Author: Gareth Roberts

Carmel College

PAPER 1 Practise questions – NAME:
Questions I struggled with | What I need to revise to improve this

Q1.          (a)     Give one feature of starch and explain how this feature enables it to act as a storage substance.
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(b)     The diagram shows part of a cellulose molecule.

(i)      Name part A.
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(ii)     Name bond B.
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(c)     The structure of cellulose is related to its role in plant cell walls. Explain how.
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(Total 7 marks)
Q2.The diagram shows part of a DNA molecule.
  
(a)     (i)      DNA is a polymer. What is the evidence from the diagram that DNA is a polymer?
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(ii)     Name the parts of the diagram labelled C, D and E.
 
| Part C | .......................................................................
| Part D | .......................................................................
| Part E | .......................................................................
(3)
(iii)    In a piece of DNA, 34% of the bases were thymine.
Complete the table to show the names and percentages of the other bases.
 
| Name of base | Percentage
| Thymine | 34
| | 34

(2)

(b)     A polypeptide has 51 amino acids in its primary structure.
(i)      What is the minimum number of DNA bases required to code for the amino acids in this polypeptide?

(1)

(ii)     The gene for this polypeptide contains more than this number of bases.
Explain why
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(Total 8 marks)

Q3.          The diagram shows one end of a cellulose molecule.
  
(a)     (i)      Name the monomers that form a cellulose molecule.
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(ii)     Name bond Y.
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(iii)    What chemical group is at position Z?
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(b)     (i)      Complete the table to show two ways in which the structure of cellulose is different from the structure of starch.
 
| Starch | Cellulose

(2)

(ii)     Explain one way in which the structure of cellulose is linked to its function.
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(Total 7 marks)

Q4.The diagram shows the structure of the cell-surface membrane of a cell.

(a)     Name A and B.
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(b)     (i)      C is a protein with a carbohydrate attached to it. This carbohydrate is formed by joining monosaccharides together. Name the type of reaction that joins monosaccharides together.
Name the type of reaction that joins monosaccharides together.
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(1)

(ii)     Some cells lining the bronchi of the lungs secrete large amounts of mucus. Mucus contains protein.
Name one organelle that you would expect to find in large numbers in a mucus-secreting cell and describe its role in the production of mucus.
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Description of role..................................................................................
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(Total 5 marks)

Q5.          (a)     The diagram shows a bacterial cell.

(i)      Name the parts labelled D and E.
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E .........................................................................................................
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(ii)     Give one function of the cell wall.
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(b)     Name two structures present in eukaryotic cells that are not present in the cells of prokaryotes.
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(c)     Several small pieces of a saprophytic fungus were placed on a starch agar plate. After 48 hours the iodine solution was poured over the starch agar. The result is shown in the diagram below. 

(i)      Explain why there is a clear area around most of the pieces of fungus.
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(ii)     Suggest why one piece of fungus has no clear area round it.
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(Total 8 marks)
 
Q6.          Read the following passage.
Human milk contains all the nutrients a young baby needs in exactly the right
proportions. It is formed in the mammary glands by small groups of milk-producing cells.
These cells absorb substances from the blood and use them to synthesise the lipids,
carbohydrates and proteins found in milk. Milk-producing cells are roughly cube-shaped
5     and have a height to breadth ratio of approximately 1.2 : 1.
The main carbohydrate in milk is lactose. Lactose is a disaccharide formed by the
condensation of two monosaccharides, glucose and galactose. (A molecule of galactose
has the same formula as a molecule of glucose – the atoms are just arranged in a different
way.)
10   Lactose is synthesised in the Golgi apparatus and transported in vesicles through the
cytoplasm. Because lactose is unable to escape from these vesicles, they increase in
diameter as they move towards the plasma membrane. The vesicle membranes fuse with
the plasma membrane and the vesicles empty their contents out of the cell.
Use the information from the passage and your own knowledge to answer the following questions.
(a)     (i)      The breadth of a milk-producing cell is 26 µm. Calculate the height of this cell.
 
Height = .......................... µm
(1)

(ii)     Describe and explain how you would expect the height to breadth ratio of an epithelial cell from a lung alveolus to differ from the height to breadth ratio of a milk-producing cell.
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(b)     How many oxygen atoms are there in a molecule of
(i)      galactose;
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(ii)     lactose?
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(c)     The lactose-containing vesicles increase in diameter as they move towards the plasma membrane of the milk-producing cell (lines 11-12). Use your knowledge of water potential to explain why.
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(d)     Suggest one advantage of milk-producing cells containing large numbers of mitochondria.
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(e)     Some substances pass through the plasma membrane of a milk-producing cell by diffusion. Describe the structure of a plasma membrane and explain how different substances are able to pass through the membrane by diffusion.
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(Total 15 marks)

Q7.          Read the following passage.
Straw consists of three main organic substances – cellulose, hemicellulose and lignin.
Cellulose molecules form chains which pack together into fibres. Hemicellulose is a small
molecule formed mainly from five-carbon (pentose) sugar monomers. It acts as a cement
holding cellulose fibres together. Like hemicellulose, lignin is a polymer, but it is not a
 5      carbohydrate. It covers the cellulose in the cell wall and supplies additional strength. In
addition to these three substances, there are small amounts of other biologically important
polymers present.
 The other main component of straw is water. Water content is variable but may be determined
         by heating a known mass of straw at between 80 and 90°C until it reaches a constant mass.
10     The loss in mass is the water content.
Since straw is plentiful, it is possible that it could be used for the production of a range of
organic substances. The first step is the conversion of cellulose to glucose. It has been
suggested that an enzyme could be used for this process. There is a difficulty here, however.
The lignin which covers the cellulose protects the cellulose from enzyme attack.

          Use information from the passage and your own knowledge to answer the following questions.
(a)     (i)      Give one way in which the structure of a hemicellulose molecule is similar to the structure of a cellulose molecule.
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(1)

(ii)     Complete the table to show two ways in which the structure of a hemicellulose molecule differs from the structure of a cellulose molecule.
 
| Hemicellulose | Cellulose
| .......................................................... .......................................................... | .......................................................... ..........................................................
| .......................................................... .......................................................... | .......................................................... ..........................................................

(2)
(b)     Name one biologically important polymer, other than those mentioned in the passage, which would be found in straw.
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(1)

(c)     Explain why the following steps were necessary in finding the water content of straw:
(i)      heating the straw until it reaches constant mass (line 9);
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(1)

(ii)     not heating the straw above 90°C (line 9).
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(d)     A covering of lignin protects cellulose from enzyme attack (line 14). Use your knowledge of the way in which enzymes work to explain why cellulose-digesting enzymes do not digest lignin.
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(e)     Describe the structure of a cellulose molecule and explain how cellulose is adapted for its function in cells.
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(Total 15 marks)

Q8.(a)     Describe the role of the enzymes of the digestive system in the complete breakdown of starch.
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(b)     Describe the processes involved in the absorption of the products of starch digestion.
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(Total 10 marks)
Q9.Omega-3 fatty acids are found in cows’ milk. Scientists investigated changes in the concentration of omega-3 fatty acids in milk when cows were moved from eating grass in fields to eating corn in cattle sheds. The following figure shows the results of one investigation.
 
(a)     The concentration of omega-3 fatty acids in milk changed when cows were fed on corn instead of grass. Describe how.
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(b)     (i)      Calculate the rate of decrease in the mean omega-3 fatty acid concentration between 0 and 40 days.
Show your working.

 
 
Answer.......................................................% per day
(2)

(ii)     The omega-3 fatty acid concentration is expressed as a percentage of total fat.
Explain the advantage of this.
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(iii)    One farmer concluded from the graph that feeding cows on corn reduces the omega-3 fatty acid content in milk. Evaluate this conclusion.
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(Total 10 marks)

Q10.(a)     In humans, the enzyme maltase breaks down maltose to glucose.
This takes place at normal body temperature.
Explain why maltase:
•        only breaks down maltose
•        allows this reaction to take place at normal body temperature.
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(b)     Scientists have investigated the effects of competitive and non-competitive inhibitors of the enzyme maltase.
Describe competitive and non-competitive inhibition of an enzyme.
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(Total 10 marks)

Q11.(a)     Describe the induced-fit model of enzyme action.
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(b)     A scientist investigated the hydrolysis of starch.He added amylase to a suspension of starch and measured the concentration of maltose in the reaction mixture at regular intervals.
His results are shown in the graph below.
 
Determine the rate of the reaction at 10 minutes.
Show how you obtained your answer.
Rate of reaction .............................................. mg dm^−3 min^−1
(2)

(c)     Explain the results shown in the graph.
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(d)     A quantitative Benedict’s test produces a colour whose intensity depends on the concentration of reducing sugar in a solution. A colorimeter can be used to measure the intensity of this colour.
The scientist used quantitative Benedict’s tests to produce a calibration curve of colorimeter reading against concentration of maltose.
Describe how the scientist would have produced the calibration curve and used it to obtain the results in the graph.
Do not include details of how to perform a Benedict’s test in your answer.
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(Total 9 marks)
Q12.(a)     Some seeds contain lipids. Describe how you could use the emulsion test to show that a seed contains lipids.
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(b)     A triglyceride is one type of lipid. The diagram shows the structure of a triglyceride molecule.
 
(i)      A triglyceride molecule is formed by condensation. From how many molecules is this triglyceride formed?
 
(1)

(ii)     The structure of a phospholipid molecule is different from that of a triglyceride.
Describe how a phospholipid is different.
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(iii)    Use the diagram to explain what is meant by an unsaturated fatty acid.
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(Total 8 marks)

Q13.(a)     The table shows some substances found in cells. Complete the table to show the properties of these substances. Put a tick in the box if the statement is correct.
 
| | Substance
| Statement | Starch | Glycogen | Deoxyribose | DNA helicase
| Substance contains only the elements carbon, hydrogen and oxygen | | | |
| Substance is made from amino acid monomers | | | |
| Substance is found in both animal cells and plant cells | | | |

(4)
(b)     The diagram shows two molecules of β-glucose.
 
On the diagram, draw a box around the atoms that are removed when the two β-glucose molecules are joined by condensation.
(2)

(c)     (i)      Hydrogen bonds are important in cellulose molecules. Explain why.
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(ii)     A starch molecule has a spiral shape. Explain why this shape is important to its function in cells.
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(Total 9 marks)
Q14.          (a)     Dietary recommendations are that lipid intake should make up 30% of energy intake. The recommended energy intake for most women aged 19-49 is 8100 kJ day^–1.The energy content of lipid is 37.8 kJ g^–1. Calculate the recommended lipid intake per day for these women. Show your working.

Answer ................................................... g
(2)

In humans, triglycerides are the main form of dietary lipids. They are digested in the gut and the products of digestion are absorbed by the small intestine.
S       (b)     Describe a biochemical test that could be performed on a sample of food to determine whether it contained triglycerides.
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(c)     The diagram shows the events that occur in the absorption of monoglycerides and fatty acids. These molecules enter the epithelial cells of the small intestine by diffusion. Once inside they are reassembled into triglycerides in organelle Q. The triglyceride molecules are formed into chylomicrons in organelle T. Chylomicrons are made from many triglyceride molecules surrounded with protein molecules. The chylomicrons leave the cell and enter vessel S.
 
S       (i)      Explain the importance of the structures labelled P.
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(1)

(ii)     Name
R; .......................................................................................................
S. ........................................................................................................
(2)

S       (iii)     Describe the role played by organelle U in the formation of chylomicrons.
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S       (iv)    Suggest how the chylomicrons leave the epithelial cell. Give a reason for your answer.
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(Total 11 marks)
 
Q15.          (a)     Amylase is an enzyme which hydrolyses starch to maltose. Some amylase and starch were mixed and the mixture incubated at 37 °C until the reaction was complete.
(i)      Sketch a curve on the axes below to show the progress of this reaction.

(1)

(ii)     Explain why the rate of the reaction decreases as the reaction progresses.
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The effect of temperature on the rate of reaction of an enzyme was investigated. A test tube containing the enzyme and a test tube containing the substrate were incubated separately at each of the temperatures being investigated. After 5 minutes, they were mixed and the rate of reaction was determined. The experiment was repeated but, this time, the enzyme and the substrate were left for 60 minutes before they were mixed. The results of the investigation are shown in the graph.

(b)     The enzyme solution used in this investigation was made by dissolving a known mass of enzyme in a buffer solution. Explain why a buffer solution was used.
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(c)     (i)      Use the graph to describe how incubation time affects the rate of the reaction.
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(ii)     The maximum rate of reaction with an incubation time of 60 minutes is less than the maximum rate of reaction with an incubation time of 5 minutes. Explain why.
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(d)     Explain how inhibitors affect the rate of enzyme-controlled reactions.
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(Total 15 marks)
Q16.          Catalase is an enzyme. It catalyses the breakdown of hydrogen peroxide in the reaction:
2H_2O_2    →    2H_2O    +    O_2
hydrogen         water      oxygen
peroxide

In an investigation, samples of different substances were added to hydrogen peroxide in a series of test tubes. The rate of reaction was measured by recording the rate at which bubbles of oxygen were produced. A scale going from 0 for no bubbles to 5 for the maximum rate of bubbling was used to measure this. The results are shown in the table.

Tube

Substance added

Rate at which bubbles of
oxygen were produced

A | Piece of liver | 4
B | Ground liver and sand | 5
C | Sand | 0
D | Piece of cooled, boiled liver | 0

(a)     Explain the difference between the rate at which bubbles were produced in.
(i)      tubes A and B;
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(ii)     tubes A and D.
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(3
(b)     Explain the purpose of tube C.
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(1)

(c)     The graph shows the energy changes which take place during the reaction in which hydrogen peroxide is converted to water and oxygen.

Use the graph to explain why
(i)      hydrogen peroxide breaks down at a lower temperature when catalase is present than when it is not present;
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(1)

(ii)     test tubes A and B became warmer when the reaction was taking place.
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(Total 9 marks)

 

Q17.          Read the following passage.
During the course of a day, we come into contact with many poisonous substances. These include industrial and household chemicals. The skin acts as a barrier and prevents many of these substances entering and harming the body.
The skin is one of the largest organs in the body. It is composed of several layers of
5        tissue. The outer layer consists of dead cells packed with keratins. Keratins are a group of proteins that differ from each other in their primary structure. Each keratin molecule consists of several polypeptide chains, each individual chain wound into a spiral or helix. The polypeptide chains include many sulphur-containing amino acids and these help to give the keratin molecules their characteristic strength.
Use information from the passage and your own knowledge to answer the questions.
(a)     What is the evidence from the passage that keratin molecules have a quaternary structure?
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(1)

(b)     Explain how sulphur-containing amino acids help to give keratin molecules their characteristic strength (lines 8–9).
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(c)     Explain why differences in primary structure result in keratins with different properties 
(line 6).
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(d)     The skin prevents poisonous substances entering and harming the body (line 3). Explain why these substances are unable to pass through the outer layer of skin cells by active transport.
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(e)     Skin cells may be studied with a transmission electron microscope or an optical microscope. Explain the advantages and limitations of using a transmission electron microscope to study cells.
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(6)
(Total 14 marks)

 
Q18.Biological washing powders contain enzymes which hydrolyse substances that cause stains on clothes.
A manufacturer tested the ability of two types of the same brand of washing powder to remove different food substances that stain clothes.
•        Type A contained an enzyme.
•        Type B was identical to A except it did not contain the enzyme.
Figure 1 shows the results.
Figure 1

A scientist worked for a company that wanted to develop a biological washing powder that was effective over a range of temperatures. He investigated the effect of temperature on the rates of the reaction catalysed by two enzymes, P and S used in biological washing powders.
Figure 2 shows his results.
Figure 2

(a)     Many of the substances causing the food stains are large, insoluble proteins.
Suggest how a biological washing powder removes this type of stain.
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(2)
(b)     The manufacturer of type A and type B washing powder claimed that these results showed that biological washing powders are better at removing stains from clothes.
Use the information in Figure 1 to evaluate this claim.
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(c)     Most customers want a washing powder which removes stains from clothes over a range of temperatures. After obtaining the results shown in Figure 2, which enzyme should the scientist recommend for use in a biological powder?
Give reasons for your answer.
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(d)     Biological washing powders often contain a number of different enzymes. This enables them to remove a wider range of stains from clothes.
Explain why a number of enzymes are required to remove a wider range of stains.
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(Total 12 marks)

Q19.Figure 1 shows one base pair of a DNA molecule.
Figure 1

(a)     Name part F of each nucleotide.
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(1)

(b)     Scientists determined that a sample of DNA contained 18% adenine.
What were the percentages of thymine and guanine in this sample of DNA?
 
| Percentage of thymine |
| Percentage of guanine |

(2)

During replication, the two strands of a DNA molecule separate and each acts as a template for the production of a new strand.
Figure 2 represents DNA replication.
Figure 2

(c)     Name the enzyme shown in Figure 2.
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(1)

The arrows in Figure 2 show the directions in which each new DNA strand is being produced.
(d)     Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions.
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(Total 8 marks)

Q20.(a)     The events that take place during interphase and mitosis lead to the production of two genetically identical cells. Explain how.
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(b)     A student cut thin sections of tissue at different distances from the tip of a root. She stained the sections and viewed them with an optical microscope.
For each section, the student counted the number of cells in mitosis and the total number of cells in each field of view. She then calculated a mitotic index for each section using the equation:
mitotic index = 
 
The student’s results arer shown in the graph.

(i)      The student cut thin sections of tissue to view with an optical microscope. 
Explain why it was important that the sections were thin.
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(2)
(ii)     What does the graph show about the growth of roots?
Use the data to explain your answer.
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(2)
(Total 8 marks)
Q21.(a)    Describe how DNA is replicated.
(6)
(b)     The graph shows information about the movement of chromatids in a cell that has just started metaphase of mitosis.

(i)      What was the duration of metaphase in this cell?
   minutes
(1)

(ii)     Use line X to calculate the duration of anaphase in this cell.
   minutes
(1)

(iii)    Complete line Y on the graph.
(2)

(c)     A doctor investigated the number of cells in different stages of the cell cycle in two tissue samples, C and D. One tissue sample was taken from a cancerous tumour. The other was taken from non-cancerous tissue. The table shows his results.
 
| | Percentage of cells in each stage of the cell cycle
| Stage of the cell cycle | Tissue sample C | Tissue sample D
| Interphase | 82 | 45
| Prophase | 4 | 16
| Metaphase | 5 | 18
| Anaphase | 5 | 12
| Telophase | 4 | 9

(i)      In tissue sample C, one cell cycle took 24 hours. Use the data in the table to calculate the time in which these cells were in interphase during one cell cycle. Show your working.

 
Time cells in interphase ...................................... hours     (2)
(ii)     Explain how the doctor could have recognised which cells were in interphase when looking at the tissue samples.
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(iii)    Which tissue sample, C or D, was taken from a cancerous tumour? 
Use information in the table to explain your answer.
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(2)
(Total 15 marks)
Q22.          Figure 1 shows a short section of a DNA molecule.
Figure 1

(a)     Name parts R and Q.
(i)      R ....................................................
(ii)     Q ....................................................
(2)

(b)     Name the bonds that join A and B.
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(1)
(c)     Ribonuclease is an enzyme. It is 127 amino acids long.
          What is the minimum number of DNA bases needed to code for ribonuclease?

(1)
(d)     Figure 2 shows the sequence of DNA bases coding for seven amino acids in the enzyme ribonuclease.
Figure 2
G  T  T  T  A  C  T  A  C  T  C  T  T  C  T  T  C  T  T  T  A
The number of each type of amino acid coded for by this sequence of DNA bases is shown in the table.
 
Amino acid | Number present
Arg | 3
Met | 2
Gln | 1
Asn | 1

Use the table and Figure 2 to work out the sequence of amino acids in this part of the enzyme. Write your answer in the boxes below.
 
Gln | | | | | |

(1)
(e)     Explain how a change in a sequence of DNA bases could result in a non-functional enzyme.
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(3)
(Total 8 marks)
Q23.          (a)     There are two forms of nitrogen. These different forms are called isotopes. ^15N is a heavier isotope than the normal isotope ^14N.
In an investigation, a culture of bacteria was obtained in which all the nitrogen in the DNA was of the ^15N form. The bacteria (generation 0) were transferred to a medium containing only the normal isotope, ^14N, and allowed to divide once. A sample of these bacteria (generation 1) was then removed. The DNA in the bacteria of generation 1 was extracted and spun in a high-speed centrifuge.
The bacteria in the ^14N medium were allowed to divide one more time. The DNA was also extracted from these bacteria (generation 2) and spun in a high speed centrifuge.
The diagram shows the results of this investigation.

