Topic 5 Class: Date: · 2019-12-19 · Chloroplasts contain chlorophyll a and chlorophyll b....

Topic 5

Name: ________________________

Class: ________________________

Date: ________________________

Time: 252 minutes

Marks: 182 marks

Comments:

of 67Woodford County High School

A student isolated chloroplasts from spinach leaves into a solution to form a chloroplastsuspension. He used the chloroplast suspension and DCPIP solution to investigate the light-dependent reaction of photosynthesis. DCPIP solution is blue when oxidised and colourlesswhen reduced.

The student set up three test tubes as follows:

• Tube 1 – 1 cm3 of solution without chloroplasts and 9 cm3 of DCPIP solution in light.

• Tube 2 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in darkness.

• Tube 3 – 1 cm3 of chloroplast suspension and 9 cm3 of DCPIP solution in light.

The student recorded the colour of the DCPIP in each of the tubes at the start and after the tubeshad been left at 20 °C for 30 minutes.

His results are shown in the table.

TubeColour of DCPIP in tube

At start After 30 minutes

1 blue blue

2 blue blue

3 blue colourless

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(a) The solution that the student used to produce the chloroplast suspension had the samewater potential as the chloroplasts.

Explain why it was important that these water potentials were the same.

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(b) Explain why the student set up Tube 1.

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(c) Explain the results in Tube 3.

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(d) The student evaluated the effectiveness of different chemicals as weed-killers by assessingtheir ability to prevent the decolourisation of DCPIP in chloroplast suspensions.

He added different concentrations of each chemical to illuminated chloroplast suspensionscontaining DCPIP. He then determined the IC50 for each chemical. The IC50 is theconcentration of chemical which inhibits the decolourisation of DCPIP by 50%.

Explain the advantage of the student using the IC50 in this investigation.

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(e) Explain how chemicals which inhibit the decolourisation of DCPIP could slow the growth ofweeds.

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(2)

(Total 9 marks)


Ecologists developed a method for estimating the biomass of trees in a plantation.The plantation consisted of trees of the same species.

They collected samples of wood from trees. For each sample they:

• determined the density of the freshly cut wood• dried the wood in an oven at 103 °C for 24 hours• determined the volume of the dried wood sample• determined the density of the dried wood.

The table below shows data about one wood sample.

Volume of freshly cutwood sample

/ dm3

Density of freshly cutwood

/ g per dm3

Volume of dried woodsample/ dm3

Density of dried woodsample

/ g per dm3

1.345 993.0 1.125 769.0

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(a) The loss of mass of the wood sample was due to loss of water. Water has a density of1 g per cm3.

Use the data in the table to calculate the percentage of water in the freshly cut woodsample. Show your working.

Percentage of water = __________________________________

(2)

(b) The ecologists dried the samples in an oven at 103 °C for 24 hours. Describe how theecologists could have determined whether or not this drying removed all the water from asample of wood.

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(c) Ecologists then investigated the relationship between the diameter of the trunk of the treesand their biomass.

The graph below shows their results. Each point is the result for one tree.

What does the graph show about the relationship between the diameter of the trunk of thetrees and their biomass?

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(2)


(d) Plantations of trees are often created to remove carbon dioxide from the atmosphere, tohelp to balance the carbon dioxide released by burning fossil fuels.

For different species of tree, information is available for:

• the relationship between diameter of trunk and freshly cut biomass• the percentage of water in fresh-cut wood• the mean dried density of wood.

Using only the information provided in part (c), suggest how the mass of carbon in thewood of a plantation of trees of a particular species could be estimated.

Start with measuring the diameter of a large number of trees.

Assume that the dry biomass of a tree consists of biological molecules that contain carbon.

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(4)

(Total 10 marks)


Chloroplasts contain chlorophyll a and chlorophyll b. Scientists found tobacco plants with amutation that caused them to make more chlorophyll b than normal tobacco plants. Theyinvestigated the effect of this mutation on the rate of photosynthesis.

The scientists carried out the following investigation.

• They grew normal and mutant tobacco plants. They grew some of each in low light intensityand grew others in high light intensity.

• They isolated samples of chloroplasts from mature plants of both types.• Finally, they measured oxygen production by the chloroplasts they had isolated from the

plants.

The figure below shows the scientists’ results.

Light intensity / μmol photons m-2s-1

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(a) Explain why the scientists measured the rate of production of oxygen in this investigation.

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In each trial, the scientists collected oxygen for 15 minutes.

(b) Calculate the difference in the oxygen produced by the chloroplasts from mutant plantsgrown in low and high light intensities at a light intensity of 500 μmol photons m–2 s–1.

Show your working.

Difference ____________________ μmol O2 mg–1 hour–1

(2)

(c) The scientists suggested that mutant plants producing more chlorophyll b would growfaster than normal plants in all light intensities.

Explain how these data support this suggestion.

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(4)

(Total 8 marks)


Read the following passage carefully.

A large and growing number of disorders are now known to be due to typesof mitochondrial disease (MD). MD often affects skeletal muscles, causingmuscle weakness.

We get our mitochondria from our mothers, via the fertilised egg cell.Fathers do not pass on mitochondria via their sperm. Some mitochondrialdiseases are caused by mutations of mitochondrial genes inside themitochondria. Most mitochondrial diseases are caused by mutations ofgenes in the cell nucleus that are involved in the functioning ofmitochondria. These mutations of nuclear DNA produce recessive alleles.

One form of mitochondrial disease is caused by a mutation of amitochondrial gene that codes for a tRNA. The mutation involvessubstitution of guanine for adenine in the DNA base sequence. Thischanges the anticodon on the tRNA. This results in the formation of anon-functional protein in the mitochondrion.

