Experimental Data Analysis Questions Marks= 109 Time ...€¦ · Calculate how much antibiotic...

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Experimental Data Analysis Questions

Marks= 109 Time allowed=141

Q1. Scientists investigated treatment of a human bladder infection caused by a species of bacterium. This species of bacterium is often resistant to the antibiotics currently used for treatment.

They investigated the use of a new antibiotic to treat the bladder infection. The new antibiotic inhibits the bacterial ATP synthase enzyme.

(a) Place a tick (✔) in the appropriate box next to the equation which represents the

reaction catalysed by ATP synthase.

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

(1)

(b) The new antibiotic is safe to use in humans because it does not inhibit the ATP synthase found in human cells.

Suggest why human ATP synthase is not inhibited and bacterial synthase is inhibited.

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(1)

(c) The scientists tested the new antibiotic on mice with the same bladder infection. They divided these mice into three groups, C, R and A.

• Group C was the control (untreated).

• Group R was treated with an antibiotic currently used against this bladder infection.

• Group A was treated with the new antibiotic.


They removed samples from the bladder of these mice after treatment and estimated the total number of bacteria in the bladder.

Their results are shown in the graph.

The antibiotics were given to the mice at a dose of 25 mg kg−1 per day.

Calculate how much antibiotic would be given to a 30 g mouse each day.

Show your working.

Answer = ____________________ mg

(2)

(d) Calculate the percentage difference in actual numbers of bacteria in group A compared with group R. The actual number of bacteria can be calculated from the

log10 value by using the 10x function on a calculator.


Show your working.

Answer = ____________________ %

(2)

(e) The scientists suggested that people newly diagnosed with this bladder infection should be treated with both the current antibiotic and the new antibiotic.

Explain why the scientists made this suggestion.

Use information from the graph in part (c) and your knowledge of evolution of antibiotic resistance in bacteria in your answer.

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(3)

(Total 9 marks)

Q2. (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)

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.


Bacteriophages are viruses that kill bacteria.

The diagram below shows drawings of a bacteriophage and a bacterium.

(a) Using the diagram and your own knowledge, put a tick (✔) in the box next to the

only correct statement about the structures of the bacteriophage and the bacterium

Both have ribosomes.

Both have a cell-surface membrane.

The bacteriophage has a capsid and the bacterium has a cell-surface membrane.

The bacteriophage has a cell wall and

the bacterium has a capsid.

(1)

(b) Using the scales in the diagram above, calculate how many times longer the bacterium is than the bacteriophage.

Use the distance between the points labelled A and B on each drawing in your calculations. Show your working.


The bacterium is ____________________ times longer

(2)

(c) Scientists investigated the use of bacteriophages to treat lung infections caused by bacteria. They infected the lungs of mice with a pathogenic species of bacterium. The mice were then divided into two groups, A and B.

• The mice in group A were not treated with bacteriophage. • The mice in group B were treated by breathing in a spray containing

bacteriophage particles.

After 3 days, the scientists killed the mice and removed their lungs. They washed out each set of lungs with a set volume of liquid. The scientists determined the number of live bacteria in the liquid.

The graph below shows the scientists’ results. The graph shows the mean and the range of the data about the mean for each group. Standard deviations of the means are not shown.

What would the scientists’ null hypothesis be for this investigation?

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(1)

(d) With some samples, the scientists decided they needed to carry out a series of dilutions of the sample before counting the bacteria.


Use evidence from the graph above to explain why dilutions were necessary for some samples but not for others.

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(2)

(e) Using only the graph above, what can you conclude from these data about the effectiveness of the bacteriophage in treating this lung infection in mice?

Do not consider statistical analyses in your answer.

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(3)

(Total 9 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) Describe the role of two named enzymes in the process of semi-conservative

replication of DNA.

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(3)

(b) Scientists investigated the function of a eukaryotic cell protein called cyclin A. This protein is thought to be involved with the binding of one of the enzymes required at the start of DNA replication.

The scientists treated cultures of cells in the following ways. C – Control cells, untreated


D – Added antibody that binds specifically to cyclin A E – Added RNA that prevents translation of cyclin A F – Added RNA that prevents translation of cyclin A and added cyclin A protein

They then determined the percentage of cells in each culture in which DNA was replicating.