(i)      Which part of the DNA molecule contains nitrogen?
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(1)

(ii)     Explain why the DNA from generation 1 is found in the position shown.
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(2)
(iii)     Complete the diagram to show the results for generation 2.
(2)

(b)     The table shows the percentage of different bases in the DNA of different organisms.
 
Organism | Adenine% | Guanine% | Thymine% | Cytosine%
Human | | 19 | |
Bacterium | 24 | 26 | 24 | 26
Virus | 25 | 24 | 33 | 18

(i)      Complete the table to show the percentages of different bases in human DNA.
(2)

(ii)     The structure of virus DNA is different from the DNA of the other two organisms. Giving evidence from the table, suggest what this difference might be.
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(2)
(Total 9 marks)
 
Q24.          (a)     Describe and explain how the structure of DNA results in accurate replication.
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(4)

(b)     Describe the behaviour of chromosomes during mitosis and explain how this results in the production of two genetically identical cells.
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(7)

(c)     A cancerous tumour is formed by uncontrolled mitotic division. This results in a mass of cells with an inadequate blood supply. Drugs are being developed which only kill cells in a low oxygen environment. Suggest how these drugs could be useful in the treatment of cancer.
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(2)
(Total 13 marks)

Q25.          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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(Extra space)  .....................................................................................
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(1)
(ii)     Large numbers of organelles X and Z are found in mucus-secreting cells.
Explain why.
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(2)
(Total 5 marks)

Q26.          The drawing shows an electron micrograph of parts of epithelial cells from the small intestine.

(a)     (i)      Name the structures labelled A.
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(1)

(ii)     Explain how these structures help in the absorption of substances from the small intestine.
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(1)

(b)     (i)      The scale bar on this drawing represents a length of 0.1μm. Calculate the magnification of the drawing. Show your working.

Magnification .............................................
(2)

(ii)     Explain why an electron microscope shows more detail of cell structure than a light microscope.
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(2)

(c)     The length of mitochondria can vary from 1.5 μm to 10 μm but their width never exceeds 1μm. Explain the advantage of the width of mitochondria being no more than 1μm.
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(1)
(Total 7 marks)

Q27.          (a)     Changes to the protein coat of the influenza virus cause antigenic variability. Explain how antigenic variability has caused some people to become infected more than once with influenza viruses.
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(2)

(b)     The drawings show the changes in a B lymphocyte after stimulation by specific antigens.

B lymphocyte before stimulation        B lymphocyte after stimulation
(i)      Describe the role of macrophages in stimulating B lymphocytes.
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S       (ii)     Explain how the changes shown in the drawings are related to the function of B lymphocytes.
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(4)
(Total 7 marks)
Q28.          The electron micrograph shows part of a chloroplast.

(a)     Name the parts labelled A and B and, for each, describe one role in the process of photosynthesis.
A Name ........................................................................................................
   Role ...........................................................................................................
(2)
B Name ........................................................................................................
   Role ...........................................................................................................
(2)

(b)     (i)      Name the main substance present in the part labelled C.
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(1)

(ii)     How is this substance formed?
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(1)
(Total 6 marks)

 

Q29.          Read the following passage.
In a human, there are over 200 different types of cell clearly distinguishable from each other.
What is more, many of these types include a number of different varieties. White blood cells, for example, include lymphocytes and granulocytes.
Although different animal cells have many features in common, each type has adaptations.
5       associated with its function in the organism. As an example, most cells contain the same
organelles, but the number may differ from one type of cell to another. Muscle cells contain many mitochondria, while enzyme-secreting cells from salivary glands have particularly large amounts of rough endoplasmic reticulum.
The number of a particular kind of organelle may change during the life of the cell. An
10      example of this change is provided by cells in the tail of a tadpole. As a tadpole matures into a frog, its tail is gradually absorbed until it disappears completely. Absorption is associated with an increase in the number of lysosomes in the cells of the tail.
Use information from the passage and your own knowledge to answer the following questions.
(a)     Explain the link between.
(i)      mitochondria and muscle cells (lines 6 - 7);
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(3)

(ii)     rough endoplasmic reticulum and enzyme-secreting cells from salivary glands 
(lines 7 - 8).
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(2)

(b)     Use information in the passage to explain how a tadpole’s tail is absorbed as a tadpole changes into a frog.
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(2)

(c)     Starting with some lettuce leaves, describe how you would obtain a sample of undamaged chloroplasts. Use your knowledge of cell fractionation and ultracentrifugation to answer this question.
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(6)
(Total 13 marks)

Q30.          (a)     A plant cell was observed with an optical microscope. Describe how the length of the cell could be estimated.
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(2)

(b)     The water potential of a plant cell is –400 kPa. The cell is put in a solution with a water potential of –650 kPa. Describe and explain what will happen to the cell.
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(3)

(c)     A group of students investigated the effect of sucrose concentration on the change in length of cylinders of tissue cut from a young carrot. They measured the initial lengths of the carrot cylinders, then placed one in each of a number of sucrose solutions. After 18 hours, they removed the carrot cylinders and measured their final lengths. Some of the results are shown in the table.
 
| Concentration of sucrose / mol dm^–3 | Percentage decrease in length of carrot cylinder
| 0.4 | 4.2
| 0.5 | 8.7
| 0.6 | 13.0
| 0.7 | 16.8
| 0.8 | 18.1
| 0.9 | 18.1
| 1.0 | 18.1

(i)      The carrot cylinders were left for 18 hours in the sucrose solutions. Explain why they were left for a long time.
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(1)

(ii)     Explain how you would use a graph to predict the concentration of sucrose that would result in no change in length of the carrot cylinders.
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(iii)     Young carrots store sugars in their tissues but, in older carrots, some of this is converted to starch. How would using cylinders of tissue from older carrots affect the results obtained for a sucrose solution of 0.6 mol dm^–3? Give a reason for your answer.
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(Total 10 marks)
Q31.          The diagrams show four stages of mitosis.

(a)     (i)      Name stage A.
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(1)

(ii)     Starting with stage C, give the stages A to D in the correct order.
C   ............   .............   .............
(1)

(iii)     Describe and explain the appearance of one of the chromosomes in stage B.
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(2)
(b)     Colchicine is a substance that prevents the formation of the spindle in mitosis. Dividing cells were treated with colchicine. This stopped them dividing. After a few hours, the colchicine was removed and the cells began to divide again. The diagram shows the chromosomes from one of the treated cells at stage B after the cell began dividing again.

(i)      What has happened to the chromosome number?
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(1)

(ii)     Suggest an explanation for the change in the chromosome number.
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(1)
(Total 6 marks)
Q32.          The diagram shows a cell cycle.
 
(a)     In prophase of mitosis, the chromosomes become visible. Describe what happens in
(i)      metaphase
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(2)

(ii)     anaphase.
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(2)

(b)     (i)      Cells lining the human intestine complete the cell cycle in a short time. Explain the advantage of these cells completing the cell cycle in a short time.
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(ii)     The time required for a cell to complete the cell cycle was 4 hours 18 minutes.
Calculate the time required in minutes for this cell to multiply to produce eight cells.
Show your working.

 

Answer .....................................................   (2)
(c)     Mikanolide is a drug that inhibits the enzyme DNA polymerase. Explain why this drug may be effective against some types of cancer.
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(Total 9 marks)
Q33.Figure 1 shows three cells, B, C and D, from tissues in the same organism. Each cell is in a stage of either mitosis or meiosis.
Figure 1
 
(a)     Complete the table with a tick if the cell shows the feature.
 
| | Cell B | Cell C | Cell D
| homologous chromosomes are present | | |
| a stage of mitosis | | |

(2)
(b)     Describe and explain the appearance of chromosome K in cell C.
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(2)
(c)     Explain what is happening at point J in cell B.
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(2)
(d)     Use information from all three cells in Figure 1 to explain how the number of chromosomes in cell D was produced.
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(1)

(e)     Figure 2 shows the mass of DNA present in cells of a population of healthy cells where mitosis is occurring.
Figure 2
 
Explain why some cells contain a mass of DNA between 1 and 2 arbitrary units.
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(1)
(Total 8 marks)

Q34.          This question should be answered in continuous prose.
Quality of Written Communication will be assessed in these answers.
(a)     Describe and explain four ways in which the structure of a capillary adapts it for the exchange of substances between blood and the surrounding tissue.
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(4)

(b)     Explain how tissue fluid is formed and how it may be returned to the circulatory system.
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(Total 10 marks)
﻿
Q35.          Tradescantia is a house plant. There are small hairs on its flowers. These hairs are made of cells. Figure 1 shows the appearance of cells from one of these hairs after 20 minutes in distilled water. Figure 2 shows cells from another hair after 20 minutes in a solution of potassium nitrate.
Figure 1 (in distilled water)

Figure 2 (in potassium nitrate solution)

(a)     What does Figure 2 suggest about the permeability of the plasma membranes surrounding these cells?
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(1)

(b)     What is present in the space labelled F? Explain your answer.
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(2)

(c)     How would the water potential of the sap in the vacuole of cell E differ from the water potential of the sap in the vacuole of cell D? Explain your answer.
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(3)
(Total 6 marks)
 Q1.          The diagram shows the human immunodeficiency virus (HIV).

(a)     Name
A ..................................................................................................................
B ..................................................................................................................
(2)

The graph shows changes in the number of T-cells and HIV particles in the blood of a person following infection.

(b)     Explain why the number of HIV particles in the blood
(i)      rises during the first few months after infection
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(2)

(ii)     remains low between 1 and 7 years after infection.
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(1)

(c)     This person developed a large number of infections about 9 years after he first became infected with HIV. Using information from the graph, explain why.
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 (4)
(Total 9 marks)

Q2.          Read the following passage.
An anti-gal antibody is a type of antibody that helps to fight infections caused
by bacteria. If a person has a bacterial infection, for example Salmonella, anti-gal
antibodies bind to antigens on the surface of the Salmonella. Not all the
anti-gal antibodies are used to fight the infection. Even after the infection, anti-gal
antibodies remain in the blood.                                                                                                    5
Scientists have made adaptor molecules to try to use the anti-gal antibodies
against viruses such as HIV. The adaptor molecules are proteins. Each adaptor
molecule had a receptor site to which the HIV binds. This receptor site was
similar to the receptor site on human cells to which the HIV binds. The
adaptor molecule has another site to which an anti-gal antibody will bind.                               10
The scientists then investigated whether adding adaptor molecules and anti-gal
antibodies can prevent HIV entering cells. They added adaptor molecules
and anti-gal antibodies to a culture of human cells. They then added HIV
to the culture. Their results showed that 90% of the virus particles failed
to infect cells.                                                                                                                             15
The scientists are hoping to develop a different type of adaptor molecule to use against MRSA.
(a)     (i)      What is an antigen? (line 3)
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(2)

(ii)     Explain why antibodies against Salmonella do not normally bind to HIV.
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(2)
(iii)     Explain how the adaptor molecule allows anti-gal antibodies to associate with HIV.
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(b)     Describe how humans produce antibodies against a pathogen such as Salmonella.
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(Extra space) ................................................................................................ (6)
(c)     (i)      HIV infects some human cells, such as T-cells, but not others. Suggest why.
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(2)
(ii)     Antibiotics are not used to treat viral infections, such as HIV. Explain why.
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(2)

(d)     (i)      When HIV, anti-gal and the adaptor molecule were added to a culture of human cells, 90% of the virus did not infect human cells. (lines 12-15). Explain why.
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(2)

(ii)     Explain why a different type of adaptor molecule will have to be made to use against MRSA. (lines 16-17)
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(2)
(Total 20 marks)

Q3.          Doctors use Zevalin to kill cancerous B-cells. Zevalin is a monoclonal antibody which has a highly radioactive substance called yttrium attached to it. The antibody binds to the surface of B-cells and the radioactivity kills the cells.
(a)     Only B-cells are killed by Zevalin.
Explain why.
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The cancerous B-cells are found mainly in the lymphatic system of patients. Before treating any patient with Zevalin containing yttrium, doctors test the patient with a different form of Zevalin. This form has radioactive indium attached to the antibody instead of yttrium. The radioactivity from indium is strong enough for doctors to detect but not strong enough to kill a patient’s cells.
The diagram shows the lymphatic systems of two patients, P and Q, after being given Zevalin with indium.  The crosses (+) show where indium was detected.

(b)     The doctors decided they could treat Patient P with Zevalin containing yttrium but not Patient Q.
Suggest why Patient P could be treated with Zevalin containing yttrium and Patient Q could not.
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(c)     Suggest one reason for the difference in distribution of the radioactivity detected in these patients.
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(d)     The antibody in Zevalin comes from mice. Patients are tested for antibodies against Zevalin before treatment for their cancer.
Suggest why.
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(Total 9 marks)
Q4.          (a)     The graph shows the number of deaths from influenza per year in a developed country.

(i)      Suggest an explanation for the change in the number of deaths from influenza during the first 10 years.
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(ii)     Suggest an explanation for the large increase in the number of deaths from influenza in year 11.
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(b)     The diagram shows some of the structures on the outside of an influenza virus.

Haemagglutinin and neuraminidase are protein molecules. Haemagglutinin binds to receptor molecules on the surface of epithelial cells in the breathing system.
Neuraminidase is an enzyme which breaks down molecules in the surface membrane of epithelial cells and allows the viruses to be released from the cells.
(i)      Describe how T lymphocytes recognise and respond to the influenza virus.
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(ii)     Describe how B lymphocytes respond to the influenza virus.
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(c)     New drugs have recently become available for treating influenza. One type is a neuraminidase inhibitor. Explain how this type of drug would act as a treatment for influenza.
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 (2)(Total 9 marks)
Q5.          People considered ‘at risk’ are offered a vaccination against influenza each year. The bar chart shows the number of people in the UK population aged 65 and over and the percentage of those who were vaccinated against influenza each winter.

(a)     Suggest one reason to explain the change in the percentage of people aged 65 and over being vaccinated.
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......................................................................................................................(1)
(b)     (i)      Calculate the change in the total number of people aged 65 and over being vaccinated between 1990/91 and 2000/01. Show your working.

 
 
Answer ................................. (2)
(ii)     A student suggested that some people aged 65 and over were being vaccinated every year. Explain how the information in the bar chart supports this suggestion.
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(iii)     Suggest why it is advisable for people to be vaccinated against influenza every year.
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(2)
(c)     An influenza virus consists of a protein coat surrounding nucleic acid. The influenza vaccine consists only of the protein coat of the virus. Explain how the influenza vaccine produces immunity in the body.
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(2)
(Total 9 marks)

Q6.          Some strains of the bacterium that causes gonorrhoea are resistant to antibiotics. This makes the disease difficult to treat. One way of testing the effectiveness of antibiotics is to use discs of paper soaked in antibiotic. These are placed in the centre of an agar plate covered by bacteria. A clear zone forms around the disc if the antibiotic is effective.
The table shows some results of an investigation into the effect of four different antibiotics on gonorrhoea bacteria.
 
Antibiotic | Diameter of clear zone / mm | Minimum diameter of clear zone if antibiotic is effective / mm
A | 47 | 52
B | 30 | 28
C | 22 | 40
D | 33 | 34

(a)     Give two reasons why it would be important to use sterile techniques during this investigation.
1 ...................................................................................................................
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(b)     (i)      The antibiotic reached the bacteria by diffusion. Suggest why an effective antibiotic may produce only a small clear zone.
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(1)
(ii)     Which antibiotic used in the investigation would be most useful for treating gonorrhoea? Explain your answer.
Antibiotic ...............................................................
Explanation .........................................................................................
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(2)
(Total 5 marks)

Q7.          (a)     What is an antigen?
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(2)

(b)     Describe how B-lymphocytes respond when they are stimulated by antigens.
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(4)

(c)     The diagram shows some components of a human immunodeficiency virus (HIV).

(i)      Suggest which labelled component of the virus is most likely to act as an antigen. Give a reason for your answer.
Component .........................................................................................
Reason ...............................................................................................
............................................................................................................. (1)
(ii)     A cell that HIV infects is 15 µm in diameter. Calculate how many times larger in diameter this cell is than an HIV particle. Show your working.

Answer ...................................... times larger
(2)
(Total 9 marks)
Q8.          (a)     Describe how B-lymphocytes respond when they are stimulated by antigens.
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(b)     The table gives information about some components of a red blood cell.
 
Component | Glycoprotein | Phospholipid | Haemoglobin
Location in cell | on outer surface of plasma membrane | within plasma membrane | in cytoplasm

Suggest which component of an intact red blood cell is most likely to act as an antigen during a blood transfusion. Explain your answer.
Component ..................................................................................................
Explanation ..................................................................................................
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(2)
(Total 6 marks)

Q9.          Different cells in the body have different functions.
(a)     Some white blood cells are phagocytic. Describe how these phagocytic white blood cells destroy bacteria.
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(b)     The epithelial cells that line the small intestine are adapted for the absorption of glucose. Explain how.
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(Total 10 marks)

Q10.(a)     Human papilloma virus (HPV) is the main cause of cervical cancer. A vaccine has been developed to protect girls and women from HPV.
Describe how giving this vaccine leads to production of antibody against HPV.
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(4)
(b)     Doctors investigated whether it was better to give two or three doses of the HPV vaccine. They determined the mean concentration of antibody against HPV in blood samples from girls who were given either two or three doses of the vaccine.
•        Girls given two doses received an initial vaccination, followed by a second at 6 months.
•        Girls given three doses received an initial vaccination, followed by a second at 1 month and a third at 6 months.
The doctors measured the concentration of antibody each month.
The results are shown below.

What do these results suggest about whether it is better to give two or three doses of the vaccine? Give reasons for your answer.
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(c)     The doctors carried out a statistical test to determine whether the antibody concentrations were significantly different in girls given two doses of the vaccine, compared with those given three doses. They determined the mean concentrations of antibody 9 months after the first dose of vaccine.
What statistical test should the doctors have used? Give the reason for your choice.
Test .....................................................................................................................
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(1)

(d)     There is genetic diversity within HPV.
Give two ways doctors could use base sequences to compare different types of HPV.
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(Total 9 marks)
Q11.(a)     The table below shows features of a bacterium and the human immunodeficiency virus (HIV) particle.
Complete the table by putting a tick (✔) where a feature is present.
 
| Feature | Bacterium | Human immunodeficiency virus (HIV) particle
| RNA | |
| Cell wall | |
| Enzyme molecules | |
| Capsid | |
(2)
(b)     When HIV infects a human cell, the following events occur.
•        A single-stranded length of HIV DNA is made.
•        The human cell then makes a complementary strand to the HIV DNA.
The complementary strand is made in the same way as a new complementary strand is made during semi-conservative replication of human DNA.
Describe how the complementary strand of HIV DNA is made.
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(c)     Contrast the structures of DNA and mRNA molecules to give three differences.
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(3)
(Total 8 marks)

Q12.          Mitochondria were isolated from the liver tissue using differential centrifugation. The tissue was chopped in cold, isotonic buffer solution. A buffer solution maintains a constant pH. The first stages in the procedure are shown in the diagram.
Tissue chopped               Homogenised                                      Centrifuged
 in cold isotonic                                                                          at low speed
  buffer solution                                                                        for 10 minutes

 Stage 1                          Stage 2                                              Stage 3

(i)      The tissue was chopped in cold, isotonic buffer solution. Explain the reason for using
a cold solution;
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an isotonic solution;
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a buffer solution.
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(3)

(ii)      Why is the liver tissue homogenised?
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(1)

(iii)     Describe what should be done after Stage 3 to obtain a sample containing only mitochondria.
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(Total 6 marks)

Q13.(a)     Describe and explain how cell fractionation and ultracentrifugation can be used to isolate mitochondria from a suspension of animal cells.
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(b)     Describe the principles and the limitations of using a transmission electron microscope to investigate cell structure.
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(Total 10 marks)

Q14.The figure below shows some cells from an onion root tip at different stages of the cell cycle.
 
