Topic 3 Assessment Booklet

Marks = 155Time Allowed 202 minutes

Q1.(a)     Explain two ways in which the structure of fish gills is adapted for efficient gas exchange.

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(2)

(b)     Explain how the counter current mechanism in fish gills ensures the maximum amount of the oxygen passes into the blood flowing through the gills.

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(3)

(Total 5 marks)

Q2.

(a)     Binding of one molecule of oxygen to haemoglobin makes it easier for a second oxygen molecule to bind.

Explain why.

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(2)

A haemocytometer is a special microscope slide which can be used to count the numbers of blood cells in a sample of blood.

•        The surface of the slide has many small, equal-sized squares marked on it.

•        The depth of the liquid under each square is 0.1 mm

•        When counting, cells that touch top or left lines are counted but cells that touch right or bottom lines are not counted.

A doctor used a haemocytometer to determine the number of red blood cells per mm3 in a blood sample. He diluted the original blood sample by a factor of 200 times before putting some on a haemocytometer.

The diagram shows the distribution of cells in a typical small square.

(b)     The doctor counted the red blood cells in many small squares.

The mean number of red blood cells per small square was 7

The original blood sample was diluted by a factor of 200 times.

Calculate the number of red blood cells per mm3 in the original blood sample.

Give your answer in standard form.

Answer = ____________________ red blood cells per mm3

(2)

(c)     When counting, cells that touch top or left lines are counted but cells that touch right or bottom lines are not counted.

Suggest two reasons for this rule.

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(2)

The doctor also wanted to know how many white blood cells per mm3 there were in a different sample of blood. To do this he first diluted the sample by a factor of 20 times. He then made the white blood cells clearly visible by using a stain that makes nuclei appear dark blue.

(d)     When counting white blood cells, the doctor only diluted the blood sample by a factor of 20 times, instead of 200 times when counting red blood cells.

Suggest why he only diluted the sample by a factor of 20 times.

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(e)     Explain how the stain allowed the doctor to count the white blood cells amongst all the red blood cells.

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(1)

(Total 8 marks)

Q3.

Under the correct conditions, new roots grow from the cut end of a plant stem. A scientist investigated the effect of substance X on the growth of new roots.

She used a ringing experiment to investigate the movement of substance X in stems taken from lemon plants. She cut out a length of stem from each plant. She then put a small block of agar on the top of each length of stem. Some agar blocks contained substance X.

The diagram below shows how she treated each length of stem.

She grew the lengths of stem in the same environmental conditions for 6 weeks, and then found the number of roots per length of stem. Roots grew at the other end of the stem from where the agar blocks were placed.

The table below shows the scientist’s results.

 

Treatment

Mean number of roots per length of stem

D

5

E

11

F

4

G

3

(a)     Treatment D is a control. Explain how the measurement obtained from this control is used by the scientist.

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(2)

(b)     Using the diagram and the table above, what can you conclude from treatments D and E about root growth?

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(3)

(c)     The mass flow hypothesis is used to explain the movement of substances through phloem.

Evaluate whether the information from this investigation supports this hypothesis.

Do not consider statistical analysis in the answer.

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(4)

(Total 9 marks)

Q4.

(a)     Describe the pathway taken by an oxygen molecule from an alveolus to the blood.

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(2)

(b)     Explain how one feature of an alveolus allows efficient gas exchange to occur.

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(2)

Carbon monoxide is a poisonous gas that is present in cigarette smoke. This carbon monoxide can be absorbed into the blood where it binds with haemoglobin.

Scientists investigated the concentration of carbon monoxide in cars in which people were smoking or not smoking. They measured the concentration with the car windows open and closed. The graph shows the scientists’ results as they presented them. A value of ± 2 standard deviations from the mean includes over 95% of the data.

(c)     In England, in October 2015, a law was introduced making it illegal to smoke in a car carrying someone who is under the age of 18.

Following the introduction of the law, a politician stated:

‘It is dangerous to smoke when a child is in the car. Higher levels of deadly toxins can build up, even on short journeys, and children breathe faster than adults, meaning they inhale more of the deadly toxins.’

Use the information provided and the data in the graph to evaluate the politician’s statements.

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(4)

(Total 8 marks)

Q5.

Scientists investigated the effect of a heat treatment on mass transport in barley plants.

•        They applied steam to one short section of a leaf of the heat-treated plants. This area is shown by the arrows in Figure 1.

•        They did not apply steam to the leaves of control plants.

•        They then supplied carbon dioxide containing radioactively-labelled carbon to each plant in the area shown by the rectangular boxes in Figure 1.

•        After 4 hours, they:

◦        found the position of the radioactively-labelled carbon in each plant. These results are shown in Figure 1.

◦        recorded the water content of the parts of the leaf that were supplied with radioactively-labelled carbon dioxide. These results are shown in the table.

Figure 1

 

Plant from which the leaf was taken

Water content of leaf / % of maximum

(± 2 standard deviations)

Heat-treated Plant

A

84.6

(±11.3)

Control Plant, not heat treated

B

92.8

(±8.6)

(a)     The scientists concluded that this heat treatment damaged the phloem.

