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Topic 2 Assessment Booklet

Marks = 134 Time Allowed 165 minutes

Q1.(a) A student prepared a stained squash of cells from the tip of an onion root and

observed it using an optical microscope.

During the preparation of the slide, he:

• cut the first 5 mm from the tip of an onion root and placed it on a glass slide• covered this tip with a drop of stain solution and a cover slip• warmed the glass slide• pressed down firmly on the cover slip.

He identified and counted nuclei in different stages of the cell cycle.

Explain why the student:

1. used only the first 5 mm from the tip of an onion root.

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2. pressed down firmly on the cover slip.

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Figure 1 shows the cells the student saw in one field of view. He used this field of view to calculate the length of time these onion cells spent in anaphase of mitosis.

Figure 1

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(b) Scientists have found the mean length of time spent by onion cells in anaphase of mitosis is 105 minutes. They also found the cell cycle of cells in the onion root shown in Figure 1 takes 1080 minutes.

32 whole cells are shown in Figure 1.

Use this information and Figure 1 to calculate the length of time the cells of this onion root are in anaphase and then calculate the percentage difference between your answer and the mean length of time found by the scientists.

Show your working.

Answer = ____________________ %(2)

(c) Tick (✓) the name given to the division of cytoplasm during the cell cycle.

A Binary fission

B Cytokinesis

C Phagocytosis

D Segregation

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(1)

(d) Describe and explain what the student should have done when counting cells to make sure that the mitotic index he obtained for this root tip was accurate.

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(e) A scientist treated growing tips of onion roots with a chemical that stops roots growing. After 24 hours, he prepared a stained squash of these root tips.

Figure 2 is a drawing showing the chromosomes in a single cell observed in the squash of one of these root tips in anaphase. This cell was typical of other cells in anaphase in these root tips.

Figure 2

Use all of this information to suggest how the chemical stops the growth of roots.

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(Total 10 marks)

Q2.(a) What is an antigen?

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(b) What is an antibody?

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Poliomyelitis is an infection caused by a virus.

A doctor vaccinated a group of patients against poliomyelitis. He gave each patient two doses of vaccine, 3 months apart.

An immunologist tested three samples of blood from each of the patients:• (sample 1) taken 2 weeks before the first dose of vaccine• (sample 2) taken 2 weeks after the first dose of vaccine• (sample 3) taken 2 weeks after the second dose of vaccine.

He measured the concentration of antibodies against the poliomyelitis virus in the patients’ blood each time. The results are shown in the graph.

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(c) Calculate the percentage increase in the mean concentration of antibodies in blood between samples 2 and 3.

Answer = ________________________ %(1)

(d) Explain the differences between the mean concentrations of antibodies in blood samples 1, 2 and 3.

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(Total 9 marks)

Q3.(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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(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

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how they reached this conclusion.

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(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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(Total 9 marks)

Q4.A bacterium is shown in the diagram.

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(a) Calculate the magnification of the image.

Magnification = _________________(1)

(b) Complete the table to show the features of a bacterium and a virus.

Put a tick (✔) in the box if the feature is shown.

Surface Bacterium Virus

Cell-surface membrane

Nucleus

Cytoplasm

Capsid

(2)

(c) DNA and RNA can be found in bacteria.

Give two ways in which the nucleotides in DNA are different from the nucleotides in RNA.

1. _________________________________________________________________

2. _________________________________________________________________(2)

(Total 5 marks)

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Q5.(a) Cells lining the ileum of mammals absorb the monosaccharide glucose by co-

transport with sodium ions. Explain how.

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A student set up the experiment shown in the diagram below.

The material from which Visking tubing is made is partially permeable.

After 15 minutes, the student removed samples from the liquid in the beaker and from the liquid inside the Visking tubing. She carried out biochemical tests on these samples. She drew the table below to record her results.

(b) Complete the table by placing a tick (✔) in each box that you expect to have shown a positive result.

Biochemical test Liquid from beaker

Liquid inside Visking tubing

Biuret reagent

Iodine in potassium iodide

Benedict’s solution

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(3)

(c) Justify your answers to part (b).