© Ed Reschke/Oxford Scientific/Getty Images
(a)     Place stages A to E in the correct order. Start with stage D.
D
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(1)

To obtain these images, the onion root tip was cut off, stained and put on a microscope slide. A cover slip was placed on top. The root tip was then firmly squashed and viewed under an optical microscope.
(b)     Complete the table below to give one reason why each of these steps was necessary.
 
| Step | Reason
| Taking cells from the root tip |
| Firmly squashing the root tip |

(2)

The figure below shows how the amount of DNA per cell changed during interphase and meiosis in an animal.
 
(c)     Explain how the behaviour of chromosomes causes these changes in the amount of DNA per cell between F and G.
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(d)     What would happen to the amount of DNA per cell at fertilisation of cell G?
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(1)
(Total 7 marks)
Q15.          The drawing shows four common plants found in the Mojave Desert.
W                       X                   Y                                 Z

(a)     Explain how three features of the plants shown in the drawing are adaptations to desert conditions.
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(3)

(b)     Resurrection plants can lose up to 95% of their water content without dying. They can survive for many years in this desiccated state and will revive within hours of rainfall. Suggest which of the plants W to Z is most likely to be a resurrection plant. Give a reason for your choice.
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(2)
(Total 5 marks)

Q16.          The drawing shows an electron micrograph of a section through part of an alveolus from a lung.

(a)     Describe the path of a molecule of oxygen from the air in the alveolus at X to the plasma membrane of cell A.
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(1)

(b)     Cell A is a eukaryotic cell. Give two features that may be found in a prokaryotic cell which are not found in cell A.
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(2)

(c)     Cells A and B are biconcave discs. Explain one advantage of a biconcave discover a spherical cell of the same volume in transporting oxygen.
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(2)
(d)     The diameter of a human red blood cell is 7 µm.
(i)      Calculate the magnification of the drawing. Show your working.

 
Magnification = ...............................
(2)

(ii)     In calculating the magnification, what assumption did you have to make about how the section was cut?
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(1)
(Total 8 marks)
Q17.          (a)     Many different substances enter and leave a cell by crossing its cell surface membrane. Describe how substances can cross a cell surface membrane.
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(b)     Describe and explain how the lungs are adapted to allow rapid exchange of oxygen between air in the alveoli and blood in the capillaries around them.
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(5)
(Total 10 marks)

Q18.Scientists investigated the effect of lipase and a 3% bile salts solution on the digestion of triglycerides. The graph below shows their results.

 

The scientists also incubated triglycerides with different concentrations of bile salts. After 30 minutes they measured the diameter of the triglyceride droplets. They used the results to calculate the mean radius of the droplets at each concentration. The table below shows their results.

| Concentration of bile salts /% | 0 | 1 | 2 | 3 | 4 | 5
| Mean radius of triglyceride droplet / μm | 6 | 5 | 4 | 3 | 2 | 1

(a)     Describe how you would use a microscope to find the mean diameter of triglyceride droplets on a slide.
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(b)     (i)      The ratio of mean radius of triglyceride droplets in bile salts at a concentration of 0% to the mean radius in bile salts at a concentration of 3% is 2 : 1.
What is the ratio of their surface areas? Show your working.
You can calculate the surface area of a droplet from the formula
A = 4πr^2
Where  A = surface area
    r = radius
   π = 3.14

 
 (2)
(ii)     Use the data in the table to explain the difference between curves Y and Z in the graph.
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(Total 8 marks)
Q19.Organic compounds synthesised in the leaves of a plant can be transported to the plant’s roots.
This transport is called translocation and occurs in the phloem tissue of the plant.
(a)     One theory of translocation states that organic substances are pushed from a high pressure in the leaves to a lower pressure in the roots.
Describe how a high pressure is produced in the leaves.
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PCMBS is a substance that inhibits the uptake of sucrose by plant cells.
Scientists investigated the effect of PCMBS on the rate of translocation in sugar beet.
The figure below shows their results.
                         
Time / minutes
(b)     During their experiment, the scientists ensured that the rate of photosynthesis of their plants remained constant.
Explain why this was important.
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(c)     The scientists concluded that some translocation must occur in the spaces in the cell walls.
Explain how the information in the figure above supports this conclusion.
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(Total 7 marks)

Q20.          (a)     Describe how air is taken into the lungs.
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(3)

The volume of air breathed in and out of the lungs during each breath is called the tidal volume. The breathing rate and tidal volume were measured for a cyclist pedalling at different speeds. The graph shows the results.

(b)     Describe the two curves.
(i)      Tidal volume
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(ii)     Breathing rate
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(c)     Calculate the total volume of air breathed in and out per minute when the cyclist is cycling at 20 km h^–1. Show your working.
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(2)
(Total 7 marks)

Q21.          Read the following passage.
When a person breathes in, the diaphragm muscle contracts and the diaphragm flattens.
This, together with movement of the ribs, leads to air being drawn into the lungs.
Breathing out is generally passive and results from the relaxation of the diaphragm
muscle and the elastic recoil of the lung tissue.
5     Two sets of intercostal muscles also play an important part in breathing in humans.
Contraction of the external intercostal muscles is associated with breathing in. During
strenuous exercise, contraction of the internal intercostal muscles helps force air out of
the lungs. In this case, breathing out is active.
Crocodiles also have lungs and breathe air. They have well developed intercostal
10   muscles but do not appear to use these during breathing. They also lack a diaphragm.
Breathing in, in crocodiles, is brought about by contraction of muscles attaching the
liver to the hip bones (see diagram). This pulls the liver back and causes air to enter
the lungs. Breathing out results from the contraction of abdominal muscles which
move the liver forwards.

Use information in the passage and your own knowledge to answer the questions.
(a)     Describe the movement of the ribs when a person breathes in (line 2).
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(1)

(b)     (i)      Explain what is meant by passive (line 3).
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(ii)     Is breathing out in crocodiles active or passive? Explain your answer.
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(c)     Explain how movement of the liver causes air to enter a crocodile’s lungs.
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(d)     Describe the difference in the composition of gases in inhaled and exhaled air. Explain how these differences are caused.
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(Total 12 marks)
Q22.(a)     Explain how oxygen is loaded, transported and unloaded in the blood.
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Midges are very small insects. The early stages of the life cycle of midges are called larvae. Midge larvae live in water. A biologist investigated the uptake of oxygen by the larvae of two species of midge. He measured the rate of uptake of oxygen by the larvae in water containing different concentrations of oxygen. The table shows his results.
 
| Concentration of oxygen in water / cm^3dm-3 | Mean rate of oxygen uptake / cm^3g^-1h-1
| | Chironomus longistylus | Tanytarsus brunnipes
| 1 | 220 | 141
| 2 | 285 | 246
| 3 | 304 | 342
| 4 | 313 | 362
| 5 | 320 | 367
| 6 | 318 | 430
| 7 | 320 | 469

(b)     The larvae in this investigation were kept at a temperature of 17 °C. Why was it important that the larvae of both species were kept at the same temperature?
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(c)     Describe the effect of an increase in oxygen concentration on the mean rate of oxygen uptake in Chironomus longistylus.
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(d)     Chironomus longistylus lives in still water whereas Tanytarsus brunnipes lives in fast running streams. The water in fast running streams has a higher concentration of oxygen than in still water. Use the table in part (a) to suggest how Chironomus longistylus is better adapted than Tanytarsus brunnipes to living in still water.
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(e)     Lungfish are freshwater fish which have gills and lungs. Scientists investigated how Australian and African lungfish use their lungs and gills for gas exchange. The graphs show the results of this investigation.

(i)      Describe the difference in the way carbon dioxide is lost from the body of an Australian lungfish and an African lungfish.
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(ii)     African lungfish are likely to survive for longer than Australian lungfish when living in pools that dry up. Explain why.
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(Total 15 marks)
Q23.Scientists investigated the effect of the water potential of soil water on plant growth. They investigated the effect of this water potential on several plant processes.
The figure below shows their results in the form they were presented. The bars show whether or not each process was occurring.
The plants stopped growing when the water potential of the soil water was below –0.7 mPa. All of the changes in the plants were related to the ability of the roots to take up water from the soil.

(a)     Describe the results in the figure.
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(b)     Explain the relationship between stomatal opening and photosynthesis.
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(c)     Although photosynthesis is still occurring, plants stop growing when the soil water potential falls below –0.7 mPa.
Use information from the figure above to suggest two reasons why.
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(Total 7 marks)
Q24.Breathing out as hard as you can is called forced expiration.
(a)     Describe and explain the mechanism that causes forced expiration.
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Two groups of people volunteered to take part in an experiment.
•        People in group A were healthy.
•        People in group B were recovering from an asthma attack.
Each person breathed in as deeply as they could. They then breathed out by forced expiration.
A scientist measured the volume of air breathed out during forced expiration by each person.
The graph below shows the results.
                         
Time breathing out / s

(b)     Forced expiration volume (FEV) is the volume of air a person can breathe out in1 second.
Using data from the first second of forced expiration, calculate the percentage decrease in the FEV for group B compared with group A.

 
Answer = ................................... %
(1)

(c)     The people in group B were recovering from an asthma attack.
Explain how an asthma attack caused the drop in the mean FEV shown in the figure below.
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(Total 9 marks)

Q25.A student investigated the distribution of stomata on leaves from two species of plant. She removed small pieces from the lower surface of the leaves of each plant species. She mounted these pieces on separate microscope slides. She then counted the number of stomata in several parts of the epidermis on each piece of leaf tissue using an optical microscope.

(a)     Suggest appropriate units the student should use to compare the distribution of stomata on leaves.
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(1)

(b)     The pieces of leaf tissue examined were very thin.
Explain why this was important.
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(c)     Give two reasons why it was important that the student counted the number of stomata in several parts of each piece of leaf tissue.
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(d)     One of the two plant species used by the student in this investigation was a xerophyte.
Other than the distribution of stomata, suggest and explain two xerophytic features the leaves of this plant might have.
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(e)     The student then compared the rate of transpiration (evaporation of water) from the two species of plant. She did this by measuring the rate of water uptake by each plant species.
Suggest two reasons why the rate of water uptake by a plant might not be the same as the rate of transpiration.
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(2)
(Total 9 marks)

Q26.          Lactose is a disaccharide found in milk. In the small intestine, it is digested into glucose and galactose by the enzyme lactase. Molecules of lactase are located in the plasma membranes of cells lining the small intestine.
(a)     What evidence in the paragraph suggests that galactose is a monosaccharide?
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(1)

(b)     (i)      Name one other digestive enzyme that is located in the plasma membranes of cells lining the small intestine.
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(1)

(ii)     Give an advantage of lactase and other digestive enzymes being located in the plasma membranes of cells lining the small intestine, rather than being secreted into the lumen of the small intestine.
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(1)

(c)     The absorption of galactose from the small intestine is reduced if the absorbing cells are treated with a respiratory inhibitor, such as cyanide. Suggest an explanation for this.
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(Total 5 marks)
Q27.(a)     Describe the difference between the structure of a triglyceride molecule and the structure of a phospholipid molecule.
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(1)

(b)     Describe how you would test for the presence of a lipid in a sample of food.
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(2)

(c)     Animal fats contain triglycerides with a high proportion of saturated fatty acids. If people have too much fat in their diet, absorption of the products of fat digestion can increase the risk of obesity. To help people lose weight, fat substitutes can be used to replace triglycerides in food.
Describe how a saturated fatty acid is different from an unsaturated fatty acid.
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(1)

The diagram shows the structure of a fat substitute.

(d)     This fat substitute cannot be digested in the gut by lipase.
Suggest why.
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(e)     This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface membranes of cells lining the gut.
Suggest why it cannot cross cell-surface membranes.
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(Total 7 marks)

Q28.          (a)     A woman takes moderate exercise. Explain what causes her heart rate to increase while she exercises.
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(6)

(b)     An electrocardiogram (ECG) measures the electrical changes occurring in cardiac muscle as a heart is beating. An ECG trace for a healthy person and an ECG trace for a person suffering from heart disease are shown.

(i)      Describe the route taken when electrical impulses are transmitted from the sinoatrial node to the muscles of the ventricles in a healthy heart.
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(ii)     Explain how information from these ECG traces suggests that the damage caused to the diseased heart is unlikely to have affected the sinoatrial node.
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(Total 10 marks)
Q29.          (a)     Describe how water is moved through a plant according to the cohesion-tension hypothesis.
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(4)
(b)     The mass of water lost from a plant was investigated. The same plant was used in every treatment and the plant was subjected to identical environmental conditions. In some treatments, the leaves were coated with a type of grease. This grease provides a waterproof barrier. The results of the investigation are given in the table.
 
Treatment | Mass lost in 5 days / g
No grease applied | 10.0
Grease applied only to the upper surface of every leaf | 8.7
Grease applied to both surfaces of every leaf | 0.1

(i)      What is the advantage of using the same plant in every treatment?
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(1)

(ii)     Why was it important to keep the environmental conditions constant?
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(iii)     What is the evidence that the grease provides a waterproof barrier?
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(1)
(c)     (i)      Calculate the mass of water lost in 5 days through the upper surface of the leaves.

 
Answer ................................................................
(1)

(ii)     Use your knowledge of leaf structure to explain why less water is lost through the upper surface of leaves than is lost through the lower surface.
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(2)
(Total 10 marks)

 

Q30.          (a)     Explain how each of the following is related to the function of xylem tissue.
(i)      Xylem tissue contains hollow tubes.
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(ii)     Lignin is present in xylem cell walls.
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(2)

(b)     In an investigation the total area of the stomatal openings and the rate of flow of water through xylem were measured in a plant over a period of 24 hours. The results are shown in the graph.

(i)      Describe the relationship between the rate of flow of water and the total area of the stomatal openings for the period of time between midday and midnight.
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(1)

(ii)     Between 8 am and midday the rate of flow of water continues to rise although the total area of the stomatal openings remains constant. Explain why the rate of flow of water rises.
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(1)
(iii)     How would the curve showing the total area of the stomatal openings differ if the investigation was repeated on a dull day?
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(1)
(c)     Some xerophytic plants have sunken stomata. Explain the advantage of this adaptation.
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(2)
(Total 7 marks)
 
Q31.          The graph shows dissociation curves for human oxyhaemoglobin at rest and during exercise.
Table 1 gives information about conditions in the body at rest and during exercise.

 
| Rest | Exercise
Plasma pH | 7.4 | 7.2
Blood temperature / °C | 37.0 | 39.0
Alveolar partial pressure of oxygen / kPa | 13.3 | 13.3
Tissue partial pressure of oxygen / kPa | 5.0 | 4.0

Table 1
(a)     What is meant by the term partial pressure?
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(1)

(b)     Use Table 1 and the graph to calculate the difference in the percentage saturation of haemoglobin in the tissues between rest and exercise.

Answer ............................ %
(1)

(c)     Explain the differences between the figures shown in Table 1 for rest and exercise.
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(d)     Explain the advantage of the difference in position of the dissociation curve during exercise.
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(2)

Table 2 shows how the oxygen concentration in the blood going to and from a muscle changes from rest to heavy exercise.

Oxygen concentration / cm^3 per 100 cm^3
blood

| Blood in arteries | Blood in veins
At rest | In solution | 0.3 | 0.2
| As oxyhaemoglobin | 19.5 | 15.0
| Total oxygen | 19.8 | 15.2
During heavy exercise | In solution | 0.3 | 0.1
| As oxyhaemoglobin | 20.9 | 5.3
| Total oxygen | 21.2 | 5.4

Table 2
(e)     By how many times is the volume of oxygen removed from the blood by the muscle in Table 2 during heavy exercise greater than the volume removed at rest?
Show your working.

 
 
Answer ..................................... times
(2)

(f)      Does enriching inspired air with oxygen have any effect on the amount of oxygen reaching the tissues? Support your answer with evidence from the graph and Table 2.
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(3)

S       (g)     The change to the dissociation curve is one of a number of ways in which the total oxygen supplied to muscles is increased during exercise. Give two other ways in which the total oxygen supplied to muscles during exercise is increased.
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(2)
(Total 15 marks)

Q32.          Relaxation of heart muscle is called diastole. Contraction is called systole. The diagram shows the periods of diastole and systole when the heart is beating.

(a)     At what time is the volume of blood in the ventricle at a maximum?
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(1)

(b)     Calculate the heart rate in beats per minute. Show your working.

 
 
Heart rate = ................................... beats per minute
(2)

(c)     The valves between the atria and the ventricles are closed between 0.1 s and 0.4 s.
(i)      Explain how pressure causes these valves to be shut.
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(ii)     Explain how closure of these valves is essential to the functioning of the heart.
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The graph shows the effect of the pressure of blood in the veins on the stroke volume of the heart.

(d)     Describe how venous blood pressure affects stroke volume.
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(2)
(e)     At a venous pressure of 2 kPa, the cardiac output is 5600 cm^3 per minute. Calculate the number of times the ventricle contracts in one minute. Show your working.

 

Answer ......................................
(2)
(f)      Explain how blood in a vein in the leg is returned to the heart.
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(Total 15 marks)
Q33.          The diagram shows a molecule of haemoglobin.

(a)     What is the evidence from the diagram that haemoglobin has a quaternary structure?
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(1)
(b)     (i)      A gene codes for the α-polypeptide chain. There are 423 bases in this genethat code for amino acids. How many amino acids are there in the α-polypeptide chain?
 
(1)
(ii)     The total number of bases in the DNA of the α-polypeptide gene is more than 423.
Give two reasons why there are more than 423 bases.
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(c)     The haemoglobin in one organism may have a different chemical structure from the haemoglobin in another organism. Describe how.
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(1)
(d)     The graph shows oxygen dissociation curves for horse haemoglobin and for llama haemoglobin. Horses are adapted to live at sea level and llamas are adapted to live in high mountains.

Use the graph to explain why llamas are better adapted to live in high mountains than horses.
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(3)
(Total 8 marks)

Q34.          (a)     Scientists measured the rate of water flow and the pressure in the xylem in a small branch. Their results are shown in the graph.
 
(i)      Use your knowledge of transpiration to explain the changes in the rate of flow in the xylem shown in the graph.
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(3)

(ii)     Explain why the values for the pressure in the xylem are negative.
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(1)

(b)     Doctors measured the thickness of the walls of three blood vessels in a large group of people. Their results are given in the table.
 
| Name of vessel | Mean wall thickness /mm (± standard deviation)
| Aorta | 5.7 ± 1.2
| Pulmonary artery | 1.0 ± 0.2
| Pulmonary vein | 0.5 ± 0.2

(i)      Explain the difference in thickness between the pulmonary artery and the pulmonary vein.
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(1)
(ii)     The thickness of the aorta wall changes all the time during each cardiac cycle.
Explain why
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(3)
(iii)    Which of the three blood vessels shows the greatest variation in wall thickness?
Explain your answer.
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(1)
(c)     Describe how tissue fluid is formed and how it is returned to the circulatory system.
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(6)
(Total 15 marks)

Q35.In mammals, the mesenteric artery connects the aorta to blood vessels of the small intestine.
Sport scientists recorded increases in blood flow in the mesenteric artery after different types of meal. The types of meal were:
•        carbohydrate only
•        fat only
•        protein only.
Typical results are shown in the table.
 
| Type of meal | Maximum percentage increase in blood flow in mesenteric artery | Time taken to reach maximum increase in blood flow in mesenteric artery / minutes
| Carbohydrate only | 64 | 15
| Fat only | 60 | 30
| Protein only | 57 | 45

(a)     The sport scientists concluded that the three types of meal had no major effect on maximum percentage increase in blood flow in the mesenteric artery.
What else can be concluded from their results?
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(2)

In another investigation, the sport scientists recorded blood flow in the mesenteric artery before and after vigorous exercise.
The graph shows their results for a large group of volunteers.

(b)     Suggest the advantage of the change in blood flow in the mesenteric artery during exercise.
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(3)

(c)     The sport scientists concluded people should not do vigorous exercise after a meal.
Does the information from the table and the graph support this conclusion?
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(4)
(Total 9 marks)

Q1.          The Solomon Islands are situated in the Pacific Ocean. The nearest large land mass is Australia, which is about 1500 km away. The biggest islands are mountainous, with large areas of tropical forest and a wide range of habitats. Some islands have a very high species diversity, and many species are endemic, that is they occur only in the Solomon Islands.
The table shows the total number of species on the islands in four vertebrate classes and the percentage which are endemic.

