1. (a)€€€€€State and explain the property of water that

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

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(2)

1.

(b) Water is used to hydrolyse ATP.

Name the two products of ATP hydrolysis.

1. _________________________________________________________________

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 usedshortening of strips of muscle tissue caused by contraction as evidence that ATP was beinghydrolysed.

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

• She then added the same volume of ATP solutions of different concentrations to the fourslides 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.

Page 1 of 57www.examqa.com

Slide

Concentration ofATP solution addedto slide / × 10–6 mol

dm–3

Final length of muscletissue 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.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

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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)

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

Explain how.

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(2)

2.


(b) A scientist measured the rate of removal of amino acids from a polypeptide with andwithout an enzyme present. With the enzyme present, 578 amino acids were released persecond. 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 thegraph.


(c) Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm–3 of ATPas 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 withoutlyxose.

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(3)

(Total 9 marks)


The diagram shows part of a DNA molecule.

(a) How many nucleotides are shown in the diagram above?

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(1)

3.

(b) Name the type of bond labelled X in the diagram.

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(1)

(c) The enzymes DNA helicase and DNA polymerase are involved in DNA replication.

Describe the function of each of these enzymes.

DNA helicase _______________________________________________________

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___________________________________________________________________

DNA polymerase _____________________________________________________

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(2)


(d) Adenosine triphosphate (ATP) is a nucleotide derivative.

Contrast the structures of ATP and a nucleotide found in DNA to give two differences.

1. _________________________________________________________________

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2. _________________________________________________________________

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(2)

(Total 6 marks)

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 antibioticinhibits the bacterial ATP synthase enzyme.

(a) Place a tick (✔) in the appropriate box next to the equation which represents the reactioncatalysed by ATP synthase.

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

(1)

4.


(b) The new antibiotic is safe to use in humans because it does not inhibit the ATP synthasefound 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. Theydivided 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 thetotal 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 comparedwith 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 betreated 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 antibioticresistance in bacteria in your answer.

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(3)

(Total 9 marks)


(a) Describe the roles of calcium ions and ATP in the contraction of a myofibril.

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(5)

5.

(b) ATP is an energy source used in many cell processes. Give two ways in which ATP is asuitable energy source for cells to use.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

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(2)

(Total 7 marks)

Cells constantly hydrolyse ATP to provide energy.

(a) Describe how ATP is resynthesised in cells.

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(2)

6.


(b) Give two ways in which the hydrolysis of ATP is used in cells.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

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(2)

(c) This is a photograph (micrograph) of a mitochondrion taken using a scanning electronmicroscope.

What is the evidence that a scanning electron microscope was used to take thisphotograph?

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(1)

(d) Name the part of the mitochondrion labelled X in the photograph.

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(1)


(e) The actual length of the mitochondrion between points A and B in the photograph is 4 μm.

What is the magnification of the mitochondrion in the photograph?

Show your working.

Magnification ____________________

(2)

(Total 8 marks)

(a) What is the role of ATP in myofibril contraction?

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

7.


(b) Scientists investigated the effect of not being able to produce creatine on the forceproduced by muscle. They used mice with a mutation that made them not able to producecreatine.The force produced when these mice gripped with their paws was compared with the forceproduced by normal mice that were able to produce creatine.

The graph shows the scientists’ results.

(i) What was the percentage fall in the mean force produced by mice not able toproduce creatine, compared with the normal mice? Show your working.

Answer ____________________ %

(2)


(ii) Suggest an explanation for these results.

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(2)

(c) The mice that were not able to produce creatine were homozygous for a recessive allele ofa gene. Mice that are heterozygous for this allele are able to produce forces similar tothose of normal mice that are homozygous for the dominant allele of the same gene.

Explain why the heterozygous mice can produce forces similar to those of normal mice.

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(2)

(Total 8 marks)

(a) ATP is useful in many biological processes. Explain why.

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(4)

8.