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(Total 9 marks)

Q6.(a) NMO is a disease that leads to damage to nerve cells in the spinal cord.

A person with NMO produces anti-AQP4 antibody that attacks only these nerve cells.

Explain why the anti-AQP4 antibody only damages these cells.

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(b) Scientists measured the concentration of anti-AQP4 antibody in the blood of people with NMO.

The spinal cord is surrounded by small bones called vertebrae. For each person, the scientists also determined the number of vertebrae surrounding damaged nerve cells.

Their results are shown in the graph.

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A scientist suggested that the concentration of anti-AQP4 antibody in a person’s blood could be used to predict the number of vertebrae surrounding damaged nerve cells they are likely to have.

Use the graph above to suggest reasons why this suggestion might not be valid.

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(c) A new treatment for NMO involves using a monoclonal antibody. The structure of the variable region of this monoclonal antibody is identical to the variable region of an anti-AQP4 antibody, but the rest of its structure is different.

Use this information and your knowledge of antigen-antibody complexes to suggest how this monoclonal antibody prevents anti-AQP4 damaging nerve cells.

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(Total 9 marks)

Q7.The cells of beetroot contain a red pigment. A student investigated the effect of temperature on the loss of red pigment from beetroot. He put discs cut from beetroot into tubes containing water. He maintained each tube at a different temperature. After 25 minutes, he measured the percentage of light passing through the water in each tube.

(a) The student put the same volume of water in each tube.

Explain why it was important that he controlled this experimental variable.

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(b) Describe a method the student could have used to monitor the temperature of the water in each tube.

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The graph shows the student’s results.

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(c) Draw a suitable curve on the graph above.(1)

(d) The decrease in the percentage of light passing through the water between 25 °C and 60 °C is caused by the release of the red pigment from cells of the beetroot.

Suggest how the increase in temperature of the water caused the release of the red pigment.

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(Total 6 marks)

Q8.(a) Describe how you would use cell fractionation techniques to obtain a sample of

chloroplasts from leaf tissue. Do not include in your answer information about any solutions.

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(b) The table shows features of a mitochondrion and a chloroplast. Complete the table with ticks where a feature is present.

Feature Mitochondrion Chloroplast

Double outer membrane

Starch grains

Diffusion of oxygen into the organelle

(3)

(c) Give the function of a mitochondrion.

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(d) Scientists investigated the effect of an exercise programme on the number and size of mitochondria in skeletal muscle. They took samples of muscle from a large number of volunteers before and after the exercise programme. From each sample, they cut thin sections and used these to determine the mean number of mitochondria per μm2 and the mean area of inner mitochondrial membranes.

Their results are shown in Graph 1 and Graph 2.

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What do the data in Graph 1 and Graph 2 suggest about the effect of the exercise programme on mitochondria?

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(Total 9 marks)

Q9.(a) Give three properties of water that are important in biology.

1. _________________________________________________________________

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A student investigated the effect of different concentrations of sucrose solution on “chips” cut from a potato. Each chip had the same dimensions.

The student:

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• weighed each chip at the start• placed each chip in a separate test tube, each containing 10 cm3 of sucrose solution

at a different concentration• left the chips in the sucrose solution for 24 hours• dried the surface of the chips and then weighed them again.

The table shows the student’s results.

Concentration of sucrose

solution / mol dm−3

Initial mass of chip / g

Final mass of chip / g

Ratio of final mass to

initial mass of chips

0.0 2.79 3.82

0.2 2.75 2.97

0.4 2.78 2.67

0.6 2.69 2.31

0.8 2.72 2.20

1.0 2.77 1.99

(b) The student produced the sucrose solutions with different concentrations from a concentrated sucrose solution.

Name the method she would have used to produce these sucrose solutions.

Name of method _____________________________________________________(1)

(c) Calculate the ratio of final mass to initial mass of potato chips and plot a suitable graph of your processed data. Express the ratios in the table in part (a) as a single number (for example 5.26:1 would be expressed as 5.26).