Vertebrate class

Total number of
species

Endemic species
/ %

Mammals
Birds
Reptiles
Amphibians

53
223
61
17

36
20
16
53

(a)     How many reptile species are endemic?
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(1)

S       (b)     Suggest an explanation for the high proportion of endemic species on the Solomon Islands.
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(Total 4 marks)

Q2.          The vegetation on a large heap of waste from an old mine was investigated. The table shows the results of the measurements of certain factors in 1m^2 frame quadrats placed on the south-facing slope.

Quadrat

Angle of
slope / °

Vegetation
cover / %

Moisture
content of
soil / %

pH of
soil

1
2
3
4
5
6

45
30
25
12
7
1

60
70
68
100
85
100

17.2
14.6
20.3
23.5
21.0
21.2

5.6
4.2
5.2
7.1
5.4
6.8

(a)     Which of the factors measured are abiotic?
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(1)
(b)     Describe how the investigators could obtain the value for vegetation cover in each quadrat.
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(2)

(c)     The correlation between vegetation cover and soil moisture content was tested statistically. These two factors were found to be positively correlated, and p < 0.05. Explain what this result means.
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(2)
S       (d)     At first the waste heap had no plants growing on it. Some of the first plants to colonise it were small herbaceous plants. Explain one way in which colonisation by herbaceous plants could change the physical environment.
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(2)
(Total 7 marks)

Q3.          (a)     Explain how crossing over can contribute to genetic variation.
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(3)

(b)     Maize seeds were an important food crop for the people who lived in Peru. The seeds could be kept for long periods. Each year, some were sown to grow the next crop. Archaeologists have found well-preserved stores. The graph shows the lengths of seeds collected from three stores of different ages.

(i)      Within each store the maize seeds showed a range of different lengths.
Explain one cause of this variation.
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(2)

(ii)     Use your knowledge of genetics and selection to explain the changes in the mean length of the seeds between AD 450 and AD 1530.
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(4)
(Total 9 marks)

Q4.          (a)     CFTR is a transmembrane regulator protein.  Its molecules have 1480 amino acids. People with cystic fibrosis produce defective CFTR protein which is missing one amino acid from its structure.
(i)      What is the minimum number of bases on DNA which would code for the normal CFTR protein? Explain your answer.
Number of bases ...............................
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(2)
(ii)     Which type of gene mutation produced the cystic fibrosis allele?
Explain your answer.
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(2)
(b)     The diagram shows part of the process of making normal and defective CFTR in a cell. A normal CFTR protein molecule has sugar molecules attached to it which make it functional.

Describe how the information on mRNA is translated into CFTR at the ribosome.
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(4)

(Total 8 marks)

Q5.          (a)     Explain the principles which biologists use to classify organisms into groups.
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(3)

Cytochrome c is a protein with about 100 amino acids and is present in all eukaryotic organisms. It has the same three-dimensional shape in all species, but only 30 of the amino acids are the same in all species. The amino acid sequence of cytochrome c has been used to construct the phylogenetic tree shown below.

(b)     Name the kingdoms represented in this phylogenetic tree.
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(c)     What does the phylogenetic tree show about the evolutionary relationship between fungi and insects?
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(d)     Suggest how information on amino acid sequences is used to construct a phylogenetic tree.
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(2

(e)     Suggest one advantage and one disadvantage of using cytochrome c to construct a phylogenetic tree. 
Advantage
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Disadvantage
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(Total 10 marks)

Q6.          Lake Malawi in East Africa contains around 400 different species of cichlids which are small, brightly coloured fish. All these species have evolved from a common ancestor.
(a)     Describe one way in which scientists could find out whether cichlids from two different populations belong to the same species.
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(2)

(b)     During the last 700 000 years there have been long periods when the water level was much lower and Lake Malawi split up into many smaller lakes. Explain how speciation of the cichlids may have occurred following the formation of separate, smaller lakes.
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(4)

(c)     Many species of cichlids are similar in size and, apart from their colour, in appearance. Suggest how the variety of colour patterns displayed by these cichlids may help to maintain the fish as separate species.
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(Total 8 marks)

Q7.          (a)     Figure 1 shows the exposed bases (anticodons) of two tRNA molecules involved in the synthesis of a protein.
Figure 1

Complete the boxes to show the sequence of bases found along the corresponding section of the coding DNA strand.
(2)

(b)     Describe the role of tRNA in the process of translation.
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(3)

(c)     Figure 2 shows the sequence of bases in a section of DNA coding for a polypeptide of seven amino acids.
Figure 2
TACAAGGTCGTCTTTGTCAAG
The polypeptide was hydrolysed. It contained four different amino acids. The number of each type obtained is shown in the table.
 
Amino acid | Number present
Phe | 2
Met | 1
Lys | 1
Gln | 3

Use the base sequence shown in Figure 2 to work out the order of amino acids in the polypeptide. Write your answer in the table below.
 
Met | | | | | |

(2)
(Total 7 marks)

Q8.          (a)     Give two factors, other than cost, that should be considered when selecting an antibiotic to treat a bacterial disease.
1 ...................................................................................................................
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(2)

S       (b)     The table describes the effects of two antibiotics on bacteria.
 
Antibiotic | Effect
Tetracycline | prevents tRNA binding
Chloramphenicol | prevents peptide bonds forming

(i)      Explain how each of these antibiotics slows down the rate of growth of bacteria.
Tetracycline ........................................................................................
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(4)

(ii)     Suggest why tetracycline has no effect on human cells.
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(Total 7 marks)

Q9.          (a)     (i)      What is the role of RNA polymerase in transcription?
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(1)

(ii)     Name the organelle involved in translation.
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(1)

(b)     Figure 1 shows some molecules involved in protein synthesis.
Figure 1

Complete Figure 1 to show
(i)      the bases on the DNA strand from which the mRNA was transcribed;
(ii)     the bases forming the anticodons of the tRNA molecules.
(2)
Figure 2 shows the effects of two different mutations of the DNA on the base sequence of the mRNA. The table shows the mRNA codons for three amino acids.
Figure 2

(c)     Name the type of mutation represented by mutation 1.
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(1)
(d)     Use the information in the table to
(i)      identify amino acid X in Figure 1;
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(1)

(ii)     explain how each mutation may affect the polypeptide for which this section of DNA is part of the code.
Mutation 1 ...........................................................................................
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Mutation 2 ...........................................................................................
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(Total 10 marks)

Q10.          (a)     Explain what is meant by stabilising selection and describe the circumstances under which it takes place.
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(5)
(b)     Some European clover plants can produce cyanide. Those plants that can produce cyanide are called cyanogenic; those that cannot produce cyanide are called acyanogenic. Cyanide is toxic to the cells of animals and plants.
When the leaves of cyanogenic plants are damaged by slugs, or exposed to low temperatures, membranes within the cells are broken. This causes the release of the enzymes that control the reactions which produce cyanide.

The proportions of cyanogenic and acyanogenic plants in clover populations were determined in different parts of Europe. These are shown in the diagram below, together with the mean minimum winter temperatures. Slugs are not usually active at temperatures below 0 °C.

Explain the proportions of cyanogenic and acyanogenic plants in clover populations growing in the area where the mean minimum winter temperature is below –4°C and in the area where it is above 10 °C.
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(5)(Total 10 marks)
Q11.          Figure 1 shows a fresh-water shrimp.
Figure 1
 
Biologists collected shrimps from a stream inside a cave and from the same stream when it was in the open.

They measured the maximum diameter of each shrimp’s eye. They also measured the length of its antenna. From these measurements they calculated the mean values for each site. Figure 2 shows their results.
Figure 2
 
| | Shrimps from the stream
| | Inside the cave | In the open
| Mean diameter of eye /mm | 0.09 | 0.24
| Mean length of antenna /mm | 8.46 | 5.81

(a)     The biologists measured the maximum diameter of each shrimp’s eye.
Explain why they measured the maximum diameter.
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(1)

(b)     A scientist working many years earlier suggested that animals which live in caves had similar adaptations. These adaptations included
•        smaller eyes
•        greater use of sense organs such as those involved in detecting touch.

(i)      Do the data in Figure 2 support this scientist’s suggestion? Explain your answer.
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(ii)     The data in Figure 2 are mean values. Explain how standard deviations of these cmean values would help you to interpret the data in Figure 2.
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(c)     The biologists investigated shrimps living in other streams. They measured the length of the antennae of these shrimps. They also measured their body length. Figure 3 shows the mean antenna length plotted against mean body length for each site.

Figure 3

(i)      What does the information in the graph suggest about the body lengths of shrimps living in caves and living in the open?
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(2)

(ii)     Do the data in the graph support the conclusion that shrimps with longer bodies have longer antennae? Give the reason for your answer.
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(1)

Other biologists investigated the genetic diversity of these shrimps. Figure 4 shows some of the data they collected.
Figure 4
 
| Gene | Allele | Percentage of shrimps with this allele in steam
| | | Inside a cave | In the open
| PGI | A | 0.9 | 2.5
| | B | 0.0 | 3.3
| | C | 98.2 | 66.4
| | D | 0.9 | 6.6
| | E | 0.0 | 21.3
| ACO2 | J | 0.0 | 5.6
| | K | 0.0 | 76.7
| | L | 100.0 | 17.8

(d)     The biologists concluded that the shrimps in the open had a higher genetic diversity than those in the cave. Explain how the data in Figure 4 support this conclusion.
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(1)

(e)     The percentage of shrimps with allele L in the cave is different from the percentage of shrimps with allele L in the open. Use your knowledge of the founder effect to suggest a reason for this difference.
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(3)

(f)      The biologists who studied these shrimps wanted to know if the shrimps living in the cave were the same species as those living in the open. They used breeding experiments to investigate this.
(i)      Describe how the biologists should carry out these breeding experiments.
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(ii)     The results of breeding experiments would help the biologists to decide whether the shrimps were the same species. Explain how.
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(3)
(Total 15 marks)

Q12.The body markings of cheetahs vary, in particular the pattern of bands on their tails. Cheetahs are solitary animals but the young stay with their mother until they are between 14 and 18 months old.
Scientists investigated the banding pattern on the tails of cheetahs living in the wild.
•        They drove a car alongside a walking cheetah and used binoculars to study the tail pattern.
•        They gave each cheetah a banding pattern score based on the width of the dark and light bands on the end of the tail.
•        They scored the width of the bands on the right and left side of the tail using a 5 point scale of width.

A typical pattern on the right side of one cheetah’s tail is shown in Figure 1.
Figure 1
 
| Band number | 1 | 2  3 | 4 | 5 | 6 | 7

 
| Band width score | 3 | 1  1 | 4 | 3 | 3 | 3

The scientists collected data from each cheetah on four separate occasions. Figure 2 shows the data for one of the cheetahs.
Figure 2
 
| Side of tail | Mean band width score (± standard deviation)
| | Band 1 | Band 2 | Band 3 | Band 4 | Band 5 | Band 6 | Band 7
| Right | 3.00 (± 0.82) | 1.00 (± 0.00) | 1.00 (± 0.00) | 3.75 (± 0.50) | 2.75 (± 0.50) | 3.00 (± 0.00) | 3.00 (± 0.00)
| Left | 3.75 (± 0.50) | 3.25 (± 0.50) | 2.00 (± 0.50) | 3.00 (± 0.00) | 2.00 (± 0.00) | 2.50 (± 0.50) | 3.00 (± 0.50)

(a)     The scientists only used data from cheetahs which were fully grown. Suggest why.
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(1)
(b)     The scientists estimated the width of the bands on the same cheetah on four separate occasions. They did not always get the same score.
(i)      Give two pieces of evidence from Figure 2 which show that the scientists sometimes obtained different scores for the same band.
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(ii)     The method the scientists used resulted in them getting different scores for the same band. Suggest why.
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(1)

(c)     What is the evidence from Figure 2 that the dark and light bands do not form rings of equal width around the tail?
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(1)

(d)     The scientists found the difference in banding pattern between
•        offspring in the same family
•        cheetahs chosen randomly.
Explain how scientists could use this information to show that some variation in tail banding was genetic.
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(Total 8 marks)
Q13.The Amazonian forest today contains a very high diversity of bird species.
•        Over the last 2 000 000 years, long periods of dry climate caused this forest to separate into a number of smaller forests.
•        Different plant communities developed in each of these smaller forests.
•        Each time the climate became wetter again, the smaller forests grew in size and merged to reform the Amazonian forest.
(a)     Use the information provided to explain how a very high diversity of bird species has developed in the Amazonian forest.
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 (5
(b)     Speciation is far less frequent in the reformed Amazonian forest. Suggest one reason for this.
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(1)
(Total 6 marks)
Q14.Farmland previously used for growing crops was left for 30 years and developed into woodland. During this period, ecologists recorded an increase in the diversity of birds in the area.
(a)     Name the process that resulted in the development of woodland from farmland.
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(1)

(b)     Explain the increase in the diversity of birds as the woodland developed.
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(Extra space) .................................................................................................
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(3)

(c)     The ecologists also investigated photosynthesis in two species of plant found in the woodland. One of the species was adapted to growing in bright sunlight (sun plant) and the other was adapted to growing in the shade (shade plant). The ecologists’ results are shown in the figure below.
 
(i)      Give two factors which could be limiting the rate of photosynthesis in the sun plant between points A and B on the figure.
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2.............................................................................................................
(1)

(ii)     Explain why CO_2 uptake is a measure of net productivity.
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(Extra space) ........................................................................................
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(1)

(iii)    Use the information in the figure to explain how the shade plant is better adapted than the sun plant to growing at low light intensities.
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(Extra space) ........................................................................................
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(2)
(Total 8 marks)
Q15.(a)     What two measurements are needed to calculate an index of diversity?
1 .....................................................................................................................
2 .....................................................................................................................
(2)

(b)     A herbicide is a chemical used to kill weeds. Ecologists investigated the effect of a herbicide on crop yield and the diversity of insects. They sprayed different fields with the same volume of different concentrations of the herbicide. At harvest, the ecologists determined the mean crop yield and the mean index of diversity of insects for fields that had received the same concentration of the herbicide.
The figure below shows their results.
 
Concentration of herbicide sprayed on field / mg dm^−3
(i)      Some fields acted as controls. They were sprayed with a solution that did not contain the herbicide. Explain the purpose of these control fields.
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(1)

(ii)     Suggest an explanation for the relationship between the concentration of herbicide and the mean crop yield.
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(2)
(iii)     Explain the relationship between the concentration of herbicide and the mean index of diversity of insects.
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(Extra space) ........................................................................................
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(3)
(Total 8 marks)
Q16.The following figure shows how some animals with hooves are classified.
 
(a)     This type of classification can be described as a phylogenetic hierarchy.
(i)      What is meant by a hierarchy?
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(Extra space) ........................................................................................
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(ii)     How many different families are shown in the figure?
 
(1)
(iii)    To which phylum does the white rhino belong?
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(1)
(b)     (i)      Explain the role of independent segregation in meiosis.
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(2)
(ii)     A zedonk is the offspring produced from breeding a mountain zebra with a donkey.
•        The body cells of a mountain zebra contain 32 chromosomes.
•        The body cells of a donkey contain 62 chromosomes.
Use this information to suggest why zedonks are usually infertile.
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(2)
(Total 8 marks)
Q17.(a)     A mutation can lead to the production of a non-functional enzyme. Explain how.
(6)
Scientists investigated the effect of a specific antibiotic on two strains of the same species of bacterium.
•        One strain, SR, shows a stringent response in the presence of this antibiotic. Part of this response involves stopping cell division. This gives this strain a greater resistance to the effects of this antibiotic.
•        The other strain, non-SR, cannot carry out a stringent response.
The scientists grew cultures of the SR strain and the non-SR strain containing the same number of bacterial cells. They then stopped each strain from dividing and exposed them to different concentrations of the antibiotic. After a fixed time, the scientists estimated the number of living bacteria remaining in the cultures.
Figure 1 shows their results.
Figure 1

(b)     Describe differences in the effect of increasing the concentration of antibiotic on the SR strain and the non-SR strain.
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[Extra space] ................................................................................................
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(2)
(c)     One way in which the stringent response gives resistance to this antibiotic is by stopping cell division.
The scientists concluded that stopping cell division is not the only way in which the stringent response gives resistance to this antibiotic.
Explain how Figure 1 supports this conclusion.
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 (2)

(d)     The stringent response involves a number of enzyme-catalysed reactions.
Explain how scientists could use this knowledge to design drugs that make the treatment of infections caused by the SR strain more successful.
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(2)
The antibiotic damages the bacterium by causing the production of substances called free radicals.
The scientists exposed the SR strain and the non-SR strain to the antibiotic. They then measured the amounts of free radicals and an enzyme called catalase in both strains.
Figure 2 shows their results.
Figure 2

(e)     Use the information provided and Figure 2 to suggest an explanation for the greater resistance of the SR strain to this antibiotic.
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[Extra space] .............................................................................. (3)(Total 15 marks)
Q18.Ecologists investigated the size of an insect population on a small island. They used a mark-release-recapture method. To mark the insects they used a fluorescent powder. This powder glows bright red when exposed to ultraviolet (UV) light.
(a)     The ecologists captured insects from a number of sites on the island. Suggest how they decided where to take their samples.
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(b)     Give two assumptions made when using the mark-release-recapture method.
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(c)     Suggest the advantage of using the fluorescent powder in this experiment.
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The ecologists did not release any of the insects they captured 1–5 days after release of the marked insects.
The table below shows the ecologists’ results.

| Days after release | Number of marked insects remaining in population | Number of insects captured | Number of captured insects that were marked
| 1 | 1508 | 524 | 78
| 2 | 1430 | 421 | 30
| 3 | 1400 | 418 | 18
| 4 | 1382 | 284 | 2
| 5 | 1380 | 232 | 9

(d)     Calculate the number of insects on this island 1 day after release of the marked insects.
Show your working.

 
Answer = ...................................
(2)
(e)     The ecologists expected to obtain the same result from their calculations of the number of insects on this island on each day during the period 1–5 days after release. In fact, their estimated number increased after day 1.
During the same period, the number of insects they caught decreased.
The method used by the ecologists might have caused these changes.
Use the information provided to suggest one way in which the method used by the ecologists might have caused the increase in their estimates of the size of the insect population.
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(2)
(Total 10 marks)
Q19.A student investigated the distribution of plants in a heathland.
The table below shows the number of plants he found in a sample area of 1 m^2.
 
| Species of plant | Number counted in 1 m^2
| Common heather | 2
| Red fescue | 14
| Vetch | 2
| White clover | 8

| (a) | What is the species richness of this sample? |
(1)
(b)     Calculate the index of diversity of this sample. Show your working.
Use the following formula to calculate the index of diversity.
d = 
where N is the total number of organisms of all species
and     n is the total number of organisms of each species

 
Index of diversity = ...................................
(2)
(c)     Suggest how this student would obtain data to give a more precise value for the index of diversity of this habitat.
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(2)
(Total 5 marks)
Q20.A student investigated the effectiveness of four different concentrations, A, B, C and D, of the same disinfectant on the growth of a bacterium. He grew a culture of this bacterium on nutrient agar (a solid growth medium) in a Petri dish. The student then cut out four filter paper discs and soaked each disc in one of the four concentrations. He then placed the discs on the nutrient agar in the Petri dish. He then left the Petri dish at 25 °C for 24 hours.
The diagram below shows the appearance of the Petri dish after 24 hours.
 
