(1)themathsandsciencetutor.co.uk/wordpress/wp-content...leaf area than did phloem-sap-feeding...

The diagram shows part of a DNA molecule.

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

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

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

Describe the function of each of these enzymes.

DNA helicase _______________________________________________________

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DNA polymerase _____________________________________________________

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Page 1 of 60Catalyst Tutors

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

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

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2. _________________________________________________________________

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(Total 6 marks)

Scientists measured the mean amino acid concentration in white wines made from grapes grownorganically and white wines made from grapes that were not grown organically.

(a) Which test could the scientists have used to identify that there are amino acids in whitewine?

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(b) All amino acids have the same general structure. The image below shows the structure ofthe amino acid isoleucine.

Draw a box around the part of the molecule that would be the same in all amino acids.

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(c) Name the chemical element found in all amino acids that is not found in triglycerides.

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(d) The scientists used a statistical test to determine whether there was a significant differencein the amino acid concentration in the two types of white wine. They obtained a value for Pof 0.04.

Name the statistical test the scientists used and give a reason for your answer.

Was the difference significant? Give a reason for your answer.

Name of statistical test ________________________________________________

Reason for choice ____________________________________________________

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Explanation of test result _______________________________________________

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(Total 6 marks)

(a) Glycogen and cellulose are both carbohydrates.Describe two differences between the structure of a cellulose molecule and a glycogenmolecule.

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(b) Starch is a carbohydrate often stored in plant cells.Describe and explain two features of starch that make it a good storage molecule.

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(c) Tick (✔) the box that identifies the test which would be used to show the presence ofstarch.

Acid hydrolysis test

Benedict’s test

Emulsion test

Iodine/potassium iodide test

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(d) The diagram shows a section through a plant tissue at a magnification of ×500.

Calculate the actual diameter of the starch grain between points A and B.

Answer = ____________________ μm(2)

(e) What type of microscope was used to obtain the image shown in the diagram above?

Give one piece of evidence to support your answer.

Type of microscope _______________________________________________

Evidence ________________________________________________________

(2)

(Total 9 marks)


(a) Most human cells contain two copies of each gene. However, there might be up to 15copies of the gene for amylase (AMY1). Scientists investigated the number of copies of theAMY1 gene in individual people in two populations. One population had a high-starch dietand the other population had a low-starch diet.

The graph below shows their results.

Describe what their results show.

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(b) Multiple copies of the AMY1 gene is an adaptation to a high-starch diet.

Use your knowledge of protein synthesis and enzyme action to explain the advantage ofthis adaptation.

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(c) Multiple copies of the AMY1 gene is an adaptation to a high-starch diet.

Suggest how this evolved through natural selection.

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(Total 9 marks)


Scientists investigated how the concentration of protein in blood plasma changes in peoplebetween the ages of 60 and 95.

The graph shows the scientists’ results. The bars show ±1 standard deviation.

(a) What is the difference between males and females in the fall in mean concentration ofprotein in blood plasma between 60 and 95 years?

Answer = ___________ g dm−3

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(b) Use the graph above to calculate the rate of change of the mean concentration of protein inthe blood plasma of males between the ages of 60 and 95.

Show your working.

Answer = ___________ g dm−3 year−1

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(c) What can you conclude from the graph above about the effect of ageing on the meanconcentration of protein in the blood plasma in males and females?

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(d) The scientists measured the absorption of each sample of blood plasma using acolorimeter. They used a calibration curve to find the concentration of protein in samples ofblood plasma.

Describe how the scientists could obtain data to produce a calibration curve and how theywould use the calibration curve to find the concentration of protein in a sample of bloodplasma.

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(e) Older people are more likely to suffer from infectious diseases.

Suggest how this may be linked to the decrease in the mean concentration of protein in theblood as people get older.

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(Total 9 marks)


(a) The genetic code is degenerate and non-overlapping.

Explain the meaning of:

Degenerate _________________________________________________________

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Non-overlapping _____________________________________________________

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The table shows a short section of a messenger RNA (mRNA) molecule and the section of apolypeptide for which it codes.

mRNA G G G G C U U C A C C G G C A A C G

Polypeptide glycine alanine serine proline alanine threonine

(b) Name the bases represented in the table by:

A _________________________________

C _________________________________

G _________________________________

U _________________________________

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(c) Use information in the table to give the sequence of bases in DNA that codes for serine.

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(Total 5 marks)


(a) Cells lining the ileum of mammals absorb the monosaccharide glucose by co-transport withsodium ions. Explain how.

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A student set up the experiment shown in the diagram below.

The material from which Visking tubing is made is partially permeable.

After 15 minutes, the student removed samples from the liquid in the beaker and from the liquidinside the Visking tubing. She carried out biochemical tests on these samples. She drew thetable below to record her results.

(b) Complete the table by placing a tick (✔) in each box that you expect to have shown apositive result.

Biochemical test Liquid from beakerLiquid inside

Visking tubing

Biuret reagent

Iodine in potassium iodide

Benedict’s solution

(3)

(c) Justify your answers to part (b).

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(Total 9 marks)


Water and inorganic ions have important biological functions within cells.