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(3)

(d) Explain the result for the chip in 0.8 mol dm−3 sucrose solution.

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(Total 9 marks)

Q10.A student investigated the effect of surface area on osmosis in cubes of potato.

• He cut two cubes of potato tissue, each with sides of 35 mm in length.• He put one cube into a concentrated sucrose solution.• He cut the other cube into eight equal-sized smaller cubes and put them into a

sucrose solution of the same concentration as the solution used for the large cube.• He recorded the masses of the cubes at intervals.

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His results are shown in the graph.

(a) Describe the method the student would have used to obtain the results in the graph. Start after all of the cubes of potato have been cut. Also consider variables he should have controlled.

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(b) The loss in mass shown in the graph is due to osmosis. The rate of osmosis between 0 and 40 minutes is faster in B (the eight small cubes) than in A (single large cube).

Is the rate of osmosis per mm2 per minute different between A and B during this time?Use appropriate calculations to support your answer.

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(Total 6 marks)

Q11.(a) Figure 1 shows part of the blood circulation in a mammal.

Figure 1

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Use Figure 1 to give the letter that represents each of these blood vessels.

Aorta

Renal vein

Vena cava

(3)

(b) Name the blood vessels that carry blood to the heart muscle.

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(c) Figure 2 shows a photograph of part of a mitochondrion from a mouse liver cell taken using a transmission electron microscope at × 62 800 magnification.

Figure 2

Produce a scientific drawing of the mitochondrion in Figure 2 in the box below.

Label the following parts of the mitochondrion on your drawing.

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• Matrix• Crista

(4)(Total 8 marks)

Q12.(a) When a person is bitten by a venomous snake, the snake injects a toxin into the

person. Antivenom is injected as treatment. Antivenom contains antibodies against the snake toxin. This treatment is an example of passive immunity.

Explain how the treatment with antivenom works and why it is essential to use passive immunity, rather than active immunity.

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The chart shows a procedure used to produce antivenom.

(b) A mixture of venoms from several snakes of the same species is used.

Suggest why.

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(c) Horses or rabbits can be used to produce antivenoms.When taking blood to extract antibody, 13 cm3 of blood is collected per kg of the

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animal’s body mass.The mean mass of the horses used is 350 kg and the mean mass of the rabbits used is 2 kg

Using only this information, suggest which animal would be better for the production of antivenoms.Use a calculation to support your answer.

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(d) During the procedure shown in the chart the animals are under ongoing observation by a vet.

Suggest one reason why.

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(e) During vaccination, each animal is initially injected with a small volume of venom. Two weeks later, it is injected with a larger volume of venom.

Use your knowledge of the humoral immune response to explain this vaccination programme.

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(Total 10 marks)

Q13.(a) What is a monomer?

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(b) Lactulose is a disaccharide formed from one molecule of galactose and one molecule of fructose.

Other than both being disaccharides, give one similarity and one difference between the structures of lactulose and lactose.

Similarity ___________________________________________________________

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Difference __________________________________________________________

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(c) Following digestion and absorption of food, the undigested remains are processed

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to form faeces in the parts of the intestine below the ileum.

The faeces of people with constipation are dry and hard. Constipation can be treated by drinking lactulose. Lactulose is soluble, but is not digested or absorbed in the human intestine.

Use your knowledge of water potential to suggest why lactulose can be used to help people suffering from constipation.

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(d) Lactulose can also be used to treat people who have too high a concentration of hydrogen ions (H+) in their blood.

The normal range for blood H+ concentration is 3.55 × 10–8 to 4.47 × 10–8 mol dm–3

A patient was found to have a blood H+ concentration of 2.82 × 10–7 mol dm–3

Calculate the minimum percentage decrease required to bring the patient’s blood H+

concentration into the normal range.

Answer = ____________________(2)

(Total 7 marks)

Q14.Figure 1 shows all the chromosomes present in one human cell during mitosis. A scientist stained and photographed the chromosomes. In Figure 2, the scientist has arranged the images of these chromosomes in homologous pairs.