(a)     Explain why there is a clear zone around each paper disc.
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(1)

(b)     The student researched information on this disinfectant prior to carrying out his investigation. On the basis of this research, the student used a maximum concentration of disinfectant of 40%.
Use the diagram to explain why.
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(1)

(c)     Suggest two variables the student should control in using the filter paper discs in this investigation.
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(2)

(d)     Use the areas of the clear zones in the diagram above to determine how many times more effective concentration D is than concentration B. Show your working.
Answer = ................................
(2)
(Total 6 marks)

 

M1.          (a)     Helical / spiral / coiled;
Compact / description e.g. ‘tightly packed’;
Feature = one mark
Explanation = one mark
Insoluble;
Prevents osmosis / uptake of water / does not affect water
potential / (starch) does not leave cell;
These must be related for both marks but can be in reverse order.
Large molecule / long chain;
Does not leave cell;
Allow idea of compact / helical / spiral / coiled due to bonding for two marks.
2 max

(b)     (i)      β / beta Glucose;
Q Reject alpha glucose
1
(ii)     Glycosidic;
1

(c)     Long / straight / unbranched chains (of glucose joined by) hydrogen bonds;
Q Ignore reference to alpha glucose
Form (micro)fibrils / (macro)fibrils;
Provide rigidity / strength / support;
Allow suitable descriptions for last point e.g. ‘prevents bursting’;
3
[7]

M2.(a)     (i)     Repeating units / nucleotides / monomer / molecules;
Allow more than one, but reject two
1

(ii)     1.      C = hydrogen bonds;
2.      D = deoxyribose;
Ignore sugar
3.      E = phosphate;
Ignore phosphorus, Ignore molecule
3

(iii)
| Name of base | Percentage
| Thymine | 34
| Cytosine / Guanine | 16
| Adenine | 34
| Cytosine / Guanine | 16

Spelling must be correct to gain MP1
First mark = names correct
Second mark = % correct, with adenine as 34%
2

(b)     (i)     153;
1

(ii)     Some regions of the gene are non-coding / introns / start / stop code / triplet / there are two DNA strands;
Allow addition mutation
Ignore unqualified reference to mutation
Accept reference to introns and exons if given together
Ignore ‘junk’ DNA / multiple repeats
1
[8]

M3.         (a)     (i)     β / Beta glucose;
Accept b / B
Reject any reference to alpha / α
1

(ii)     Glycosidic;
Reject references to α(1-4) glycosidic bond, but allow beta 1-4, or unspecified reference to 1-4 (1,4)
1

(iii)    OH / hydroxyl / HO;
Reject hydroxide
Reject OH / HO molecule
Ignore alcohol
1

 
(b)     (i) | Starch | Cellulose
| 1. (1,4 and) 1,6 bonds / contains 1,6 bonds / branching | 1. 1,4 bonds / no 1,6 bonds / unbranched / straight;
| 2. All glucoses / monomers same way up | 2. Alternate glucoses /     monomers upside down;
| 3. Helix /     coiled / compact | 3. Straight;
| 4. Alpha glucose | 4. Beta glucose;
| 5. No (micro / macro) fibrils / fibres | 5. Micro / macro fibrils / fibres;

1 mark per pair of contrasts, both starch and cellulose required
Accept other comparable differences eg hydrogen bonds within starch but between cellulose molecules
2 max

(ii)     1.      H-bonds / micro / macro fibrils / fibres;
Reject strong hydrogen bonds
2.      Strength / rigidity / inelasticity;
‘Strong hydrogen bonds’ = 0 but ‘Strong hydrogen bonds give strength (to the molecule)’ = 1
2
[7]

M4.(a)     1.       A: phospholipid (layer);
1. Reject hydrophobic / hydrophilic phospholipid
2.      B: pore / channel / pump / carrier / transmembrane / intrinsic / transport protein;
2. Ignore unqualified reference to protein
2

(b)     (i)      Condensation (reaction);
1

(ii)     Organelle named; Function in protein production / secretion;
Function must be for organelle named 
Incorrect organelle = 0
eg
1.      Golgi (apparatus);
1. Accept smooth endoplasmic reticulum
2.      Package / process proteins;
OR
3.      Rough endoplasmic reticulum / ribosomes;
3. Accept alternative correct functions of rough endoplasmic reticulum. ER / RER is insufficient
3. Accept folding polypeptide / protein
4.      Make polypeptide / protein / forming peptide bonds;
OR
5.      Mitochondria;
6.      Release of energy / make ATP;
6. Reject produce / make energy
6. Accept produce energy in the form of ATP
OR
7.      Vesicles;
8.      Secretion / transport of protein;
2
[5]

M5.          (a)     (i)      D plasmid / ribosome(s) / cytoplasm / storage granules;
(accept any sensible structure)
E (slime / mucous) capsule
OR
slime / mucous layer; 
2
(ii)     protection / maintain shape / prevent lysis / strength / support;
1

(b)     two of the following:
nucleus;
OR
nuclear envelope / mitochondria / chloroplasts / sER / rER / 
golgi apparatus / 80s ribosomes
linear DNA / chromosomes / lysosomes / vacuole / vescicles / 
cellulose cell wall;
2 max

(c)     (i)      starch digested / broken down;
by amylase / carbohydrase;
2
(ii)     any sensible suggestion e.g. no secretion of amylase / 
functional amylase / 
piece of fungus might have died;
(accept carbohydrase / enzyme for amylase)
(reject “no digestion” without qualification)
1
[8]

M6.          (a)     (i)      31 / 31.2;
1
(ii)     Ratio would be less / smaller;
Cell is thin / has large surface area / (adapted) for diffusion;
Accept converse. Must relate to concept of ratio.
2
(b)     (i)      6;
1
(ii)     11;
1

(c)     Water potential inside vesicle more negative / lower; 
Water moves into vesicle by osmosis / diffusion;
2
(d)     Mitochondria supply energy / ATP;
For active transport / absorption against concentration
gradient / synthesis / anabolism / exocytosis / pinocytosis;
Do not credit references to making,
creating or producing energy.
2

(e)     1   Phospholipids forming bilayer / two layers;
2   Details of arrangement with “heads” on the outside;
3   Two types of protein specified;
     e.g.   passing right through or confined to one layer / 
              extrinsic or intrinsic / 
              channel proteins and carrier proteins / 
              two functional types
4   Reference to other molecule e.g. cholesterol or glycoprotein;
5   Substances move down concentration gradient / from high to low
     concentration;
Reject references to across or along a gradient
6   Water / ions through channel proteins / pores;
7   Small / lipid soluble molecules / examples pass between phospholipids / 
through phospholipid layer;
8   Carrier proteins involved with facilitated diffusion;
Ignore references to active transport.
Credit information in diagrams.
max 6
[15]

M7.          (a)     (i)      both are polymers / polysaccharides / built up from many sugar units / both contain glycosidic bonds / contain (C)arbon, (H)ydrogen and (O)xygen;
1
(ii)     hemicellulose shorter / smaller than cellulose / fewer carbons;
hemicellulose from pentose / five-carbon sugars and cellulose from
hexose / glucose / six-carbon sugars;
(only credit answers which compare like with like.)
2

(b)     protein / nucleic acid / enzyme / RNA / DNA / starch / amylose / amylopectin polypeptide;
1
(c)     (i)      to make sure that all the water has been lost;
1
(ii)     only water given off below 90 °C;
(above 90°C) other substances straw burnt / oxidised / broken down; and lost as gas / produce loss in mass;
2

(d)     enzymes are specific;
shape of lignin molecules will not fit active site (of enzyme);
OR
shape of active site (of enzyme);
will not fit molecule;
2 max

(e)     1. made from β-glucose;
2. joined by condensation / removing molecule of water / glycosidic bond;
3. 1 : 4 link specified or described;
4. “flipping over” of alternate molecules;
5. hydrogen bonds linking chains / long straight chains;
6. cellulose makes cell walls strong / cellulose fibres are strong;
7. can resist turgor pressure / osmotic pressure / pulling forces;
8. bond difficult to break;
9. resists digestion / action of microorganisms / enzymes;
(allow maximum of 4 marks for structural features)
6 max
[15]

M8.(a)     Amylase;
(Starch) to maltose:
Maltase;
Maltose to glucose;
Hydrolysis;
(Of) glycosidic bond;
Q Do not penalise incorrect site for digestion or incorrect site of enzyme production.
5 max

(b)     Glucose moves in with sodium (into epithelial cell);
Via (carrier / channel) protein / symport;
Sodium removed (from epithelial cell) by active transport / sodium- potassium pump;
Into blood;
Maintaining low concentration of sodium (in epithelial cell) / maintaining sodium concentration gradient (between lumen and epithelial cell);
Glucose moves into blood;
By (facilitated) diffusion;
Q Only allow diffusion mark in context of movement of glucose into the blood.
5 max
[10]

M9.(a)     (Omega-3 concentration) falls more rapidly at first;
Levels out at 140 days / concentration of 0.4%;
2

(b)     (i)      Two marks for correct answer of 0.04 or 0.043;;
One mark for incorrect answer which clearly identifies total fall of 1.7;
2

(ii)     To take into account variation in fat content of milk / fat content varies from cow to cow;
Allows comparison;
2

(iii)    The graph shows a decrease with time feeding on corn;
No control group;
Might have fallen anyway / might decrease with time rather than with time spend feeding on corn;
Other factors / other named factor might also have changed;
Only one investigation so might not be representative;
4 max
[10]

M10.(a)     1.      Tertiary structure / 3D shape of enzyme (means);
Accept references to active site
2.      Active site complementary to maltose / substrate / maltose fits into active site / active site and substrate fit like a lock and key;
Idea of shapes fitting together
3.      Description of induced fit;
4.      Enzyme is a catalyst / lowers activation energy / energy required for reaction;
Accept “provides alternative pathway for the reaction at a lower energy level”
5.      By forming enzyme-substrate complex;
Accept idea that binding stresses the bonds so more easily broken
Do not award point 5 simply for any reference to E-S complex
5

(b)     1.      Inhibitors reduce binding of enzyme to substrate / prevent formation of ES complex;
Max 3 if only one type of inhibition dealt with. Accept maltase and maltose as examples of enzyme and substrate (and others)
Only once, for either inhibitor
(Competitive inhibition),
2.      Inhibitor similar shape (idea) to substrate;
3.      (Binds) in to active site (of enzyme);
Accept allows max rate of reaction to be reached / max product will eventually be formed
Accept complementary to active site
4.      (Inhibition) can be overcome by more substrate;
(Non-competitive inhibition),
5.      Inhibitor binds to site on enzyme other than active site;
6.      Prevents formation of active site / changes (shape of) active site;
Accept does not allow max rate of reaction to be reached / max product will not be formed
7.      Cannot be overcome by adding more substrate;
5 max
[10]

M11.(a)     1.      (before reaction) active site not complementary to/does 
not fit substrate;
2.      Shape of active site changes as substrate binds/as 
enzyme-substrate complex forms;
Note. Points 1 and 2 may be made in one statement and ‘complementary’ introduced at any point.
Points 1&2 – active site mentioned once applies for both points
Point 2 – Ignore references to how shape change is caused
3.      Stressing/distorting/bending bonds (in substrate leading to reaction);
2 max

(b)     1.      Tangent to curve drawn;
Tangent drawn at about 10 minutes
2.      Value in range of 8 to 11;
1 mark only for correct answer
2

(c)     1.      (Rate of) increase in concentration of maltose slows as substrate/starch is used up
OR
High initial rate as plenty of starch/substrate/more E-S complexes;
Reject ref. to amylase being used up
2.      No increase after 25 minutes/at end/levels off because no substrate/starch left;
Accept ‘little’
Ignore references to substrate a limiting factor
2

(d)     1.      Make/use maltose solutions of known/different concentrations 
(and carry out quantitative Benedict’s test on each);
2.      (Use colorimeter to) measure colour/colorimeter value of each 
solution and plot calibration curve/graph described;
Axes must be correct if axes mentioned, concentration on x-axis and colorimeter reading on y-axis
3.      Find concentration of sample from calibration curve;
3
[9]

M12.(a)     1.      Crush / grind;
2.      With ethanol / alcohol;
3.      Then add water / then add to water;
2. Water must be added after ethanol for third mark.
4.      Forms emulsion / goes white / cloudy;
4. Do not accept carry out emulsion test.
3

(b)     (i)      4 / four;
1

(ii)     1.      Phosphate / PO_4;
“It” refers to phospholipid.
2.      Instead of one of the fatty acids / and two fatty acids;
1. Accept minor errors in formula. Do not accept phosphorus / phosphorus group.
2

(iii)    1.      Double bonds (present) / some / two carbons with only one hydrogen / (double bonds) between carbon atoms / not saturated with hydrogen;
Answer refers to unsaturated unless otherwise clearly indicated.
May be shown in appropriate diagram.
2.      In (fatty acid) C  / 3;
2
[8]

M13.(a)

One mark for each correct column
Mark ticks only and ignore crosses
4

(b)     1.      Two marks for box round two hydrogens and one of the oxygens from OH groups on carbons 1 and 4;;
2.      One mark from incorrect answer involving any two hydrogens and an oxygen from carbons 1 and 4;
Do not award marks if all atoms concerned are on same carbon atom or are on carbon atoms other than 1 and 4 or where the answer does not have two hydrogen and one oxygen
2

(c)     (i)      1.      Holds chains / cellulose molecules together / forms cross links between chains / cellulose molecules / forms microfibrils, providing strength / rigidity (to cellulose / cell wall);
2.      Hydrogen bonds strong in large numbers;x
Principles here are first mark for where hydrogen bonds are formed and second for a consequence of this.
Accept microfibres
2

(ii)     Compact / occupies small space / tightly packed;
Answer indicates depth required. Answers such as “good for storage”, “easily stored” or “small” are insufficient.
1
[9]

M14.          (a)     Two marks for correct answer of 64.285 / 64.3 / 64;
(allow 1 mark for (8100 / 100 × 30) / 37.8)
2
(b)     dissolve in / add ethanol then mix with water;
emulsion / white colour indicates triglycerides present;
2

(c)     (i)      increase the surface area for absorption;
(ignore wrong ref. to name)
1
(ii)     R = tissue fluid / interstitial fluid / extracellular fluid / intercellular space;
S = lymph(atic) vessel / lymph capillary / lacteal;
2

(iii)     proteins are synthesised by U;
involvement of ribosomes;
protein isolation / transport (inside RER);
vesicle formation;
2 max
(iv)    exocytosis / description of;
because of size / too large to leave by other methods;
2
[11]

M15.          (a)     (i)      Curve rising and levelling out;
1
(ii)     Substrate becomes limiting / falls / gets less;
Fewer collisions / complexes formed; 
2

(b)     To keep pH the same / optimum pH / so change in pH does not affect reaction;
1

(c)     (i)      For temperature up to 40 – 50 °C has no effect;
Over temperature (of 40 – 50 °C) reduces rate of reaction;
Note. Award one mark for general statement about the
longer the incubation time, the slower the rate of reaction.
2
(ii)     Bonds (holding tertiary structure) broken;
More enzyme denatured / tertiary structure destroyed / 
active sites lose shape / no longer fit;
Fewer enzyme-substrate complexes formed;
Note. Award marks if clearly in the context of more denaturation. Allow credit here for converse relating to exposure for 5 minutes.
3

(d)     Competitive
2 Similarity of shape of inhibitor and substrate;
3 Inhibitor can enter / bind with active site (of enzyme);
Non-competitive
4 Affect / bind to enzyme other than at active site;
5 Distorts shape of active site;
Inhibitors
6 Prevent entry of / binding of substrate to active site;
7 Therefore fewer / no enzyme-substrate complexes formed;
6
[15]

M16.          (a)     (i)      (Grinding) breaks open cells / increases surface area (of liver);
Releases catalase / enzyme / more catalase / allows more hydrogen peroxide into liver;
2
(ii)     Heating causes bonds (maintaining tertiary structure) to break;
Denatures / changes tertiary structure so active site changed;
Substrate no longer fits / ES complex not formed;
3

(b)     (Control) to show that sand did not affect reaction (with ground liver);
1
(c)     (i)      Lower activation energy / less energy required to bring about reaction;
1
(ii)     Energy in products / water and oxygen less than energy in substrate / reactants / hydrogen peroxide;
(Difference) given out as heat / exothermic;
2
[9]

M17.          (a)     Several / more than one polypeptide chain in molecule;
Evidence must only relate to 4º structure
1
(b)     Chemical bonds formed between sulphur-containing groups / 
R-groups / form stronger disulphide bonds;
Bind chain(s) to each other;
2

(c)     Different number / sequences of amino acids;
Bonds in different places which gives different shape;
2
(d)     Outer layer of skin cells are dead / do not respire
Do not contain mitochondria / do not produce ATP / release energy;
Cells do not have required proteins / carriers;
3

(e)     Advantages:
1       Small objects can be seen;
2       TEM has high resolution as wavelength of electrons shorter;
Accept better
Limitations:
3       Cannot look at living cells as cells must be in a vacuum;
4       must cut section / thin specimen;
5       Preparation may create artefact
6       Does not produce colour image;
6
[14]

M18.(a)     1.      Enzyme hydrolyses / breaks down protein to amino acids;
2.      Products are soluble / can be washed away;
2

(b)     Arguments for biological washing powder:
3 max if only arguments against biological washing powder are referred to
1.      More effective with all stains;
Accept different ways of expressing ‘effective’ e.g. higher % of stain removed
2.      Greater improvement with salad dressing / chocolate milkshake / chocolate pudding;
Arguments against biological washing powder:
3.      Little / less improvement with raspberry sorbet / raspberry smoothie;
4.      Only tested 5 / a small number of stains;
5.      Only chose stains that would work / didn’t select stains that wouldn’t work;
6.      Only included results that did work / didn’t show results that didn’t work;
7.      Only one set of results / not repeated;
8.      Only compared against one washing powder / may not be true for other washing powders;
Ignore references to unknown masses of powder, temperature of washes or other aspects of technique or different fabrics
4 max

(c)     1.      Enzyme S effective across a wider range of temperatures;
2.      Enzyme S more active above 50 °C / active up to 80 °C / active above 60 °C;
3.      Enzyme S more active below (about) 37 °C temperature;
4.      (Although) Enzyme P has higher rate of reaction at optimum / 40 – 50 °C;
5.      Enzyme P denatured above 50 °C;
Answers should be in the context of choosing enzyme S but, if P is chosen, points 4 and 5 may still be awarded, if described
In points 2 and 3, a temperature must be stated. Allow ± 5 degrees of values shown
3 max

(d)     1.      Stains caused by different substances;
2.      Enzymes are specific;
3.      Active site specific to substrate / other substrates cannot fit active site;
This could be expressed in other ways e.g. ‘other substrates are not complementary to the active site’
3
[12]

M19.(a)     Deoxyribose.
1

(b)     1.      Thymine 18 (%);
2.      Guanine 32 (%).
2

(c)     DNA polymerase.
1

(d)     1.      (Figure 1 shows) DNA has antiparallel strands / described;
2.      (Figure 1 shows) shape of the nucleotides is different / nucleotides aligned  differently;
3.      Enzymes have active sites with specific shape;
4.      Only substrates with complementary shape / only the 3’ end can bind with active site of enzyme / active site of DNA polymerase.
4
[8]

M20.(a)     1.      DNA replicated;
Reject: DNA replication in the wrong stage
2.      (Involving) specific / accurate / complementary base-pairing;
Accept: semi conservative replication
3.      (Ref to) two identical / sister chromatids;
4.      Each chromatid / moves / is separated to (opposite) poles / ends of cell.
Reject: meiosis / homologous chromosomes / crossing over
Note: sister chromatids move to opposite poles / ends = 2 marks for mp 3 and mp 4
Reject: events in wrong phase / stage
4

(b)     (i)      1.      To allow (more) light through;
Accept: transparent
2.      A single / few layer(s) of cells to be viewed.
Accept: (thin) for better / easier stain penetration
2

(ii)     1.      More / faster mitosis / division near tip / at 0.2 mm;
Neutral: references to largest mitotic index
2.      (Almost) no mitosis / division at / after 1.6 mm from tip;
Accept: cell division for mitosis
Penalise once for references to meiosis
3.      (So) roots grow by mitosis / adding new cells to the tip.
Accept: growth occurs at / near / just behind the tip (of the root)
Accept: converse arguments
2 max
[8]

M21.(a)     1.      Strands separate / H-bonds break;
1. Q Neutral: strands split
1. Accept: strands unzip
2.      DNA helicase (involved);
3.      Both strands / each strand act(s) as (a) template(s);
4.      (Free) nucleotides attach;
4. Neutral: bases attach
4. Accept: nucleotides attracted
5.      Complementary / specific base pairing / AT and GC;
6.      DNA polymerase joins nucleotides (on new strand);
6. Reject: if wrong function of DNA polymerase
7.      H-bonds reform;
8.      Semi-conservative replication / new DNA molecules contain one old strand and one new strand;
8. Reject: if wrong context e.g. new DNA molecules contain half of each original strand
6 max

(b)     (i)      18;
Do not accept 17.5
1

(ii)     10;
1

(iii)    1.      Horizontal until 18 minutes;
Allow + / - one small box
2.      (Then) decreases as straight line to 0 μm at 28 minutes;
2. Allow lines that start from the wrong place, ending at 0 at 28 minutes
2

(c)     (i)      Two marks for correct answer of 19.68 or 19.7;;
Accept 19hrs 41mins
One mark for incorrect answers in which candidate clearly multiplies by 0.82;
Allow one mark for incorrect answers that clearly show 82% of 24 (hours)
2

