OxfordAQA International AS and A-level Biology: Skeletal muscle and sliding filaments (9610) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions test section 3.4.4, skeletal muscles as effectors, from OxfordAQA International AS and A-level Biology (9610, Version 5.1). Explanations are in the study guide and the revision notes. See also the course hub, the printable checklist and the free diagnostics.

Questions

1. A person bends the elbow to lift a book. Explain why the arm needs two muscles, the biceps and the triceps, and why the skeleton must be incompressible. [3]

2. (a) Describe the gross structure of a skeletal muscle. [2] (b) Describe the microscopic structure of a skeletal muscle fibre. [3]

3. An electron micrograph of one myofibril is at ×24 000. A relaxed sarcomere measures 60.0 mm on the image and a contracted sarcomere measures 46.8 mm. The A band measures 37.2 mm. (a) Calculate the actual lengths of the relaxed and contracted sarcomeres, in µm. [2] (b) Calculate by what percentage the sarcomere shortens. [1] (c) Calculate the length of the I band in the relaxed and contracted sarcomeres. [2]

4. Four images (W to Z) show the same myofibril at different stages. Each thin filament is 0.95 µm and the A band is 1.55 µm. The sarcomere lengths are: W 1.85 µm, X 2.50 µm, Y 2.15 µm, Z 3.05 µm. (a) Put the images in order from most contracted to most relaxed. [1] (b) For image Y, calculate the I band and the H zone. [2] (c) Explain what the data show about the filaments. [3]

5. Describe how the filaments in a sarcomere change position and how the interaction of myosin, actin, calcium ions and ATP brings this about. [6]

6. A chemical stops ATP binding to myosin heads. It does not affect calcium ions. A muscle fibre in calcium-rich fluid is treated with it. Predict the state of the fibre and explain your answer. [4]

7. The table shows the tension developed by a muscle preparation at different calcium ion concentrations (data supplied).

Calcium ions / µmol dm⁻³	0.05	0.2	0.5	1.0	2.0	5.0
Tension / % of maximum	0	8	40	72	96	100

(a) Describe the relationship shown. [2] (b) Calculate the mean increase in tension per µmol dm⁻³ between 0.5 and 1.0. [1] (c) Explain the tension at 5.0 µmol dm⁻³ compared with 0.05 µmol dm⁻³. [3]

8. A muscle holds 7.2 mmol of ATP and 21.6 mmol of phosphocreatine per kg. In an all-out effort it uses 2.4 mmol of ATP per kg each second. Take it that each phosphocreatine molecule regenerates one ATP. (a) Calculate the time for which the ATP store alone would supply the muscle. [1] (b) Calculate how long both stores last together. [2] (c) Calculate the percentage of the stored ATP supply that comes from phosphocreatine. [1]

9. The table shows the phosphocreatine and ATP content of a muscle during 10 s of maximal effort, as a percentage of the resting value (data supplied).

Time / s	0	2	4	6	8	10
Phosphocreatine	100	70	45	25	12	8
ATP	100	99	97	95	92	90

(a) Calculate the mean rate of fall of phosphocreatine between 0 and 4 s. [1] (b) Explain why the ATP content falls so little. [2] (c) Between 6 s and 8 s the phosphocreatine falls at a lower rate than between 0 s and 2 s. Suggest why. [2] (d) Describe how the phosphocreatine store is restored after the effort. [2]

10. Data supplied for two fibre types from one muscle.

	Fibre J	Fibre K
Diameter / µm	48	76
Mitochondria / % of fibre volume	7.5	2.1
Time to peak force / ms	110	38
Time for force to fall to 50% / s	540	24

(a) Identify the slow fibre, using two pieces of evidence. [2] (b) Calculate how many times greater the mitochondrial volume of J is than K, to 1 decimal place. [1] (c) Explain how the properties of J suit a postural muscle. [3]

11. Bending the elbow lifts a forearm. Describe and explain the changes in the sarcomeres of the biceps and of the triceps as the elbow bends, referring to antagonistic action. [6]

Answers

1. Muscles can only pull when they contract, not push [1]; the biceps contracts to bend the elbow while the triceps relaxes, and the triceps contracts to straighten it, so two muscles are needed to move the bone in both directions [1]; an incompressible skeleton does not squash, so the pull is transmitted through the tendon to the bone, which acts as a lever [1] [3] Examiner insight: Say “pull” and “relaxes” explicitly. An answer that says the muscles “work together” does not show the opposing action.

2. (a) Many bundles of muscle fibres wrapped in connective tissue [1]; supplied with blood vessels and nerves, and attached to bone by tendons [1] [2] (b) Each fibre is one long cell with many nuclei [1]; the sarcoplasm contains mitochondria and sarcoplasmic reticulum, and is full of parallel myofibrils [1]; a surface membrane (sarcolemma) surrounds the fibre and the myofibrils give cross-striations in a longitudinal section [1] [3] Examiner insight: “Gross” describes the whole organ and “microscopic” describes the cells. Give features at the right scale or the points are lost.

3. (a) Relaxed: 60.0 ÷ 24 000 = 0.0025 mm = 2.5 µm [1]; contracted: 46.8 ÷ 24 000 = 0.00195 mm = 1.95 µm [1] [2] (b) (2.5 − 1.95) ÷ 2.5 × 100 = 22.0% [1] [1] (c) A band = 37.2 ÷ 24 000 = 1.55 µm; relaxed I band = 2.5 − 1.55 = 0.95 µm [1]; contracted I band = 1.95 − 1.55 = 0.40 µm [1] [2] Examiner insight: Show the unit conversion and give the answer in the unit asked for.