Explain how the results in Figure 1 support this conclusion.

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(2)

(b)     The scientists also concluded that this heat treatment did not affect the xylem.

Explain how the results in the table support this conclusion.

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(2)

(c)     The scientists then investigated the movement of iron ions (Fe3+) from the soil to old and young leaves of heat-treated barley plants and to leaves of plants that were not heat treated. Heat treatment was applied half way up the leaves. The scientists determined the concentration of Fe3+ in the top and lower halves of the leaves of each plant.

Their results are shown in Figure 2.

Figure 2

What can you conclude about the movement of Fe3+ in barley plants?

Use all the information provided.

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(4)

(Total 8 marks)

Q6.

(a)     Sodium ions from salt (sodium chloride) are absorbed by cells lining the gut. Some of these cells have membranes with a carrier protein called NHE3.

NHE3 actively transports one sodium ion into the cell in exchange for one proton (hydrogen ion) out of the cell.

Use your knowledge of transport across cell membranes to suggest how NHE3 does this.

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(3)

(b)     Scientists investigated the use of a drug called Tenapanor to reduce salt absorption in the gut. Tenapanor inhibits the carrier protein, NHE3.

The scientists fed a diet containing a high concentration of salt to two groups of rats, A and B.

•   The rats in Group A were not given Tenapanor (0 mg kg−1).

•   The rats in Group B were given 3 mg kg−1 Tenapanor.

One hour after treatment, the scientists removed the gut contents of the rats and immediately weighed them.

Their results are shown in the table.

 

Concentration of Tenapanor / mg kg−1

Mean mass of contents of the gut / g

0

2.0

3

4.1

The scientists carried out a statistical test to see whether the difference in the means was significant. They calculated a P value of less than 0.05.

They concluded that Tenapanor did reduce salt absorption in the gut.

Use all the information provided and your knowledge of water potential to explain how they reached this conclusion.

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(4)

(c)     High absorption of salt from the diet can result in a higher than normal concentration of salt in the blood plasma entering capillaries. This can lead to a build-up of tissue fluid.

Explain how.

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(2)

(Total 9 marks)

Q7.

The graph shows the oxyhaemoglobin dissociation curves for fetal haemoglobin (HbF) and adult haemoglobin (HbA).

 

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

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(2)

(b)     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.

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(2)

(c)     Sickle cell disease (SCD) is caused by production of faulty HbA. This results in a reduced ability to transport oxygen to tissues. Scientists investigated the use of a substance called hydroxyurea to treat babies with SCD. Hydroxyurea changes the concentration of HbF in the blood.

The scientists carried out an investigation with 122 babies who had SCD. Each baby was given hydroxyurea for 41 months. The scientists then found the mean change in the concentration of HbF in the babies’ blood.

Their results are shown in the table.

 

Mean concentration of HbF in the babies’ blood / arbitrary units

Before treatment with hydroxyurea (± 1 standard deviation)

After treatment with hydroxyurea (± 1 standard deviation)

7.6

(± 4.5)

19.1

(± 6.5)

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.

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(3)

(Total 7 marks)

Q8.

A student used a potometer to measure the movement of water through the shoot of a plant. The potometer is shown in Figure 1. As water is lost from the shoot, it is replaced by water from the capillary tube.

 

(a)     In one experiment, the air bubble moved 7.5 mm in 15 minutes. The diameter of the capillary tube was 1.0 mm.

Calculate the rate of water uptake by the shoot in this experiment.

Give your answer in mm3 per hour. Show your working. (The area of a circle is found using the formula, area = πr 2)

____________________ mm3 hour−1

(2)

(b)     The student wanted to determine the rate of water loss per mm2 of surface area of the leaves of the shoot in Figure 1.

Outline a method she could have used to find this rate. You should assume that all water loss from the shoot is from the leaves.

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(3)

(c)     The rate of water movement through a shoot in a potometer may not be the same as the rate of water movement through the shoot of a whole plant.

Suggest one reason why.

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(1)

(d)     Aquaporins are channel proteins that allow the diffusion of water across membranes. One type of aquaporin, called PIP1, can also transport carbon dioxide molecules across membranes.

Figure 2 shows the structure of a water molecule and of a carbon dioxide molecule. They are drawn to the same scale.

 

Suggest two reasons why water molecules and carbon dioxide molecules can both pass through PIP1.

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(2)

(e)     The scientists first produced transgenic poplar trees. These trees all had a length of foreign DNA inserted into them. This DNA led to the production of single-stranded RNA that specifically inhibited expression of the gene for PIP1.

The scientists then measured the difference in the amount of PIP1 in leaves of transgenic poplars and in leaves of wild type poplars without the foreign DNA. The amount of PIP1 in the transgenic poplars was approximately 15% of that in the wild type poplars.

Using this information, what can you conclude about the effect of the foreign DNA in the transgenic poplar trees?

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(3)

(f)      The transgenic poplars still produced some PIP1.

Suggest why.

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(1)

(g)     The scientists investigated the importance of PIP1 in the movement of water and carbon dioxide through the tissues of leaves of poplar trees.