(b) Describe how ATP is made in mitochondria.

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(6)

(c) Plants produce ATP in their chloroplasts during photosynthesis. They also produce ATPduring respiration. Explain why it is important for plants to produce ATP during respirationin addition to during photosynthesis.

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(5)

(Total 15 marks)


(a) The table contains statements about three biological processes.

Complete the table with a tick if the statement in the first column is true, for each process.

Photosynthesis Anaerobic

respirationAerobic respiration

ATP produced

Occurs inorganelles

Electron transportchain involved

(3)

9.

(b) Write a simple equation to show how ATP is synthesised from ADP.

___________________________________________________________________

(1)

(c) Give two ways in which the properties of ATP make it a suitable source of energy inbiological processes.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

___________________________________________________________________

(2)

(d) Humans synthesise more than their body mass of ATP each day. Explain why it isnecessary for them to synthesise such a large amount of ATP.

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(2)

(Total 8 marks)


(a) Describe the part played by the inner membrane of a mitochondrion in producing ATP.

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(3)

10.

(b) A scientist investigated ATP production in a preparation of isolated mitochondria. Hesuspended the mitochondria in an isotonic solution and added a suitable respiratorysubstrate together with ADP and phosphate. He bubbled oxygen through the preparation.

(i) Why was the solution in which the mitochondria were suspended isotonic?

______________________________________________________________

______________________________________________________________

(1)

(ii) Explain why the scientist did not use glucose as the respiratory substrate.

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(2)

(iii) Explain why the oxygen concentration would change during this investigation.

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(1)

(Total 7 marks)


In the light-dependent reaction of photosynthesis, light energy generates ATP.

Describe how.

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(Total 5 marks)

11.

The diagram represents two of the stages of aerobic respiration that take place in amitochondrion.

(a) Name substance X.

___________________________________________________________________

(1)

12.

(b) Which stage of aerobic respiration takes place inside a mitochondrion and is notrepresented on the diagram?

___________________________________________________________________

(1)


(c) Explain why oxygen is needed for the production of ATP on the cristae of themitochondrion.

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(3)

(Total 5 marks)


(a) The flow chart shows the main stages in aerobic respiration.

13.


(i) Complete the flow chart by writing, in the appropriate boxes, the number of carbonatoms in substance P and the name of substance Q.

(2)

(ii) Some ATP is formed in the cytoplasm and some in the mitochondria. Use theinformation given to calculate the number of molecules of ATP formed in amitochondrion from one molecule of glucose in aerobic respiration. Show how youarrived at your answer.

Answer____________________

(2)

(iii) In the presence of oxygen, respiration yields more ATP per molecule of glucose thanit does in the absence of oxygen. Explain why.

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(3)


(b) Anabaena is a prokaryote found inside the leaves of a small fern. Anabaena can produceammonia from nitrogen (nitrogen fixation). This reaction only takes place in the anaerobicconditions found in cells called heterocysts. Heterocysts are thick-walled cells that do notcontain chlorophyll. The drawing shows the relationship between Anabaena and the fern.


(i) Suggest how the features of the heterocysts improve the efficiency of the process ofnitrogen fixation.

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(3)

(ii) In China, the fern is cultivated and ploughed into fields to act as an organic fertiliser.Explain how ploughing the fern plants into the soil results in an improvement in thegrowth of the rice crop grown in these fields.

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(5)

(Total 15 marks)


The diagram summarises the process of anaerobic respiration in yeast cells.

(a) (i) In anaerobic respiration, what is the net yield of ATP molecules per molecule ofglucose?

______________________________________________________________

(1)

14.

(ii) Give two advantages of ATP as an energy-storage molecule within a cell.

1. ____________________________________________________________

______________________________________________________________

2. ____________________________________________________________

______________________________________________________________

(2)

(b) Describe how NAD is regenerated in anaerobic respiration in yeast cells.