Cell treatment Percentage of cells where

DNA was replicating

C Control

91

D Antibody that binds specifically to cyclin A

11

E RNA that prevents translation of cyclin A

10

F RNA that prevents translation of cyclin A and added cyclin A protein

92

Suggest explanations for the results in the table.

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(3)


(Total 6 marks)

Q7. (a) Formation of an enzyme-substrate complex increases the rate of reaction.

Explain how.

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(2)

(b) A scientist measured the rate of removal of amino acids from a polypeptide with and without an enzyme present. With the enzyme present, 578 amino acids were released per second. Without the enzyme, 3.0 × 10–9 amino acids were released per second.

Calculate by how many times the rate of reaction is greater with the enzyme present. Give your answer in standard form.

Answer = ____________________ times faster

(2)

Another scientist investigated an enzyme that catalyses the following reaction.

ATP → ADP + Pi


The scientists set up two experiments, C and L.

Experiment C used • the enzyme • different concentrations of ATP.

Experiment L used • the enzyme • different concentrations of ATP • a sugar called lyxose.

The scientists measured the rate of reaction in each experiment. Their results are shown in the graph.

(c) Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm–3 of ATP as a percentage of the maximum rate shown with lyxose.

Answer = ____________________ %

(2)

(d) Lyxose binds to the enzyme.

Suggest a reason for the difference in the results shown in the graph with and


without lyxose.

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(3)

(Total 9 marks)

Q8. Figure 1 shows a faulty form of meiosis that can occur in some plants.

Figure 1


(a) Complete Figure 2 to show the chromosome content of the cells that would result from a normal meiotic division of the diploid parent cell shown in Figure 1.

Figure 2


(2)

(b) If two diploid (2n) gametes fuse at fertilisation, it can result in the growth of a tetraploid plant which has 4 copies of each chromosome.

Red clover is a plant grown to produce cattle feed. Tetraploid red clover plants produce a higher yield than diploid red clover plants.

Whether a red clover plant produces 2n gametes is genetically controlled.

Scientists investigated the possibility of breeding red clover plants that only produced 2n gametes.

• In breeding cycle 0, they grew red clover plants and identified plants that produced 2n gametes.

• In breeding cycle 1, they used the plants producing 2n gametes to produce offspring.

• In breeding cycles 2 and 3, they identified plants producing 2n gametes and used these to produce offspring.


Observed Expected

Breeding cycle

Number of plants that did not produce

Number of plants that did

produce 2n

Number of plants that did not produce

Number of plants that did

produce 2n


2n gametes gametes 2n gametes gametes

0 50 4 50 4

1 14 42

2 2 44

3 0 56

The scientists used the following null hypothesis.

‘The proportion of plants that produce 2n gametes will not change from one breeding cycle to the next.’

Complete the table to show the expected number of plants that did not produce 2n gametes and the expected number of plants that did produce 2n gametes after 1 cycle. Give each answer to the nearest whole number.

(2)

(c) The scientists tested their null hypothesis using the chi-squared statistical test. After 1 cycle their calculated chi-squared value was 350 The critical value at P=0.05 is 3.841

What does this result suggest about the difference between the observed and expected results and what can the scientists therefore conclude?

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(2)

(d) Use your knowledge of directional selection to explain the results shown in the table.

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(3)

(Total 9 marks)

Q9. Figure 1 represents the phylogenetic classification of four different species of fruit fly.

Figure 1

(a) Figure 1 shows a hierarchy. Explain how.

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(2)

(b) Name the taxon to which Drosophilidae belongs.

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Drosophila fruit flies display courtship behaviour. One of the stages of courtship is singing by males. Normally a male will produce a ‘sine song’, in which continual noise is made, and a ‘pulse song’, in which there is continual noise with some louder peaks.


Scientists showed fruit flies a visual stimulus that made them sing. They made recordings of these songs.

Figure 2 shows the recordings of the songs of three flies over the same time period.

Figure 2

(c) D. erecta and D. willistoni are closely related species but different species.

Describe evidence from Figure 2 that supports this statement.