There are a number of ways to try to diagnose whether someone has amitochondrial disease. One test involves measuring the concentration oflactate in a person’s blood after exercise. In someone with MD, theconcentration is usually much higher than normal. If the lactate testsuggests MD, a small amount of DNA can be extracted from mitochondriaand DNA sequencing used to try to find a mutation.

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

(a) Mitochondrial disease (MD) often causes muscle weakness (lines 1–3). Use yourknowledge of respiration and muscle contraction to suggest explanations for this effect ofMD.

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Two couples, couple A and couple B, had one or more children affected by a mitochondrialdisease. The type of mitochondrial disease was different for each couple.

None of the parents showed signs or symptoms of MD.

• Couple A had four children who were all affected by an MD.• Couple B had four children and only one was affected by an MD.

(b) Use the information in lines 5–9 and your knowledge of inheritance to suggest why:

• all of couple A’s children had an MD• only one of couple B’s children had an MD.

Couple A ___________________________________________________________

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Couple B ___________________________________________________________

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(c) Suggest how the change in the anticodon of a tRNA leads to MD (lines 10–13).

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(d) If someone has MD, the concentration of lactate in their blood after exercise is usuallymuch higher than normal (lines 15–17). Suggest why.

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(e) A small amount of DNA can be extracted from mitochondria and DNA sequencing used totry to find a mutation (lines 18–19).

From this sample:

• how would enough DNA be obtained for sequencing?• how would sequencing allow the identification of a mutation?

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(2)

(Total 15 marks)

The figure below shows the apparatus used for measuring the rate of oxygen consumption inaerobic respiration by seeds.

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(a) For the first 10 minutes, the tap attached to tube A was left open and the syringe from tubeB was removed.

Suggest three reasons why the apparatus was left for 10 minutes.

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(b) Suggest and explain why the chosen temperature was 20 °C for this experiment.

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After 10 minutes, the tap attached to tube A was closed and the syringe was attached to tube B.Every minute, the syringe plunger was moved until the levels in the U-tube were the same. Thereading on the syringe volume scale was then recorded.

The results are shown in the table below.

Time / minutes Reading on syringevolume scale / cm3

0 0.84

1 0.81

2 0.79

3 0.76

4 0.73

5 0.70

6 0.68

7 0.66

8 0.63

9 0.62

10 0.58

(c) During the experiment, the coloured liquid in the tubing moved towards tube B.Explain what caused this.

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(3)


(d) The mass of the seeds was 1.6 g. Use the information in the table above to calculate therate of oxygen consumption in cm3 g–1 hour–1 by the seeds.

Show your working.

Rate = ____________________ cm3 g–1 hour–1

(2)

(Total 10 marks)


Herbicides can be used to reduce the growth of weeds.

Scientists completed seven studies to determine how the use of the herbicide Atrazine affectedthe yield of sugarcane. In each study, some plots were treated with Atrazine and some plots werenot treated with Atrazine.

The graph below shows the scientists’ results. (1 hectare = 10 000 m 2)

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(a) Calculate the percentage decrease in yield caused by the use of Atrazine in study G.

Answer = _________________ %

(1)


(b) A teacher studying these data with her students told her class that no definite conclusionscould be drawn when comparing the mean values in the graph.

Suggest why the teacher said this.

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(c) Atrazine binds to proteins in the electron transfer chain in chloroplasts of weeds, reducingthe transfer of electrons down the chain.

Explain how this reduces the rate of photosynthesis in weeds.

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(d) When treated with Atrazine, weeds have been shown to give off small amounts of heat.

Suggest an explanation for this observation.

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(Total 8 marks)

Scientists have investigated the use of different types of stem cell to treat damage to the heartafter a myocardial infarction. During a myocardial infarction, a number of different cell types in theheart die. This includes cardiomyocytes which are heart-muscle cells.

Embryonic pluripotent stem cells (ESCs) can divide and differentiate into a wide range of differentcell types.

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(a) Using the information given, suggest one reason why ESCs might be suitable to treatdamage to the heart.

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(b) ESCs have not yet been used to treat people who have had a myocardial infarction. This isbecause of concern that the use of ESCs might lead to more harm to the person. One waythat ESCs might lead to more harm is by differentiating into the wrong types of cells.

Suggest one other way that putting ESCs into a person’s heart might lead to more harm tothe person.

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(c) Transplants of cardiomyocytes have been shown to increase the repair of heart tissuedamaged by myocardial infarction.

One group of scientists investigated the hypothesis that these transplants work bystimulating growth of new blood vessels into damaged heart tissues. They obtained threegroups of mice, A, B and C that had suffered myocardial infarctions.

• Group A were operated on but no transplant was given.• Group B were operated on and given transplants containing cardiomyocytes and two

other types of heart cell.• Group C were operated on and given transplants containing the two other types of

heart cells but no cardiomyocytes.

After a suitable time, the scientists measured the mean number of capillaries per mm2 insections taken from areas of the hearts of the mice affected by myocardial infarction.

Their results are shown in the graph below. The bars show ±2 standard deviations, whichincludes 95.4% of the data.

Group A was a control group. Explain two ways in which Group A acts as a control.

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(d) What can you conclude from these data about the stimulation by cardiomyocytes on growthof new blood vessels into damaged heart tissues?

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(e) Suggest how the growth of new blood vessels into damaged heart tissues could increasethe rate of repair of tissues.

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(f) The scientists used an optical microscope to measure the number of capillaries in thinsections cut from samples of heart muscle.

Describe the method they would have used to find the mean number of capillariesper mm2.

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(4)

(Total 15 marks)


The image below shows a transmission electron micrograph of a longitudinal section of skeletalmuscle.