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(1)

(Total 4 marks)


The diagram shows some of the stages in two processes that produce ATP.

Process 1

15.

Process 2

(a) In Process 1, what causes substance X to lose electrons (e–)?

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___________________________________________________________________

(1)

(b) Where precisely, within a cell, does electron transport take place in Process 2?

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(1)

(Total 2 marks)


The diagram shows the structure of a chloroplast.

(a) Label the diagram with an X to show where the light-dependent reactions take place andwith a Y to show where the light-independent reactions take place.

(1)

16.

(b) The photolysis of water is an important part of the process of photosynthesis. Describewhat happens in the photolysis of water.

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___________________________________________________________________

___________________________________________________________________

(2)

(c) ATP and reduced NADP are two products of the light-dependent reactions. Describe onefunction of each of these substances in the light-independent reactions.

ATP _______________________________________________________________

___________________________________________________________________

Reduced NADP ______________________________________________________

___________________________________________________________________

(2)

(Total 5 marks)


The diagram gives an outline of the process of aerobic respiration.

17.


(a) Name substances X, Y and Z.

X ________________________________________________________________

Y ________________________________________________________________

Z ________________________________________________________________

(3)

(b) Give the location of each of the following in a liver cell.

(i) Glycolysis ____________________________________________________

(ii) The Krebs cycle _______________________________________________

(2)

(c) (i) Write the letter A on the diagram to show one step where ATP is used.

(ii) Write the letter B on the diagram at two steps where ATP is produced.

(3)

(d) Apart from respiration, give three uses of ATP in a liver cell.

1. _________________________________________________________________

2. _________________________________________________________________

3. _________________________________________________________________

(3)

(e) Human skeletal muscle can respire both aerobically and anaerobically. Describe whathappens to pyruvate in anaerobic conditions and explain why anaerobic respiration isadvantageous to human skeletal muscle.

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(4)

(Total 15 marks)


(a) The table contains some statements relating to biochemical processes in a plant cell.Complete the table with a tick if the statement is true or a cross if it is not true for eachbiochemical process.

Statement Glycolysis Krebs cycle Light-dependentreaction of

photosynthesis

NAD is reduced

NADP is reduced

ATP is produced

ATP is required

(4)

18.

(b) An investigation was carried out into the production of ATP by mitochondria. ADP,phosphate, excess substrate and oxygen were added to a suspension of isolatedmitochondria.

(i) Suggest the substrate used for this investigation.

______________________________________________________________

(1)

(ii) Explain why the concentration of oxygen and amount of ADP fell during theinvestigation.

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(2)


(iii) A further investigation was carried out into the effect of three inhibitors, A, B and C,on the electron transport chain in these mitochondria. In each of three experiments, adifferent inhibitor was added. The table shows the state of the electron carriers, W–Z,after the addition of inhibitor.

Inhibitoradded

Electron carrier

W X Y Z

A oxidised reduced reduced oxidised

B oxidised oxidised reduced oxidised

C reduced reduced reduced oxidised

Give the order of the electron carriers in this electron transport chain. Explain youranswer.

Order _______ _______ _______ _______

Explanation ____________________________________________________

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(2)

(Total 9 marks)

(a) Explain how three features of a plasma membrane adapt it for its functions.

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(6)

19.


(b) ATP breaks down to ADP and phosphate releasing energy. The graph shows the rate of ionmovement and the rate of ATP production in an investigation carried out on a suspensionof cells. At a certain point in the investigation, a respiratory poison was added to the cellsuspension. Later, ATP was added to the same cell suspension.

Describe and explain the changes in the rate of ion movement.

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(4)

(Total 10 marks)


(a) The diagram shows the banding pattern observed in part of a relaxed muscle fibril.

(i) Describe what causes the different bands seen in the muscle fibril.

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(2)

20.

(ii) Describe how the banding pattern will be different when the muscle fibril iscontracted.