Figure 1 Figure 2

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(a) Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis. Explain your answers.

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2. _________________________________________________________________

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(b) Tick (✓) one box that gives the name of the stage of mitosis shown in Figure 1.

A Anaphase

B Interphase

C Prophase

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D Telophase

(1)

(c) When preparing the cells for observation the scientist placed them in a solution that had a slightly higher (less negative) water potential than the cytoplasm. This did not cause the cells to burst but moved the chromosomes further apart in order to reduce the overlapping of the chromosomes when observed with an optical microscope.

Suggest how this procedure moved the chromosomes apart.

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(d) The dark stain used on the chromosomes binds more to some areas of the chromosomes than others, giving the chromosomes a striped appearance.

Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas.

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(e) In Figure 2 the chromosomes are arranged in homologous pairs.What is a homologous pair of chromosomes?

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(f) Give two ways in which the arrangement of prokaryotic DNA is different from the arrangement of the human DNA in Figure 1.

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2. _________________________________________________________________

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(Total 9 marks)

Q15.(a) Give two similarities in the movement of substances by diffusion and by osmosis.

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2. _________________________________________________________________

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A scientist measured the rate of uptake of a monoglyceride and a monosaccharide by epithelial cells of the small intestine of mice. A monoglyceride is a molecule of glycerol with one fatty acid attached. She did this for different concentrations of monoglyceride and monosaccharide.

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Her results are shown in the graph.

(b) Use your knowledge of transport across membranes to explain the shape of the curve in the graph for uptake of monosaccharides between concentrations:

A and B ____________________________________________________________

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C and D ____________________________________________________________

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(c) The graph is evidence for monoglycerides being lipid-soluble molecules.

Suggest how.

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(Total 7 marks)

Q16.(a) Structures A to E are parts of a plant cell.

A Cell WallB ChloroplastC NucleusD MitochondrionE Golgi apparatus

Complete the table by putting the correct letter, A, B, C, D or E in the box next to each statement.

Statement Letter

Has stacked membranes arranged in parallel and contains DNA.

Is made of polysaccharide.

Is an organelle and is not surrounded by two membranes.

(3)

(b) Human breast milk is produced and secreted by gland cells. These gland cells have adaptations that include many mitochondria and many Golgi vesicles. The milk contains a high concentration of protein.

Explain the role of these cell adaptations in the production and secretion of breast milk.

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(Total 5 marks)

Q17.The diagram shows part of a prokaryotic cell.

(a) Name the structures labelled W to Z in the diagram.

W ________________________________________________________________

X _________________________________________________________________

Y _________________________________________________________________

Z _________________________________________________________________(2)

(b) Name the main biological molecule in:

W ________________________________________________________________

X _________________________________________________________________(2)

(c) Name the process by which prokaryotic cells divide.

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(d) Some prokaryotic cells can divide every 30 minutes. A liquid culture contained a starting population of 1.35 × 104 cells.

Assuming each cell divides every 30 minutes, calculate how many cells there will be after 3 hours. Assume no cells die during this time.

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Answer = ____________________(2)

(Total 7 marks)

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Mark schemes

Q1.(a) 1. Where dividing cells are found / mitosis occurs;

ORNo dividing cells / mitosis in tissue further away / more than 5 mm from tip;ORTo get (soft) tissue that will squash;ORLength that will fit under cover slip;

Accept most dividing cells

2. Single / thin layer of cells / spread out cells so light passes through (making cells / nuclei visible);

Accept thin layer of tissueIgnore to see cells clearly

2

(b) 3.57 / 3.6 / 3.7 / 3.71 / 3.8 (%);;If the answer includes additional decimal places, award the marks if it would round to a correct answerThere are 3 cells in anaphaseAccept for 1 mark, 101.25 / 101 (students estimate in minutes)OR3.75 (difference between scientist estimate and student’s estimate in minutes)Ignore plus or minus signs

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(c) Cytokinesis;1

(d) Description;

Explanation;

