Q1.(a)     Describe how you could use cell fractionation to isolate chloroplasts from leaf tissue.

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The figure below shows a photograph of a chloroplast taken with an electron microscope.

 

© Science Photo Library

(b)     Name the parts of the chloroplast labelled A and B.

Name of A ________________________________________________________

Name of B ________________________________________________________(2)

(c)     Calculate the length of the chloroplast shown in the figure above.

 

 

 

 

Answer ____________________(1)

(d)     Name two structures in a eukaryotic cell that cannot be identified using an optical microscope.

1. _________________________________________________________________

2. _________________________________________________________________(1)

(Total 7 marks)

Q2.The drawing shows part of a human cell.

 

(a)     Name organelles

X _________________________________________________________________

Y _________________________________________________________________(2)

(b)     (i)      The organelles labelled X all have very similar shapes in this cell.Explain why they appear to have different shapes in this drawing.

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(ii)     Large numbers of organelles X and Z are found in mucus-secreting cells.Explain why.

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(Total 5 marks)

Q3.A technician investigated the effect of temperature on the rate of an enzyme-controlled reaction. At each temperature, he started the reaction using the same volume of substrate solution and the same volume of enzyme solution.

The figure below shows his results.

                Time after start of reaction / s

(a)     Give one other factor the technician would have controlled.

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(b)     Calculate the rate of reaction at 25 °C.

 

 

 

Answer ____________________(2)

(c)     Describe and explain the differences between the two curves.

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(Total 8 marks)

Q4.The diagram below represents one process that occurs during protein synthesis.

 

(a)     Name the process shown.

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(b)     Identify the molecule labelled Q.

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(c)     In the diagram above, the first codon is AUG. Give the base sequence of:

the complementary DNA base sequence __________________________________

the missing anticodon _________________________________________________(2)

The table below shows the base triplets that code for two amino acids. 

Amino acid Encoding base triplet

Aspartic acid GAC, GAU

Proline CCA, CCG, CCC, CCU

(d)     Aspartic acid and proline are both amino acids. Describe how two amino acids differ from one another. You may use a diagram to help your description.

 

 

 

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(e)     Deletion of the sixth base (G) in the sequence shown in the diagram above would change the nature of the protein produced but substitution of the same base would not. Use the information in the table and your own knowledge to explain why.

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(Total 8 marks)

Q5.A group of students carried out an investigation to find the water potential of potato tissue.

The students were each given a potato and 50 cm3 of a 1.0 mol dm−3 solution of sucrose.

•        They used the 1.0 mol dm−3 solution of sucrose to make a series of different concentrations.

•        They cut and weighed discs of potato tissue and left them in the sucrose solutions for a set time.

•        They then removed the discs of potato tissue and reweighed them.

The table below shows how one student presented his processed results. 

Concentration of sucrose solution / mol

dm−3

Percentage change in mass of potato

tissue

0.15 +4.7

0.20 +4.1

0.25 +3.0

0.30 +1.9

0.35 −0.9

0.40 −3.8

(a)     Explain why the data in the table above are described as processed results.

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(b)     Describe how you would use a 1.0 mol dm−3 solution of sucrose to produce 30 cm3 of a 0.15 mol dm−3 solution of sucrose.

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(c)     Explain the change in mass of potato tissue in the 0.40 mol dm−3 solution of sucrose.

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(d)     Describe how you would use the student’s results in the table above to find the water potential of the potato tissue.

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(Total 8 marks)

Q6.(a)     The letters A, B, C, D and E represent stages in mitosis.

•        A – anaphase•        B – interphase•        C – metaphase•        D – prophase•        E – telophase

Write one of the letters, A to E, in the box to complete the following statement.

Chromosomes line up on the equator of the mitotic spindle in    (1)

(b)     Scientists looking for treatments for cancer are investigating the use of substances called kinesin inhibitors (KI). These inhibitors prevent successful mitosis. Some kinesin inhibitors cause the development of a monopolar spindle in mitosis.

The diagram below shows chromosomes attached to a normal mitotic spindle and to a monopolar mitotic spindle.

 

Suggest why the development of a monopolar mitotic spindle would prevent successful mitosis.

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(c)     Scientists investigated the effect of different concentrations of a kinesin inhibitor (KI) on mitosis of human bone-cancer cells grown in a culture.

The following table shows the scientists’ results. 

Concentration of kinesin inhibitor / nmol dm−3

Percentage of dividing human bone-cancer cells

showing a monopolar mitotic spindle

  0   0

  1   0

10   8

 100  93

 1000   100

10 000   100

A student who saw these results concluded that in any future trials of this kinesin inhibitor with people, a concentration of 100 nmol dm−3 would be most appropriate to use.

Do these data support the student’s conclusion? Give reasons for your answer.

