement of the leg away from the midline of the body. [1]

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(a) P is the femoral head (the ball) (1); Q is cartilage (1); R is a ligament, because it joins bone to bone (1); S is a tendon, because it joins muscle to bone (1); T is synovial fluid (1). (b) Moving the leg away from the midline of the body is abduction (1).

Final answer

(a) P femoral head; Q cartilage; R ligament; S tendon; T synovial fluid. (b) Abduction.

Question B10

HL[5 marks]

The internal and external intercostal muscles are an antagonistic pair that move the ribs. (a) Explain why the muscles are described as antagonistic. [2] (b) Outline the role of titin in the stretched layer. [2] (c) State which layer contracts to lift the ribs. [1]

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(a) The two muscles pull the ribs in opposite directions (1), so when one contracts the other relaxes and is stretched (1); a muscle can only pull, so one muscle cannot return the ribs by itself. (b) The stretched muscle stores elastic energy in its titin (1), and the titin recoils and helps the next movement (1). (c) The external intercostals contract in inhalation (1).

Final answer

(a) The two layers pull the ribs in opposite directions: when one contracts the other relaxes and is stretched. (b) The stretched layer stores elastic energy in titin, which recoils it afterwards. (c) The external layer contracts to lift the ribs.

Advanced practice

Question C1

HL[8 marks]

After death a muscle becomes rigid within a few hours, then softens again. (a) Outline the role of ATP in muscle contraction. [3] (b) Explain why a muscle becomes stiff when ATP is no longer produced. [2] (c) Explain how a stretched relaxed muscle returns to its resting length in a living person, naming the role of titin. [3]

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(a) ATP binds to the myosin head and releases it from actin (1); splitting ATP re-sets the head (1); the head can then bind further along and pull again, so the filaments keep sliding (1). (b) After death no more ATP is made (1), so the heads cannot let go of actin and the cross-bridges stay attached, making the muscle stiff (1). (c) The antagonist contracts and stretches the relaxed muscle (1); the titin in the stretched sarcomere stores elastic energy and recoils the sarcomere to its resting length (1), and it also stops the sarcomere being overstretched (1).

Final answer

(a) ATP binds the head and releases it from actin, then its splitting re-sets the head for another pull. (b) No ATP is made, so the heads stay attached to actin and the filaments cannot slide: the muscle stays stiff. (c) An antagonist stretches the sarcomere; titin recoils the sarcomere and stops overstretching, returning it to its resting length.

Question C2

HL[8 marks]

Twenty volunteers were divided at random into group A, who did no exercise, and group B, who did hip-stretching exercises for six weeks. Maximum hip abduction was measured with a goniometer. The table gives mean angles with the standard error of the mean (SE, n=10 in each group). Group A (no exercise): start 33 (SE 1.3), end 34 (SE 1.4). Group B (stretching): start 32 (SE 1.4), end 48 (SE 1.5). Angles in degrees. (a) Calculate the percentage change in the mean angle for each group. [2] (b) Use the SE values to evaluate whether the groups differ after six weeks. [2] (c) Design a procedure to measure hip abduction reliably in each volunteer. [4]

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(a) Group B: \dfrac{48-32}{32}\times100=50\,\% (1); group A: \dfrac{34-33}{33}\times100\approx3\,\% (1). (b) After exercise: group B gives 48\pm1.5=46.5 to 49.5 and group A gives 34\pm1.4=32.6 to 35.4 (1); the ranges do not overlap, so the difference between the groups is probably real (1). (c) The independent variable is the exercise given and the dependent variable the angle in degrees (1); the same start position, the same tester and device and the same time of day are controlled (1); the goniometer centre is placed over the joint and the arms along the body segments (1); each volunteer is measured at least three times and the mean taken (1).

Final answer

(a) (48-32)/32\times100=50\,\% for group B and (34-33)/33\times100\approx3\,\% for group A. (b) Group B: 48\pm1.5 gives 46.5 to 49.5; group A: 34\pm1.4 gives 32.6 to 35.4; they do not overlap, so probably a real difference. (c) Same start position, goniometer centred on the joint, same tester, repeated readings; compare means with SE.

