Required practical 5: dissection of an organ system - AQA A-Level Biology (Paper 3) Exam Questions

Practise 15 exam-style questions on Required practical 5: dissection of an organ system for AQA A-Level Biology (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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15 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A student is going to dissect a set of pig lungs, with the trachea attached, obtained from a butcher. She will push a length of rubber tubing into the trachea and use it to inflate the lungs.

Give two safety precautions she should take when inflating the lungs, other than wearing a lab coat.

Model answer

1. Inflate the lungs with a foot pump, never by blowing into the tubing.
2. Inflate the lungs inside a large clear plastic bag so that any fluid or droplets are contained.

Explanation

Lungs from a butcher may carry microorganisms, so the main risks are infection and splashes of fluid. A pump keeps the student's mouth away from the tissue, and a plastic bag, eye protection and gloves stop splashes reaching the skin or eyes; hands and surfaces are cleaned afterwards. Vague answers such as 'be careful' earn no credit.

Mark scheme

1 mark
Inflate the lungs with a foot pump / bellows / hand pump, not by blowing into the tubing (by mouth)
Max 2. List rule: if more than two precautions are given, each wrong extra answer cancels one correct answer. Ignore 'take care' and 'wear a lab coat'.
1 mark
Put the lungs inside a (large, clear) plastic bag while inflating them (to contain splashes / droplets)
1 mark
Wear eye protection / goggles (in case fluid from the lungs splashes)
1 mark
Wear (disposable) gloves / cover any cuts on the hands (with waterproof dressings)
1 mark
Do not over-inflate the lungs (so that they do not burst and spray fluid)
1 mark
Wash hands with soap / disinfect the bench and equipment afterwards
Accept 'use antibacterial hand wash'.

Question 2

3 marks

A student inflated a pair of pig lungs with a pump attached to the trachea.

When she disconnected the pump, the lungs deflated without being squeezed. Explain why.

She then cut a small piece of tissue from the edge of one lung and put it into a beaker of water. The piece of tissue floated. Explain why.

Another student suggested that the lungs deflated only because air leaked out through small holes in their surface. Suggest how this suggestion could be tested on a pair of inflated lungs before any cuts are made.

Model answer

The alveolar walls contain elastic fibres, which were stretched when the lungs were inflated. When the pump was removed, the fibres recoiled and pushed the air out.

The piece of lung tissue contains air in its alveoli, so it is less dense than water and floats.

To test the suggestion, inflate the lungs and then clamp the tube in the trachea shut. If the lungs stay inflated, no air is escaping through their surface, so when the tube is open the lungs must deflate because elastic recoil pushes the air out through the trachea.

Explanation

Observations made during a dissection can be explained from structure. Elastic fibres stretch during inspiration and recoil during expiration, so a lung empties without being squeezed. Lung tissue is mostly air-filled alveoli, so it floats. A good test of a suggestion changes one thing and states the result that would support or reject it: clamping the trachea closes the exit used by recoil, so only a leak could then let the lungs deflate.

Mark scheme

1 mark
Elastic fibres / elastic tissue (in the walls of the alveoli) were stretched when the lungs were inflated AND recoiled, pushing the air out
Max 3. Points 1 and 2 explain the observations; points 3 and 4 are ways of testing the suggestion (max 1). Accept 'elastic recoil of the alveoli / lungs'. Insufficient: 'the air escaped' with no idea of elasticity.
1 mark
The tissue contains air (in the alveoli), so it is less dense than water
Accept 'air is trapped in the alveoli / air sacs'.
1 mark
Inflate the lungs and then close / clamp / tie off the tube or the trachea: if the lungs stay inflated, no air is leaking through the surface (so deflation is due to elastic recoil pushing air out through the trachea)
Points 3 and 4: 1 mark each, max 1. The method AND what result would show a leak (or no leak) are both needed. Insufficient: 'look for holes' with no method of detecting escaping air.
1 mark
Brush soapy water over the surface of the inflated lungs / hold the inflated lungs under water, and look for bubbles of air escaping (bubbles would show a leak)
Accept 'listen for air escaping' only with the trachea closed.

Question 3

3 marks

A student cut 2 cm off the bottom of a celery stalk while holding the cut end under water. Without letting the cut end come out of the water, she moved the stalk into a beaker of water containing a dye that moves through the xylem. After 30 minutes she cut slices about 2 mm thick from the bottom of the stalk until she could no longer see any dye. The dye had moved 84 mm up the stalk.

Calculate the mean rate of movement of the dye. Give your answer in mm per minute.

The student concluded that water moves up the xylem of celery at 2.8 mm per minute. Give two reasons why this value may not be an accurate measure of the rate of water movement in the xylem.

Model answer

Rate = 84 ÷ 30 = 2.8 mm per minute

The dye molecules may bind to the walls of the xylem, so the dye moves more slowly than the water. Also, the dye at the leading edge may be too faint to see, so the distance it moved, and therefore the rate, would be underestimated.

Explanation

Rate = distance ÷ time. A dye is only a marker for the water: if it moves more slowly than the water, or its leading edge cannot be seen, the measured rate is lower than the true rate of water movement. Other limitations are the thickness of the slices, the use of a single stalk with no repeats, and the assumption that the rate was constant for 30 minutes.