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(2)


(b) There is an increase in the activity of the enzyme ATPase during muscle contraction. Aninvestigation into muscle contraction involved measuring the activity of ATPase in solutionscontaining ATP, myosin and different muscle components. The table shows the results.

Solution ContentsATPase activity /

arbitrary units

A ATP, myosin and actin 1.97

B ATP, myosin, actin and tropomyosin 0.54

C ATP, myosin, actin, tropomyosin and calcium ions 3.85

(i) Explain the importance of ATPase during muscle contraction.

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______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(ii) Using your knowledge of muscle contraction, explain the difference in the resultsbetween

A and B;

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(2)

B and C.

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(2)

(Total 10 marks)


Pea plants are leguminous and have nodules on their roots which contain bacteria that are ableto fix nitrogen. The diagram shows some of the processes involved in nitrogen fixation by thesebacteria.

21.

(a) Name

(i) substance X;

______________________________________________________________

(1)

(ii) substance Y.

______________________________________________________________

(1)

(b) Pea plants respire aerobically, producing ATP which can be used for amino acid synthesis.Describe the role of oxygen in aerobic respiration.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


(c) The bacteria respire anaerobically. This produces hydrogen and ATP used in nitrogenfixation. The hydrogen comes from reduced NAD. Explain how the regeneration of NAD inthis way allows ATP production to continue.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(d) The enzyme nitrogenase is specific to the reaction shown. Explain how one feature of theenzyme would contribute to this specificity.

Feature

___________________________________________________________________

Explanation

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(e) Sodium ions act as a non-competitive inhibitor of the enzyme nitrogenase. Explain how thepresence of a non-competitive inhibitor can alter the rate of the reaction catalysed bynitrogenase.

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___________________________________________________________________

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(3)

(Total 11 marks)


The diagram shows the stages in one cycle that results in movement of an actin filament in amuscle sarcomere.

22.

(a) Describe how stimulation of a muscle by a nerve impulse starts the cycle shown in thediagram.

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___________________________________________________________________

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(3)


(b) Each cycle requires hydrolysis of one molecule of ATP and moves one actin filament 40 nm. During contraction of a muscle sarcomere, a single actin filament moves 0.6 µm.Calculate how many molecules of ATP are required to produce this movement.

Answer ___________________________________

(2)

(c) After death, cross bridges between actin and myosin remain firmly bound resulting in rigormortis. Using information in the diagram, explain what causes the cross bridges to remainfirmly bound.

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___________________________________________________________________

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(2)

(Total 7 marks)


The diagram shows a summary of the light-independent reaction of photosynthesis.

(a) (i) Complete the boxes to show the number of carbon atoms in the molecules.

(2)

23.

(ii) In which part of a chloroplast does the light-independent reaction occur?

______________________________________________________________

(1)

(iii) Which process is the source of the ATP used in the conversion of glycerate3-phosphate (GP) to triose phosphate?

______________________________________________________________

(1)

(iv) What proportion of triose phosphate molecules is converted to ribulose bisphosphate(RuBP)?

______________________________________________________________

(1)


(b) Lowering the temperature has very little effect on the light-dependent reaction, but it slowsdown the light-independent reaction. Explain why the light-independent reaction slowsdown at low temperatures.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 7 marks)


Mark schemes

(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;ORTakes a lot of heat / energy to change temperature;

Accept due to H bonding between water molecules2

1.

(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 phosphate1

(c) 1. Species / organism the muscle tissue came from;ORThickness / 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 hydrolase2 max

(d) Description1. As concentration of ATP increases, length of muscle decreases;

Accept negative correlation

Explanation2. More ATP (hydrolysed by ATP hydrolase), so more energy released, so moremuscle contraction / shortening of muscle;

Accept more ATP available for correct/named aspect of musclecontraction

Idea of more is required once.

Reject energy produced2


(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 answer3

[10]

(a) 1. Reduces activation energy;

Accept ‘reduces E a’.