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(a) Name structures C, D and E.

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(3)

(b) Give the name of the structure shown between points A and B.

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(c) Calculate the actual distance between points A and B. Give your answer inmicrometres (µm).

Answer = __________________ µm

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(d) The image shows glycogen granules present in skeletal muscle.

Explain their role in skeletal muscle.

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(e) During vigorous exercise, the pH of skeletal muscle tissue falls. This fall in pH leads to areduction in the ability of calcium ions to stimulate muscle contraction.

Suggest how.

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(3)

(Total 10 marks)

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

The scientists results are shown in the table.

Temperature / °C Mean mass / gRate of carbon

dioxide release /µdm3 minute−1

Rate of carbon dioxiderelease per gram /µdm3 g−1 minute−1

10 0.047 0.12

20 0.046 0.33

30 0.048 0.56

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

(3)


The body temperature of the insects was largely determined by the temperature they were keptat. At each temperature, the scientists recorded rate of carbon dioxide release by individualinsects 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 and20 °C.

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

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

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(3)

(Total 8 marks)

Read the following passage.

Plants require phosphate ions that they get from soil. These ions areoften in poor supply and this results in poor growth of the plants. Mostplants have mycorrhizae that help the plants to obtain nitrates.Mycorrhizal networks can connect the roots of plants growing next to eachother. The use of fertilisers containing phosphate and nitrates in farminginhibits the growth of mycorrhizae. As a result, intensively farmed cropplants do not have mycorrhizae.

Plants can defend themselves by producing defensive enzymes thatdestroy pathogens such as bacteria. Some plants express the genes fordefensive enzymes in response to signal proteins secreted by other plantsthat are being attacked by a pathogen. These signal proteins can bereleased into the air.

Scientists have discovered that tomato plants increase production ofdefensive enzymes if plants next to them become infected with apathogen. These tomato plants were connected by a mycorrhizal networkthat can carry signal proteins between them. The largest increase indefensive enzyme secretion that the scientists found in a tomato plantin response to the signal protein was by 122.6 per cent.

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

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(a) Suggest and explain two reasons why a poor supply of phosphate ions results in poorgrowth of plants (lines 1–2).

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(b) Suggest how defensive enzymes produced by plants destroy bacteria (lines 8–9).

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(c) The signal proteins secreted into the air by a plant being attacked by a pathogen act asstimuli leading to the expression of genes for defensive enzymes in other plants (lines9–12).

Suggest how they lead to the expression of these genes.

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(d) Suggest and explain the advantage to tomato plants of transmitting signal proteins throughmycorrhizal networks, rather than releasing them into the air (line 11–12 and lines 14–16).

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(e) The largest increase in defensive enzyme secretion that the scientists found in a tomatoplant in response to the signal protein was by 122.6 percent (lines 16–18).

The rate of secretion of the defensive enzymes before the signal protein was produced was450 µmol dm−3 g−1 hour−1.

Calculate the rate of secretion per second after the response to the signal protein.

Answer = _________________________ µmol dm−3 g−1 second−1

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(f) A student who read this passage concluded that farmers should not use fertilisers toincrease yields when growing tomato plants.

Evaluate his conclusion.

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(4)

(Total 15 marks)

Farmers use artificial fertilisers to maintain or increase yield from grain-producing crop plantssuch as wheat.

(a) Artificial fertiliser is used to replace mineral ions removed from the land when crops areharvested. One of the mineral ions is nitrate.

Give two examples of biological molecules containing nitrogen that would be removedwhen a crop is harvested.

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(2)

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(b) Scientists investigated changes in the use of artificial fertiliser in India between 1970 and2005. They also investigated changes in the fertiliser response ratio. This ratio showshow many kg of grain are produced for each kg of fertiliser used.

The graph shows their results in the form the scientists presented them. (A hectare is a unitof area commonly used in agriculture)

Use these data to calculate the difference in the mass of grain produced per hectare in1970 compared with 2005.

Show your working.

Difference ____________________ kg hectare−1

(2)


(c) Use the data in the graph above to evaluate the use of artificial fertilisers on grain-producing crops in India.

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(2)

(Total 6 marks)

Scientists investigated the effect of regular exercise on skeletal muscle fibres in mice. Thescientists compared the muscle fibres of mice after six weeks of regular exercise (trained mice)with those of mice that had not exercised (control mice). The scientists stained the muscle fibresfrom both sets of mice to show succinic acid dehydrogenase activity. The darker the stain thegreater the succinic acid dehydrogenase activity.

The diagram below shows a typical set of results they obtained.

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(a) Succinic acid dehydrogenase is an enzyme used in the Krebs cycle.

Suggest one reason for the difference in the staining between the muscle fibres of thecontrol mice and the trained mice.

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(b) The scientists then compared the length of time that the control mice and the trained micecould carry out prolonged exercise. The trained mice were able to exercise for a longertime period than control mice.

Explain why.

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(c) The scientists determined the mean diameter of muscle fibres in trained mice using anoptical microscope to examine sections of muscle tissue. The circular area (πr2) of onefield of view was 1.25 mm2. The diameter of this area was equal to the diameter of 15muscle fibres.

Using this information, calculate the mean diameter in μm (micrometres) of muscle fibres inthis section of tissue.

Answer ____________________ μm(2)


(d) The scientists also compared the diameter of samples of muscle fibres taken from youngmice and adult mice.

Some of their results are shown in the graph.

Describe two differences between these samples of muscle fibres.

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(Total 8 marks)


(a) Exercise causes an increase in heart rate.

Describe the role of receptors and of the nervous system in this process.