(ii)     1.      No visible chromosomes / chromatids / visible nucleus;
1

(iii)    D (no mark)
1.      Lower % (of cells) in interphase / higher % (of cells) in mitosis / named stage of mitosis;
1. Accept: ‘less’ or ‘more’ instead of ‘%’ 
1. Do not accept: higher % (of cells) in each / all stage(s)
2.      (So) more cells dividing / cells are dividing quicker;
2. Accept: uncontrolled cell division
2. Do not award if Tissue C is chosen
2
[15]

M22.          (a)     (i)      Deoxyribose;
pentose / 5C sugar = neutral
1
(ii)     Phosphate / Phosphoric acid;
phosphorus / P = neutral
1

(b)     Hydrogen (bonds);
1
(c)     381 / 384 / 387;
1
(d)     (Gln) Met Met Arg Arg Arg Asn;
1

(e)     Change in (sequence of) amino acids / primary structure;
Change in hydrogen / ionic / disulfide bonds leads to change in tertiary structure / active site (of enzyme);
Substrate cannot bind / no enzyme-substrate complexes form;
Q Reject = different amino acids are formed
3
[8]

M23.          (a)     (i)      base / named bases;
reject nucleotide or uracil
1
(ii)     it has been produced by semi-conservative
replication / one old strand and one new; 
One strand has ^15N bases and the other ^14N;   
Accept light / heavy N (therefore) it is less dense / lighter;
2
(iii)     one band is in same position as generation 1;
one band higher;
accept a line. N.B. need a visible gap
2

(b)     (i)      A = 31 and JT = 31;
C = 19;
2
(ii)     viral DNA single-stranded / not double-stranded;
evidence from table e.g. not equal amount of A and T
/ C and G / all different;
2
ignore no base-pairing In this Question assume It’ means viral DNA
[9]

M24.          (a)     1       two strands therefore semi-conservative replication (possible);
2       base pairing / hydrogen bonds holds strands together
3       hydrogen bonds weak / easily broken, allow strands to separate;
4       bases (sequence) (exposed so) act as template / can be copied;
5       A with T, C with G / complementary copy;
6       DNA one parent and one new strand; 
4 max

(b)     1       chromosomes shorten / thicken / supercoiling;
2       chromosomes (each) two identical chromatids / strands / copies 
(due to replication);
3       chromosomes / chromatids move to equator / middle of the spindle / cell;
4       attach to individual spindle fibres;
5       spindle fibres contract / centromeres divide / repel;
6       (sister) chromatids / chromosomes (separate) 
move to opposite poles / ends of the spindle;
7       each pole / end receives all genetic information / 
identical copies of each chromosome;
8       nuclear envelope forms around each group of chromosomes / 
chromatids / at each pole;
7 max

(c)     cancer cells killed, normal body cells survive;
cancer cells low oxygen (as blood supply cannot satisfy demand);
2
[13]

 

M25.         (a)     X = mitochondria;Y = (rough) endoplasmic reticulum;
Accept ribosomes/ER/RER for Y
Reject 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 X
Ignore names of organelles if function correct
2
[5]

M26.          (a)     (i)      microvilli; (reject brush border)
1
(ii)     increased surface area (for diffusion);
1

(b)     (i)       principle of ;
(15 –17 tolerance)
160000;
(correct answer award 2 marks)
2

(ii)     electron microscope has a greater resolving
power / objects closer 
together can be distinguished; 
electron (beams) have a shorter wavelength;
2

(c)     short diffusion pathway / short pathway to the centre / large SA:V ratio
for faster, more diffusion;
1
[7]

 

M27.          (a)     memory B / T cells do not recognise (new antigens);
antibodies previously produced are not effective
as shape not complementary to new antigen;
2

(b)     (i)      antigen in membrane presented to lymphocytes / 
produce cytokinins;
1
(ii)     mitochondria provide (more) ATP / energy;
(more) RER / ribosomes synthesise proteins;
(more) Golgi body secretes / modifies or packages proteins / 
produces glycoproteins;
(B lymphocytes) produces antibodies;
4
[7]

M28.          (a)     A – granum / thylakoid;
chlorophyll molecules to trap light / light absorbing pigments / 
light dependent reaction / part of light dependent reaction;
2
B – stroma;
(contains enzymes for) carbon dioxide fixation / light-independent reaction / 
part of light-independent reaction;
(allow ribosome role of protein in photosynthesis)
2

(b)     (i)      C – starch;
1
(ii)     from glucose in a condensation / polymerisation reaction / many
glucose molecules joined together;
1
[6]

M29.          (a)     (i)      Mitochondria site of respiration;
Production of ATP / release of energy;
For contraction;
Do not award credit for making or producing energy.
3
(ii)     Enzymes are proteins;
Proteins synthesised / made on ribosomes;
2

(b)     Lysosomes produce / contain enzymes;
Which break down / hydrolyse proteins / substances / cells of tail;
2

(c)     1. Chop up (accept any reference to crude breaking up);
2. Cold;
3. Buffer solution;
4. Isotonic / same water potential;
5. Filter and centrifuge filtrate;
6. Centrifuge supernatant;
7. At higher speed;
8. Chloroplasts in (second) pellet;
max 6
[13]

M30.          (a)     Measure diameter of field with ruler; And proportion taken up by the cell; or Measure length with (eyepiece) graticule / eyepiece scale;
Calibrated against stage micrometer / something of known length;
Reject divide apparent length by magnification
2

(b)     Membrane / cytoplasm shrinks / pulls away from cell wall / cell plasmolysed / goes flaccid; Water moves down water potential gradient / to lower / more negative water potential; By osmosis;
3

(c)     (i)      Reaches equilibrium / no further / maximum change in length;
Reject osmosis takes time
1
(ii)     Line / curve of best fit; Extrapolate (and read off) / find where it crosses x-axis;
2
(iii)     Greater decrease / length smaller; More water removed;
Greater difference in water potential / cell with higher / less negative water potential; Starch is insoluble / has no effect on osmosis
max 2
[10]

M31.          (a)     (i)      A anaphase;
1
(ii)     (C) B,A,D;
1
(iii)     (original) chromosome / DNA has been replicated;
each chromosome consists of two chromatids / 
chromatids attached at centromere;
(accept reference to condensed state of chromosomes)
2

(b)     (i)      it has doubled / now 8;
1
(ii)     chromosome / DNA replication but no separation
/ anaphase / cell division;
1
[6]

M32.          (a)     (i)      Spindle formed / chromosome / centromere / chromatids
attaches to spindle;
Chromosomes / chromatids line up / move to middle / equator
(of cell);
Do not award second mark for answers referring to chromosomes ‘pairing up’.
Ignore reference to homologous chromosomes unless context suggests pairing which negates second mark.
Neutral: Details on nuclear membrane.
Accept: Diagram for second marking point.
2

(ii)     Chromosome / centromere splits / chromatids / ‘chromosomes’ separate / pulled apart;
To (opposite) sides / poles / centrioles (of cell);
Reject: Homologous chromosomes separate for first marking point.
Accept: Diagram for second marking point.
Chromatids / ‘chromosomes’ move to poles / sides / centrioles = 2 marks.
2

(b)     (i)      Form / replace cells quickly / rapidly / divide / multiply / replicate rapidly;
Neutral: Repair cells.
Answers must convey idea of ‘speed’.
1

(Ii)     Correct answer = 774 minutes / 12 hours 54mins = 2 marks;;
Incorrect answer but indicates 3 cell cycles involved = one mark;
2

(c)     Prevents / slows DNA replication / doubling / prevents / slows mitosis;
New strand not formed / nucleotides (of new strand) not joined
together / sugar-phosphate bonds not formed;
First marking point must be in context of DNA replication not cell replication.
Do not negate first marking point if role of DNA polymerase is described incorrectly e.g. Reject: ‘joins bases / strands together’.
Role of DNA polymerase must be correct for last marking point.
2
[9]

M33.(a)
| | Cell B | Cell C | Cell D
| homologous chromosomes are present | ✔ | ✔ |
| a stage of mitosis | | ✔ |

Mark horizontally
1 mark for each correct row
2

(b)     Mark as pairs, do not mix and match
1.      (Chromosomes consist of) two chromatids connected at centromere;
Accept: sister chromatids for two chromatids
2.       (Because) DNA has replicated;
OR
3.       K is on equator of spindle;
Ignore: ‘middle’
4.       (because) attached at centromere;
Ignore reference to meiosis / bivalents / homologous pairs
2

(c)     1.      Crossing over / exchange of alleles /lengths of DNA / recombination;
Accept: description of crossing over eg sections of chromatids break and re-join
Accept: reference to chiasma/ chiasmata
2.      Between (chromatids of) homologous chromosomes;
Accept: ‘between non-sister chromatids’
Accept: ‘bivalent’ for homologous
Ignore: genes exchanged
2

(d)     Separation/segregation of pairs/homologous chromosomes;
Accept: result of meiosis I / result of division of cell B
Accept: pulled to opposite poles for ‘separation’
Ignore ref to chromatids
1

(e)     (DNA) replication taking place/not finished;
Accept: they are cells in S phase
1
[8]

M34.          (a)     1.      permeable capillary wall / membrane;
2.      single cell thick / thin walls, reduces diffusion distance;
3.      flattened (endothelial) cells, reduces diffusion distance;
4.      fenestrations, allows large molecules through;
5.      small diameter / narrow, gives a large surface area to volume / short diffusion distance;
6.      narrow lumen, reduces flow rate giving more time for diffusion;
7.      red blood cells in contact with wall / pass singly, gives short diffusion distance / more time for diffusion;
(allow 1 mark for 2 features with no explanation)
4 max

(b)     1.      (hydrostatic) pressure of blood high at arterial end;
2.      fluid / water / soluble molecules pass out (reject plasma);
3.      proteins / large molecules remain;
4.      this lowers the water potential / water potential becomes more negative;
5.      water moves back into venous end of capillary (reject tissue fluid) by osmosis / diffusion;
6.      lymph system collects any excess tissue fluid which returns to blood / circulatory system / link with vena cava / returns tissue fluid to vein;
6
[10]
QWC 1

M35.          (a)     partially / selectively permeable accept semi-permeable
allows water to pass through but not potassium nitrate / solute;
1
(b)     potassium nitrate (solution);
cell wall permeable;
2
(c)     water potential more negative / lower in cell E; water removed;
greater solute / sap concentration (in cell);
3
[6]

M1.         (a)     A = envelope/membrane/phospholipid (bilayer);B = capsid / nucleocapsid / capsomere / protein;
2
(i)      (HIV is) invading cells which make new viruses;Cells release viruses into blood;
2
(ii)     Virus remains dormant/exists as provirus/exists as DNA in host DNA;
Accept virus stays in cells
1

(c)     HIV destroys T cells;More (free) viruses produced leads to fall in T-cells;(So fewer) T-cells activate B-cells/memory cells;
Reduced/no antibody production;Immune system not working properly/inability to fight infection;Opportunistic infections;
4 max
[9]

M2.          (a)     (i)      Molecule/protein/glycoprotein;
Stimulates immune response;
(That causes) production of antibodies;
2 max

(ii)     Antigens on HIV are different (shape);
So, antibody will not ‘fit’/not complementary (to antigen);
Receptor sites on antibody specific to one antigen;
2 max

(iii)     (Has site with) same shape as salmonella antigen so binds
to anti-gal antibodies;
(Has site with) same shape as receptor molecule so that HIV will bind;
Binds to both molecules;
2 max

(b)     Salmonella pathogen has specific antigen on surface;
Salmonella pathogen engulfed by macrophage;
T-cells activate B-cells;
B-cell with complementary/specific receptor antibody activated/
clonal selection;
B-cells divide/form clone/clonal expansion;
Plasma cells make antibodies;
Specific to antigen/bind to salmonella bacterial antigen;
Accept macrophage presents antigen to T/B cells;
Accept T-cells release factors;
6 max

(c)     (i)      HIV binds to specific receptor;
Only present on certain cells / T-cells;
2
(ii)     Antibiotics stop metabolism, viruses don’t have metabolism;
Viruses hide in cells, antibiotics can’t reach;
Two suitable cell components antibiotics work against that
viruses don’t have;
e.g. some antibiotics work against ribosomes, that viruses don’t have
2

(d)     (i)      Adaptor molecule binds to HIV;
(This) prevents the HIV binding to the receptor;
Therefore few HIV available to infect cells;
2 max
(ii)     Would need to be complementary to MRSA (antigens);
MRSA has different antigens;
But would still need to have binding site for anti-gal;
2 max
[20]

M3.          (a)     Zevalin/antibody binds to specific receptor/cell surface protein/antigen;
(Only found) on B-cells;
2

(b)     Patient P treated with Zevalin/yttrium (no mark);
Assume ‘Zevalin’ means ‘with yttrium’ unless they state
otherwise
Where indium/antibody (only) on lymphatic system/groin and armpits;
So only (cancerous) B-cells killed;
In patient P high concentration of radioactivity/antibodies high enough
to kill cancer cells;
Patient Q ‒ radioactivity in places where other body cells could be killed/
organs damaged/named example;
Could harm patient more than cancer;
Patient Q cancer has spread;
So too late to treat;
3 max

(c)     Patient Q ‒ (cancerous) B-cells outside of lymphatic system/metastasis;
So antibody bound in other parts of the body (as well);
Patient Q ‒ has different receptors/distribution of receptors compared
to patient P;
Other body cells (than B-cells) have receptors for antibody;
2 max

(d)     Might be allergic to mouse antibody/protein;
(Mouse) antibody acts as an antigen;
Causes an immune response/antibody production;
Antibody destroys Zevalin;
Releases radioactivity into body/prevents activity against the cancer;
2 max
[9]

M4.          (a)     (i)      fall in deaths due to rise in number of people with immunity / better care / targeting vaccination at vulnerable;
1
(ii)     mutation of virus / new strain;
mutant form not recognised by memory cells (allow antibodies);
2 max

(b)     (i)      T lymphocyte receptors recognise shape of haemagglutinin / 
neuraminidase / viral antigen;
clone (once only);
destroy virus;
2 max

(ii)     clone (once only);
produce antibodies;
effect of antibody e.g. stimulation of phagocytosis / 
precipitation of toxins;
2

(c)     alter shape of active site of neuraminidase / block active site;
virus unable to leave host cells;
2
[9]

M5.          (a)     Publicity about vaccination / better health education / risks of ‘flu epidemics;
(Accept: now free on NHS (though only since 2000) / better awareness / more commonly available)
1

(b)     (i)      1990: 26% of 7.4million = 1.92million and 2000: 64% of 7.8 million = 4.99million;
increase = 3.07 million;
2
(Correct reading of all 4 figures from graph = 1)
(Correct answer but no ‘millions’ = 1)
(Correct method resulting from wrong graph reading = 1)

(ii)     Over 50% of population being vaccinated;
But only from 2000 onwards;
(Principle of more people being vaccinated each year = 1)
2

(iii)     Different strain / type of virus each year / virus mutates;
With different antigens;
Influenza antibodies / memory cells (rapidly) destroyed / need replacing;
max 2

(c)     (Protein coat) carries antigens which stimulates B-cells / production of antibodies;
Production of memory cells;
2
[9]

M6.          (a)     To prevent contamination of apparatus with other microorganisms / bacteria;
To prevent personal contact with bacteria;
To prevent release of bacteria into air;
max 2

(b)     (i)      Diffuses slowly;
1
(ii)     B;
Produces inhibition zone greater than the minimum diameter;
2
[5]

M7.          (a)     molecule / part of molecule / protein / glycoprotein / named molecule;
that stimulates an immune response / eq;
2
(b)     divide by mitosis / form clones; produce plasma cells; (plasma cells) 
make antibodies;
(plasma cells) produce memory cells;
4

(c)     (i)      glycoprotein AND
different shape to body proteins / RNA and reverse transcriptase
inside virus / phospholipids same as body’s / on the surface
of the virus;
1
(ii)     187.5;;
Accept 187 – 188
1 mark for HIV = 80nm;
2 max
[9]

M8.          (a)     divide by mitosis / form clones;
produce plasma cells; 
(plasma cells) make antibodies;
(plasma cells) produce memory cells;
4
(b)     glycoprotein;
different shape to body proteins / body phospholipids are the same / 
located on the outside of the cell / the haemoglobin is located
inside the cell;
2
[6]

M9.          (a)     1.      Phagocyte attracted to bacteria by chemicals / recognise antigens on bacteria as foreign;
2.      Engulf / ingest bacteria;
3.      Bacteria in vacuole / vesicle;
4.      Lysosome fuses with / empties enzymes into vacuole;
5.      Bacteria digested / hydrolysed;
1. Accept names chemical e.g. toxin
2. Allow description of engulfing
3. Accept: bacteria in phagosome
5. Neutral: Break down
5. Accept digestive enzymes destroy bacteria
5. Do not accept “destroy bacteria” as it is in question stem
4 max

(b)     1.      Microvilli provide a large / increased surface area;
2.      Many mitochondria produce ATP / release or provide energy (for active transport);
3.      Carrier proteins for active transport;
4.      Channel / carrier proteins for facilitated diffusion;
5.      Co-transport of sodium (ions) and glucose or symport / carrier protein for sodium (ions) and glucose;
6.      Membrane-bound enzymes digest disaccharides / produce glucose;
1. Reject villi on epithelial cells
1. Accept brush border
2. Accept large SA:vol ratio
3. Need idea of “lots”
4. Reject: energy produced
5. Accept Na^+K^+ pump
6. Neutral: Channel proteins
7. Accept named example
6
[10]

M10.(a)     1.      Vaccine/it contains antigen (from HPV);
Term ‛antigen’ may be first mentioned with point 2
2.      Displayed on antigen-presenting cells;
Accept named example, e.g. macrophage/phagocyte/B cells
3.      Specific helper T cell (detects antigen and) stimulates specific B cell;
Accept ‛helper T cell with receptor on surface’ for ‛specific’ and B cells with receptor/antibody on surface that bind to antigen for ‛specific’
4.      B cell divides/goes through mitosis/forms clone to give plasma cells;
5.      B cell/plasma cell produces antibody;
4 max

(b)     1.      Two (doses) because got more antibody;
Accept more effective in producing antibody
2.      With three doses, second dose/dose at 1 month doesn’t lead to production of any more antibody (than the two-dose group)/get same/similar response;
3.      Three doses would be more expensive/less popular with parents/girls (and serves no purpose);
Accept ‘less painful’
2 max

(c)     t-test, because comparing two means;
Mark for correct test and explanation correct
Accept ‘comparing the mean’
Reject ‘to show that the results/means are significant’
1

(d)     1.      Compare (base sequences of) DNA;
2.      Look for mutations/named mutations (that change the base sequence);
3.      Compare (base sequences of) (m)RNA;
1 and 3 accept triplet/codon sequences for comparisons
Ignore references to ‘introns/non-coding DNA’
2 max
[9]

M11.(a)
| Feature | Bacterium | Human immunodeficiency virus (HIV) particle
| RNA | ✔ | ✔
| Cell wall | ✔ |
| Enzyme molecules | ✔ | ✔
| Capsid | | ✔

1 mark for each correct vertical column
2

(b)     1.      (Complementary) nucleotides/bases pair
OR
A to T and C to G;
Ignore ‘(DNA polymerase) forms base pairs/nucleotide pairs’
2.      DNA polymerase;
3.      Nucleotides join together (to form new strand)/phosphodiester bonds form;
Ignore ‘(DNA polymerase) forms base pairs/nucleotide pairs’
If clearly writing rote answer about DNA replication 2 max e.g. helicase or separating strands
3

(c)     1.      DNA double stranded/double helix and mRNA single-stranded;
Contrast requires both parts of the statement
2.      DNA (very) long and RNA short;
Accept ‛RNA shorter’ or ‛DNA bigger/longer’
3.      Thymine/T in DNA and uracil/U in RNA;
4.      Deoxyribose in DNA and ribose in RNA;
R Deoxyribonucleic/ ribonucleic acid
Ignore ref. to histones
Ignore ref. to helix and straight chain alone
5.      DNA has base pairing and mRNA doesn’t/ DNA has hydrogen bonding and mRNA doesn’t;
6.      DNA has introns/non-coding sequences and mRNA doesn’t;
Ignore ref to splicing
3 max
[8]