E.g,

1. Examine large number of fields of view / many cells;Mark as pairs onlyAccept large number / 20 or more for many

2. To ensure representative sample;Accept typical / reliable

OR

3. Repeat count;

4. To ensure figures are correct;OR

5. Method to deal with part cells shown at edge /count only whole cells;

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6. To standardise counting;2 max

(e) 1. Stops anaphase / cell division / mitosis;Accept prevents telophase / cytokinesis

2. (By) stopping / disrupting / spindle fibres forming / attaching / pulling;Ignore affects anaphase

3. Preventing separation of (sister) chromatids;Ignore chromosomes separate / splitAccept chromatids split

4. (So) no new cells added (to root tip);3 max

[10]

Q2.(a) 1. Foreign protein;

Accept glycoprotein / glycolipid / polysaccharide

2. (that) stimulates an immune response / production of antibody;2

(b) 1. A protein / immunoglobulin specific to an antigen;

2. Produced by B cells

OR

Secreted by plasma cells;2

(c) 1750(%);1

(d) 1. Sample 1 / before vaccination no antibody released because patients not yet encountered vaccine / antigen / virus;

Accept ‘produced’ for ‘released’

2. (Sample 2 / primary response / after first dose) activation / clonal selection / expansion of B cells into plasma cells;

3. Plasma cells release antibodies;

4. (Sample 3 / secondary response / after second dose) memory cells produce more antibodies / produce antibodies more quickly;

4[9]

Q3.(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

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membrane);3. Accept ‘Na + and H + bind to protein’ but do not allow

incorrect chemical symbols3 max

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

ORLess water absorbed into cells by osmosisORWater 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’

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(c) 1. (Higher salt) results in lower water potential of tissue fluid;2. (So) less water returns to capillary by osmosis (at venule end);OR3. (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 pairs3. Accept higher hydrostatic pressure.

2[9]

Q4.(a) × 20 000

Accept range from 18 000 to 22 0001

(b)

✔

✔

✔1 mark for each correct column

2

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(c) 1. DNA contains thymine and RNA contains uracil;

2. DNA contains deoxyribose and RNA contains ribose.2

[5]

Q5.(a) 1. Sodium ions actively transported from ileum cell to blood;

2. Maintains / forms diffusion gradient for sodium to enter cells from gut (and with it, glucose);

3. Glucose enters by facilitated diffusion with sodium ions;3

(b)

Biochemical test

Liquid from beaker

Liquid inside Visking tubing

Biuret reagent ✔I2/KI ✔ or blank

Benedict’s ✔ ✔1 mark for each correct row

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(c) 1. Biuret: protein molecules too large to pass through tubing;Neutral: enzyme molecules

2. Iodine in potassium iodide solution: starch molecules too large to pass through tubing;

If no tick in 04.2, allow no starch hydrolysed

3. Benedict’s: starch hydrolysed to maltose, which is able to pass through tubing.Reject: glucose

3[9]

Q6.(a) 1. (Anti-AQP4) antibody has a (specific) tertiary structure;

2. Has binding site / variable region that only binds to / complementary to one antigen;

3. Antigen to this antibody (only) found on these nerve cells;4. So, antibody (only) binds to / forms antigen-antibody complex with these

nerve cells (causing damage);Reject “active site” (only penalise once if it occurs throughout)3. / 4. Accept ‘receptor’ for antigen

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(b) 1. Only 20 in the study;OR

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Only one study;2. For some concentrations of antibody there is a range in the number of

vertebrae surrounding damaged nerve cells;3. No statistical test used;4. Correlation is weak;

1. Accept small sample2. Accept suitable use of data2. Accept converse

3 max

(c) 1. The monoclonal antibody binds to nerve cell antigen so less / no anti-AQP4 can bind;ORThe monoclonal antibody forms antigen-antibody complex with nerve cell antigen so less / no anti-AQP4 can bind;

2. When monoclonal antibody binds it doesn’t cause damage to nerve cell;It = monoclonal antibody1. Reject “active site”Ignore “competitive inhibitor”Accept receptor for antigenDo not credit responses in the context of enzymes