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(b) AMP-activated protein kinase (AMPK) is an enzyme that regulates a number of cellularprocesses. Exercise leads to activation of AMPK.

The diagram shows one effect of activation of AMPK during exercise.


CPT1 is a channel protein that transports fatty acids into mitochondria.

Using the diagram above, explain the benefit of activation of AMPK during exercise.

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(3)

(Total 7 marks)

Arbuscular mycorrhiza fungi (AMF) are fungi which grow on, and into, the roots of plants. AMFcan increase the uptake of inorganic ions such as phosphate.

(a) Suggest one way in which an increase in the uptake of phosphate could increase plantgrowth.

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(b) Suggest one way in which AMF may benefit from their association with plants.

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(c) Scientists investigated the effects of different AMF species on the productivity of the plantcommunity of a prairie grassland ecosystem when growing in/on soil containing differentphosphate concentrations.

The scientists set up identical plots of prairie grassland soil containing seeds of the plantspecies found in the ecosystem. The scientists added different AMF species and differentconcentrations of phosphate to particular plots. Control plots without AMF species werealso set up. After 20 weeks the scientists determined the shoot biomass for each plot.

The results the scientists obtained are shown in the graph.

Explain why an increase in shoot biomass can be taken as a measurement of net primaryproductivity.

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(d) Using the data from the graph in part (c), evaluate the effect on plant productivity of addingAMF species and adding phosphate to the soil.

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(4)

(e) Using the ex button on your calculator, determine the rate of shoot biomass production ingrams per day for the control plot in soil with normal phosphate concentration.

Answer = ____________________ g day−1

(2)

(Total 10 marks)

Heat stress is a condition that often occurs in plants exposed to high temperatures for aprolonged period of time. Heat stress is a major factor in limiting the rate of photosynthesis.15

(a) Heat stress decreases the light-dependent reaction of photosynthesis.

Explain why this leads to a decrease in the light-independent reaction.

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(b) Another effect of heat stress is a decrease in the activity of the enzyme rubisco. Adecrease in the activity of an enzyme means that the rate of the reaction it catalysesbecomes slower.

A decrease in the activity of the enzyme rubisco would limit the rate of photosynthesis.

Explain why.

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(c) Where precisely is rubisco found in a cell?

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Scientists investigated the effect of temperature on the activity of two enzymes isolated from theleaf cells of cotton plants.• Rubisco• Rubisco activase – an enzyme that activates rubisco

Figure 1 and Figure 2 show their results.


(d) The scientists concluded that heat stress reduces the activity of rubisco in plant leaves byaffecting rubisco activase.

Use all the information to evaluate their conclusion.

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(4)

(Total 9 marks)


Ammonia in soil is oxidised to nitrites and nitrates by species of nitrifying bacteria.

Scientists investigated whether two soils with a different pH contained different communities ofnitrifying bacteria. These communities consist of all the nitrifying bacteria of different species ineach soil. They took samples of soil from two sites, A and B.

They measured the pH of the samples and found that

• the soil from site A had a pH of 6.9• the soil from site B had a pH of 4.3

The scientists measured the concentration of ammonia in soil samples over 20 days. Eachsample contained the same concentration of ammonia at the start and had the same mass. Theyrecorded the concentration of ammonia in

• soil A with a pH of 6.9• soil B with a pH of 4.3• a mixture of equal masses of soils A and B with its pH adjusted to 6.9

Their results are shown in Figure 1.

Figure 1

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(a) The scientists used units of μg g–1 for the concentration of ammonia in soil.

Suggest why, in this investigation, the scientists used these units.

μg ________________________________________________________________

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g–1 ________________________________________________________________

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(2)

(b) Calculate the difference in the rate of breakdown of ammonia per day between day 0 andday 2 in soil A and soil B.

Show your working and the units for your answer.

Difference in rate = ____________________

(2)


(c) The scientists concluded that the soil mixture experiment showed there were differentcommunities of bacteria in soils A and B.

What evidence from Figure 1 supports their conclusions? Give reasons for your answer.

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(3)


The oxidation of ammonia by nitrifying bacteria involves the enzyme ammonia monooxygenase.Each species of nitrifying bacteria has its own specific amoA gene that codes for production ofammonia monooxygenase.

In a second investigation, the scientists determined the expression of the amoA gene in twospecies of bacteria, S and T. Species S was from acid soil and species T was from soil with aneutral pH.

The scientists grew cultures of each species separately in soils of different pH. They determinedthe amount of mRNA from the amoA gene in each culture.

Their results are shown in Figure 2.

Figure 2


(d) In which species was the number of copies of mRNA more affected by changes in soil pHfrom 4.9 to 7.5? Use a calculation to support your answer.

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(2)

(e) This method allowed the scientists to estimate the expression of the amoA gene in eachculture but not the growth of the bacterial population in each culture.

Explain why.

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(4)


(f) The scientists set up their cultures in sterile glass bottles.

Suggest one suitable method for sterilising the bottles and explain why it was necessary tosterilise them.

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(2)

(Total 15 marks)

Yeast cells can respire aerobically or anaerobically. A student used the apparatus shown inFigure 1 to measure the rate of respiration in yeast.

She:

• positioned the flask in a water bath so that the yeast culture reached a constanttemperature

• then left the apparatus for one hour before starting her investigation.

Figure 1

17


(a) Suggest one reason why it was important that the student left the apparatus for one hourafter the yeast culture reached a constant temperature.

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(1)

(b) During her investigation, the coloured liquid moved to the right.

Explain why it moved to the right.

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(2)

(c) The student found that the coloured liquid moved 1.5 cm in 24 hours. The diameter of thelumen (hole) of the capillary tubing was 1 mm.

