3. Mid UNIT

Question 1

The adult damselfly uses a tracheal system for gas exchange.

Explain three ways in which an insect’s tracheal system is adapted for efficient gas exchange.

Answer 1

1. Tracheoles have thin walls so short diffusion distance to cells;
2. Highly branched / large number of tracheoles so short diffusion distance to cells;
3. Highly branched / large number of tracheoles so large surface area (for gas exchange);
4. Tracheae provide tubes full of air so fast diffusion (into insect tissues);
5. Fluid in the end of the tracheoles that moves out (into tissues) during exercise so faster diffusion through the air to the gas exchange surface;

OR

Fluid in the end of the tracheoles that moves out (into tissues) during exercise so larger surface area (for gas exchange);

6. Body can be moved (by muscles) to move air so maintains diffusion / concentration gradient for oxygen / carbon dioxide;
7. Do not accept unqualified references to thin membranes.

Max 2 if any reference to blood

Ignore references to spiracles

5. Accept ‘water’ for fluid.

Accept ‘cells’ and ‘tissues’ as interchangeable words.

Question 2

The damselfly larva is a carnivore that actively hunts prey. It has gills to obtain oxygen from water.

Some other species of insect have larvae that are a similar size and shape to damselfly larvae and also live in water. These larvae do not actively hunt prey and do not have gills.

Explain how the presence of gills adapts the damselfly to its way of life.

Answer 2

1. Damselfly larvae has high(er) metabolic / respiratory (rate);
2. (So) uses more oxygen (per unit time / per unit mass);

Idea of ‘more / high’ is needed for both mark points.

2. Accept ‘needs’ for ‘uses’
3. Ignore references to absorbing / obtaining / uptake of more oxygen

Question 3

A scientist measured the size of each gill lamella of the gills of 40 damselfly larvae.

His results are shown in the table.

Mean width / mm (± uncertainty / mm) 1.61 (± 0.19)

Mean width / mm

(± uncertainty / mm) 6.12 (± 0.41)

Calculate the mean surface area of one side of one gill lamella. Assume that a gill lamella is rectangular and give your answer to an appropriate number of significant figures.

Include the percentage error (uncertainty) of surface area in your answer.

Answer 3

Mean SA = 9.85 mm 2 / 9.9 mm 2 ;

Percentage uncertainty of SA = 18.5 / 18.7 / 19;

If both answers incorrect 1 mark for

Percentage uncertainty of dimensions 11.8 / 12 and 6.70 / 6.7

Surface area correctly calculated with correct units but not rounded to appropriate sf (9.8532 mm 2)

Surface area correct (with appropriate sf) but no / incorrect unit given

Question 4

Suggest two ways the student could improve the quality of her scientific drawing of this gill.

Answer 4

1. Don’t use shading;
2. Only use single lines / don’t use sketching (lines) / ensure lines are continuous / connected;
3. Add further labels / annotations;
4. Don’t cross label lines;
5. Add magnification / scale (bar

Question 5

Describe and explain how the countercurrent system leads to efficient gas exchange across the gills of a fish.

Answer 5

1. Water and blood flow in opposite directions;

Accept: diagram if clearly annotated

2. Maintains concentration / diffusion gradient / equilibrium not reached / water always next to blood with a lower concentration of oxygen;

Must have the idea of ‘maintaining’ or ‘always’ in reference to concentration / diffusion gradient

Accept: constant concentration / diffusion gradient

3. Along whole / length of gill / lamellae;

Accept: gill plate / gill filament

Question 6

Amoebic gill disease (AGD) is caused by a parasite that lives on the gills of some species of fish. The disease causes the lamellae to become thicker and to fuse together.

AGD reduces the efficiency of gas exchange in fish. Give two reasons why.

Answer 6

1. (Thicker lamellae so) greater / longer diffusion distance / pathway;

Q Neutral: ‘thicker’ diffusion pathway

2. (Lamellae fuse so) reduced surface area;

Accept: reduced SA:VOL

Question 7

The table below shows some features of gas exchange of a fish at rest.

Volume of oxygen absorbed by the gills from each dm3 of water / cm3 7

Mass of fish / kg 0.4

Oxygen required by fish / cm3 kg–1 hour–1 90

(i) Calculate the volume of water that would have to pass over the gills each hour to supply the oxygen required by the fish. Show your working.

Answer 7

Correct answer of 5.1 or 5.14(2857) (dm3) = 2 marks;;

Allow 1 mark max for an answer of 5 if the correct answer of 5.1 or 5.14(2857) is not shown

One mark for incorrect answers that show 36 or 0.4 × 90 or 90 ÷ 7;

Question 8

The volume of water passing over the gills increases if the temperature of the water increases. Suggest wh

Answer 8

1. Increased metabolism / respiration / enzyme activity;

Accept: enzymes work more efficiently

2. Less oxygen (dissolved in water);

Question 9

Explain how an arteriole can reduce the blood flow into capillaries.

Answer 9

Muscle contracts;

2. Constricts/narrows arteriole/lumen;

Accept decreases for constricts/narrows

Accept vasoconstriction for 1 mark

Question 10

What can you conclude from the appearance of valves in the image above about heart muscle activity and blood movement between:

1. ventricles and arteries?

Answer 10

1. Ventricle (muscles) relaxed
2. No (blood) backflow (into ventricles)
3. Atria (muscle) contracted;
4. Blood movement from atria (into ventricles);

Accept flow/pumped for movement

Question 11

Explain how changes in the shape of haemoglobin result in the S-shaped (sigmoid) oxyhaemoglobin dissociation curve for HbA.

Answer 11

First oxygen binds (to Hb) causing change in shape;

(Shape change of Hb) allows more O2 to bind (easily) / greater saturation with O2

Question 12

At birth 98% of the haemoglobin is HbF. By the age of 6 months, the HbF has usually completely disappeared from the baby’s blood and been replaced by HbA.

Use the graph above to explain why this change is an advantage for the baby.

Answer 12

1. (HbA has) lower affinity for O2 at low partial pressures;

OR

(HbA has) lower affinity for oxygen at pp found in tissues;

2. Easier unloading of O2 for (aerobic) respiration;

Question 13

The scientists concluded that treatment with hydroxyurea would increase the concentration of oxygen in the blood of babies with SCD.

Suggest how the graph and table above support this conclusion.

Answer 13

1. A large/significant increase in HbF;
2. (HbF has) higher affinity for O2 (than faulty HbA);
3. Higher proportion of HbF in blood so more oxygen carried;

OR

More oxygen carried after treatment;