2 max[9]

Q7.(a) 1. (If) too much water the concentration of pigment (in solution) will be lower /

solution will appear lighter / more light passes through (than expected);OR(If) too little water the concentration of pigment (in solution) will be greater / solution will appear darker / less light passes through (than expected);

2. So results (from different temperatures) are comparable;1. Ignore reference to too much water so red pigment /

solution too weak to measure2

(b) (Take) readings (during the experiment) using a (digital) thermometer / temperature sensor;

1

(c) Point-to-point line drawn between co-ordinates (with a ruler);ORSmooth s-shaped line of best fit;

Reject any extrapolations below 20 °C or above 80 °CAny line should look smooth (not ‘sketchy’)

1

(d) 1. Damage to (cell surface) membrane;

2. (membrane) proteins denature;

3. Increased fluidity / damage to the phospholipid bilayer;2 max

[6]

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Q8.(a) 1. Macerate / homogenise / blend / break tissues / cells (in solution);

2. Centrifuge;3. At different / increasing speeds until chloroplast fraction obtained;

1. Accept any suitable method to break tissues / cells / release organelles

2. and 3. Allow ‘perform differential centrifugation until chloroplasts obtained.’ for 2 marks

3

(b)

Feature Mitochondrion Chloroplast

Double outer membrane ✔ ✔

Starch grains ✔Diffusion of oxygen into the organelle

✔

1 mark for each correct rowCrosses = blank space

3

(c) The site of aerobic respiration (reactions)ORATP is made / ADP is phosphorylated;

Reject ‘energy is produced’1

(d) 1. Training made no difference to number (of mitochondria per μm2);2. Training led to an increase in the area (of inner mitochondrial

membrane);1. Accept Graph 1 as mean number of mitochondria per

μm 2

2. Accept Graph 2 as area of inner mitochondrial membrane

2[9]

Q9.(a) Accept any three suitable properties e.g.:

• Is a metabolite• Is a solvent• Has a (relatively) high heat capacity• Has a (relatively) large latent heat of vaporisation / evaporation• Has cohesion / hydrogen bonds between molecules;

No explanations are neededHowever do not accept ‘polar’ unqualified

3 max

(b) Dilution series;Accept serial dilution

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1

(c) 1. Axes correct way round with linear scales;2. Axes labelled with mol dm−3 and ratio without units;3. Correct values correctly plotted and suitable curve drawn;

3. Accept point to point or smooth curve but no extrapolation

NFP – 3. Graph starts just below 1.4 and finishes just above 0.7 and looks right.

3

(d) 1. (0.8 mol dm−3 sucrose) solution has a more negative / lower water potential than potato (cytoplasm);ORpotato (cytoplasm) has a less negative / higher water potential than (0.8 mol dm−3 sucrose) solution;

2. (therefore) water moves out (of potato) into the (sucrose) solution by osmosis (so cells decrease in mass);

1. Accept sucrose solution is hypertonic / potato cytoplasm is hypotonic

2. Accept water moves down a water potential gradient2

[9]

Q10.(a) 1. Method to ensure all cut surfaces of the eight cubes are exposed to the

sucrose solution;Credit valid method descriptions to fulfil mp1, 2 and 3 (no explanation is required).

2. Method of controlling temperature;Accept ‘at room temperature’ for method

3. Method of drying cubes before measuring;

4. Measure mass of cubes at stated time intervals;Accept time intervals between every 5 minutes with maximum of every 40 minutes.Accept ‘weigh the cubes at stated time intervals’

3 max

(b) Yes or No (no mark)

Calculation of rate per mm2 for both sets of data, accept answers in the range1.6 × 10–5 to 1.8 × 10–5 and1.5 × 10–5 to 1.6 × 10–5;;; Both correct = 3

One correct = 2

Neither correct – look below for max 2

Allow 1 mark for calculation of surface area of two (sets of) cubes 7350 (mm2) and 14700 (mm2)

Allow 1 mark for calculation of both rates of osmosis shown in first 40 minutes –

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between 0.12 and 0.13 and between 0.22 and 0.23

If surface area and/or rate of osmosis is incorrect then, allow 1 mark for (their) calculated rate divided by (their) calculated surface area

Accept answers not given in standard form or to any number of significant figures ≥2sf as long as rounding correct.