2. Due to bending bondsORWithout enzyme, very few substrates have sufficient energy for reaction;

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

Ignore references to ‘breaking bonds’.2

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]

(a) 8;

Accept eight1

3.

(b) Phosphodiester (bond);

Accept phonetic spellings1


(c) 1. DNA helicase – (unwinding DNA and) breaking hydrogenbonds / bonds between chains / bases / strands;

2. DNA polymerase – joins (adjacent) nucleotides OR formsphosphodiester bond / sugar-phosphate backbone;

1. Accept H bonds.

1. Accept hydrolyses for breaks

2. Reject forms hydrogen bonds (between nucleotides / bases)2

(d) 1. ATP has ribose and DNA nucleotide has deoxyribose;2. ATP has 3 phosphate (groups) and DNA nucleotide has 1

phosphate (group);3. ATP – base always adenine and in DNA nucleotide base can

be different / varies;

Both parts of each MP needed

3. Reject Uracil / U

3. Accept C, T or G for different bases

Accept annotated diagram for any of the three marks2 max

[6]

(a)

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

1

4.

(b) 1. Human ATP synthase has a different tertiary structure to bacterial ATP synthaseORHuman ATP synthase has a different shape active site to bacterial ATPsynthaseORAntibiotic cannot enter human cells/mitochondriaORAntibiotic 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%;ORAnswer in range 3288 − 4368%;

2. 1 mark for correct log10 readings from graph converted toactual 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 / oldantibiotic;

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

ORUnlikely that bacteria are resistant to both the new and the old antibiotic;ORUse 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 – ‘somebacteria are killed’ or ‘the bacteria number is reduced ’ isinsufficient.

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

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

[9]

(a) 1. Calcium ions diffuse into myofibrils from (sarcoplasmic) reticulum;2. (Calcium ions) cause movement of tropomyosin (on actin);3. (This movement causes) exposure of the binding sites on the actin;4. Myosin heads attach to binding sites on actin;5. Hydrolysis of ATP (on myosin heads) causes myosin heads to bend;6. (Bending) pulling actin molecules;7. Attachment of a new ATP molecule to each myosin head causes myosin heads

to detach (from actin sites).5 max

5.

(b) 1. Releases relatively small amount of energy / little energy lost as heat;

Key concept is that little danger of thermal death of cells

2. Releases energy instantaneously;

Key concept is that energy is readily available

3. Phosphorylates other compounds, making them more reactive;4. Can be rapidly re-synthesised;5. Is not lost from / does not leave cells.

2 max

[7]


(a) 1. From ADP and phosphate;

Accept

Reject P/Phosphorus

Reject use of water in the reaction

2. By ATP synthase;3. During respiration/photosynthesis;

2 max

6.

(b) 1. To provide energy for other reactions/named process;

Reject ‘produce’ energy2. To add phosphate to other substances and make them more reactive/change their

shape;2

(c) (Can see) 3D image;1

(d) Crista/cristae;

Ignore matrix1

(e) Value between 20,750 (83mm) and 21,250 (85mm) two marks;;Formula given/used but calculation wrong, award 1 mark

Magnification = image size Object size

(Large number divided by 4)2

[8]

(a) 1. (Reaction with ATP) breaks/allows binding of myosin to actin/ actinomyosin bridge;2. Provides energy to move myosin head;

1. Credit ‘breaks’ or ‘allows’ binding to actin (because cyclical)

2. Allow in context of ‘power stroke’ or ‘re-cocking’ (becausecyclical)

2. Ignore contraction on its own2

7.