They measured the mean rates of movement of carbon dioxide and water through the tissues of leaves of transgenic poplars and through the tissues of leaves of wild type poplars.

Their results are shown in the graph below.

 

Using only the graph above, evaluate the importance of PIP1 in the movement of carbon dioxide and water through leaves of poplar trees.

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(3)

(Total 15 marks)

Q9.

(a)     Describe the cohesion-tension theory of water transport in the xylem.

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(5)

(b)     Describe how mRNA is produced in a plant cell.

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(5)

(Total 10 marks)

Q10.

The diagram outlines the digestion and absorption of lipids.

 

(a)     Tick (✔) the box by the name of the process by which fatty acids and glycerol enter the intestinal epithelial cell.

 

Active transport

Diffusion

Endocytosis

Osmosis

(1)

(b)     Explain the advantages of lipid droplet and micelle formation.

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(3)

(c)     Name structure Q in the diagram above and suggest how it is involved in the absorption of lipids.

Name _____________________________________________________________

How it is involved ____________________________________________________

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(4)

(Total 8 marks)

Q11.

Haemoglobin transports oxygen around the body of many animals.

(a)     Haemoglobin is a protein with a quaternary structure.

Explain the meaning of quaternary structure.

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(b)     When fully saturated, each molecule of haemoglobin is bound to four molecules of oxygen.

The graph shows the percentage saturation of haemoglobin with oxygen at different partial pressures.

 

Give the formula for calculating the percentage saturation of haemoglobin with oxygen.

 

Percentage saturation of haemoglobin with oxygen

 

=

 

(1)

(c)     The first molecule of oxygen to bind causes a change in the shape of the haemoglobin molecule.

This change of shape makes it easier for other oxygen molecules to bind to the haemoglobin molecule.

Explain how the graph provides evidence for this.

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(2)

(d)     Suggest one advantage of this change in the affinity of haemoglobin for oxygen.

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(1)

(e)     Tests on the man whose blood was used to construct the graph gave the following data.

•        Concentration of haemoglobin in blood = 150 g dm−3.

•        Volume of oxygen carried by fully saturated haemoglobin = 1.35 cm3 g−1.

•        Resting heart rate = 60 beats minute−1.

•        Volume of blood pumped out of left ventricle each beat = 60 cm3.

Use these data and information from the graph to calculate the volume of oxygen released to the man’s tissues per minute whilst he was at rest.

Show your working.

Answer =______________cm3 minute−1

(3)

(Total 8 marks)

Q12.

Boron is an element that is needed in very small amounts for normal plant growth.

One group of scientists tested a hypothesis that boron combines with sucrose to produce a sucrose-borate complex that is translocated more effectively than sucrose molecules.

They grew tomato plants in nutrient-poor sand. Prior to starting their experiment, they left the mature plants in a dark room for 48 hours.

For each plant, the scientists put one of its leaves into a solution of sucrose that was radioactively labelled. These leaves were left attached to the plants. They used two radioactively labelled sucrose solutions:

•        solution A contained boron at a concentration of 10 parts per million.

•        solution B contained no boron.

After a period of time, the scientists removed samples from parts of the plants, dried them in an oven and ground each into a powder. They then measured the radioactivity in each powdered sample. The scientists’ results are shown in the table.

 

 Part of plant

Mean radioactivity / counts minute−1 g−1

Plants with leaf immersed in solution A (with boron)

Plants with leaf immersed in solution B (no boron)

 Stem tip

14.2

  1.7

 First leaf above treated leaf

  3.3

  0.0

 Upper stem

31.2

  8.3

 Lower stem

28.3

13.3

 First leaf below treated leaf

21.7

  0.0

 Roots

  3.5

  1.7

(a)     Explain the following steps in the scientists’ method.

They grew tomato plants in nutrient-poor sand.

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They left the mature plants in the dark for 48 hours before starting their experiment.

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(2)

(b)     The scientists dried the plant samples in an oven at 100 °C.

Give two reasons why they used this temperature.

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(2)

(c)     Do the scientists’ results support their hypothesis?

Use evidence from the table to support your answer.

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(4)

(d)     Suggest how the scientists could adapt their method to determine which tissue carried the radioactively labelled sucrose.

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(2)

(Total 10 marks)

Q13.

The graph shows the volume changes in the left ventricle of a human heart during two cardiac cycles. The numbers 1, 2, 3 and 4 represent times when heart valves open or close.

 

(a)     Use information from the graph to complete the table in part (a). Place the number 1, 2, 3 or 4 in the appropriate box.

 

 

Valve opens

Valve closes

Semi-lunar valve

 

 

Atrioventricular valve

 

 

(2)

(b)     Use the diagram above to calculate the volume of blood pumped per minute by the left ventricle.

Answer = ____________________ cm3 min−1

(2)

(c)     Explain the role of the heart in the formation of tissue fluid.

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(2)

(d)     Lymphoedema is a swelling in the legs which may be caused by a blockage in the lymphatic system.

Suggest how a blockage in the lymphatic system could cause lymphoedema.

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(1)

(Total 7 marks)

Q14.

Scientists measured the rate of carbon dioxide release by three groups of insects of the same species at 10 °C, 20 °C and 30 °C. They also determined the mean mass of each group of insects.