Mark scheme

1 mark
2.8 (mm per minute) (a correct answer of 2.8 with no working scores this mark)
Max 3. Point 1 is the calculation: a correct answer of 2.8 with no working scores this mark (84 ÷ 30). Do not accept 0.36 (30 ÷ 84) or 0.047 (a rate in mm per second). Points 2 to 6 are reasons, max 2.
1 mark
The dye may move more slowly than the water (e.g. dye molecules bind to / are absorbed by the xylem walls), so the rate of water movement is underestimated
Points 2 to 6: 1 mark each, max 2. List rule: if more than two reasons are given, each wrong or contradictory extra answer cancels one correct answer. Ignore references to light, temperature or humidity (these change the rate but do not make the measurement inaccurate).
1 mark
The dye at the leading edge may be too faint / dilute to see, so the distance moved (and the rate) is underestimated
1 mark
The slices were about 2 mm thick, so the distance moved is only known to the nearest slice (about 2 mm)
Accept 'the resolution of the method is limited by the thickness of the slices'.
1 mark
It is a mean over 30 minutes: the rate may not have been constant (e.g. faster at the start)
1 mark
Only one stalk was used / there were no repeats, so the value may not be representative (stalks vary)
Accept 'water may move at different rates in different xylem vessels'.

Question 4

6 marks

During the dissection of a sheep's heart, a student compared the volumes of the two ventricles. She tied off the aorta and the pulmonary artery with string and cut away the atria just above the valves between the atria and the ventricles. She then filled each ventricle with water from a burette until it was full, and recorded the volume of water added. The left ventricle held 82 cm³ and the right ventricle held 88 cm³.

In a living sheep, each ventricle pumps out about 60% of the blood it contains during each beat, and the resting heart rate is 80 beats per minute. Use the volume of the left ventricle to estimate the sheep's resting cardiac output. Give your answer in dm³ per minute.

The student had expected the left ventricle to hold much more water than the right ventricle, because its wall is much thicker. Explain why, in a living sheep, the two ventricles pump out the same volume of blood in each beat.

Give two reasons why the volumes she measured may be different from the volumes of the ventricles in a living sheep.

Model answer

Stroke volume = 82 × 0.60 = 49.2 cm³
Cardiac output = stroke volume × heart rate = 49.2 × 80 = 3936 cm³ per minute = 3.9 dm³ per minute

The right ventricle pumps blood through the lungs, and all of it returns to the left side of the heart, which pumps it round the body and back to the right side. The two circulations are in series, so the two ventricles must pump out the same volume in each beat; otherwise blood would build up in the lungs or in the body. The thicker wall of the left ventricle produces a higher pressure, not a larger volume.

The heart muscle may have stiffened after death, so the ventricles did not expand as they do when they fill with blood. Water may also have leaked out around the string ties, so more water was added than the ventricle holds.

Explanation

Cardiac output = stroke volume × heart rate: first find the stroke volume (60% of 82 cm³), then multiply by the heart rate and convert cm³ to dm³ by dividing by 1000. Because the pulmonary and systemic circulations are joined in series, both ventricles have the same stroke volume; wall thickness reflects the pressure each ventricle must produce, not the volume it holds. Measurements on an organ from a dead animal are limited by changes after death and by the method, such as leaks and trapped air.

Mark scheme

1 mark
Correct method: 82 × 0.60 × 80 (÷ 1000); or a correct final answer of 3.9 (accept 3.9 to 3.94) with no working
Max 6. Points 1 and 2 are the calculation: a correct answer of 3.9 dm³ per minute (accept 3.9 to 3.94) with no working scores 2 marks. Accept for 1 mark: 3936 (cardiac output in cm³ per minute, not converted) OR evidence of a stroke volume of 49.2 cm³ OR 6.6 (accept 6.56; 60% not used) OR 4.2 (accept 4.22; the volume of the right ventricle used). Points 3 to 5 explain the equal volumes, max 2. Points 6 to 9 are reasons, max 2.
1 mark
3.9 (dm³ per minute) (accept 3.9 to 3.94)
Accept 3.936.
1 mark
The pulmonary and systemic circulations are in series: all the blood pumped by the right ventricle (through the lungs) returns to the left side of the heart, and all the blood pumped by the left ventricle returns to the right side
Points 3 to 5: 1 mark each, max 2. Accept 'the same blood passes through both ventricles in turn'.
1 mark
If one ventricle pumped more blood in each beat than the other, blood would build up in one circulation (e.g. in the lungs) and the other circulation would run short
Accept 'the outputs of the two sides of the heart must balance'.
1 mark
The thicker wall of the left ventricle produces a higher pressure (to pump blood to the whole body), not a larger volume
Accept 'wall thickness affects the force of contraction, not the volume of the chamber'.
1 mark
The heart muscle has stiffened / contracted after death (or with cold storage), so the ventricles do not expand as they would when filling with blood
Points 6 to 9: 1 mark each, max 2. List rule: if more than two reasons are given, each wrong or contradictory extra answer cancels one correct answer. Ignore 'measurement error', 'human error' and 'the heart was dead' with no effect on the volume.
1 mark
Water leaked out (around the string ties / through the cut vessels or valves), so more water was added than the ventricle holds
1 mark
Air bubbles were trapped in the ventricle, so it held less water than its volume
1 mark
Cutting away the atria / tying the arteries distorted or damaged the ventricles, changing their volume
Accept 'the cut may not have been made at the top of each ventricle'.

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