Gas exchange

Describe the roles of ribosomes, tRNA and ATP in the process of translation.

Describe and explain the adaptations of leaves that maximise gas exchange.

Do not refer to organelles in your answer.

Describe the pathway taken by blood between a kidney and the coronary arteries.

Include names of the major blood vessels (arteries and veins) and heart valves involved.

A student completed a dissection of an animal gas exchange system.

Figure 5 shows the scientific drawing he produced from his observations.

Suggest two ways the scientific drawing in Figure 5 could be improved.

Do not refer to adding labels or using a sharp pencil in your answer.

Describe the role of structures A and B in increasing the volume in the thoracic cavity.

Lung volume can be determined with a spirometer using the helium dilution method.

The lung volume (V2) is calculated using the equation:

where C1 = the initial concentration of helium in the spirometer C2 = the helium concentration in the spirometer when equal with the lungs V1 = volume of gas in the spirometer.

A student used a spirometer with a volume of 3250 cm³ and an initial concentration of 10% helium. They found C2 was 3.78%

Use the equation to calculate this student’s lung volume in dm³

Show your working.

Asbestos is a fibrous mineral. If asbestos is inhaled, it can cause a lung disease

called asbestosis.

The latency period of a disease is the time between exposure to the cause of the disease and the symptoms occurring. The latency period for asbestosis is 15 to 40 years.

Asbestosis can be diagnosed using a biopsy. This is where a sample of lung cells is removed from a patient and examined with a microscope.

Asbestosis can also be diagnosed without a biopsy. To diagnose asbestosis without a biopsy, five conditions must be met. Three of these are given below.

1. A scan of the lungs showing fibrous scar tissue.

2. Reduced gas exchange.

3. Shortness of breath.

These three conditions are also present in many other lung diseases.

Using the information provided, suggest the other two conditions that must be met before a diagnosis of asbestosis can be made.

Suggest two advantages of diagnosing asbestosis without a biopsy.

Idiopathic pulmonary fibrosis (IPF) is a lung disease similar to asbestosis. Both

diseases cause:

• the death of alveolar epithelial cells due to DNA damage

• dead cells to be replaced by inelastic, fibrous scar tissue.

Currently, there is no cure for asbestosis and no effective treatment to slow its progression. There is also no cure for IPF. However, the drugs nintedanib and pirfenidone can slow the formation of scar tissue and slow the progression of IPF.

In two separate clinical trials, doctors investigated the effect of nintedanib and pirfenidone on 263 IPF patients.

• 124 patients were treated with nintedanib. 7 of these patients had to stop treatment due to side effects.

• 139 patients were treated with pirfenidone. 14 of these patients had to stop treatment due to side effects.

The doctors recorded how many days each patient lived after treatment.

Figure 6 shows their results.

Suggest why neither clinical trial had a control group.

The doctors concluded that both drugs increased life expectancy in patients with IPF compared with survival data for previous patients with IPF.

Using all the information, evaluate whether nintedanib and pirfenidone should be considered as treatments for asbestosis.

Turn over for Section B

Section B

one question.

You are advised to spend no more than 45 minutes on this section.

Describe how we breathe in.

A scientist prepared alveolar tissue to view using an optical microscope. The scientist cut very thin slices of the alveolar tissue.

Explain why the scientist used very thin slices of alveolar tissue with the optical microscope.

Figure 2 is an image of the lung tissue observed using an optical microscope.

Identify the tube labelled A.

The scientist used a ruler to measure the diameter of some of the alveoli.

Table 2 shows the scientist’s results.

The magnification of the image in Figure 2 is × 40

Use this information and Table 2 to calculate the mean diameter, in μm, of the alveoli.

Show your working.

Give the uncertainty associated with taking a measurement using a ruler with 1 mm graduations.

Calculate the percentage error for a measurement using the ruler of 4 mm

Describe and explain the relationship between surface area to volume ratio of the human body and metabolic rate.

Table 2 shows the height and mass of two adults.

The surface area of a person is estimated using the following formula:

The volume of a person is estimated using the following formula:

Using suitable calculations, deduce which person has the smaller surface area to volume ratio.

Show your working and complete the sentence below.

Give your answer to 3 significant figures.

Which is not a possible explanation for the difference in surface area to volume ratio between person A and person B ?