The scientists results are shown in the table.

 

 Temperature / °C

Mean mass / g

Rate of carbon dioxide release / µdm3 minute−1

Rate of carbon dioxide

release per gram /

µdm3 g−1 minute−1

 10

0.047

0.12

 

 20

0.046

0.33

 

 30

0.048

0.56

 

(a)     Complete the table above and plot a graph of your calculated values against temperature on the graph paper. Express your calculated rates with the appropriate number of significant figures.

 

(3)

The body temperature of the insects was largely determined by the temperature they were kept at. At each temperature, the scientists recorded rate of carbon dioxide release by individual insects over time. This rate depends upon spiracles opening or closing.

The graphs below show results for three insects.

 

(b)     Calculate the change in the rate per hour of opening of the spiracles between 10 °C and 20 °C.

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(1)

(c)     Explain how you could determine the total amount of carbon dioxide secreted at 30 °C during the period of recording.

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(1)

(d)     Suggest an explanation for the effect of temperature on the rate of carbon dioxide release.

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(3)

(Total 8 marks)

Q15.

Many humans are unable to digest lactose. A scientist investigated the production of lactose-free milk. He produced gel beads containing the enzyme lactase and placed the beads in a column. He poured milk (Milk A) into the column and collected the milk (Milk B) after it had moved through the column over the beads. This is shown in the diagram below.

 

(a)     Milk A contains no glucose. Milk B contains glucose. Explain why Milk B contains glucose.

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(b)     The enzyme was trapped within the gel beads. Suggest one advantage of trapping the enzyme within the gel beads.

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The scientist varied the flow rate of the milk through the column. The effect of flow rate on the concentration of glucose in Milk B is shown in the table below.

 

Flow rate of milk through the column / cm3 minute−1

Concentration of glucose in Milk B / arbitrary units

50

45

100

6

(c)     Explain the difference in the results in the table.

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(1)

(d)     The gel beads were all similar sizes. Use the formula below to calculate the volume of one of the beads with a 3.0 mm diameter.

Volume of sphere = πr3

Volume = _____________________ mm3

(1)

(e)     Galactose has a similar structure to part of the lactose molecule.

Explain how galactose inhibits lactase.

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(2)

(Total 6 marks)

Q16.

Scientists investigated the effect of humidity on the movement of the insect, Tenebrio molitor.

The insects were placed in choice chambers with one side kept at 100% humidity and the other side kept at 30% humidity.

The insects were used one at a time and the path the insect followed recorded on paper.

Figure 1 shows a typical result. The solid dot shows the final recorded position of the insect.

Figure 1

 

(a)     What type of behaviour was shown by the insect in Figure 1?

Give evidence from Figure 1 to support your answer.

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(2)

(b)     The scientists found that the insects moved for 94% of the time in the more humid side, but in the drier side they moved only 20% of the time. The scientists concluded that reduced movement in the drier side was an adaptation that reduced water loss.

Use your knowledge of gas exchange in insects to explain how this behaviour would reduce water loss in the insects.

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(2)

(c)     Tenebrio molitor has two antennae on its head. These are sense organs.

The scientists found that one insect stopped when it reached the boundary between the two sides of the choice chamber and seemed to perform various movements with its antennae. The insect then moved to the drier side.

This behaviour can be seen in Figure 2. The points marked with a Q indicate where the insect showed this behaviour.

Figure 2

 

What type of behaviour did the scientists conclude that the insect in Figure 2 was showing?

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(1)

(d)     After observing the behaviour of the insect in Figure 2, the scientists hypothesised that if an insect had one of its antennae removed it would have a tendency to turn to one side and move in circles. The scientists tested this hypothesis by cutting one antenna off another insect and observing its movement.

The result of this experiment can be seen in Figure 3.

Figure 3

 

Does the movement observed in Figure 3 support the scientists’ hypothesis?

Give the reason for your answer.

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(2)

(e)     The scientists then investigated the effect of a range of humidities on the activity of the insects. Figure 4 shows their results. The triangles represent the number of insects still moving after 15 minutes.

Figure 4

 

A student studying Figure 4 concluded that as humidity increases, so does movement of the insects.

Evaluate the student’s conclusion.

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(2)

(Total 9 marks)

Q17.

Read the following passage.

 

Some insect species feed on the leaves of plants. These leaf-chewersbite off pieces of leaves. Other insect species feed on sap from phloemor xylem. These sap-feeders have sharp, piercing mouthparts that theyinsert directly into either xylem or phloem. Leaf-chewers and insectsthat feed on xylem sap are active feeders; this means they use theirjaw muscles to obtain their food. In contrast, insects that feed on phloemsap are passive feeders; this means they do not use their jaw musclesto take up sap from phloem.

Feeding on phloem sap presents two problems. Firstly, phloem sap hassa high sugar concentration. This could lead to a high pressure of liquidin the insect’s gut because of water entering the gut from the insect’sbody tissues. A phloem-sap-feeder polymerises some of these sugarsinto polysaccharides which are passed out of its anus as ‘honey dew’.The secondproblem is that phloem sap has a low concentration ofamino acids. Phloem-sap-feeding insects rely on bacteria in their guts toproduce amino acids. Each phloem-sap-feeding insect receives a few ofthese bacteria from its parent. This has resulted in a reduction in thegenetic diversity of the bacteria found within these insects.