Question C3

HL[7 marks]

Actin filaments are 1.0\ \mum long and each myosin filament is 1.6\ \mum long. The sarcomere lengths in four micrographs of one myofibril are M1 2.6\ \mum, M2 2.2\ \mum, M3 1.8\ \mum and M4 1.6\ \mum. (a) Identify the most contracted sarcomere. [1] (b) Calculate the length of the H zone in M1. [2] (c) Calculate the length of the light band in M2. [1] (d) Explain why no H zone is seen in M3. [2] (e) State which length is the same in all four micrographs. [1]

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(a) M4, which has the shortest sarcomere (1). (b) The H zone is the sarcomere length minus the two actin filaments: 2.6-2\times1.0=0.6\ \mum (2). (c) The light band is the sarcomere length minus the dark band: 2.2-1.6=0.6\ \mum (1). (d) At 1.8 \mum two actin filaments of 1.0 \mum would measure 2.0 \mum, so they overlap in the middle and no part of the dark band holds myosin only; the H zone is absent (2). (e) The dark band, because it is as long as the myosin filaments, which do not shorten (1).

Final answer

(a) M4. (b) H zone =2.6-2\times1.0=0.6\ \mum. (c) Light band =2.2-1.6=0.6\ \mum. (d) In M3 the thin filaments from the two Z-lines have met in the middle, so no stretch of the dark band holds only myosin. (e) The dark band: 1.6 \mum in every micrograph.

Question C4

HL[7 marks]

Three diving mammals were compared. A humpback whale also swims every year from cold feeding waters to warm breeding waters. Species A: myoglobin 6 mg per g, typical dive 5 min. B: 25 and 20. C: 55 and 40. (a) Describe the relationship between myoglobin content and dive time. [2] (b) Suggest how a greater oxygen store allows longer dives. [2] (c) Describe two features of a whale that reduce the energy needed for swimming. [2] (d) State the reason for locomotion shown by the whale's yearly trip. [1]

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(a) The longer dive times go with the higher myoglobin content (1); quote values, such as 5 minutes at 6 mg per g and 40 minutes at 55 mg per g (1). (b) Muscle with more myoglobin holds a larger oxygen store (1), so aerobic respiration can supply ATP for the muscles for longer during a dive (1). (c) Two linked features, such as a streamlined body that reduces drag (1) and a fluke moved up and down that gives thrust (1); one mark each. (d) The yearly trip between feeding and breeding waters is migration (1). Maximum 7.

Final answer

(a) Dive time rises as myoglobin rises: 5 minutes at 6, 20 at 25 and 40 at 55 mg per g. (b) More stored oxygen lets muscle make ATP aerobically for longer, so the animal can stay under water longer. (c) A streamlined body reduces drag; a fluke moved up and down gives thrust. (d) Migration.

Question C5

HL[7 marks]

The angle between the ribs and the muscle fibres was measured in the 8th intercostal space of six volunteers. The table gives the mean angle between the rib and the muscle fibres, in degrees, measured on the front side of the fibres. Mean angle in degrees: external intercostal back 38, front 52; internal intercostal back 122, front 135. (a) State the mean angle of the internal muscle at the front. [1] (b) Compare and contrast the angles in the two muscles. [3] (c) Explain how the arrangement lets the pair move the ribs in both directions. [3]

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(a) 135^\circ (1; units needed). (b) Similarity: in both muscles the front value is larger than the back (1); difference: the external angles are below 90^\circ and the internal above (1); the internal angles are larger at both positions (1). (c) The fibres run in opposite directions, so the layers pull the ribs in opposite directions (1); the external layer contracts to lift the ribs and the internal layer to lower them, with the other layer relaxed (1); the layer that is stretched stores elastic energy in its titin (1), which recoils to help the next movement.