(b) (i) Any value between 68.5 and 69.49 (%);;

If get difference of 0.9 but calculation of percentage incorrect,then award 1 mark;

2

(ii) (Mutant mice)

1. Unable to make phosphocreatine/ less phosphateavailable to make/recycle ATP;

2. So less energy/so less ATP available for contraction/fastmuscle fibres;

1 and 2. Reject production/creation of energy once

2. Accept less energy for grip

2. Accept no energy/no ATP for contraction/fast muscle fibres2

(c) 1. (Heterozygous) have one dominant/normal allele (for creatineproduction);

2. (This) leads to production of enough/normal amount ofcreatine;

1. Accept has one allele/one copy of the gene for/that is makingcreatine

2[8]

(a) 1. Releases energy in small / manageable amounts;

1. Accept less than glucose

2. (Broken down) in a one step / single bond broken immediate energy compound/ makes energy available rapidly;

2. Accept easily broken down

3. Phosphorylates / adds phosphate makes (phosphorylated substances) morereactive / lowers activation energy;

3. Do not accept phosphorus or P on its own

4. Reformed / made again;

4. Must relate to regeneration4

8.


(b) 1. Substrate level phosphorylation / ATP produced in Krebs cycle;

Accept alternatives for reduced NAD

2. Krebs cycle / link reaction produces reduced coenzyme / reduced NAD /reduced FAD;

2. Accept description of either Krebs cycle or link reaction

3. Electrons released from reduced / coenzymes / NAD / FAD;

4. (Electrons) pass along carriers / through electron transport chain / throughseries of redox reactions;

5. Energy released;

5. Allow this mark in context of electron transport or chemiosmosis

6. ADP / ADP + Pi;

6. Accept H+ or hydrogen ions and cristae

7. Protons move into intermembrane space;

7. Allow description of movement through membrane

8. ATP synthase;

8. Accept ATPase. Reject stalked particles6 max

(c) 1. In the dark no ATP production in photosynthesis;

1. In context of in photosynthetic tissue / leaves

2. Some tissues unable to photosynthesise / produce ATP;

3. ATP cannot be moved from cell to cell / stored;

4. Plant uses more ATP than produced in photosynthesis;

5. ATP for active transport / synthesis (of named substance);5

[15]


(a)

Photosynthesis

Anaerobicrespiration

Aerobicrespiration

ATP produced

Occurs inorganelles

Electrontransport chaininvolved

1 mark per column

Mark ticks only. Ignore anything else if different symbols such ascrosses are used as well.

If crosses are used instead of ticks allow cross as equivalent to atick.

Reject tick with a line through3

9.

(b) ADP + Pi → ATP;

Both sides correct, but allow other recognised symbols or words forphosphate ion. Reject P unless in a circle.

Accept = as equivalent to arrow

Accept reversible arrow

Ignore any reference to kJ / water1

(c) 1. Energy released in small / suitable amounts;

2. Soluble;

3. Involves a single / simple reaction;

1. In context of release, not storage. Ignore producing energy /manageable amounts.

2. Reject "broken down easily / readily". Reject "quickly / easilyresynthesised".

2 max

(d) 1. ATP cannot be stored / is an immediate source of energy;

2. ATP only releases a small amount of energy at a time;2

[8]


(a) Electrons transferred down electron transport chain;

Provide energy to take protons / H+ into space between membranes;

Protons / H+ pass back, through membrane / into matrix / throughATPase;

Energy used to combine ADP and phosphate / to produce ATP;

Accept: alternatives for electron transport chain.3 max

10.

(b) (i) Prevent damage to mitochondria caused bywater / osmosis / differences in water potential;

Accept: other terms that imply damage e.g. shrink / burst1

(ii) Glucose is used / broken down during glycolysis in cytoplasm / not inmitochondria;

Accept: ‘glucose is converted to pyruvate’ for description ofbreakdown

Glucose cannot cross mitochondrial membrane / does notenter mitochondria;

Accept: only pyruvate can2

(iii) Terminal / final acceptor (in electron transport chain) / used tomake water;

Could be shown by symbols1

[7]

1. Light (energy) excites / raises energy level of electrons in chlorophyll;

2. Electrons pass down electron transfer chain;

Q Accept any reasonable alternative for electron transfer chain.