3 max[6]

Q11.(a) D;

G;F;

3

(b) Coronary arteries;Accept coronary arteryIgnore aorta, arteriole and capillaryReject coronary veinsDo not accept coronary by itselfAccept phonetic spelling

1

(c) 1. No sketched / hanging / crossing lines / shading;Ignore stippling

2. Must look similar;3. Matrix and crista correctly labelled;

Ignore any other labels4. Correct scale stated (x 62 800);

Accept other suitable scale given4

[8]

Q12.(a) 1. (Antivenom/Passive immunity) antibodies bind to the toxin/venom/antigen and

(causes) its destruction;For ‘bind’ accept ‘attach’, ignore ‘attack’.For ‘destruction of toxin’ accept agglutination or phagocytosis.Ignore reference to antibodies ‘neutralising toxin/stopping damage’Reject reference to ‘killing’ toxin/venom.

2. Active immunity would be too slow/slower;Accept ‘passive immunity is faster’, not simply ‘passive immunity is fast’.

2

(b) 1. May be different form of antigen/toxin (within one species)ORSnakes (within one species) may have different mutations/alleles;

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2. Different antibodies (needed in the antivenom)OR(Several) antibodies complementary (to several antigens);

No mark points are available for answers related to collecting venom from different species of snake.

2 max

(c) 1. Horses because more antivenom/antibodies could be collected (as more blood collected);

2. 4550 (cm3) v 26 (cm3) (blood collected);Accept 175 rabbits needed to (collect the volume of blood from) one horse.

2

(d) 1. (So) the animal does not suffer from the venom/vaccine/toxin;

2. (So) the animal does not suffer anaemia/does not suffer as a result of blood collection;

3. (So) the animal does not have pathogen that could be transferred to humans;Accept ‘To fulfil licence/legal requirements’.Accept ‘(So) the animal does not have pathogen that could result in it producing other antibodies (not wanted in the antivenom)’.For ‘pathogen’ accept correct form of pathogen.

1 max

(e) 1. B cells specific to the venom reproduce by mitosis;Accept in context of primary or secondary immune response.Credit idea of specificity if given once in relation to T or B cell.Accept a description for specificity.Accept ‘clone’ for ‘reproduce by mitosis’.‘Clonal selection of B cells’ = MP1.

2. (B cells produce) plasma cells and memory cells;

3. The second dose produces antibodies (in secondary immune response) in higher concentration and quicklyORThe first dose must be small so the animal is not killed;

Accept ‘a lot of antibody’ for ‘higher concentration of antibody’.

3[10]

Q13.(a) (a monomer is a smaller / repeating) unit / molecule from which larger molecules /

polymers are made;Reject atoms / elements / ’building blocks’ for units / moleculesIgnore examples

1

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(b) Similarity1. Both contain galactose / a glycosidic bond;

Ignore references to hydrolysis and / or condensation

Difference2. Lactulose contains fructose, whereas lactose contains glucose;

Ignore alpha / beta prefix for glucoseDifference must be stated, not implied

2

(c) 1. (Lactulose) lowers the water potential of faeces / intestine / contents of the intestine;

Accept Ψ for water potential

2. Water retained / enters (due to osmosis) and softens the faeces;Accept descriptions of soft faeces, eg faeces is less dry / less hard

2

(d) (-) 84.1(%);;Accept (-) 84.15(%)Allow 1 mark for84OR

OR

2[7]

Q14.(a) 1. The (individual) chromosomes are visible because they have condensed;

Both parts of each answer are required – evidence and explanation.For ‘they’ accept ‘chromosomes/chromatin/DNA’Accept ‘tightly coiled’ or ‘short and thick’ for condensed but do not accept ‘contracted’.Ignore references to nucleus/nucleolus/nuclear membrane.