A scientist investigated the effect of three different insects on the growthof a plant called the goldenrod. He found that leaf-chewing insects andxylem-sap-feeding insects caused a much greater reduction in totalleaf area than did phloem-sap-feeding insects.

5

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20

Use the information from the passage and your own knowledge to answer the following questions.

(a)     Phloem-sap-feeders are passive feeders (lines 6–7).

Phloem-sap-feeders do not use their jaw muscles to take up sap from phloem.

Explain why they can take up sap without using their jaw muscles.

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(3)

(b)     A phloem-sap-feeder polymerises some of these sugars into polysaccharides (line 12-13).

Suggest the advantage of this.

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(2)

(c)     Each phloem-sap-feeding insect receives a few of these bacteria from its parent.

(lines 16–17).

Suggest how this has caused a reduction in genetic diversity of the bacteria.

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(2)

(d)     A scientist found that leaf-chewers and xylem-sap-feeders had a greater effect on plant growth than phloem-sap-feeders (lines 20–22).

Other than environmental factors, give two features the scientist would have controlled in his experiment to ensure this conclusion was valid.

1. _________________________________________________________________

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2. _________________________________________________________________

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(2)

(e)     The scientist used the reduction in total leaf area of the experimental plants as an indicator of plant growth.

Outline a method by which you could find the area of a plant leaf.

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(1)

(Total 10 marks)

Q18.

(a)     Explain four ways in which the structure of the aorta is related to its function.

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(4)

Figure 1 shows the oxyhaemoglobin dissociation curves for two different species, A and B.

Figure 1

 

(b)     Species B is more active than species A. Use Figure 1 to explain how the haemoglobin of species B allows a greater level of activity.

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(4)

(c)     An electrocardiogram (ECG) shows the electrical activity of the heart. Figure 2 shows an ECG for an animal of species B at rest. Each large spike represents a contraction of the ventricles.

Figure 2

 

For species B, the mean volume of blood leaving the left ventricle during each contraction is 0.03 cm3.

Calculate the mean volume of blood leaving the left ventricle per minute.

Volume of blood = ___________________________ cm3 minute−1

(2)

(Total 10 marks)

Page 1 of 52

Mark schemes

Q1.

(a)     1.      Many lamellae / filaments so large surface area;

2.      Thin (surface) so short diffusion pathway;

1 & 2 must each have a feature and a consequence

2

(b)     1.      Water and blood flow in opposite directions;

Allow diagram showing counter-flow

2.      Blood always passing water with a higher oxygen concentration;

3.      Diffusion gradient maintained throughout length (of gill)

OR

Diffusion occurs throughout length of gill

OR

If water and blood flowed in same direction equilibrium would be reached;

3

[5]

Q2.

(a)     1.      Binding of first oxygen changes tertiary / quaternary (structure) of haemoglobin;

Ignore ref. to ‘positive cooperativity’ unqualified

Ignore ref. to named bonds

Accept conformational shift caused

2.     Creates / leads to / uncovers second / another binding site

OR

Uncovers another iron / Fe / haem group to bind to;

Reject ref. to active site

2

(b)     5.6 × 106 (red blood cells per mm3);;

Award 1 max for one of

2.8 × 104 (standard form but ignoring dilution)

OR

5 600 000 (correct but not standard form)

OR

5.6 × 105 (failure to use depth of liquid on slide);

2

(c)     1.      To avoid dealing with parts of cells;

2.      To avoid counting same cells twice / more than once;

3.      To be consistent / get comparable results;

Accept more accuracy

Ignore reliability / repeatability / reproducibility / precision / validity

2 max

(d)     There are fewer white cells, so no need to dilute (further to see enough);

Accept converse of too few to see if greater dilution / at 200 times

Do not accept ref. to numbers of red and white cells unqualified

Ignore ref. to white cells larger

1

(e)     White cells have a nucleus (that stains but red cells do not);

Accept converse for red cells

1

[8]

Q3.

(a)     1.      Used to compare effect of other treatments / as a baseline;

Accept for 2 marks, substance (X) and not agar / block / water that caused the difference in the number of roots.

Do not accept unqualified reference to “compare results”.

2.      Shows / Measures effect of substance (X);

OR

Accounts for effect of substances produced naturally;

Accept measures effect of independent variable

2

(b)     1.      (D shows) substance (X) is not required for (some) root growth / production of roots;

OR

Substances (already) present in stem cause (some) root growth;

2.      Substance X moves through plant;

Accept X moves through stem / phloem

3.      (E shows) substance (X) causes / increases / doubles number of roots / root growth;

3

(c)     In support of mass flow hypothesis

1.      (F shows) phloem is involved;

2.      (G shows) respiration / active transport is involved (in flow / movement);

3.      Because 4 °C / cooling reduces / slows / stops flow / movement;

4.      The agar block is the source;

5.      Roots are the sink;

Against the mass flow hypothesis

6.      No bulge above ringing (in F);

7.      No (role for) osmosis / hydrostatic pressure / water movement;

Accept no turgor pressure

8.      Movement could be due to gravity;

9.      Roots still grow without (intact/functioning) phloem;

10.    No leaves / sugars / photosynthesis to act as a source;

Each point must be clearly made in the context of support or against.