M12.          (i)      cold - no / reduced enzyme action / e.g. stops autolysis;
(reject “cell activity reduced”)
isotonic - stops osmotic effects / description of effect on
cells or organelles;
buffer - prevents damage to enzymes / proteins;
3
(ii)      break open the cells / release the cell contents;
1
(iii)     supernatant / liquid above the pellet;
spun at a high(er) speed;
(mark as independent points)
2
[6]

M13.(a)     Any five from:
1.      Cell homogenisation to break open cells;
1. Accept suitable method of breaking open cells.
2.      Filter to remove (large) debris / whole cells;
2. Reject removes cell walls.
3.      Use isotonic solution to prevent damage to mitochondria / organelles;
3. Ignore to prevent damage to cells.
4.      Keep cold to prevent / reduce damage by enzymes / use buffer to prevent protein / enzyme denaturation;
5.      Centrifuge (at lower speed / 1000 g) to separate nuclei / cell fragments / heavy organelles;
5. Ignore incorrect numerical values.
6.      Re-spin (supernatant / after nuclei / pellet removed) at higher speed to get mitochondria in pellet / at bottom.
6. Must have location
Reject ref to plant cell organelles only once
5 max

(b)     Principles:
1.      Electrons pass through / enter (thin) specimen;
2.      Denser parts absorb more electrons;
3.      (So) denser parts appear darker;
4.      Electrons have short wavelength so give high resolution;
Principles:
Allow maximum of 3 marks
Limitations:
5.      Cannot look at living material / Must be in a vacuum;
6.      Specimen must be (very) thin;
7.      Artefacts present;
8.      Complex staining method / complex / long preparation time;
9.      Image not in 3D / only 2D images produced.
Limitations:
Context of limitation must be clear, not simply explaining how TEM works
E.g “allows you to see organelles as a thin section is used” is not a limitation
Allow maximum of 3 marks
Ignore ref to colour
5 max
[10]

M14.(a)      (D)CBEA.
1

(b)     
| Step | Reason
| (Taking cells from the root tip) | Region where mitosis / cell division occurs;
| (Firmly squashing the root tip) | To allow light through / make tissue layer thin;

2

(c)     (Increase)
1.      Chromosomes / DNA replicates;
(First decrease)
2.      Homologous chromosomes separate;
(Second decrease)
3.      Sister chromatids separate.
3

(d)     1.      (DNA would) double / go to 2 (arbitrary units).
1
[7]

M15.          (a)     shallow roots enable rapid uptake of rainfall (in X and / or Z);
widespread / shallow roots allow collection of larger volume
water / over a larger area / rapid uptake of water (in Z);
swollen stem for water storage (in X);
deep roots for accessing deep groundwater (in Y);
small / no leaves so little transpiration;
3

(b)     Z;
wide spread of roots for rapid water absorption;
(accept X; if linked to leaves channelling water to roots)
(ignore references to water storage abilities)
(accept other responses if justified)
2
[5]

M16.          (a)     Epithelium of alveolus, capillary wall / epithelium / endothelium, plasma;
1
(b)     Cell wall;
Capsule;
Flagellum;
Mesosomes;
Plasmid;
Genetic material / DNA / nucleoid;
Ribosomes;
Accept references to size only if some idea of range is given
max 2

(c)     Large (surface) area;
For diffusion;
     or
Short distance to centre of cell / to all haemoglobin;
For diffusion;
2

(d)     (i)      Correct answer of approximately 7800 / 8000 = 2 marks
Incorrect answer but clearly derived by
dividing diameter of cell A by 7                        = 1 mark
2
(ii)     Idea of cut through maximum diameter / middle;
1
[8]

M17.          (a)     1.      (Simple / facilitated) diffusion from high to low concentration / down concentration gradient;
Q Do not allow across / along / with concentration gradient
2.      Small / non-polar / lipid-soluble molecules pass via phospholipids / bilayer;
Reject: named molecule passing through membrane by an incorrect route
Accept: diagrams if annotated
OR
Large / polar / water-soluble molecules go through proteins;
3.      Water moves by osmosis / from high water potential to low water potential / from less to more negative water potential;

4.      Active transport is movement from low to high concentration / against concentration gradient;
Only penalise once if active transport is not named
e.g. ‘movement against the concentration gradient involves proteins and requires ATP’ = 2 marks
5.      Active transport / facilitated diffusion involves proteins / carriers;
Accept: facilitated diffusion involves channels
Reject: active transport involves channels
6.      Active transport requires energy / ATP;
7.      Ref. to Na^+ / glucose co-transport;
Credit ref. to endo / exocytosis as an alternative
5 max

(b)     1.      Many alveoli / alveoli walls folded provide a large surface area;
Neutral: alveoli provide a large surface area
2.      Many capillaries provide a large surface area;
3.      (So) fast diffusion;
Neutral: greater / better diffusion
Neutral: fast gas exchange
Allow ‘fast diffusion’ only once
4.      Alveoli or capillary walls / epithelium / lining are thin / short distance between alveoli and blood;
Reject: thin membranes / cell walls
Accept: one cell thick for ‘thin’

5.      Flattened / squamous epithelium;
Accept: endothelial
6.      (So) short diffusion distance / pathway;
7.      (So) fast diffusion;
8.      Ventilation / circulation;
Accept: descriptions for ventilation / circulation
9.      Maintains a diffusion / concentration gradient;
10.    (So) fast diffusion;
Do not double penalise if description lacks detail
e.g. thin membranes so a short diffusion distance = 1 mark
5 max
[10]

M18.(a)     Measure with eyepiece graticule / scale;
Calibrate with stage micrometer / scale on slide / object of known size; 
Repeats and calculate the mean;
OR
Use a ruler to estimate the field diameter under microscope;
How many droplets go across the field;
Repeats and calculate mean;
Accept references to radius
3

(b)     (i)      Two mark for correct answer of 4 : 1;;
One mark for incorrect answer but working shows that candidate has clearly attempted to compare values of r^2 / 6^2 and 3^2 / 36 and 9;
Idea of comparing ratios
A ratio of 1 : 4 should gain 1 mark
2
(ii)     Small droplets have a larger surface area to volume ratio;
More surface for lipase (to act), leading to faster digestion of triglycerides;
Fatty acids are produced more quickly so pH will drop more quickly in curve Y / with bile salts / less fatty acids in curve Z / without bile salts so pH drop more slowly;
3
[8]

M19.(a)     1.      Water potential becomes lower / becomes more negative (as sugar enters phloem);
2.      Water enters phloem by osmosis;
3.      Increased volume (of water) causes increased pressure.
3

(b)     1.      Rate of photosynthesis related to rate of sucrose production;
2.      Rate of translocation higher when sucrose concentration is higher.
2

(c)     1.      Rate of translocation does not fall to zero / translocation still occurs after 120 minutes;
2.      But sucrose no longer able to enter cytoplasm of phloem cells.
2
[7]

M20.          (a)     contraction of (diaphragm) muscles flattens diaphragm;
contraction of intercostal muscles raises ribcage;
increase in volume decreases pressure;
3

(b)     (i)      tidal volume increases steeply, then increase slows down after 10 to 15 km h^-1;
1
(ii)     breathing rate increases slowly then steeply after 10 to 15 km h^-1;
(max 1 if no reference to speed where change occurs in either (i) or (ii))
1

(c)     20 × 2.75 = 55 dm^2;
(award 1 mark for correct method i.e. tidal volume × rate);
2
[7]

M21.          (a)     up and out;
1
(b)     (i)      does not require work / effort / involve muscle contraction / energy 
expenditure;
1
(ii)     active as it involves contraction of muscles;
1

(c)     liver moves back;
increases volume of lungs; 
pressure lower (in lungs than outside);
3 max

(d)     maximum of three marks for description, points 1 to 4
1       inhaled air contains more oxygen than exhaled air;
2       inhaled air contains less carbon dioxide than exhaled air;
3       inhaled air contains less water (vapour);
4       relative amount / percentage of nitrogen also changes;
5       respiration results in lower blood oxygen / higher blood carbon dioxide;
6       oxygen enters blood / carbon dioxide leaves blood in alveoli;
7       by diffusion;
8       water vapour diffuses from moist surface;
6 max
[12]

M22.(a)     1.      Haemoglobin carries oxygen / has a high affinity for oxygen / oxyhaemoglobin;
2.      Loading / uptake / association in lungs;
3.      at high p.O_2;
4.      Unloads / dissociates / releases to respiring cells / tissues;
5.      at low p.O_2;
6.      Unloading linked to higher carbon dioxide (concentration);
6. Ignore reference to incorrect pH in relation to effect of higher carbon dioxide concentrations for marking point
6

(b)     1.      Allows comparison;
Do not credit 'temperature affects results' on its own;
2.      (Different temperature) affects enzymes;
2. Allow reference to denaturation of enzymes.
3.      (Different temperature) affects respiration / metabolism;
4.      (Different temperature) affects amount of dissolved oxygen;
2 max

(c)     1.      Increases then levels out / stops increasing / fluctuates slightly;
2.      At 5 (cm^3 dm^-3) / 320 (cm^3 g^-1h^-1);
Allow description of 'fluctuates slightly' in terms of candidate quoting figures after 320.
2

(d)     1.      Chronimus longistylus has higher uptake at low (oxygen) concentrations;
Chronimus longistylus has higher uptake to (oxygen concentration of) 2 / lower uptake after 2; (= 2 marks)
2.      (Higher uptake) up to 2 cm^3 dm^-3;
2. Award mark if candidate uses figures from table e.g. higher at concentration 1 (220) or concentration 2 (285).
Higher uptake at concentration 1 or 2 = 2 marks.
2

(e)     (i)      More (than in African) lost via gills in Australian lungfish / less (than African) lost via lungs in Australian lungfish;
1
(ii)     1.      More / most exchange is via lungs (in African lungfish);
1. Allow converse for first point.
2.      Gills will not function / function less efficiently (in air);
2. Allow water is required for gills to function.
2
[15]

M23.(a)     1.      Protein synthesis and cell wall synthesis and cell expansion 
         stop at −0.7 / at a higher water potential than other two;
If all 3 are correctly identified in marking point 1, accept ‘the others / the other two’ in marking point 2, and vice versa
2.      Photosynthesis and stomatal opening stop at -1.5 / at a lower water potential than other three;
Correct processes must be named in at least one of marking point 1 or marking point 2
Where reference to water potential differences are made, they must be comparative, eg ‘higher’
2

(b)     1.      Stomata allow uptake of carbon dioxide;
2.      Carbon dioxide used in / required for photosynthesis;
2

(c)     1.      Growth involves cell division / cell expansion / increase in mass;
Marking point 1 is for the principle
2.      Protein synthesis stops so no enzymes / no membrane proteins / no named protein (for growth / division);
Marking points 2, 3 and 4 require appreciation of ‘why’ before credit can be awarded
‘named’ protein must relate to proteins involved in growth or cell division
3.      Cell wall synthesis stops so no new cells can be made;
Full credit is possible without a statement of the principle 
(marking point 1)
4.      No cell expansion / increase in mass because (cells) stop taking up water;
3 max
[7]

M24.(a)     1.      Contraction of internal intercostal muscles;
2.      Relaxation of diaphragm muscles / of external intercostal muscles;
3.      Causes decrease in volume of chest / thoracic cavity;
4.      Air pushed down pressure gradient.
4

(b)     19(%);
1

(c)     1.      Muscle walls of bronchi / bronchioles contract;
2.      Walls of bronchi / bronchioles secrete more mucus;
3.      Diameter of airways reduced;
4.      (Therefore) flow of air reduced.
4
[9]

M25.(a)     Stomata per mm^2 or cm^2
OR
Number per mm^2 or cm^2;
Accept: mm^−2 or cm^−2.
Reject: per μm^2 or μm^−2.
Reject: the use of a solidus / as being equivalent to per.
Ignore: ‘amount’.
1

(b)     1.      Single/few layer(s) of cells;
Accept: more/too many/overlapping.
‛Single layer’ without reference to cells/tissue should not be credited.
2.      So light can pass through;
2

(c)     1.      Distribution may not be uniform
OR
So it is a representative sample;
Accept: more/fewer stomata in different areas.
Ignore: anomalies/random/bias.
2.      To obtain a (reliable) mean;
Accept: ‘average’.
2

(d)     1.      Hairs so ‘trap’ water vapour and water potential gradient decreased;
2.      Stomata in pits/grooves so ‘trap’ water vapour and water potential gradient decreased;
3.      Thick (cuticle/waxy) layer so increases diffusion distance;
4.      Waxy layer/cuticle so reduces evaporation/transpiration.
5.      Rolled/folded/curled leaves so ‘trap’ water vapour and water potential gradient decreased;
6.      Spines/needles so reduces surface area to volume ratio;
1, 2 and 5. Accept: humid/moist air as ‛water vapour’ but not water/moisture on its own.
1, 2 and 5. Accept: diffusion gradient as equivalent to water potential gradient.
1, 2 and 5. Accept: less exposed to air as an alternative to water potential gradient.
6. Accept: spines/needles so ‘reduce area’.
2 max

(e)     1.      Water used for support/turgidity;
2.      Water used in photosynthesis;
3.      Water used in hydrolysis;
4.      Water produced during respiration;
2 max
[9]

M26.          (a)     Digestion / hydrolysis / breakdown of a disaccharide into monosaccharides;
OR
(glucose and galactose form lactose) glucose is a monosaccharide;
max 1

(b)     (i)      Dipeptidase / disaccharidase / named disaccharidase;
1
(ii)     Enzymes not lost (with gut contents) / more effective absorption
of products formed by these enzymes;
1

(c)     No ATP formed / no energy released by respiration;
[reject “making” energy]
Link ATP to active transport (of galactose) into cells;
2
[5]

M27.(a)     1.      In phospholipid, one fatty acid replaced by a phosphate;
Ignore references to saturated and unsaturated
Accept 
Reject P/Phosphorus
Accept annotated diagrams
1

(b)     1.      Add ethanol, then add water;
Reject ethanal/ethonal
Accept ‘Alcohol/named alcohol’
2.      White (emulsion shows lipid);
Accept milky – Ignore ‘cloudy’
Sequence must be correct
If heated then DQ point 1
Reject precipitate
2

(c)     Saturated single/no double bonds (between carbons)
OR
Unsaturated has (at least one) double bond (between carbons);
Accept hydrocarbon chain/R group for ‘between carbons’ for either
Accept Sat = max number of H atoms bound
‘It’ refers to saturated
1

(d)     1.      (Fat substitute) is a different/wrong shape/not complementary;
OR
Bond between glycerol/fatty acid and propylene glycol different 
(to that between glycerol and fatty acid)/no ester bond;
2.      Unable to fit/bind to (active site of) lipase/no ES complex formed;
If wrong bond name given (e.g. peptide/glycosidic), then penalise once
2

(e)     It is hydrophilic/is polar/is too large/is too big;
Ignore ‘Is not lipid soluble’
1
[7]

M28.          (a)     1.      rate of respiration increases (in muscle cells);
2.      carbon dioxide concentration increases / pH falls / H^+ increases / acidity increases;
3.      chemoreceptors in aortic / carotid bodies / medulla (accept reference to aorta / carotid arteries not sinus);
4.      (impulses to) medulla / cardioaccelerator centre;
5.      increased frequency of impulses (award only once);
6.      along sympathetic pathway to sinoatrial node / SAN (not pacemaker);
6

(b)     (i)      through cardiac muscle;
to atrioventricular node;
along bundle of His / Purkyne fibres;
2 max
(ii)     sinoatrial node in the (right) atrium;
trace from healthy person is identical to the trace for the diseased heart in the region of the atria / only differences seen in trace for ventricles;
2
[10]

M29.          (a)     1. water evaporates / transpires from leaves;
2. reduces water potential in cell / water potential / osmotic gradient across
    cells (ignore reference to air space);
3. water is drawn out of xylem;
4. creates tension (accept negative pressure, not reduced pressure);
5. cohesive forces between water molecules;
6. water pulled up as a column;
4 max

(b)     (i)      same surface area of leaf / number of leaves / age / thickness of
cuticle;
1
(ii)     (environmental conditions) affect rate of transpiration / evaporation;
1
(iii)     presence of grease reduces water loss;
1

(c)     (i)      1.2 / 1.3g;
1
(ii)     more stomata on the lower surface;
(thicker) waxy cuticle on the upper surface;
2
[10]

M30.          (a)     (i)      unrestricted / free / quick / easy water flow / continuous column / maintains transpiration stream;
1
(ii)     resists tension in water (column) / provides support / strength / 
maintains column of water / adhesion / prevents water loss
(allow waterproofing in correct context i.e. not absorbing);
1

(b)     (i)      as total area of stomata decreases the rate of water flow decreases / 
decrease is proportional;
(reject proportional, ‘as one goes up the other goes up’ and ‘same shape’)
1

(ii)     increasing / higher temperature causes increasing / higher rate of evaporation / transpiration;
(not water loss)
1
(iii)     lower plateau (start and finish at same point);
(allow if curve sketched on original graph, reject ‘curve is lower’)
1

(c)     conserves water / reduces water loss / reduces transpiration / reduces evaporation;
high humidity (in pit) / reduced water potential gradient / less water blown away  /  increased diffusion pathway;
2
[7]

M31.          (a)     It is a measure of the concentration of a gas
(in a mixture of gases or a liquid);
1
(b)     37-38%
Accept 36 – 39

(c)     muscle contraction causes increased respiration;
increased CO_2 production lowering blood pH / lactate released
lowering blood pH;
increased heat released therefore increased temperature;
increased O_2 consumption lowering tissue PO_2;
4

(d)     haemoglobin has a lower affinity for oxygen;
more O_2 for respiration;
2
(e)     3.4 times = 2 marks
(incorrect answer in which candidate shows amount of oxygen removed at rest is 4.6 and amount removed during exercise is 15.8 = 1 mark)
2

(f)      Nearly all O_2 is transported by haemoglobin / v. little transported in plasma;
EITHER
Haemoglobin is (nearly) fully saturated with O_2 at the alveoli both at
rest and when exercising;
Therefore no (very little) further increase is possible;
OR
Haemoglobin is only 95% saturated with oxygen at the alveoli;
Therefore enriching inspired / air with oxygen will raise this to 100%;
3

(g)     increased depth / rate / pulmonary ventilation;
increase stroke volume / heart rate / Q increases blood flow rate;
arterioles [Accept artery] supplying the muscles
dilate / vasodilation / greater proportion of blood flow to the muscles;
max 3
[15]

M32.          (a)     0.1 / 0.9 (s);
1
(b)     Two marks for correct answer of 75 (beats per minute);
One mark for incorrect answer based on cardiac cycle taking 0.8 seconds;
2

(c)     (i)      Pressure in ventricle higher than pressure in atria;
1
(ii)     Prevents backflow of blood / prevents flow from ventricles to atria;
1

(d)     Increase (in stroke volume) as blood pressure increases, remains constant / plateaus; after 3 kPa / when stroke volume = 82cm^3
2
(e)     Two marks for correct answer of 80;
One mark for incorrect answer recognising that ventricle contracts once every cardiac cycle / stroke volume = 70 cm^3
2

(f)      1       Muscles (surrounding veins) contract and press on (walls of) vein and squeezes blood along veins;
2       Valves prevent backflow / ensure flow in one direction;
3       Systole / contraction of heart pumps blood (through arteries) into veins / residual arterial pressure / negative pressure in chest due to inspiration;
4       Recoil of heart muscle during diastole / after contraction;
5       Draws blood from veins into atria;
Accept sucks
6       Wide lumen little resistance / friction
6
[15]

M33.          (a)     More that one polypeptide / chain;
Ignore references to haem / other groups
1
(b)     (i)      141;
1

(ii)     1.      Stop / start sequences;
2.      Non coding DNA (in the gene) / introns / multiple repeats / junk DNA;
Do not credit “some bases repeated”
3      Two chains / a non-coding strand / complementary base pairs;
4.      Addition of base by mutation;
2 max

(c)     Different primary structure / amino acids / different number of polypeptide chains;
Question is about haemoglobin so do not credit differences in DNA
1

(d)     1.      Low partial pressure of oxygen in lungs;
2.      (Llama) haemoglobin able to load more oxygen / (llama)
haemoglobin saturated (at low / particular partial pressure of oxygen);
3.      Higher affinity for oxygen;
The terms used in the graph (or near approximations) should be used in this answer.
Ignore references to unloading
The answer must relate to llamas
3
[8]