Tick () one box.

Name two structural features of the gills of a fish that increase their surface area.

Chitin is a polysaccharide. The chitin monomer is a β-glucose molecule with one OH group replaced by an NHCOCH₃ group. NHCOCH₃ can be represented by N(Ac).

Figure 1 shows the monomer that forms chitin and the chitin polymer.

Chitin has a similar structure to cellulose.

Use Figure 1 to describe three ways the structure of chitin is similar to the structure of cellulose.

Chitin keeps the tracheae open in the tracheal system of gas exchange in an insect. Gas exchange does not occur in the tracheae.

Explain the importance of one adaptation of the gas exchange surface in the tracheal system of an insect.

Lignin is a polymer found in the walls of xylem vessels in plants. Lignin keeps the xylem vessel open as a continuous tube.

Explain the importance of the xylem being kept open as a continuous tube.

Figure 11 shows images of gills from two fish as seen through an optical microscope.

Image C shows gills from a fish with healthy gills.

Image D shows gills from a fish with damaged gills.

Figure 11

Figure 11 not reproduced here due to third-party copyright restrictions

To observe the fish gills with the optical microscope, the scientists used two different stains. The first stain binds to DNA; the second stain binds to the red blood cells.

Explain why a second stain would be needed to stain the red blood cells. Suggest which molecule the stain could bind to in the red blood cells.

Using Figure 11, the scientists calculated the surface area to volume ratios for each gill filament in these two fish. Some of their results are shown in Table 4.

Complete Table 4. State your calculated volume and surface area:volume ratio to 2 significant figures.

The damage to the gills causes uncontrolled cell division in the cells around the capillaries in the gill filaments.

Other than surface area:volume ratio, describe one way this uncontrolled cell division changes the gills, as shown in Figure 11.

Explain how this difference would affect gas exchange.

Figure 12 shows the general pattern of blood circulation in fish.

Use Figure 12 to complete Table 5 to show two differences between the circulation of blood in fish and the circulation of blood in a mammal.

Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake.

Figure 2 shows two models of oxygen uptake found in animals.

Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B.

Figure 3 shows changes in concentration of oxygen in two gas exchange systems.

A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system.

Use Figure 3 to justify this conclusion.

Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system.

Table 1 shows features of two mammals.

Bats are flying mammals; shrews are ground-living mammals.

Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew.

Give your answer to an appropriate number of significant figures.

Give one suggestion to explain this difference.

Particulate matter is solid particles and liquid particles suspended in air. Polluted air contains more particulate matter than clean air.

A high concentration of particulate matter results in the death of some alveolar epithelium cells. If alveolar epithelium cells die inside the human body they are replaced by non-specialised, thickened tissue.

Explain why death of alveolar epithelium cells reduces gas exchange in human lungs.

Scientists grew alveolar epithelium cells and exposed the epithelium cells to different concentrations of particulate matter. They calculated the percentage of these alveolar epithelium cells that died after 24 hours of exposure to particulate matter. Their results are shown in Figure 6.

Do the data in Figure 6 show a linear relationship between concentration of particulate matter and percentage of dead cells?

Use suitable calculations to justify your answer.

Space for your calculations:

Scientists investigated stomatal density on leaves of one species of tree.

Figure 9 shows three examples of the square fields of view the scientists used to calculate a mean stomatal density.

Calculate the mean stomatal density in the three fields of view in Figure 9.

Give your answer as number of stomata per mm²

Show your working.

The scientists used leaves from individual trees that had grown in different areas of

the world in different years. Each tree had grown in an area and year with known carbon dioxide concentration.

Their results are shown in Figure 10.

Give a null hypothesis for this investigation and name a statistical test that would be appropriate to test your null hypothesis.

From 1910 to 2000, the carbon dioxide concentration in the atmosphere increased from 300 parts per million to 365 parts per million.

Use Figure 10 to calculate the mean rate of change in stomatal density from 1910 to 2000.

Give your answer as number of stomata per mm² per 10-year period.

Show your working.

A journalist saw Figure 10 and suggested that future increases in atmospheric carbon dioxide concentration could result in less transpiration.

Evaluate his suggestion.

Describe the advantage of the Bohr effect during intense exercise.