4. (a) W, Y, X, Z [1] [1] (b) I band = 2.15 − 1.55 = 0.60 µm [1]; H zone = 2.15 − 2 × 0.95 = 0.25 µm [1] [2] (c) The A band is the same in every image [1], so the thick filaments do not change length [1]; the I band and H zone shrink as the sarcomere shortens (H zone is 0.60 µm in X but 0 in W), so the thin filaments are moving between the thick ones rather than getting shorter [1] [3] Examiner insight: Quote values from the data when you explain, and say which band stays constant.

5. Thin filaments are drawn inwards between the thick filaments, so the Z lines approach each other and the sarcomere shortens [1]; calcium ions released from the sarcoplasmic reticulum move tropomyosin off the binding sites on actin [1]; myosin heads attach to those sites, forming actomyosin bridges [1]; the heads change shape and drag the actin filaments towards the middle of the sarcomere [1]; ATP binds to the heads so they detach [1]; ATP is hydrolysed and the energy returns the heads to their original position, so the cycle repeats [1] [6] Examiner insight: Follow a sequence and say what each molecule does. A list of the four names, with no roles, shows little.

6. The fibre stays contracted, rigid and unable to relax [1]; calcium ions keep binding sites exposed so myosin heads bind and tilt [1]; without ATP the heads cannot detach [1]; so the bridges stay attached and the fibre cannot lengthen or cycle further [1] [4] Examiner insight: Predict first, then explain. “No contraction” is a poor prediction, since binding does not need ATP.

7. (a) Tension rises as calcium ion concentration rises [1]; the rise is steepest between about 0.2 and 1.0 µmol dm⁻³ and levels off near 100% above 2.0 [1] [2] (b) (72 − 40) ÷ (1.0 − 0.5) = 64% per µmol dm⁻³ [1] [1] (c) At 0.05 µmol dm⁻³ tropomyosin still covers the binding sites, so no bridges form and tension is zero [1]; at 5.0 µmol dm⁻³ calcium ions have moved tropomyosin and exposed binding sites [1]; so many myosin heads bind and tension is at its maximum [1] [3] Examiner insight: “Describe” needs the pattern with values, including where it levels off. Explanations should name tropomyosin and the exposed binding sites.

8. (a) 7.2 ÷ 2.4 = 3.0 s [1] [1] (b) (7.2 + 21.6) = 28.8 mmol per kg [1]; 28.8 ÷ 2.4 = 12.0 s [1] [2] (c) 21.6 ÷ 28.8 × 100 = 75.0% [1] [1] Examiner insight: Show the total before dividing. The assumption of one ATP per phosphocreatine is what allows the two stores to be added.

9. (a) (100 − 45) ÷ 4 = 13.75%, about 13.8% per s [1] [1] (b) Phosphocreatine passes its phosphate to ADP, regenerating ATP as fast as it is used [1]; so the ATP concentration stays nearly constant while phosphocreatine falls [1] [2] (c) Less phosphocreatine is left to use [1]; some ATP is now being supplied by glycolysis and respiration (so less is needed from phosphocreatine) [1] [2] (d) During rest, ATP from aerobic respiration [1] is used to add phosphate to creatine, rebuilding phosphocreatine [1] [2] Examiner insight: Explain the small fall in ATP by regeneration, not by saying the muscle is not using it. In recovery, state the source of the ATP.

10. (a) J is the slow fibre: it has a higher proportion of mitochondria (7.5% against 2.1%) [1]; it has a longer time to peak force and force falls much more slowly, so it resists fatigue [1] [2] (b) 7.5 ÷ 2.1 = 3.6 [1] [1] (c) Postural muscles must contract for long periods [1]; many mitochondria allow sustained aerobic respiration [1]; slower fatigue, with a rich oxygen supply from myoglobin and capillaries, means the muscle keeps holding posture without tiring [1] [3] Examiner insight: Link each property to function. Naming “many mitochondria” earns little unless you say what it allows.

11. The biceps contracts and the triceps relaxes [1]; in the biceps, myosin heads form bridges and pull thin filaments inwards [1]; the sarcomere, I band and H zone get shorter while the A band stays the same [1]; the triceps is stretched by the contracting biceps, with no bridges holding the filaments, so its thin filaments are pulled back out [1]; its sarcomeres get longer, with wider I bands and H zones [1]; the tendons pass the pull to the forearm, a lever on the incompressible skeleton [1] [6] Examiner insight: Cover both muscles. Answers about the biceps alone leave half of the question unanswered.

Where marks are usually lost

• Stating that the filaments, A band or myosin get shorter in contraction.
• Leaving out calcium ions or tropomyosin when explaining how binding sites become available.
• Describing ATP as “energy for the muscle” with no mention of detachment or resetting of the head.
• Forgetting that removal of calcium ions, and ATP for the pumps, is needed for relaxation.
• Dividing by magnification without converting mm to µm.
• Mixing up the I band and H zone when calculating from sarcomere length.
• Treating phosphocreatine as a long-term store.
• Explaining only one muscle of an antagonistic pair.
• Giving fibre properties without naming the fibre type.

Next steps

• Review the condensed revision notes.
• Read the full study guide.
• Go to the OxfordAQA A-level Biology hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

Source: OxfordAQA International AS and A-level Biology (9610), specification Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, published by OxfordAQA. Section 3.4.4, Skeletal muscles as effectors.