The volume of a capillary tubing is given by πr2l, where π is 3.14 and l = length.

Calculate the volume of gas produced in cm3 hour–1.Show your working.

Answer = ____________________ cm3 hour–1

(2)


Figure 2 shows a typical population growth curve for yeast under laboratory conditions.

Figure 2

(d) Explain why a log scale is used to record the number of cells.

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(1)

(e) Many yeast cells die during the death phase.

Suggest one reason why.

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(1)

(f) The following equation can be used to make predictions of the growth in the population ofyeast cells under ideal laboratory conditions.

Xt = X0 ert

Xt = the population after a certain timeX0 = the population at the starte = 2.72 (base of natural logarithm)r = growth ratet = time period in hours over which r applies


A population of 2000 yeast cells was left for 10 hours.The value for the growth rate was 0.5

Assuming no yeast cells died, calculate the predicted size of the population after 10 hours.Show your working.

Answer = ____________________

(2)

(Total 9 marks)

Scientists investigated the process of succession on sand dunes. They measured the percentagecover of different species of plants on sand dunes of different ages.Some of the results the scientists obtained are shown in Figure 1.

Figure 1

18


(a) Describe how you would determine the mean percentage cover for beach grass on a sanddune.

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(3)


(b) The scientists concluded that the results shown in Figure 1 were due to succession takingplace.

Use Figure 1 to explain why the scientists reached this conclusion.

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(4)


The scientists also investigated how the proportion of sunlight reaching the ground changedduring succession. Some of the results the scientists obtained are shown in Figure 2.

Figure 2

(c) Use Figure 1 to explain the results in Figure 2.

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(1)

(d) Using evidence from Figure 2, what can you conclude about the net primary productivity(NPP) in the sand dunes that are older than 1000 years?

Explain your answer.

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(2)

(Total 10 marks)


Mark schemes

(a) 1. Osmosis does not occur;2. Chloroplast / organelle does not burst / lyse / shrivel / shrink;

1. Accept: osmosis would occur if water potentials were not thesame.

1 and 2, Accept: correct reference to osmotic lysis for 2 marks.

2. Accept: chloroplast would burst / lyse / shrivel / shrink if waterpotentials were not the same.

2. Reject: ‘ cell bursts/shrivels’

2. Ignore: damage to chloroplasts on its own is not enough for amark.

3. Reject: becomes turgid / flaccid.2

1

(b) 1. To show light does not affect DCPIP;2. To show chloroplasts are required;

Ignore: comparison with other tubes.2

(c) 1. Reduction of DCPIP by electrons;2. (From) chlorophyll / light dependent reaction;

1. Accept: hydrogen / H for electrons but not protons / hydrogenions / H* on their own.

2. Accept: from chloroplasts / photosystems / water.2

(d) Provides a standard / reference pointORCan compare different chemicals/weed-killersORCan compare different concentrations of chemicals / weed-killers;

Accept: decolourises quicker than 100% or saves time waiting forcomplete decolourisation.

Note: comparisons must be qualified.

Accept: find the most effective weed-killer or the most effectiveconcentration.

Accept: answers relating to cost effectiveness.1

(e) 1. Less / no ATP produced;2. Less / no reduced NADP produced;3. Less / no GP reduced / converted to TP;

2, Accept: less / no NADPH / NADPH2 / NADPH + H2 max

[9]

(a) 1. 35.22 or 35.23% = 2 marks;

Award 1 mark if only fresh and dry masses correct, 1335.59 and 865.13 g;2

2


(b) (After 24 hours)

1. Record mass and reheat;

Accept return to oven = reheat

2. Until constant mass recorded;2

(c) 1. Positive correlation (between diameter and biomass);

2. Not linear / geometric / exponential / gradient gets steeper;2

(d) 1. Calculate a mean diameter;

2. (Use this to) estimate / determine the mean fresh biomass of trees;

3. Use the percentage water content to find the dried biomass;

4. Use the dried density to calculate the mass of tree;

5. Count / estimate the number of trees in plantation and multiply by (mean) carboncontent (to find total carbon);

4 max

[10]

(a) 1. Oxygen produced in light-dependent reaction;2. The faster (oxygen) is produced, the faster the light-dependent reaction.

2

3

(b) 35–36 μmol Oxygen per mg chlorophyll.

Correct difference at 500 μmol photons m–2 s–1 or incorrectdifference but division by 4 shown = 1 mark.

2

(c) At all light intensities, chloroplasts from mutant plants:1. Have faster production of ATP and reduced NADP;2. (So) have faster / more light-independent reaction;3. (So) produce more sugars that can be used in respiration;4. (So) have more energy for growth;5. Have faster / more synthesis of new organic materials.

Accept converse points if clear answer relates to non-mutant plants4 max

[8]

(a) 1. Reduction in ATP production by aerobic respiration;2. Less force generated because fewer actin and myosin interactions in muscle;3. Fatigue caused by lactate from anaerobic respiration.

3

4


(b) Couple A,1. Mutation in mitochondrial DNA / DNA of mitochondrion affected;2. All children got affected mitochondria from mother;3. (Probably mutation) during formation of mother’s ovary / eggs;

Couple B,4. Mutation in nuclear gene / DNA in nucleus affected;5. Parents heterozygous;6. Expect 1 in 4 homozygous affected.

4 max

(c) 1. Change to tRNA leads to wrong amino acid being incorporated into protein;2. Tertiary structure (of protein) changed;3. Protein required for oxidative phosphorylation / the Krebs cycle, so less / no

ATP made.3

(d) 1. Mitochondria / aerobic respiration not producing much / any ATP;2. (With MD) increased use of ATP supplied by increase in anaerobic respiration;3. More lactate produced and leaves muscle by (facilitated) diffusion.