3. (Electrons) reduce carriers / passage involves redox reactions;

4. Electron transfer chain / role of chain associated with chloroplastmembranes / in thylakoids / grana;

Example such as chemiosmosis;

5. Energy released / carriers at decreasing energy levels;

6. ATP generated from ADP and phosphate / Pi / phosphorylation of ATP;

[5]

11.


(a) pyruvate;1

(b) Krebs cycle;1

(c) ATP formed as electrons pass along transport chain;oxygen is terminal electron acceptor / accepts electrons from electrontransport chain / electrons cannot be passed along electron transportchain if no O2 to accept them;

forms H2O / accepts H+ from reduced NAD / FAD / oxidises reducedNAD / FAD;

3

[5]

12.

(a) (i) P = 3;Q = acetylcoenzyme A;

2

(ii) 36 ATP, however derived = 2 marks30 ATP, however derived = 1 mark

2

13.

(iii) Correct statement in the context of aerobic respiration oranaerobic respiration concerning:Oxygen as terminal hydrogen / electron acceptor allowing operation of electrontransport chain / oxidative phosphorylation;Fate of pyruvate;Significance of ATP formed in glycolysis;

3

(b) (i) Thick walls exclude oxygen;Produced by photosynthetic cells (of fern and Anabaena);Contain no chlorophyll so do not photosynthesise;Do not produce oxygen;Oxygen would inhibit nitrogen fixation process;

max. 3

(ii) Decomposers / bacteria / fungi / saprobionts (in fields);

Convert protein / organic nitrogen (in cells of fern) intoammonium ions (allow ammonia);Ammonium ions (ammonia) converted to nitrite, then converted to nitrate;

Allow 1 mark for NH3 / NHBy nitrifying bacteria / correctly named;Nitrate used to form protein / amino acids in rice;

5

[15]

NO3


(a) (i) 2 (molecules)1

(ii) Cannot pass out of cell;Quickly / easily broken down (hydrolysed) / brokendown in a on-step reaction / immediate source of energy;Stores / releases small amounts of energy;Do not credit “producing energy”

max 2

14.

(b) Formed when reduced NAD used to reduce / donate H ionsto pyruvate / convert pyruvate to ethanol;

1

[4]

(a) (Absorption of) light;1

(b) Inner membrane / cristae / stalked particles of mitochondria;1

[2]

15.

(a) On diagram, correctly labelled:

Light-dependent: granum / thylakoid membranes – labelled ‘X’ANDLight-independent: stroma – labelled ‘Y’;

1

16.

(b) Any two from:

(Water) forms H+ / hydrogen ions and electrons / e– ;

O2 / oxygen formed; [NOT ‘O’, NOT ‘O –’]

(Light) excites electrons / raises energy level of electrons / electrons tochlorophyll / to photosystem;

max 2

(c) (ATP) Provides energy for GP → TP / provides P for RuP / TP → RuBP;

(Reduced NADP) Provides H / electrons for GP → TP / reduces GP to TP;2

[5]

(a) X = Carbon dioxide;Y = Acetyl coenzyme A;

(ACCEPT Acetyl CoA)

Z = Water;3

17.


(b) (i) Cytoplasm;1

(ii) Mitochondrion;

(IGNORE named part)1

(c) On the diagram:

(i) ‘ A’ (ATP used) – between glucose and triose phosphate;1

(ii) ‘ B’ Any two from:

(ATP produced) – between triose phosphate and pyruvate; in Krebs cycle; from electron carriers (to right of bracket & not below grey box);

max 2

(d) Any three from:

Source of energy / of phosphate;Active transport;Phagocytosis / endo- / exocytosis / pinocytosis;Bile production;Cell division / mitosis;Synthesis of: glycogen; protein / enzymes; DNA / RNA; lipid / cholesterol; urea;

max 3

(e) Any four from:Forms lactate; [extras – C2H5OH / CO2 – CANCEL]

Use of reduced NAD / NADH;

Regenerates NAD;

NAD can be re-used to oxidise more respiratory substrate / correct e.g. /allows glycolysis to continue;Can still release energy / form ATPwhen oxygen in short supply / when no oxygen;

max 4

[15]


(a) x;

x x ;

x x4

18.