2. (Each) chromosome is made up of two chromatids because DNA has replicated;

Both parts of each answer are required – evidence and explanation.Accept ‘sister chromatids’ for ‘two chromatids’.Ignore references to nucleus/nucleolus/nuclear membrane.

3. The chromosomes are not arranged in homologous pairs, which they would be if it was meiosis;

Both parts of each answer are required – evidence and

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explanation.Accept not meiosis because bivalents/chiasmata/crossing over not seen.Ignore references to nucleus/nucleolus/nuclear membrane.

2 max

(b) Automarked q – ✔ prophase1

(c) 1. Water moves into the cells/cytoplasm by osmosis;Reject water moving into chromosomes/nucleus.

2. Cell/cytoplasm gets bigger;Accept idea of cell/cytoplasm has greater volume/swells/expands.Ignore references to pressure changes, turgidity and chromosomes being more dilute.Ignore references to changing water/fluid contents of the cell.Allow ECF for ‘nucleus expands’ but not for ‘chromosomes expand’.

2

(d) Differences in base sequences

OR

Differences in histones/interaction with histones

OR

Differences in condensation/(super)coiling;Answer must be in context of differences in arrangement of chromosomes not just related to the properties of the stain.Accept spec section 8 ideas e.g. different methylation/acetylationAccept different genesReject different alleles

1

(e) (Two chromosomes that) carry the same genes;Reject ‘same alleles’Accept ‘same loci’ (plural) or ‘genes for the same characteristics’

1

(f) (Prokaryotic DNA) is

1. Circular (as opposed to linear);

2. Not associated with proteins/histones ;

3. Only one molecule/piece of DNAORpresent as plasmids;

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Max 1 if prokaryotic DNA only found as plasmids OR if prokaryotic DNA is single stranded.Ignore references to nucleus, exons, introns or length of DNA. Do not credit converse statements.Ignore descriptions of eukaryotic DNA alone.

2 max[9]

Q15.(a) 1. (Movement) down a gradient / from high concentration to low concentration;

Ignore along / across gradientReject movement from gradient to gradient

2. Passive / not active processes;ORDo not use energy from respiration / from ATP / from metabolism;ORUse energy from the solution;

Reject do not use energy unqualified2

(b) 1. Movement through carrier proteins;ORFacilitated diffusion;Between A and B

Accept MP1 in either sectionIgnore co-transport / active transportAccept channel proteins

2. Rate of uptake proportional to (external) concentration;Between C and D

Accept description of proportional

3. All channel / carrier proteins in use / saturated / limiting;Accept used upAccept transport proteins

3

(c) 1. Rate of uptake is proportional / does not level off (so diffusion occurring);Accept as one increases the other increases

2. (Lipid-soluble molecules) diffuse through / are soluble in phospholipid (bilayer);2

[7]

Q16.(a) B;

A;

E;3

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(b) 1. (Many mitochondria) release energy / ATP for movement of vesicles / synthesis of protein / active transport;

Must include function of organelle and use in context of milk production.Ignore reference to lipid / triglycerideReject reference to mitochondria undergoing anaerobic respirationReject “produce energy”.Reject “energy for respiration”

2. (Many Golgi) vesicles transport protein / glycoprotein / milk to cell membrane / out of cell;

Must include function of organelle and use in context of milk production.Ignore reference to lipid / triglycerideAccept exocytosis as transport and releaseIgnore references to protein synthesis

2[5]

Q17.(a) W – (cell surface) membrane

X – cell wallY – capsuleZ – flagellum

Four correct = 2 marks.Three or two correct = 1 mark.Y - Ignore references to slime/mucusY - Reject capsidZ - accept flagella

2

(b) W - Phospholipids;X - Murein / glycoprotein;

X - Accept peptidoglycans.Accept phonetic spellings

2

(c) Binary fission;Reject binary fusion

1

(d) 8.64 × 105;;Accept 864 000 however expressed, e.g. 864 × 103

Allow one mark for26 = 64OR64 / 26 × (1.35 × 104)

2[7]

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