3

(e) 1. Enough DNA using PCR;2. Compare DNA sequence with ‘normal’ DNA.

2

[15]

(a) 1. Equilibrium reached.

Accept equilibrate

2. Allow for expansion / pressure change in apparatus;3. Allow respiration rate of seeds to stabilise.

Ignore seeds acclimatise3

5

(b) 1. Optimum temperature / temperature for normal growth of seeds;2. (Optimum temperature) for enzymes involved in respiration.

2

(c) 1. Oxygen taken up / used by seeds;2. CO2 given out is absorbed by KOH (solution);3. Volume / pressure (in B) decreases.

3

(d) 0.975 / 0.98.

If incorrect,

0.26 × 6 / or incorrect numbers divided by 1.6 for 1 mark2

[10]


(a) 7.7(%);1

(b) 1. No error bars / SD;

2. To show if overlap occurs so difference (in means) is not significant / due to chanceORTo show if no overlap occurs so difference (in means) is significant / is not due tochance.

Do not accept ‘no statistical test performed’ as Chi squared /Spearman’s rank would be inappropriate.

Ignore references to sample size as it can be assumed thatscientists completed the study using appropriate methodology.

2

6

(c) 1. Reduced transfer of protons across thylakoid membraneORReduced chemiosomotic gradient / proton gradient across thylakoid membrane;

2. (So) less ATP produced;

3. (So) less reduced NADP produced;

Accept NADPH / NADPH2 / NADPH+

Reject reduced NAD

4. (So) light-independent reaction slows / stops;ORLess reduction of GP to triose phosphate.

4

(d) Idea that energy is released from high energy / excited electron/s (that were lost fromchlorophyll)

1

[8]

(a) 1. (ESCs) can replace any type of (heart) cell;

Accept named type of cell, e.g. heart muscle cell1

7

(b) 1. Might divide out of control;

2. Leading to tumour / cancer;2

(c) 1. Shows the effects of surgery;

2. Allows effects of transplants / treatment to be seen;

Allow in either order2


(d) 1. Other cell types cause some increase but most of increase due to cardiomyocytes;

2. Large SD, so some not much increase / no better than control;

3. Overlap of SDs indicates / suggests no significant difference;3

(e) 1. Greater blood supply (to damaged areas);

2. Bringing more oxygen / glucose for respiration;

3. Brings more amino acids for protein synthesis;

4. For cell repair / mitosis / division;3 max

(f) 1. Measure diameter of field of view and calculate area;

2. Using micrometer slide and eyepiece graticule;

Accept descriptions

3. Count number of capillaries in large number of fields of view and calculate mean;

4. Select fields of view randomly4

[15]


(a) C = M line / M disc / myosin filament

D = mitochondrion

E = myofibril3

(b) Sarcomere1

(c) Answer in range 1.14–1.181

(d) 1. As a store of glucose

Ignore provide energyOR

To be hydrolysed to glucose;

2. For respiration / to provide ATP;2

(e) 1. Low pH changes shape of calcium ion receptors

Do not accept tropomyosin does not move

2. Fewer calcium ions bind to tropomyosin;

Accept troponin

3. Fewer tropomyosin molecules move away;

4. Fewer binding sites on actin revealed;

5. Fewer cross-bridges can form

OR

Fewer myosin heads can bind

Must include idea of fewer at least once3 max

[10]

8

(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 noextrapolation;

3

9

(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 / getmore oxygen;

Note – explanation required3

[8]

(a) 1. (Required to) make ATP / glucose phosphate, so less respiration / less energy forgrowth;

2. (Required to) make nucleotides, so less DNA / mRNA / tRNA for cell division /production of protein (for growth);

3. (Required to) make RuBP / NADP, so less CO2 fixed / reduced into sugar;

4. (Required to) make phospholipids for membranes;2 max

10

(b) 1. Hydrolyse;

Accept digest

2. murein / glycoprotein (in cell wall);2

(c) 1. Bind to receptor (on target plant);

2. Acts as / leads to production of a transcription factor;

3. (Which) binds to promoterORstimulates transcription of genesORproduction of mRNA (for defensive enzymes);

3

(d) 1. Direct plant-to-plant transmission;

2. (So) localised responseORfaster responseORno dilution of signal protein;

2


(e) 0.278;

Accept 1 mark for 1001.7 or

× 100

= 122.62

(f) Should not use:

1. Fertilisers prevent development of mycorrhizae;

2. Mycorrhizae help plants to defend themselves (causing an increase in crop yield);

3. Mycorrhizae help plants to take up nitrates / phosphates (causing an increase in cropyield);

Should use:

4. Fertilisers containing phosphate and nitrate increase gross primary production soincrease yield;

5. Most soil is poor in phosphate so without fertiliser (tomato) plant might not get enoughphosphate;

4 max

[15]

(a) Two suitable examples;Examples1. amino acid / protein / polypeptide / peptide;2. nucleic acid / nucleotide / base;3. DNA;4. RNA;5. ATP / ADP;6. NAD / NADP (reduced or not);7. Cyclic AMP / cAMP;8. Chlorophyll;

List rule applies

Reject for either point nitrates / nitrites / ammonia / ammonium /urea

4. Accept pre-mRNA / mRNA / rRNA / tRNA2 max

11

(b) Correct answer in the range 90 to 133.2 scores 2 marks;1 mark for answers where yield calculated correctly for 1970 OR 2005;(1970 in range) 170.8 to 176.4OR(2005 in range) 266.4 to 304.0;

Accept positive or negative values2


(c) 1. Using more but getting less response over time;2. The graph shows correlation but doesn’t prove changes in yield due to fertiliser