Ignore sugar / sucrose

3 max for “support” and 3 max for “against”

4 max

[9]

Q4.

(a)     1.      (Across) alveolar epithelium;

2.      Endothelium / epithelium of capillary;

Incorrect sequence = maximum of 1 mark

2

(b)     1.      (The alveolar epithelium) is one cell thick;

Reject thin membrane

2.      Creating a short diffusion pathway / reduces the diffusion distance;

2 max

(c)     For

1.      Significantly higher concentrations of CO (compared with no smoking) with closed window (as no overlap in 2 × SD);

Accept higher concentrations of CO with closed window are not due to chance

Idea of higher is required, not just difference

2.      Any increase in CO could be dangerous;

OR

CO causes less oxygen to be carried / provided (which could be deadly in children);

3.      (significantly) higher levels after (just) 5 minutes (with closed windows supporting short journey statement);

Idea of higher is required, not just difference

Against

4.      No idea if (roughly) 5ppm is ‘deadly’;

5.      No significant difference with open window (as 2 × SD overlaps);

Accept difference with open window could be due to chance

6.      No data on child breathing rates;

OR

Idea that children breathe faster but have smaller lung volume, so overall volume of CO inhaled could be similar;

4 max

[8]

Q5.

(a)     EITHER

1.      The radioactively labelled carbon is converted into sugar/organic substances during photosynthesis;

For ‘organic substance’ accept named organic substance, eg glucose, sucrose, amino acid.

2.      Mass flow/translocation in the phloem throughout the plant only in plants that were untreated/B/control

OR

Movement of sugar/organic substances in the phloem throughout the plant only in plants that were untreated/B/control;

Accept ‘translocation/mass transport in the phloem past the heat treatment only in the untreated plant/B/control’.

Accept converse for heat-treated plant/A ie Movement of sugar/organic substances/mass flow/translocation in the phloem stops (beyond the heat treatment) in treated plants/A.

OR

3.      Movement in phloem requires living cells/respiration/active transport/ATP;

4.      Heat treatment damages living cells so transport in the phloem throughout the plant only in plants that were untreated/B/control

OR

Heat treatment stops respiration/active transport/ATP production so transport in the phloem throughout the plant only in plants that were untreated/B/control;

Do not mix and match – award either mp1 and mp2 or mp3 and mp4.

2

(b)     1.      (The water content of the leaves was) not different because (means ± 2) standard deviations overlap;

For ‘not different’ accept ‘difference is not significant’ or ‘difference due to chance’.

2.      Water is (therefore) still being transported in the xylem (to the leaf)

OR

Movement in xylem is passive so unaffected by heat treatment;

2

(c)     1.      Heat treatment has a greater effect on young leaves than old;

Accept description of no/little/(slight) increase effect in old leaves and change in young leaves.

2.      Heat treatment damages the phloem;

3.      Fe3+ moves up the leaf/plant;

4.      (Suggests) Fe3+ is transported in the xylem in older leaf;

5.      In young leaf, some in xylem, as some still reaches top part of leaf;

6.      (Suggests) Fe3+ is (mostly) transported in phloem in young leaf

OR

Xylem is damaged in young leaf

OR

Xylem is alive in young leaf;

7.      Higher ratio of Fe3+ in (all/untreated) old leaves than (all/untreated) young;

Accept ‘more at the top’ for ‘higher ratio’.

8.      All ratios show there is less Fe3+ in the top than the lower part of leaves;

9.      (But) no statistical test to show if the difference(s) is significant;

Accept ‘(But) no standard deviations to show if the difference(s) is significant’.

4 max

[8]

Q6.

(a)     1.      Co-transport;

2.      Uses (hydrolysis of) ATP;

3.      Sodium ion and proton bind to the protein;

4.      Protein changes shape (to move sodium ion and / or proton across the membrane);

3.      Accept ‘Na + and H + bind to protein’ but do not allow incorrect chemical symbols

3 max

(b)     1.      Tenapanor / (Group)B / drug causes a significant increase;

OR

There is a significant difference with Tenapanor / drug / between A and B;

2.      There is a less than 0.05 probability that the difference is due to chance;

3.      (More salt in gut) reduces water potential in gut (contents);

4.      (so) less water absorbed out of gut (contents) by osmosis

OR

Less water absorbed into cells by osmosis

OR

Water moves into the gut (contents) by osmosis.

OR

(so) water moves out of cells by osmosis.

1. and 2. Reject references to ‘results’ being significant / due to chance once only.