Final answer

(a) 135^\circ. (b) Both are larger at the front than at the back; the internal angles are larger, with all internal angles above 90^\circ and all external below. (c) The layers run in opposite directions, so they pull the ribs in opposite directions: the external layer raises them, the internal lowers them; each stretches the other, and its titin stores energy that helps the next movement.

Connected questions

Question X1

Connected questions · HL[6 marks]

B3.3 + B3.1 + C1.2. During a 400 m race a sprinter's breathing rate and the strength of her muscle contractions both rise. (a) Explain how the intercostal muscles and diaphragm cause air to enter the lungs. [2] (b) Explain why her muscle fibres need oxygen at a higher rate. [2] (c) Suggest why anaerobic respiration is used in the last metres. [2]

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(a) The external intercostal muscles contract and move the ribs up and out (1); the diaphragm contracts and flattens, which increases the volume of the thorax and lowers the pressure, so air enters (1). (b) The myosin heads need ATP to release actin and be re-set, so ATP use rises (1); aerobic respiration in the mitochondria uses oxygen from the lungs to make most of the ATP (1). (c) When oxygen runs short, anaerobic respiration makes ATP quickly (1) but yields little ATP per glucose and produces lactate (1).

Final answer

(a) The external intercostals contract and lift the ribs and the diaphragm flattens, so thorax volume rises and pressure falls, drawing air in. (b) Her muscle fibres use ATP faster; aerobic respiration in mitochondria makes ATP, using oxygen. (c) Anaerobic respiration makes ATP faster but less of it per glucose and produces lactate.

Question X2

Connected questions · HL[6 marks]

B3.3 + C2.2 + B2.3. Botulinum toxin stops motor neurons releasing their neurotransmitter at neuromuscular junctions. (a) Outline the events at a neuromuscular junction. [2] (b) Explain the effect of the toxin on a motor unit. [2] (c) Explain two ways a muscle fibre is specialised for contraction. [2]

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(a) An action potential arrives at the axon terminal (1); neurotransmitter is released from vesicles by exocytosis, diffuses across the cleft and binds to receptors on the muscle fibre, starting an action potential in the fibre (1). (b) No neurotransmitter is released, so the fibres receive no signal and do not contract (1); the whole motor unit is paralysed (1). (c) A fibre has many nuclei, so a very long cell can make enough protein for all its myofibrils (1), and many mitochondria, which supply the ATP that the myosin heads use (1).

Final answer

(a) An action potential reaches the terminal, vesicles release acetylcholine by exocytosis, it diffuses across the cleft and binds receptors on the fibre, which starts an action potential in the fibre. (b) With no neurotransmitter, no signal reaches the fibres, so the muscle is paralysed. (c) The fibre is a long cell with many nuclei, so many genes are active, and it has many mitochondria for ATP.

Question X3

Connected questions · HL[6 marks]

B3.3 + B3.2 + C1.2. A Weddell seal can dive for more than 20 minutes. Its blood carries more haemoglobin than a human's, and its muscles hold much myoglobin. (a) Explain how haemoglobin helps to supply the muscle with oxygen. [2] (b) Explain how a high myoglobin content helps the diving seal. [2] (c) Suggest why a dive cannot continue once the oxygen stores are used up. [2]

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(a) Haemoglobin in the red blood cells binds oxygen in the lungs (1) and carries it in the blood to the muscle, where it is released (1). (b) Myoglobin stores oxygen inside the muscle fibre (1), so aerobic respiration can supply ATP to the myosin heads for longer (1). (c) When the oxygen runs out the muscle relies on anaerobic respiration, which makes little ATP (1) and produces lactate, so the dive must end (1).

Final answer

(a) Haemoglobin in red blood cells binds oxygen at the lungs and carries it in the blood to the muscles. (b) More myoglobin gives a bigger store of oxygen inside the muscle for aerobic respiration. (c) When the oxygen is used up, the muscle must use anaerobic respiration, which makes little ATP and lactate, so the dive ends.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Muscle and motility from the core ideas to full exam questions, with question types, worked examples and graded practice.

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