/ but there could be other factors;3. Becomes less cost effective with time;

Idea of over time is important

1. accept fertiliser becomes less effective over time

1. Accept use of figures from graph

1. Accept the idea of less grain / crop over time

2. Ignore whether correlation is positive or negative2 max

[6]

(a) Increase in aerobic respirationORIncrease in / more mitochondriaORIncrease in / more slow muscle fibres;

Ignore: reference to Krebs cycle as this in the stem of the question.1 max

12

(b) 1. (More aerobic respiration) produces more ATP;2. Anaerobic respiration delayed;3. Less or no lactate;

1. Accept: produces ATP faster.

2. Accept: aerobic respiration can continue.

2. Accept : no anaerobic respiration.

3. Accept: lactic acid.3

(c) 1. Correct answer in range 84 to 84.2 = 2 marks;2. For one mark accept incorrect answer but shows r (radius) = 0.63 (mm)

ORd (diameter) = 1.26 (mm);

2. Ignore: numbers after 0.63 and 1.26.2


(d) 1. A numerical comparison of range = 2 marks i.e.Young (fibres) range 14/15 – 47/48 (μm) and adult (fibres) 17/18 – 86/87/88(μm)ORYoung (fibres) range 32/33/34 and adult (fibres) range 68/69/70/71;

2. Comparison of range without numbers = one mark i.e.Adult (fibres) greater range / spread / variation (of diameters)ORYoung (fibres) smaller range / spread (of diameters);

3. Comparison of mode = one mark i.e.Adult (fibres) peak/most common/frequent/mode at 50 (μm) and young (fibres)peak/most common/frequent/mode at 30 (μm);

1. Accept: one mark for comparison of minimum values i.e.

14/15 compared to 17/18

Allow one mark for comparison of maximum values i.e.

47/48 compared to 86/87/88.

1. Note: comparison of both maximum and minimum values = 2marks.

3. Accept: adult (fibres) peaks at higher diameter or young(fibres) peak / most frequent at lower diameter.

3. Reject: reference to mean / average.2 max

[8]

(a) 1. Chemoreceptors detect rise in CO2 / H+ / acidity / carbonic acid / fall in pHORBaro / pressure receptors detect rise in blood pressure;

2. Send impulses to cardiac centre / medulla;3. More impulses to SAN;4. By sympathetic (nervous system for chemoreceptors / CO2)

ORBy parasympathetic (nervous system for baro / pressure receptors / bloodpressure);

1. Ignore: location of receptors.

1. Ignore: chemoreceptors detect oxygen.

2 and 3. Accept: action potentials.

2. Reject: ‘messages’, ‘signals’, ‘an impulse’ or an ‘actionpotential’.

3. Ignore: messages’, ‘signals’, ‘an impulse’ or an ‘actionpotential’ as emphasis here is on increase in frequency.

4

13


(b) 1. Less / no malonyl-CoA;2. (More) fatty acids transported / moved into mitochondria;3. Respiration / oxidation of fatty acids provides ATP;

1. ‘Inhibition of malonyl-CoA’ on its own is not enough but acceptproduction of malonyl-CoA is inhibited.

2. Accept: ‘transport of fatty acids into mitochondria is notinhibited’.

2. Ignore: method of entry.

3. Accept: for respiration any stage of aerobic respiration e.g.Krebs (cycle), link (reaction) etc.

3. Reject: production of energy, but accept production of energyin the form of ATP.

3. Accept: acetyl CoA can enter Krebs cycle / mitochondria toprovide ATP.

3

[7]

(a) Used to produce named phosphate compound in cells;e.g. ATP / ADP / phospholipids / DNA / RNA / RuBP / TP /GP etc.

1

14

(b) Example of a carbon-containing biological compound e.g. carbohydrate / amino acid /vitamin;

Accept: sugars / organic (compounds).

Ignore: products of photosynthesis.

Ignore: starch.1

(c) 1. Represents dry mass / mass of carbon;2. Represents gross production minus respiratory losses;

2. Accept: NPP = GPP −R.2. Accept: Chemical energy minus respiratory losses.

1 and 2. Chemical energy store minus respiratory losses = 2 marks.2

(d) 1. For the control an increase in phosphate increases (plant) growth;2. For Entrophospora an increase in phosphate reduces (plant) growth;3. Scutellospora reduces (plant) growth (compared to control);4. Entrophospora and Glomus increases (plant) growth (compared to control);5. No SD / statistical test to determine significance;6. Only 20 weeks of growth;7. Underground / root growth not known;

5. Accept: no error bars.

7. Accept: only shows shoot growth.4 max


(e) 1. Answer in range 0.07 to 0.09 = 2 marks;2. Answer in range 9.97 to 12.2

ORShows division by 140 or 20 × 7 = 1 mark;

2

[10]

(a) 1. (Less/no) ATP;

2. (Less/no) reduced NADP;

Accept NADPH, NADPH + H, NADPH2 NADPH + H+

Reject reduced NAD, NADH etc,2

15

(b) 1. (Less/no) carbon dioxide (reacts) with RuBP;

2. (Less/no) GP;2

(c) 1. Stroma (of/in chloroplast);

Reject: stoma

Reject stroma of cytoplasm/chlorophyll

Reject stroma of mitochondrion

Ignore references to Calvin cycle or the light-independent reaction1

(d) 1. Rubisco activity increases with temperatureORRubisco optimum temperature is above (rubisco activase);

2. (Rubisco) activase activity decreases at high temperatures (allow any temperatureabove 25 ºC.)OR(Rubisco) activase optimum (allow in range) 25 to 30 ºC.;