2.      Do not credit suggestion that probability is 0.05% or 5.

2.      Accept ‘There is a greater than 0.95 / 95% probability that any difference between observed and expected is not due to chance’

4

(c)     1.      (Higher salt) results in lower water potential of tissue fluid;

2.      (So) less water returns to capillary by osmosis (at venule end);

OR

3.      (Higher salt) results in higher blood pressure / volume;

4.      (So) more fluid pushed / forced out (at arteriole end) of capillary;

For ‘salt’ accept ‘sodium ions’.

Do not allow mix and match of points from different alternative pairs

3.      Accept higher hydrostatic pressure.

2

[9]

Q7.

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

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

OR

Cooperative binding;

2

(b)     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;

2

(c)     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;

3

[7]

Q8.

(a)     Correct answer 23.55 – 24 two marks;

For one mark

5.9

OR

94.2;

2

(b)     1.      Method for measuring area;

e.g. draw round (each) leaf on graph paper and count squares;

2.      Of both sides of (each) leaf;

3.      Divide rate (of water loss / uptake from potometer) by (total) surface area (of leaves);

3

(c)     Plant has roots

OR

xylem cells very narrow;

Ignore references to air bubbles / mass flow / photosynthesis

Accept xylem damaged when cut

1

(d)     1.      Both small / similar size (so fit channel);

2.      Have a similar shape (so bind to / fit channel);

1.      Accept same height and width

Ignore refs to polar / non-polar

2.      Accept Aquaporin complementary to oxygen(s)

2

(e)     1.      Single-stranded RNA (has base sequence) complementary to PIP1 mRNA;

2.      Binds to mRNA (of PIP1) / leads to destruction of mRNA;

3.      Prevents / reduces translation (of PIP1);

4.      Reduces photosynthesis/named process that uses water;

3.      Less made is insufficient

3 max

(f)      Not all of mRNA bound to single-stranded RNA / there is more mRNA than interfering RNA

OR

Not all mRNA destroyed / disabled;

Accept mutations in transgene,

Accept not all cells with transgenes

1

(g)     1.      Loss of PIP reduces water and carbon dioxide movement;

2.      Differences significant because SDs don’t overlap

OR

Need stats test to see whether significant differences (or not);

3.      Greater (proportional) effect on carbon dioxide transport;

4.      Not all movement through PIP;

1.      Accept converse for wild type

2.      Reject references to results significant or not significant

2.      Accept error bars for SDs

3 max

[15]

Q9.

(a)     1.      Water lost from leaf because of transpiration / evaporation of water (molecules) / diffusion from mesophyll / leaf cells;

OR

Transpiration / evaporation / diffusion of water (molecules) through stomata / from leaves;

2.      Lowers water potential of mesophyll / leaf cells;

3.      Water pulled up xylem (creating tension);

4.      Water molecules cohere / ‘stick’ together by hydrogen bonds;

5.      (forming continuous) water column;

6.      Adhesion of water (molecules) to walls of xylem;

2.      Accept Ψ or WP

5 max

(b)     1.      The DNA strands separate by breaking the H bonds;

OR

H bonds broken between (complementary) (DNA) bases;

2.      (Only) one of the strands/template strand is used (to make mRNA/is transcribed);

3.      (Complementary) base pairing so A ⟶ U, T ⟶ A, C ⟶ G, G ⟶ C;

4.      (RNA) nucleotides joined by RNA polymerase;

5.      pre-mRNA formed;

6.      Splicing / introns removed to form mRNA;

1.      Ignore ‘hydrolysis’ of bonds

1.      Accept DNA “unzips” by breaking the H bonds

6.      Accept ‘non-coding’ sections for introns

5 max

[10]

Q10.

(a)    Diffusion

Automarker

1

(b)     1.      Droplets increase surface areas (for lipase / enzyme action);

2.      (So) faster hydrolysis / digestion (of triglycerides / lipids);

3.      Micelles carry fatty acids and glycerol / monoglycerides to / through membrane / to (intestinal epithelial) cell;

1.      Context is important

1.      Reject micelles increase surface area

2.      Ignore ‘breakdown’

3.      Ignore ‘small enough’

3.      Accept description of membrane

3.      Reject any movement through membrane proteins

3

(c)     1.      Golgi (apparatus);

2.      Modifies / processes triglycerides;

3.      Combines triglycerides with proteins;

4.      Packaged for release / exocytosis

OR

Forms vesicles;

Ignore ‘processes and packages’ unqualified

2.      Reject synthesises triglycerides

3.      Accept ‘forms / are lipoproteins’

4

[8]

Q11.

(a)     (Molecule contains) more than one polypeptide (chain).

Accept: has four polypeptides

1

(b)     

1

(c)     1.      At low partial pressure of oxygen, little increase in saturation as oxygen increases;

2.      (then) rapid rise as it gets easier for oxygen to bind.

Accept use of appropriate numbers from graph

2

(d)     Ensures rapid / more intake of oxygen in lungs / release of oxygen in tissues.

1

(e)     Volume of blood leaving heart = (0.6 × 0.6) dm3 minute−1 = 3.6 dm3 minute-1

Mass of haemoglobin in this volume of blood = (3.6 × 150) g = 540g

Volume of oxygen at 100% saturation of this haemoglobin = (540 × 1.35) = 729 cm3

The graph shows 60% of this volume of oxygen has been released to the tissues, so final answer is (729 × 0.6) = 437.4 cm3 minute −1

3

[8]

Q12.