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(2)

(d) The scientists repeated their experiments, using female fruit flies as the visual stimulus. When a male and female D. willistoni were together, their songs led to mating.


When two female D. willistoni were together, their songs did not lead to any attempt to mate.

Use information from Figure 2 to suggest why the two females did not attempt to mate.

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(2)

(Total 7 marks)

Q10. (a) Give two similarities in the movement of substances by diffusion and by osmosis.

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2. _________________________________________________________________

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(2)

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.

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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(3)

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

Suggest how.

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(2)

(Total 7 marks)

Q11. (a) Describe the relationship between size and surface area to volume ratio of

organisms.

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(1)

(b) A scientist calculated the surface area of a large number of frog eggs. He found that the mean surface area was 9.73 mm2. Frog eggs are spherical.

The surface area of a sphere is calculated using this equation

Surface area = 4πr2

where r is the radius of a sphere

π = 3.14

Use this equation to calculate the mean diameter of a frog egg.

Show your working.


Diameter = ____________________ mm

(2)

The scientist calculated the ratio of surface area to mass for eggs, tadpoles and frogs. He also determined the mean rate of oxygen uptake by tadpoles and frogs.

His results are shown in the table.

Stage of frog

development

Ratio of surface

area to mass

Mean rate of oxygen uptake / μmol g–1 h–1

Egg 2904 : 1 no information

Tadpole 336 : 1 5.7

Adult 166 : 1 1.3

(c) The scientist used units of μmol g–1 h–1 for the rate of oxygen uptake.

Suggest why he used μmol in these units.

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(1)

(d) The scientist decided to use the ratio of surface area to mass, rather than the ratio of surface area to volume. He made this decision for practical reasons.

Suggest one practical advantage of measuring the masses of frog eggs, tadpoles and adults, compared with measuring their volumes.

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(e) Explain why oxygen uptake is a measure of metabolic rate in organisms.

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(1)

(f) A student who looked at these results said that they could not make a conclusion about the relationship between stage of development and metabolic rate.

Use information in the table to explain reasons why they were unable to make a conclusion.

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(3)

(Total 9 marks)

Q12. (a) State and explain the property of water that can help to buffer changes in

temperature.

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(2)

(b) Water is used to hydrolyse ATP.

Name the two products of ATP hydrolysis.

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2. _________________________________________________________________

(1)

Hydrolysis of ATP is catalysed by the enzyme ATP hydrolase.

A student investigated the effect of ATP concentration on the activity of ATP hydrolase. She used shortening of strips of muscle tissue caused by contraction as evidence that ATP was being hydrolysed.

• She took four slides A, B, C and D, and added strips of muscle tissue of the same length to each slide.

• She then added the same volume of ATP solutions of different concentrations to the four slides and left each slide for five minutes.

• She then recorded the final length of each strip of muscle tissue.

Her results can be seen in the table.

Slide

Concentration of ATP solution added to slide / × 10–6 mol dm–3

Final length of muscle tissue after 5 minutes

/ mm

A 2 36

B 4 31

C 6 29

D 8 26

(c) Other than those given, name two variables the student should have controlled.

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(2)

(d) Describe and explain the pattern shown by the data in the table.

Description __________________________________________________________

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Explanation _________________________________________________________

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(2)

(e) The hydrolysis of 1 dm3 of a 1 mol dm–3 solution of ATP releases 30 500 J of energy.

60% of the energy released during the hydrolysis of 1 mol dm–3 of ATP is released as heat; the rest is used for muscle contraction.

The student added 0.05 cm3 of ATP solution to slide D.

Calculate the energy available from ATP for contraction of the muscle on this slide.