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(d)     At the start of their investigation, the scientists made a solution of kinesin inhibitor (KI) with a concentration of 10 000 nmol dm−3. They used this to make the other concentrations by a series of dilutions with water.

Describe how they made 100 cm3 of 1000 nmol dm−3 solution of kinesin inhibitor.

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(Total 9 marks)

Q7.The figure below summarises the process of meiosis. The circles represent cells and the structures within each cell represent chromosomes.

          

(a)     Describe and explain the appearance of one of the chromosomes in cell X.

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(b)     Describe what has happened during division 1 in the figure above.

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(c)     Identify one event that occurred during division 2 but not during division 1.

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(d)     Name two ways in which meiosis produces genetic variation.

1. _________________________________________________________________

2. _________________________________________________________________(2)

(Total 8 marks)

Q8.Many insects release carbon dioxide in short bursts even though they produce it at a constant rate. The diagram shows how this is achieved in one particular insect.

 

(a)     Using information from the diagram, suggest what stimulates the spiracles to open.

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(b)     Explain what causes the oxygen concentration in the tracheae to fall when the spiracles are closed.

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(c)     The insect lives in dry conditions. Suggest an advantage of the pattern of spiracle movements shown in the diagram.

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(Total 5 marks)

 

Q9.Scientists investigated the effect of lipase and a 3% bile salts solution on the digestion of triglycerides. The graph below shows their results.

 

(a)     Describe what curve Y shows about the effect of lipase and bile salts on the pH of the mixture.

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(b)     The concentration of lipase did not change during the course of the investigation.Explain why.

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(c)     One of the scientists decided to repeat the investigation at a temperature 10°C below the original temperature.Describe how you would expect his plotted curve to be different from curve Z.

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(Total 4 marks)

Q10.(a)    The table shows three statements about some biological molecules. Complete the

table with a tick in each box if the statement is true for haemoglobin, cellulose or

starch. 

Statement Haemoglobin Cellulose Starch

Has a quaternary structure

Formed by condensation reactions

Contains nitrogen

(3)

The graph shows oxygen dissociation curves for the haemoglobin of a mother and her fetus.

 

Partial pressure of oxygen (pO2) / kPa

(b)     What is the difference in percentage saturation between the haemoglobin of the mother and her fetus at a partial pressure of oxygen (pO2) of 4 kPa?

 (1)

(c)     The oxygen dissociation curve of the fetus is to the left of that for its mother. Explain the advantage of this for the fetus.

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(d)     After birth, fetal haemoglobin is replaced with adult haemoglobin. Use the graph to suggest the advantage of this to the baby.

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(e)     Hereditary persistence of fetal haemoglobin (HPFH) is a condition in which production of fetal haemoglobin continues into adulthood. Adult haemoglobin is also produced.

People with HPFH do not usually show symptoms. Suggest why.

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(Total 9 marks)

Mark schemes

Q1.(a)     1.      How to break open cells and remove debris;

2.      Solution is cold / isotonic / buffered;3.      Second pellet is chloroplast.

3

(b)     1.      A stroma;2.      B granum.

Accept thylakoid2

(c)      μm1

(d)     Two of the following for one mark:Mitochondrion / ribosome / endoplasmic reticulum / lysosome / cell-surface membrane.

1 max[7]

Q2.(a)     X = mitochondria;

Y = (rough) endoplasmic reticulum;Accept ribosomes/ER/RER for YReject smooth endoplasmic reticulum for Y

2

(b)     (i)      (Sections cut at) different angles/in different planes;Ignore name given to organelle

1

(ii)     Z modifies/packages/transports/secretes mucus/ Z adds sugars to proteins;X provides ATP/energy (for this);

Accept makes in relation to Z but not XIgnore names of organelles if function correct

2[5]

Q3.(a)     Concentration of substrate solution / of enzyme solution / pH.

1

(b)     1.      2.5 / 0.04;1 mark for correct value

2.      g dm–3 minute–1 / g dm–3 s–1;1 mark for related unit

2

(c)     1.      Initial rate of reaction faster at 37 °C;2.      Because more kinetic energy;3.      So more E–S collisions / more E–S complexes formed;

4.      Graph reaches plateau at 37 °C;5.      Because all substrate used up.

Allow converse for correct descriptions and explanations for curve at 25 °C

5[8]

Q4

(a) benedicts

heat

Blue to red/orange 2

(b) Know concerntrations of maltose (range

Benedicts

Examine colour in colorimeter (abs or trans)

Plot graph of conc vs abs/trans

Read off graph 3

Q5.(a)     Translation.

1(b)     Transfer RNA / tRNA.