A cyclist completed a fitness test on an exercise bike. The intensity of the exercise was increased every 10 seconds. The test finished when he was unable to cycle any further. The partial pressure of oxygen (pO₂) and of carbon dioxide (pCO₂) in air breathed out was measured.

Figure 3 shows the results of the cyclist’s fitness test.

Ventilatory threshold (VT) is a measure of the point when anaerobic respiration increases because aerobic respiration alone can no longer maintain muscle contraction.

VT can be identified as the first point when there is an increase in pO₂ breathed out, without an equivalent increase in pCO₂ breathed out.

Use Figure 3 to determine the time after the exercise started when the cyclist reached VT.

Calculate the ratio of pO₂ to pCO₂ in breathed-out air at this time.

Show your working.

An increase in the intensity of exercise produces an increase in the volume of carbon dioxide produced.

However, Figure 3 shows that the pCO₂ in air breathed out did not show a large increase during the exercise.

Suggest one physiological change that would cause this result. Explain how the physiological change would allow for the removal of the increase in the volume of carbon dioxide produced.

Physiological change

When muscle fibres have very low concentrations of ATP, they may get ATP in the

following ways.

• AMPK (an enzyme) oxidises fatty acids.

• Phosphocreatine donates phosphate to ADP in anaerobic conditions.

Figure 4 shows how these chemicals work.

At more than 80% of maximum muscle effort, ATP can only be made for a limited time.

Use Figure 4 to suggest one reason why.

Tick () the correct box.

GW1516 is a performance-enhancing drug. GW1516 activates AMPK and develops slow muscle fibres at rest.

Use Figure 4 to justify why professional athletes are not allowed to take GW1516.

Do not include details of chemiosmotic theory in your answer.

EPO is another performance-enhancing drug. It can increase the haematocrit (the percentage of red blood cells in blood).

A heart attack is caused by a lack of glucose and oxygen being delivered to cardiac muscle via the coronary arteries. The overuse of EPO can increase the risk of a heart attack.

Suggest how.

The normal haematocrit for human males is 47(±5)%. For professional male cyclists, the maximum haematocrit allowed is 50%.

A student suggested that professional male cyclists should be allowed to use EPO until their haematocrit is 50%.

Give two reasons why this suggestion is not valid.

Section B

one question.

You are advised to spend no more than 45 minutes on this section.

The water potential of leaf cells is affected by the water content of the soil.

Scientists grew sunflower plants. They supplied different plants with different volumes of water.

After two days, they determined the water potential in the leaf cells by using an instrument that gave a voltage reading.

The scientists generated a calibration curve to convert the voltage readings to water potential.

Figure 8 shows their calibration curve.

The scientists needed solutions of known water potential to generate their calibration curve.

Table 5 shows how to make a sodium chloride solution with a water potential of

Complete Table 5 by giving all headings, units and volumes required to make 20 cm³ of this sodium chloride solution.

Table 6 shows some of the concentrations of sodium chloride solution the scientists used and the water potential of each solution.

There is a linear relationship between the water potential and the concentration of sodium chloride solution.

Use the data in Table 6 to calculate the concentration of sodium chloride solution with a water potential of 3.41MPa

In addition to determining the water potential in the leaf cells, the scientists measured the growth of the leaves.

They recorded leaf growth as a percentage increase of the original leaf area.

Their results are shown in Figure 9.

One leaf with an original area of 60cm² gave a voltage reading of 7 µV

Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area. Give your answer in cm²

Sunflowers are not xerophytic plants. The scientists repeated the experiment with xerophytic plants.

Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 9.

Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow only slowly.

Describe the gross structure of the human gas exchange system and how we breathe in and out.

Mucus produced by epithelial cells in the human gas exchange system contains triglycerides and phospholipids.

Compare and contrast the structure and properties of triglycerides and phospholipids.

Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with the sugar, lactose, attached.

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein.

Figure 4 shows the stages of development of an insect called a damselfly.

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.

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.

A scientist measured the size of each gill lamella of the gills of 40 damselfly larvae. His results are shown in Table 5.

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. Show your working.

A student used an optical microscope to observe part of a damselfly larva gill.

Figure 5 shows the drawing the student produced.

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

Write an essay on one of the topics below.

EITHER

The importance of nitrogen-containing substances in biological systems.

OR

The importance of diffusion in organisms.