(b) (i) pyruvate / succinate / any suitable Krebs cycle substrate;1

(ii) ADP and phosphate forms ATP;oxygen used to form water / as the terminal acceptor;

2

(iii) Y X W Z;order of carriers linked to sequence of reduction / reducedcarriers cannot pass on electrons when inhibited;

2

[9]

(a) feature and adaption; for example

1 phospholipid bilayer (as a barrier);2 forms a barrier to water soluble / charged substances /

allows non-polar substances to pass

OR

maintains a different environment on each side / compartmentalisation;

3 bilayer is fluid;4 can bend to take up different shapes for phagocytosis /

form vesicles / self repair;5 channel proteins (through the bilayer) / intrinsic protein;6 let water soluble / charged substances through / facilitated diffusion;7 carrier proteins (through the bilayer);8 allow facilitated diffusion / active transport;9 surface proteins / extrinsic proteins, glycoproteins / glycolipids;10 cell recognition / act as antigens / receptors;11 cholesterol;12 regulates fluidity / increases stability;

6 max

principle mark (only for 5, 6, 7, 8)proteins transport material across the membrane

3 features max

19.


(b) curve description:1 Curve goes down when the poison is added and rises when

ATP added;1

explanation:

2 Ion movement is by active transport;3 ATP / energy needed for active transport;4 respiration provides ATP / energy;5 poison inhibits / stops respiration / ATP production;

3 max

[10]

(a) (i) A / dark band is mainly due to myosin filaments;H zone only myosin filaments;darker band has both types of filament;light band has only actin filaments;

2 max

(ii) H zone narrows;light band narrows;outer darker regions of A / dark band widen;

2 max

20.

(b) (i) breaks down ATP yielding energy;used to form / break actomyosin bridges;

2

(ii) A and Btropomyosin covers binding site on actin;no cross bridges formed /ATPase activity on myosin head reduced;

2

B and Ccalcium ions remove tropomyosin;binding / calcium ions increase ATPase activity;

2

[10]

(a) (i) ammonia / ammonium ions / compound;1

(ii) glucose;1

(b) final acceptor for hydrogen:to form water;

2

21.

(c) glycolysis can continue;NAD can accept more hydrogen;

2


(d) secondary / tertiary structure;produces particular shape of active site;or(shape of) active site;complementary to shape of substrate;

2

(e) sodium ions / non-competitive inhibitor binds to enzymeat a site other than active site;resulting in change of shape of active site / no longer complementary;substrate can no longer bind with the enzyme / enzyme-substratecomplexes no longer formed;

3

[11]

(a) calcium ions;bind to / displace tropomysin; (allow troponin)reveal binding site on actin;myosin binds to exposed sites on actin / actomyosin formed /cross bridges form between actin and myosin;activates ATPase;

3 max

22.

(b) distance single actin filament moves divided by distance movedusing 1 ATP;15 ATP;

2

(c) respiration stops / no ATP produced;ATP required for separation of actin and myosin / cross bridges;

2

[7]

(a) (i) RuBP – 5; GP – 3; TP – 3; Glucose – 6;

(all correct = 2 marks; 3 or 2 correct = 1 mark)2

(ii) stroma;1

(iii) light-dependent reaction / (photo)phosphorylation;

(accept photolysis)1

(iv) 5 out of 6 / 83% / equivalent;1

23.

(b) enzymes involved / not a photochemical reaction;slow rate of enzyme / chemical reaction at low temperature /less kinetic energy / fewer collisions;

2

[7]


Documents

1. (a)€€€€€State and explain the property of water that