M34.         (a)     (i)      1.      Stomata open;
Allow converse
2.      Transpiration highest around mid-day as middle of day warmer / lighter;
2. Allow ‘Sun is at its hottest’
3.      (Increased) tension / water potential gradient;
Ignore ‘pull, suck’
3

(ii)     (Inside xylem) lower than atmospheric pressure / (water is under) tension;
Accept cohesion tension. Ignore vacuum
1

(b)     (i)     High pressure / smoothes out blood flow / artery wall contains more collagen / muscle / elastic (fibres) / connective tissue;
Accept converse for pulmonary vein
Incorrect function of artery disqualifies mark
1

(ii)     1.      (Aorta wall) stretches because ventricle / heart contracts / systole / pressure increases;
1. Allow expand
2.      (Aorta wall) recoils because ventricle relaxes / heart relaxes / diastole / pressure falls;
2. Allow spring back
Reject any reference to contract / relax in MP1 and 2
3.      Maintain smooth flow / pressure;
3

(iii)    Aorta 1.2 / largest SD;
Allow pulmonary vein provided candidate relates standard deviation to mean
1

(c)     Formation
1.      High blood / hydrostatic pressure / pressure filtration;
2.      Forces water / fluid out;
2. Reject plasma, ignore tissue
3.      Large proteins remain in capillary;
Return
4.      Low water potential in capillary / blood;
5.      Due to (plasma) proteins;
6.      Water enters capillary / blood;
7.      (By) osmosis;
7. Osmosis must be in correct context 
8.      Correct reference to lymph;
6 max
[15]

M35.(a)     1.       Time taken to reach maximum blood flow varied widely/significantly;
Must be emphasis on idea of ‘widely’. Mention only of ‘vary’ is insufficient. Ignore use of numbers unless a comparison is given
Ignore any mention of a correlation between maximum percentage increase in blood flow and time taken to reach maximum increase in blood flow
2.       Quickest after a carbohydrate-only meal;
OR
Slowest after a protein-only meal;
2
(b)     1.      More blood flows to (skeletal) muscles (during exercise);
2.      (supplying) more oxygen / glucose / removing more carbon dioxide/ lactic acid/ heat;
1 and 2. Idea of ‘more’ is needed
More blood to muscles delivering oxygen = 2 marks
3.      For high (rate of) respiration / to meet increased demand for energy/ATP;
OR
Prevents anaerobic respiration/lactic acid build up;
Accept: reduces/delays for prevent
3

(c)     Immediate effect of exercise after meal
1.      Meal increases blood flow in (mesenteric) artery AND exercise decreases blood flow in (mesenteric) artery;
1. Will relate to information given in the tables
Overall effect on blood circulation
2.      Insufficient blood (flow to small intestines / muscles);
2. Accept: blood diverted away/shunted
Ignore references to ‘strain on heart’, ‘heart disease’, ‘cardiovascular diseases'
Ignore references to controlling variables and reliability
Effect on blood flow of type of meal
3.      Carbohydrate meal quick(er) / during exercise;
OR
Protein/fat meal slow(er) / after exercise;
Effect of reduced blood flow on cells
4.      (More) anaerobic (respiration) / lactic acid produced;
OR
less aerobic respiration;
Consequence for person of changed blood flow
5.      Less absorption (of digested food) / faeces contains digested food;
6.      Cramp / indigestion / discomfort / fatigue;
Look for ideas in each of 5 areas
MP1 might be spread throughout the answer
6. Ignore references to digestion
Max 4

(c)     1.      (blood flows from kidney along) renal vein to vena cava;
2.      (along) vena cava to right atrium/side of heart;
3.      (along) pulmonary artery to lungs;
4.      (along) capillaries to pulmonary vein;
5.      (along) pulmonary vein to left atrium/side of heart;
6.      (along) aorta to renal artery (to kidney);
7.      Blood may pass through several complete circuits before returning to kidney;
Reject: ‘blood vessel pumps’ only once
Ignore references to valves
Ignore references to heart action/cardiac cycle
Accept labelled diagram must include directional arrows
Max 6
[15]

M1.          (a)     10
(reject: 9.76)
1
(b)     isolation (on islands);
variety of habitats / conditions different from origin / other islands;
differing pathways of natural selection;
leading to organisms too different to interbreed.
3 max
[4]

M2.          (a)     angle, moisture and pH
(all required)
1
(b)     system for subdividing quadrat into, e.g. many squares;
method of estimating cover in small squares, e.g. counting those where cover over 50%, or cover at points (of intersection);
(not just ‘count squares with vegetation’ unless very small)
2

(c)     increasing vegetation cover is related to increasing moisture content
(allow ‘affects‘ moisture content or vice versa, not ‘causes); 
correlation is significant / not due to chance / can reject null hypothesis
/ only 1 in 20 / 5% probability that the correlation is due to chance;
2

(d)     factor; and linked effect e.g.
wind-blown particles trapped;
accumulation of soil;
OR
accumulation of organic / dead / decomposed matter / humus;
increase in mineral ions / improved water retention / improved
soil structure;
OR
nitrogen fixation;
increased nitrate concentration / improved soil fertility;
2 max
[7]

M3.          (a)     sections of chromatids exchanged;
sections have different alleles;
new combinations of (linked) alleles;
(allow 1 mark for idea that ‘genes’ are exchanged,
if no other marks gained)
3

(b)     (i)      length controlled by many genes / polygenes;
each gene may have different alleles / idea of additive effects;
OR
environmental factors / or named factor;
how named factor may affect growth of seeds;
2 max

(ii)     1.      selection of large seeds for sowing;
2.      higher proportion of alleles for long length / loss of alleles for short seeds from population;
3.      (possible appearance of) new alleles through mutation;
4.      process repeated over many generations;
(G - allow 1 mark idea for that ‘largeness’ selected, survives and inherited)
4
[9]
QWC 1

M4.          (a)     (i)      number of bases = 4440
allow 4446 if they refer to start / stop
each amino acid coded for by triplet / three bases
(so three times more bases than amino acids);
2

(ii)     deletion;
(deletion) of three bases;
because substitution / addition would change amino acid(s);
2 max

(b)     codon on mRNA;
specific / complementary base pairing with;
anti-codon on tRNA;
specific tRNA for each amino acid;
protein formed by condensation reactions / 
peptide bonds formed;
4 max
[8]

M5.          (a)     large groups are divided into smaller groups;
(not just ‘hierarchical’)
members of a group have features in common based on anatomy
/ fossils / embryology / DNA / specific aspect of cell biology
/ homologous structures; 
reflects evolutionary history;
3

(b)     fungi and animals;
1
(c)     (insects and fungi) have common ancestor;
they diverged a long time ago / before others referred to in phylogenetic tree;
2

(d)     those with similar sequences put in same groups / are more closely related;
the greater difference in amino acid sequence the longer ago the groups
diverged;
2

(e)     A - present in all (eukaryotic) species or organisms / quantifiable;
D - extinct species not considered / no timing of events available / only limited number of amino acid sequences / can’t include prokaryotic species
2
[10]

M6.          (a)     breed together;
if fertile offspring, then same species;
2
(b)     isolation of two populations;
variation already present due to mutations;
different environmental conditions / selection pressures leading to
selection of different features and hence different alleles;
different frequency of alleles;
separate gene pools / no interbreeding;
4

(c)     selection of mate dependent on colour pattern;
prevents interbreeding / keeps gene pools separate;
2
[8]

M7.          (a)     AGC; TTC;
2

(b)     anticodon complementary to codon / reads message on mRNA;
specific amino acid;
carried / transferred (to ribosome);
correct sequence of amino acids along polypeptide;
3 max

(c)     
(Met) | Phe | Gln | Gln | Lys | Gln | Phe

2
(three / four / five correct 1 mark; six correct 2 marks)
[7]

M8.          (a)     side effects / allergic reactions / low toxicity to cells;
interaction with other drugs / effective in conditions of use / reasonably stable;
should only act on the problem bacteria / narrow spectrum;
how much resistance the bacteria have built up;
2 max

(b)     (i)      tetracycline
prevents tRNA binding to ribosomes / amino acid / mRNA;
1
amino acids not available / brought / picked up;
1
chloramphenicol
prevents amino acids being joined / prevents primary structure forming;
1
no enzymes / no structural proteins formed;
(accept cell wall formation if qualified) (prevents protein synthesis gains one mark in either section, once only)
1

(ii)     only prevents tRNA binding to 70S / prokaryotic / bacterial
ribosomes / human ribosomes are different sizes / shapes / structure;
1
[7]

M9.          (a)     (i)      join / attach nucleotides, to form a strand / along backbone / phosphodiester bonds;
(reject reference to H bonds, complementary base pairing)
1
(ii)     ribosome / RER;
1

(b)     (i)      CGTTACCAA;
1
(ii)     CGU UAC CAA;
1
(c)     substitution;
1

(d)     (i)      alanine;
1
(ii)     (mutation 1)
no change(to sequence of amino acids);
codon for alanine / degenerate codon / same amino acid coded for;
2
(mutation 2)
(change in sequence) valine replaced by alanine / codon for alanine;
folding / shape / tertiary structure / position of bonds may change;
(reject peptide bonds)
2
[10]

M10.         (a)     1.      Occurs in an unchanging environment;
1
+
2.      Selection against extremes / selection for the mean / mean / median / mode unaltered
3.      Range / S.D is reduced
4.      Increasing proportion of populations becomes well adapted to environment;
4

(b)     1.      All plants are acyanogenic below –4 °C and (most) cyanogenic above +10 °C;
2.      Cyanogenic plants’ cells freeze below –4°;
3.      Releasing cyanide (into their own tissues) / damaging / killing plants / disrupting metabolism;
4.      Selective advantage not to produce cyanide at –4 °C;
5.      Slugs present at higher temperatures / not usually present / inactive at lower temperatures and cyanide production kills / deters slugs;
5
[10]

M11.          (a)     (So results) can be compared / so measurement is the same each time / because eye is not perfectly round / uniform;
Accept eye opens to different amounts
1

(b)     (i)      1.      Eye (diameter) is smaller and antennae longer;
2.      Antennae detecting touch;
3.      Data only refers to shrimps / data may not apply to all animals / only in one area;
The principle here is that candidate has recognised that both features confirm suggestion. Exact wording does not matter.
2 max

(ii)     1.      Standard deviation gives a measure of spread / variation;
2.      More standard deviations overlap, the less likely it is that differences are real / significant / the more likely they are caused by chance;
Do not accept range
Accept converse.
Although we are looking for the idea of significance, we cannot require this term.
2

(c)     (i)      Qualitative statement about
          difference in size / 
          difference in variation / 
          overlap in size;
Quantitative statement about
          difference in size / 
          difference in variation / 
          overlap in size;
Supported by relevant two sets of figures from graph;;
Note simplistic answer involving a quantitative statement gains 1 mark.
More specific answer involving quantitative information gains 2 marks.
2

(ii)     (No) for same body length, antenna are longer / antenna are shorter / some with longer body have short antennae / some with shorter body length have longer antennae;
OR
(Yes) positive correlation in open / in cave;
Habitat not critical as a term.
Must refer to idea of same habitat
Accept description
1

(d)     More alleles of each gene / shrimps in open have all the alleles;
Candidates are required to use the information from the table. Must therefore refer to alleles.
1

(e)     1.      A small number of shrimps were / went into the cave;
2.      All / high proportion of shrimps had allele L;
3.      Cave population descended from these / these reproduce;
3

(f)      (i)      1.      Cross shrimps from two sites / watch courtship;
2.      Breed young together / observe mating;
3.      Allow 1 mark for any method of improving quality of results e.g. carry out reciprocal crosses / large number of crosses / isolate beforehand;
Other valid equivalent suggestions should be accepted.

(ii)     If same species the shrimps would breed, producing fertile young / courtship species specific;
Accept any form of evidence – mating / laying eggs / giving birth to young.
3
[15]

M12.(a)     Banding pattern changes as cheetah gets older / difficult to judge as tail is short / fluffy;
1

(b)     (i)      Mean not (always) a whole number;
Standard deviation not (always) zero;
2

(ii)     Movement of tail / angle of sight / confused it with another band / subjective estimation;
Accept reference to Figure 1
E.g. Bands 2 and 3 have same thickness but look different
1

(c)     Band width not the same on both sides of tail;
1

(d)     Offspring of the same family will be more similar genetically;
As have same mother (and father) / parent;
Expect to see more differences in randomly chosen cheetahs;
3
[8]

M13.(a)     1.      No interbreeding / gene pools are separate / geographic(al) isolation;
Accept: all marks if answer written in context of producing increased diversity of plants
1 Do not award this mark in context of new species being formed and then not interbreeding
1 Accept reproductive isolation as an alternative to no interbreeding
2.      Mutation;
2 Accept: genetic variation
3.      Different selection pressures / different foods / niches / habitats;
3 Accept: different environment / biotic / abiotic conditions or named condition
3 Neutral: different climates
4.      Adapted organisms survive and breed / differential reproductive success;
5.      Change / increase in allele frequency / frequencies;
5

(b)     Similar / same environmental / abiotic / biotic factors / similar / same selection pressures / no isolation / gene flow can occur (within a species);
Accept: same environment
1
[6]

M14.(a)     Succession;
Ignore any word in front of succession e.g. secondary / ecological succession.
Neutral ‘forestation’.
1

(b)     1.      Greater variety / diversity of plants / insects / more plant / insect species;
Neutral: more plants.
2.      More food sources / more varieties of food;
Neutral: more food / more / greater food source (singular).
3.      Greater variety / more habitats / niches;
Accept: more nesting sites.
Q Neutral: more homes / shelters.
3

(c)     (i)      Temperature and carbon dioxide;
Neutral: water, chlorophyll.
1

(ii)     Shows (gross) photosynthesis / productivity minus respiration / more carbon dioxide used in photosynthesis than produced in respiration;
Correct answers are often shown as: net productivity = (gross) photosynthesis – (minus) respiration.
1

(iii)    1.      (Shade plant) has lower (rate of) respiration / respiratory losses / less CO2 released at 0 light intensity / in dark;
Accept use of figures.
Accept: lower compensation point.
2.      Greater (net) productivity / less sugars / glucose used / more sugars / glucose available; 
Neutral: any references to rate of photosynthesis.
2
[8]

M15.(a)     1.      Number of (individuals of) each species;
Accept: ‘population’ for ‘number’
2.      Total number of individuals / number of species;
Accept: ‘species richness’
MP2 allows for other types of diversity index
2

(b)     (i)      (Shows) results are due to the herbicide / are not due to another factor / (to) compare the effect of using and not using the herbicide / shows the effect of adding the herbicide;
Neutral: allows a comparison
Neutral: ensures results are due to the independent variable
Reject: ‘insecticide’
Accept: ‘pesticide’
1

(ii)     1.      (More) weeds killed so more crops / plants survive / higher yield / less competition;
2.      High concentrations (of herbicide) harm / damage / kill / are toxic to crops / plants;
Accept: ‘pesticide’
Neutral: ‘insecticide’
Accept: use of figures (eg 400+)
2

(iii)    1.      Reduced plant diversity / fewer plant species / fewer varieties of plant;
Accept: ‘weed’ for ‘plant’
Neutral: fewer plants
Accept: only one crop species remains
2.      Fewer habitats / niches;
Q Neutral: fewer homes / shelters
3.      Fewer food sources / varieties of food;
Neutral: less food
3
[8]

M16.(a)     (i)      1.      Groups within groups;
Accept: idea of larger groups at the top or smaller groups at the bottom
2.      No overlap (between groups);
2

(ii)     3;
1

(iii)    Chordata;
Accept: if phonetically correct eg ‘Cordata’
1

(b)     (i)      1.      (To provide) genetic variation;
Genetic variation must be directly stated and not implied
2.      (Allows) different combinations of maternal and paternal chromosomes / alleles;
Accept: any allele of one gene can combine with any allele of another gene
2

(ii)     1.      (Zedonk has) 47 / odd / uneven number of chromosomes;
Accept: diploid number would be odd
Reject: if wrong number of chromosomes is given
2.      Chromosomes cannot pair / are not homologous / chromosome number cannot be halved / meiosis cannot occur / sex cells / haploid cells are not produced;
Accept: cannot have half a chromosome
Q Reject: meiosis cannot occur in sex cells
2
[8]

M17.(a)      1.      Change / mutation in base / nucleotide sequence (of DNA / gene);
Q.
Ignore: references to changing base-pairing
Accept: affect for change, if in correct context
Accept: changes triplets / codons
2.      Change in amino acid sequence / primary structure (of enzyme);
Accept: different amino acid(s) coded for
Q Reject: different amino acids produced / formed / made
3.      Change in hydrogen / ionic / disulfide bonds;
Accept: references to sulfur bonds
4.      Change in the tertiary structure / shape;
Neutral: alters 3D structure / 3D shape
5.      Change in active site;
6.      Substrate not complementary / cannot bind (to enzyme / active site) / no enzyme-substrate complexes form.
Accept: no E S complexes form
6

(b)     1.      Non-SR strain falls more / SR strain falls less / up to 10(μg / cm−3);
Must include 10 but only required once in either MP1 or MP2
Ignore: units or absence of
This must be a comparative statement
2.      Above 10(μg / cm^−3), SR strain levels out / off and non-SR strain continues to decrease;
3.      Greater difference between strains with increasing concentration of antibiotic.
This must be a comparative statement
2 max

(c)     1.      Division stopped (of both strains by scientist);
Reject: references to mitosis stopping
2.      SR strain still more resistant / fewer die / none die (at higher concentrations of antibiotic).
Accept: SR strain and non-SR strain would be similar if resistance is due to only stopping division
Need some comparison with non-SR
2

(d)     1.      Make a competitive / non-competitive inhibitor;
Mark in pairs
either MP1 and MP2 OR MP3 and MP4
2.      Competitive competes with / blocks active site / non-competitive inhibitor affects / changes active site;
Do not mix and match
OR
3.      (Make a drug) that inhibits / denatures / destroys enzyme / stringent response;
Accept: drug that ‘knocks out’ / destroys enzyme
4.      Give at the same time as / before an antibiotic.
2 max

(e)     (SR strain)
1.      Fewer free radicals (than non-SR);
Note: has to be comparative statement
2.      Produces more catalase (than non-SR);
Accept converse statements for non-SR.
3.      Catalase (might be) linked to production of fewer free radicals / breaking down / removing free radicals.
Accept: hydrolysis of radicals by catalase.
3
[15]

M18.(a)     1.      Draw grid over (map of) area;
2.      Select squares / coordinates at random.
2

(b)     1.      No emigration / immigration;
2.      No losses to predation;
3.      Marking does not affect survival;
4.      Birth rate and death rate equal;
5.      (In this case) all belong to one population.
2 max

(c)     1.      Only glows brightly with UV, so doesn’t make insects more visible;
2.      So doesn’t affect / increase predation;
OR
1.      Glows brightly with UV marking visible;
2.      So makes it easy to pick out labelled insects.
2

(d)     10 130.
Tolerance of ±1
 = 1 marks
2

(e)     1.      Scientists removed large numbers of insects (which were not returned) from same area / same population;
2.      Affecting ratio of marked to unmarked.
2
[10]

M19.(a)      4:
1

(b)     2.68(6).
If answer incorrect:
Σn(n-1) = 242 = 1 mark
N(N-1) = 650 = 1 mark
2

(c)     1.      Take more samples and find mean;
2.      Method for randomised samples described.
Allow larger area = 1 mark
2
[5]

M20.(a)     Bacteria killed;
Ignore: no growth or growth of bacteria prevented.
Accept: bacteria destroyed.
Accept: no living bacteria.
1

(b)     Clear zone would be too large
OR
Clear zones would overlap/merge
OR
Could kill all bacteria (on the plate);
Must convey idea of too large.
1

(c)     1.      (Same) size;
Accept: any measure of size e.g. thickness, area, diameter.
Ignore: ‘same shape’ as shape shown on the diagram.
2.      (Same) material/absorbency;
3.      In solution for same time period;
Ignore: reference to volume of disinfectant.
2 max

(d)     Any number between 2.5 to 3.2 = two marks;;
Allow one mark for an incorrect answer but shows method of calculating how many times more effective D is than B
e.g. 22 divided by 13/14
or 11 divided by 6.5/7
or 1.57/1.6/1.69/1.7.
2
[6]