Answer = ____________________ J

(3)

(Total 10 marks)

Q13. (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)


Mark schemes

Q1. (a)

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

1

(b) 1. Human ATP synthase has a different tertiary structure to bacterial ATP synthase OR Human ATP synthase has a different shape active site to bacterial ATP synthase OR Antibiotic cannot enter human cells/mitochondria OR Antibiotic not complementary (to human ATP synthase);

1

(c) 0.75; One mark for showing 30 g = 0.03 kg; One mark for showing 0.025 mg g−1

2

(d) Answer in range 97.0 − 97.8%; OR Answer in range 3288 − 4368%;

2. 1 mark for correct log10 readings from graph converted to actual numbers

(16.98 − 19.50 and 660.7 − 758.6) 2

(e) 1. (From the graph in part c) New / old antibiotic does not kill all bacteria; OR (From the graph in part c) Some bacteria are resistant to the new / old antibiotic;

2. Resistant bacteria will reproduce to produce (more) resistant bacteria; 3. (Use of both) one antibiotic will kill bacteria resistant to the other

antibiotic; OR Unlikely that bacteria are resistant to both the new and the old antibiotic; OR Use of both antibiotics (likely to) kill all / most bacteria;


Accept use of ‘A’ for ‘new antibiotic’ and ‘R’ for ‘old antibiotic’.

1. Must relate to the bacteria that are still present – ‘some bacteria are killed’ or ‘the bacteria number is reduced ’ is insufficient.

2. Accept ‘resistant bacteria reproduce to pass on resistance gene / allele’

3. ‘Use of both antibiotics will be more effective’ is insufficient.

3

[9]

Q2. (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]

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)

The bacteriophage has a capsid and the bacterium has a cell-surface membrane;

Third box down

Reject if more than one box with tick. Ignore crossed-out ticks

Accept tick to right or left of correct box 1

(b) Correct number of times between 13.0 / 12.96 and 13.9 / 13.92 scores 2 marks; One mark if correct sizes in ranges of 150.7nm to 154.4nm / 0.151μm to 0.154μm and 1953.5nm to 2097.6nm / 1.954μm to 2.098μm;

Both lengths required for 1 mark credit

Accept refs to 150 / 0.15 and 2000 / 2

Ignore number of sig fig 2

(c) The bacteriophage makes no difference to the number of (living) bacteria / there will be no difference in the number of (live) bacteria in treated and untreated mice / there will be no difference in the number of (live) bacteria in A and B;

Ignore no difference between mice 1

(d) 1. Log scale (on graph) shows big range in number of bacteria OR Use of suitable data from log scale to give the range in number of bacteria;

2. Some samples too many to count (so dilute) but some countable (so don’t dilute) OR Use of figures from graph relating to ease (or otherwise) of counting Example. 631 000 000 bacteria would be too big to count (without serial dilution), 100 000 bacteria is small enough to count;

1 and 2 Do not accept simple statements of log10 values from graph

Look for answers in standard form

Group A

Mean between 79 million and 100 million

Range 4 million to 631 million

Group B

Mean between 100 thousand and 126 thousand

Range 3.98/4 up to 251 thousand 2


(e) 1. (Bacteriophage) reduces number of bacteria; 2. (In all cases / mice because) ranges don’t overlap; 3. But big range of effect/some mice a big reduction/a few

mice with big falls in number OR Doesn’t bring bacteria down to 0 in any / works for some (mice) but not for all;

Do not accept just quotes of log10 figures direct from graph

Ignore refs to significance

Range 3.98/4 up to 251 thousand

2. Reject ref to SD / SE 3

[9]

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. (DNA) helicase causes breaking of hydrogen/H bonds (between DNA strands);

Reject ‘helicase hydrolyses hydrogen bonds’.

2. DNA polymerase joins the (DNA) nucleotides;

Reject if suggestion that DNA polymerase joins the complementary nucleotides or forms H bonds.

Reject if joining RNA nucleotides or forming RNA.

3. Forming phosphodiester bonds; 3

(b) 1. (Treatment D Antibody binds to cyclin A so) it cannot bind to DNA/enzyme/initiate DNA replication;

For ‘bind to enzyme’ accept ‘activate’.

Idea of ‘initiate DNA replication’ must be linked to start not just less replication.

For ‘enzyme’ accept named enzyme.

2. (Treatment E) RNA interferes with mRNA/tRNA/ribosome/polypeptide formation (so cyclin A not made);


3. In Treatment F added cyclin A can bind to DNA/enzyme (to initiate DNA replication) OR Treatment F shows that it is the cyclin A that is being affected in the other treatments OR Treatment F shows that cyclin A allows the enzyme to bind (to DNA) OR (Some cells in D or E) can continue with DNA replication because they have a different cyclin A allele OR (Some cells in D or E) can continue with DNA replication because the antibody/RNA has not bound to all the cyclin A protein/mRNA OR (Some cells in E) can continue with DNA replication because they contain previously translated cyclin A;

Context needed for Treatment F but it does not need to be named.