1(c)     TAC;

UAC.2

(d)     Have different R group.Accept in diagram

1

(e)     1.      Substitution would result in CCA / CCC / CCU;2.      (All) code for same amino acid / proline;3.      Deletion would cause frame shift / change in all following codons /

change next codon from UAC to ACC.3

[8]

Q6.(a)     Calculations made (from raw data) / raw data would have recorded initial and final

masses.1

(b)     Add 4.5 cm3 of (1.0 mol dm–3) solution to 25.5 cm3 (distilled) water.If incorrect, allow 1 mark for solution to water in a proportion of 0.15:0.85

2

(c)     1.      Water potential of solution is less than / more negative than that of potato tissue;

Allow Ψ as equivalent to water potential

2.      Tissue loses water by osmosis.2

(d)     1.      Plot a graph with concentration on the x-axis and percentage change in mass on the y-axis;

2.      Find concentration where curve crosses the x-axis / where percentage change is zero;

3.      Use (another) resource to find water potential of sucrose concentration (where curve crosses x-axis).

3[8]

Q7.(a)     C

Auto mark1

(b)     1.      No separation of chromatids/chromosomes/centromeres;Accept anaphase preventedAccept nondisjunctionReject homologous pairs

2.      Chromatids/chromosomes all go to one pole/end/sides of cell/not pulled to opposite poles;

3.      Doubles chromosome number in cell/one daughter cell gets no chromosomes or chromatids;

Accept DNA for chromosomesAccept ploidyIgnore references to ‘genetic information’Ignore simple descriptions of what normally happens in mitosis

2 max

(c)     1.      (No, because) at 100 there are still some (7%) cancer cells dividing/undergoing mitosis;

Accept idea that all division stops only at 10002.      So, cancer not destroyed/may continue to

grow/spread/form tumours;Must refer to cancer spreading not cells dividing

3.      Best concentration may be between 100 and

1000/need trials between 100 and 1000;4.      This research in culture, don’t know effect of KI on

people;Reject ‘not tested on humans’Reject ‘done in animals’

5.      (Yes, because) above 100 produces little increase in % of cells not dividing/undergoing mitosis/at 100, most (93%) cancer cells unable to divide/dead;

Must clearly link lack of monopolar mitotic spindles with cell division

6.      Above 100 may be harmful (to body);Accept ‘above 100/high concentrations produce harmful side effects/named effects'

7.      Higher concentrations more expensive;8.      (Above 100) will have more effect on (rapidly dividing)

cancer cells;Must relate to 100

4 max

(d)     1.      10 cm3 of 10 000 nmol dm−3/ (original) solution;2.      90 cm3 of water;

If ratio correct but make wrong volume e.g. 1 litre, award 1 mark

2[9]

Q8.(a)     1.      Chromosome is formed of two chromatids;

2.      (Because) DNA replication (has occurred);3.      (Sister) chromatids held together by centromere.

3

(b)     1.      Chromosomes in homologous pair;2.      One of each into daughter cells / haploid number.

2

(c)     Separation of (sister) chromatids / division of centromere.1

(d)     1.      Independent segregation (of homologous chromosomes);Accept random assortment

2.      Crossing over / formation of chiasmata.2

[8]

Q9.(a)     increasing carbon dioxide concentration / partial pressure;

(decrease in oxygen negates)1

(b)     (oxygen is used in) respiration therefore diffuses (from tracheae) to tissues;oxygen unable to enter organism;

2

(c)     spiracles not open all the time;therefore there is less water loss(by diffusion through spiracles);

2[5]

Q10.(a)     pH goes down and levels out;

after 30 min / pH 6.5;2

(b)     Enzyme not used up in reaction;1

(c)     Curve will be less steep:Only accept answers relating to curve not rate of reaction

1[4]

Q11.(a)

 One mark for each correct row

3

(b)     16;1

(c)     1.      Higher affinity / loads more oxygen at low / same / high partial pressure / pO2;

2.      (Therefore) oxygen moves from mother / to fetus;2

(d)     1.      Low affinity / oxygen dissociates;Assume ‘it’ is adult haemoglobin1. Accept: converse if ‘fetal haemoglobin’ is clearly stated

2.      (Oxygen) to respiring tissues / muscles / cells;2. Q: Neutral ‘respirate’

2

(e)     Enough adult Hb produced / enough oxygen released / idea that curves /

affinities / Hb are similar / more red blood cells produced;Neutral: ‘adult Hb is also produced’ as in the question stemReject: curves / affinities / Hb are the same

1[9]

Q12

(a) Injected with small amount/inactive pathogen/antigenB-Lymphocytes detect and produce specific antibodyB-lymphocytes also make memory cells by mitosisMemory cell produce antibody quickly in second infection

(b) Variation in bacterial population, gives resistanceThese survive antibiotic treatment More space/food/resourcesReproduce by mitosis Pass on resistance

(c) In bacterial cell: No nucleus/loop of DNAPlasmidCapsulenon-9:2 (eukaryotic) flagellumPilliNo membrane bound organelles (examples)Small/70s ribosomesSmall size (figures)