Accept denatures at high temperature (allow any temperatureabove 25 ºC)

3. (Results/graphs suggest) activase cannot/does not affect activity of rubisco;

4. (Results are) only for cotton;

Accept may not be the same in other species/types of plant

Ignore: only one study

5. (Results are) for isolated enzymes;

6. No stats test;4 max

[9]


(a) 1. (μg because) very little ammonia (in soil);

2. (μg because) avoids use of (lots of) decimal places (in their results) / avoids the useof powers of 10 / avoids the use of standard form;

Accept makes numbers more manageable

Accept makes easier to plot graph

3. (g–1) to allow comparisons (between samples);2 max

16

(b) Answer between 4.5 and 4.6 μg g–1 day–1;;

Award 1 mark for correct number but wrong / no units

Ignore plus or minus signs

Accept ‘per gram’ AND / OR ‘per day’2

(c) 1. pH 4.3 / B has fastest rate of breakdown (of ammonia);

2. A + B / mixture at pH 6.9 slowest / slower (than A or B);

Not just ref. to A and then B on their own

3. Suggests (community / bacteria at) pH 4.3 / B doesn’t work (well) at pH 6.9 / pH ofmixture;

Accept converse that only (community / bacteria at) pH 6.9 / A isworking in the mixture

3

(d) (Species S because) no mark

1. Species S change of 990,000 (per gram of soil);

Award MP1 and 2 OR MP3 and 4

Accept standard forms 9.9 × 105 and 9.9 × 103 for either

Accept for 1 mark for 100 times greater in correct context with noother calculations shown

2. Species T change of 9,900 (per gram of soil);

Accept standard forms 9.9 × 105 and 9.9 × 103 for either

OR

(Species T because) no mark

3. Species S has 99% change;

4. Species T has 9900% change;2


(e) 1. They didn’t count bacteria / cells / population(s);

Ignore ref. to other factors / other named factors affecting growth

2. Copies / number of mRNA related to amount of enzyme / amoA produced /translated;

3. Don’t know how much mRNA / amoA produced by each cell;

Accept some bacteria produce more mRNA / amoA than others

4. Don’t know if amoA (mRNA / enzyme) is linked to cell division / growth (ofpopulation);

Amount of amoA does not show cell division / growth

Reject references to mitosis / meiosis4

(f) 1. Suitable method;eg in boiling water / steam / autoclave / wash in disinfectant / wash in alcohol

Ignore heat unqualified

Ignore flaming of bottle

Accept radiation

2. (Reason) to remove / kill other bacteria / organisms that might break down ammonia;

Ignore ref to removing bacteria that ‘affect the result’

Accept other bacteria producing amoA

Accept other bacteria compete with / kill bacteria that produceamoA

Ignore contamination unqualified2

[15]

(a) (So the) oxygen is used/absorbed/respired;117

(b) 1. Anaerobic respiration produces carbon dioxide;

2. Increase in pressure/volume (of gas);

Reference to either volume or pressure required for the mark2

(c) 1. Correct answer in range of4.9 × 10–4 to 4.91 × 10–4 = 2 marks;;

Accept any equivalent mathematical representation of this answer

2. Incorrect answer buts shows division by 24 = 1 markORIncorrect answer but shows a number from 1175 to 1178 (ignore position of decimalpoint, standard form and any numbers that follow) = 1 mark;ORIncorrect answer but show the number 49 (ignore position of decimal point, standardform and any numbers after 49) = 1 mark;

2


(d) Large range/difference/increase in numbers;

Accept reference to exponential (increase)

Ignore if the answer only refers to numbers being high

Ignore to ‘fit on the scale’1

(e) Decrease/no glucose/substrateORIncrease in ethanol/carbon dioxide/acidity;

Accept decrease/no oxygen as Figure 2 is not linked to Figure 1.

Accept competition for glucose/oxygen.

Accept any named sugar

Accept decrease in pH

Accept increase in toxins

Ignore food/nutrients1

(f) 1. Correct answer of 298000 or 297766 or 297765.59 or 296826 = 2 marks;;

Accept: any equivalent answer with appropriate rounding

e.g. 2.98 × 105,

29.78 × 104 etc.

2. Incorrect answer but working shows 2000 × 2.72 = 1 mark;ORIncorrect answer but working shows 2.720.5 × 10 / 2.725 / e0.5 × 10 / = 1 mark

2

[9]

(a) 1. Method of randomly determining position (of quadrats) e.g. random numberstable/generator;

Ignore line/belt transect

2. Large number/sample of quadrats;

Accept many/multiple

Ignore point quadrat

If a specified number is given, it must be 20 or more

3. Divide total percentage by number of quadrats/samples/readings;3

18


(b) 1. Beach grass is the pioneer (species);

2. Pioneers/named species change the (abiotic) environment/habitat/conditions/factors;

Must convey idea of change being caused by a species

Accept example of change e.g. more humus

3. (So) less hostile for named speciesOR(So) more suitable for named species;

4. Conifer/hardwood trees represent climax community;4

(c) Trees block/reduce (sun)light;

Reject ‘blocks’ all of the light1

(d) 1. (NPP) remains constant;

2. GPP/photosynthesis and respiration constant;

Accept GPP/photosynthesis equals respiration

OR

3. (NPP) low/decreases;

4. Less light so less photosynthesis/GPP;

Reject no photosynthesis

Mark in paired statements

1 and 2 or 3 and 42

[10]


Topic 5 Class: Date: · 2019-12-19 · Chloroplasts contain chlorophyll a and chlorophyll b....

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Transcript of Topic 5 Class: Date: · 2019-12-19 · Chloroplasts contain chlorophyll a and chlorophyll b....