For ‘enzyme’ accept named enzyme. 3

[6]

Q7. (a) 1. Reduces activation energy;

Accept ‘reduces Ea’.

2. Due to bending bonds OR Without enzyme, very few substrates have sufficient energy for reaction;

Accept ‘Due to stress/pressure/tension on bonds’ OR ‘Due to weakening bonds’.

Ignore references to ‘breaking bonds’. 2

(b) 1.93 × 1011;;

Allow 1 max for

578/3.0 × 10–9

1.93 × 10x when x ≠11

Correct answer with incorrect standard form e.g. 19.3 × 1010

Accept any number of significant figures ≥2, if rounding

correct (1.926• × 1011). Same principle applies to one max

answers. 2

(c) 31.4;;

Allow 1 max for

0.44 and 1.4


32.8

33.1

30

29.3

Accept any number of significant figures ≥2, if rounding correct (31.4284714).

Same principle applies to 1 max answers.

32.8 = Both readings at 2.5 mmol dm–3 (0.44/1.34)

33.1 = Both readings at 2.5 mmol dm–3 (0.44/1.33)

30 = Incorrect reading for C (0.42/1.4)

29.3 = Incorrect reading for C (0.41/1.4) 2

(d) 1. (Binding) alters the tertiary structure of the enzyme ;

Max 1

if lyxose acting as an inhibitor

OR if answer linked to lower rate of reaction

OR if lyxose used an energy source/respiratory substrate

2. (This causes) active site to change (shape);

3. (So) More (successful) E-S complexes form (per minute)

OR

E-S complexes form more quickly

OR

Further lowers activation energy;

Accept ‘acts as a co-enzyme’

Accept description for E-S complexes. 3

[9]

Q8. (a) 1. 1 long and 1 short chromosome, each made up of 2 chromatids held (by

centromere), in each cell of 1st division;

2. 1 long and 1 short (separate) chromosome in each cell of 2nd division;

Allow ECF for correct chromosomes shown in each cell from candidate’s 1st division cells.

Ignore drawing of centromere. 2

(b)

52 4


;;

Allow 1 mark for numbers totalling 56 except 14/42 - repetition of observed values.

If table is blank, award 1 mark for evidence of 56.

Both 52 and 4 required in table for two marks, do not credit 52 or 4 for one mark.

Award 1 max for answers not given as whole numbers. 2

(c) 1. There is a less than 0.05/5% probability that the difference(s) (between observed and expected) occurred by chance;

Reject ‘results (without reference to difference) occurring by chance’. Overall max 1 with this statement.

Accept ‘there is a greater than 0.95/95% probability that the difference did not occur by chance’.

Ignore ‘difference is significant’

2. Calculated value is greater than critical value so the null hypothesis can be rejected;

Ignore ‘difference is significant’

Do not accept ‘P value’ for ‘critical value’.

3. (The scientists can conclude that) the proportion of plants that produce 2n gametes does change from one breeding cycle to the next;

2 max

(d) 1. The scientists selected/used for breeding plants that produced 2n gametes;

Answer must be in context of the scientists selecting plants to breed. Accept ‘artificial selection’ or ‘selectively bred’.

2. (So these plants) passed on their alleles (for production of 2n gametes to the next generation);

Both mark points can be awarded if one correct reference is made to alleles (in either context).

3. The frequency of alleles for production of 2n gametes increased (in the population).

Both mark points can be awarded if one correct reference is made to alleles (in either context).

For ‘production of 2n gametes’ accept ‘abnormal meiosis’.

Do not accept ‘number’ for frequency.

Accept converse answers linked to plants that produce n gametes.

3

[9]

Q9. (a) 1. (It shows) smaller groups within larger groups / larger groups containing

smaller groups;

Accept groups within groups


2. With no overlap (between groups); 2

(b) Family;

Accept phonetic spellings 1

(c) 1. Sine song is (very) similar / same length (for both, so closely related).