(a)     1.      Sand: to ensure no boron provided;

2.      Dark: to ensure no sucrose produced in leaves / produced by photosynthesis.

2

(b)     1.      Evaporates all water

2.      (But) does not burn (organic compounds).

2

(c)     Yes because in the presence of boron

1.      Uptake of sucrose greater;

2.      Transport to other parts of plant greater;

3.      Correct use of data that supports MP1 or MP2.

No because:

4.      No evidence that boron reacts with sucrose / that a sucrose-borate complex is formed.

4

(d)     1.      Take thin (horizontal) sections of plant tissue / stem;

2.      Place against photographic film in dark for several hours / carry out autoradiography.

2

[10]

Q13.

(a)

 

open

closed

Semi-lunar valves

2

3

Atrioventricular valves

4

1

One mark for each correct column

General marker

2

(b)    (Acceptable range is) 6315.79 to 6400;

Allow one mark for (SV = 120 − 40 =) 80 (cm 3)

OR

(1 cycle = 1.24 − 0.48 =) 0.76 (s)

OR

79 / 80 (beats minute −1)

2

(c)     1.      Contraction of ventricle(s) produces high blood / hydrostatic pressure;

2.      (This) forces water (and some dissolved substances) out (of blood capillaries);

1.      Do not accept contraction / pumping of the heart

1.      Reject blood / plasma / tissue fluid forced out

2

(d)     Excess tissue fluid cannot be (re)absorbed / builds up;

The idea of excess is important

Accept ‘drained’ for absorbed

1

[7]

Q14.

(a)     1.      Line graph with rate on y axis and temperature on x axis and linear scales;

2.      Values calculated to appropriate sf;

3.      Rates correctly calculated and plotted, with ruled line connecting points and no extrapolation;

3

(b)     8 or 9;

1

(c)     1.      Determine the area under the curve;

1

(d)     1.      Enzymes / metabolism faster;

2.      Higher rate of respiration and carbon dioxide production / release;

3.      Spiracles open more often / remain open to excrete / get rid of carbon dioxide / get more oxygen;

Note – explanation required

3

[8]

Q15.

(a)     Lactase hydrolyses lactose in to glucose (and galactose);

1

(b)     No lactase in the milk

OR

Enzyme can be reused.

1

(c)     100 cm3 minute–1 is too fast to bind to active site / converse for 50 cm3 minute–1;

1

(d)     14.1(4);

1

(e)     1.      Galactose is a competitive inhibitor / attaches to the active site (of lactase);

2.      Fewer enzyme substrate complexes formed.

2

[6]

Q16.

(a)     1.      Kinesis;

Ignore any prefix

2.      Movement is random / non-directional

OR

Insect is not moving towards a particular stimulus;

2

(b)     1.      Less respiration so less gas exchange;

2.      (So) spiracles open less so less water loss;

2

(c)     Taxis;

Ignore any prefix

1

(d)     No (no mark), the insect does not move in circles;

Shows kinesis / results similar to Figure 1;

2

(e)     For (max 1)

1.      The data show a positive correlation;

Must state this as the description is given in the stem.

2.      Large sample / number of insects so valid / reliable / representative;

Against (max 1)

3.      (however) there are overlaps in individual experiments at all humidities;

4.      70–90% humidity there is little / no change in movement / movement only increases after 76% humidity;

Accept any value in this range

2 max

[9]

Q17.

(a)     1.      Contents of phloem vessel pushed into insect’s mouth by high pressure;

2.      (High pressure in phloem vessel) caused by loading of sugars into phloem in leaf;

3.      And (resulting) entry of water by osmosis.

3

(b)     1.      Polysaccharides are insoluble;

2.      So do not affect water potential of gut.

2

(c)     1.      (Only few bacteria passed from parent, so) only a few (copies of) genes passed on (in bacteria);

2.      May not / does not include all alleles (of genes, so diversity reduced)

OR

Small number of bacteria transmitted means unrepresentative sample.

2

(d)     1.      Number / mass / density of insects per plant;

2.      Stage of development / size of plants / insects;

Ignore any abiotic factor

2

(e)     Draw around leaf on graph paper and count squares;

1

[10]

Q18.

(a)     1.      Elastic tissue to allow stretching / recoil / smoothes out flow of blood / maintains pressure;

2.      (Elastic tissue) stretches when ventricles contract

OR

Recoils when ventricle relaxes;

3.      Muscle for contraction / vasoconstriction;

4.      Thick wall withstands pressure OR stop bursting;

5.      Smooth endothelium reduces friction;

6.      Aortic valve / semi-lunar valve prevents backflow.

4 max

(b)     1.      Curve to the right so lower affinity / % saturation (of haemoglobin);

2.      Haemoglobin unloads / dissociates more readily;

3.      More oxygen to cells / tissues / muscles;

4.      For greater / more / faster respiration;

Idea of a higher rate of respiration

4

(c)     16.5–18 (cm3 minute–1);

Allow 1 mark if heart rate wrongly calculated but then multiplied by 0.03

2

[10]