2. (But) have different peaks / pulses (in pulse song);

Must give a difference, not just state they are different

Accept suitable differences eg number / length / amplitude / interval

2

(d) 1. (Three) peaks (in pulse song) occur at the same time (since both female) / songs identical / male peaks are different;

Accept suitable differences in male peaks eg number / length / amplitude / interval

2. (Therefore) no male (song) to stimulate / cause mating; OR Nothing to stimulate / cause mating;

2

[7]

Q10. (a) 1. (Movement) down a gradient / from high concentration to low concentration;

Ignore along / across gradient

Reject movement from gradient to gradient

2. Passive / not active processes; OR Do not use energy from respiration / from ATP / from metabolism; OR Use energy from the solution;

Reject do not use energy unqualified 2

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

Accept MP1 in either section

Ignore co-transport / active transport

Accept 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 up


Accept 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]

Q11. (a) As size increases, ratio (of surface area to volume) decreases;

Accept converse.

Comparison required, e.g., smaller organisms have a larger ratio

1

(b) Two marks for correct answer in range of 1.75 to 1.76032;;

Accept for 1 mark, incorrect answer using radius 0.87 / 0.88 / 0.880 / 0.8802 / 0.88015;

OR

Accept for 1 mark, incorrect answer with correct rearranged equation, e.g.,

Radius =

OR

=

OR

=

OR

r2 = surface area ÷ 4 π

OR

r2 = 9.73 ÷ 12.56

OR

r2 = 0.77 / 0.774 / 0.775 2

(c) (Measures) small uptake / amount / quantity / volume / concentration / rate (of oxygen uptake);

OR Avoids use of powers of ten / standard form / many decimal places;

Ignore weight / accuracy 1

(d) More accurate / less error (in measuring mass); OR Causes less distress / damage to animal (to measure mass); OR Easier / quicker (to find mass) because irregular shapes; OR Fewer measurements / calculations;


Ignore references to human error

Accept converse if reference made to volume

Reject if comparison is made with surface area. 1

(e) (Oxygen used in) respiration, which provides energy / ATP; OR (Oxygen is used in) respiration, which is a metabolic process / chemical reaction;

Reject produces energy

Reject references to anaerobic respiration 1

(f) 1. No information about egg;

2. So cannot compare all stages (in Table 2);

Idea of comparing all three stages needed

3. No statistical information / test / t-test / comparison of standard deviations; OR No measure of significant differences;

Reject statements that “results” are not significant

Reject references to chi squared or correlation coefficient 3

[9]

Q12. (a) 1. (water has a relatively) high (specific) heat capacity;

Ignore numbers relating to heat capacity

2. Can gain / lose a lot of heat / energy without changing temperature; OR Takes a lot of heat / energy to change temperature;

Accept due to H bonding between water molecules 2

(b) Adenosine diphosphate and (inorganic) phosphate;

Accept ADP for adenosine diphosphate

Accept Pi / PO43– / P in a circle for inorganic phosphate

Reject adenine diphosphate

Reject phosphorus / P for phosphate 1

(c) 1. Species / organism the muscle tissue came from; OR Thickness / type / source of the muscle tissue;

Ignore surface area of muscle tissue

2. Temperature of the muscle tissue / ATP solution / slides;

Need to be qualified

3. pH of the ATP solution;


Need to be qualified

Reject concentration / volume of ATP hydrolase 2 max

(d) Description 1. As concentration of ATP increases, length of muscle decreases;

Accept negative correlation

Explanation 2. More ATP (hydrolysed by ATP hydrolase), so more energy released, so more muscle contraction / shortening of muscle;

Accept more ATP available for correct/named aspect of muscle contraction

Idea of more is required once.

Reject energy produced 2

(e) 4.88 × 10–6 ;;;

If answer incorrect

EITHER

Allow 1 mark for 0.244

Allow 1 mark for 1.22 × 10–5

OR

Allow 1mark for 12200 / 1.525

Allow 1 mark for 0.61

Accept 5 × 10–6

Accept correct answer however expressed

Max 2 for incorrect final answer 3

[10]

Q13. (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]

Experimental Data Analysis Questions Marks= 109 Time ...€¦ · Calculate how much antibiotic...

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