(a) Give two properties of water that are important in the cytoplasm of cells.For each property of water, explain its importance in the cytoplasm.

Property 1__________________________________________________________

Biological importance within cells________________________________________

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Property 2__________________________________________________________

Biological importance within cells________________________________________

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(b) Other than sodium, name one inorganic ion and give one example of its biologicalimportance in a cell.

Name of inorganic ion_________________________________________________

Biological importance__________________________________________________

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(c) Compare and contrast the processes by which water and inorganic ions enter cells.

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(Total 9 marks)


A biochemist isolated a protease from a bacterium. He investigated the effect of temperature onthe rate of hydrolysis of a protein by this protease. He measured the mass of protein hydrolysedin 5 minutes at each temperature.

The results are shown in the table below.

Temperature / °CMass of proteinhydrolysed / g

Rate of hydrolysis /

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5 0.48

10 1.11

15 1.23

20 1.05

30 0.78

45 0.12

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(a) Process the data in the table. Plot the processed data on the graph paper.

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(b) A student concluded from a graph of the data in the table that the bacterium lives at 15 °C.

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

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(c) Suggest two variables the biochemist controlled when investigating the effect oftemperature on the rate of breakdown of a protein by the protease.

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2. _________________________________________________________________

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(Total 9 marks)


Read the following passage.

Some insect species feed on the leaves of plants. These leaf-chewersbite off pieces of leaves. Other insect species feed on sap from phloemor xylem. These sap-feeders have sharp, piercing mouthparts that theyinsert directly into either xylem or phloem. Leaf-chewers and insectsthat feed on xylem sap are active feeders; this means they use theirjaw muscles to obtain their food. In contrast, insects that feed on phloemsap are passive feeders; this means they do not use their jaw musclesto take up sap from phloem.

Feeding on phloem sap presents two problems. Firstly, phloem sap hassa high sugar concentration. This could lead to a high pressure of liquidin the insect’s gut because of water entering the gut from the insect’sbody tissues. A phloem-sap-feeder polymerises some of these sugarsinto polysaccharides which are passed out of its anus as ‘honey dew’.The second< >problem is that phloem sap has a low concentration ofamino acids. Phloem-sap-feeding insects rely on bacteria in their guts toproduce amino acids. Each phloem-sap-feeding insect receives a few ofthese bacteria from its parent. This has resulted in a reduction in thegenetic diversity of the bacteria found within these insects.

A scientist investigated the effect of three different insects on the growthof a plant called the goldenrod. He found that leaf-chewing insects andxylem-sap-feeding insects caused a much greater reduction in totalleaf area than did phloem-sap-feeding insects.

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10

15

20

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Use the information from the passage and your own knowledge to answer the followingquestions.

(a) Phloem-sap-feeders are passive feeders (lines 6–7).Phloem-sap-feeders do not use their jaw muscles to take up sap from phloem.

Explain why they can take up sap without using their jaw muscles.

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(b) A phloem-sap-feeder polymerises some of these sugars into polysaccharides (line 12-13).Suggest the advantage of this.

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(c) Each phloem-sap-feeding insect receives a few of these bacteria from its parent.(lines 16–17).

Suggest how this has caused a reduction in genetic diversity of the bacteria.

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(d) A scientist found that leaf-chewers and xylem-sap-feeders had a greater effect on plantgrowth than phloem-sap-feeders (lines 20–22).

Other than environmental factors, give two features the scientist would have controlledin his experiment to ensure this conclusion was valid.

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2. _________________________________________________________________

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(e) The scientist used the reduction in total leaf area of the experimental plants as an indicatorof plant growth.

Outline a method by which you could find the area of a plant leaf.

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(Total 10 marks)

Many humans are unable to digest lactose. A scientist investigated the production of lactose-freemilk. He produced gel beads containing the enzyme lactase and placed the beads in a column.He poured milk (Milk A) into the column and collected the milk (Milk B) after it had movedthrough the column over the beads. This is shown in the diagram below.

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(a) Milk A contains no glucose. Milk B contains glucose. Explain why Milk B contains glucose.

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(b) The enzyme was trapped within the gel beads. Suggest one advantage of trapping theenzyme within the gel beads.

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The scientist varied the flow rate of the milk through the column. The effect of flow rate on theconcentration of glucose in Milk B is shown in the table below.

Flow rate of milk through the column /cm3 minute−1

Concentration of glucose in Milk B /arbitrary units

50 45

100 6

(c) Explain the difference in the results in the table.

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(d) The gel beads were all similar sizes. Use the formula below to calculate the volume of oneof the beads with a 3.0 mm diameter.

Volume of sphere = πr3

Volume = _____________________ mm3

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(e) Galactose has a similar structure to part of the lactose molecule.Explain how galactose inhibits lactase.

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(Total 6 marks)


A bacterium is shown in the diagram.

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(a) Calculate the magnification of the image.

Magnification = _________________

(1)

(b) Complete the table to show the features of a bacterium and a virus.

Put a tick (✔) in the box if the feature is shown.

Surface Bacterium Virus

Cell-surface membrane

Nucleus

Cytoplasm

Capsid

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(c) DNA and RNA can be found in bacteria.

Give two ways in which the nucleotides in DNA are different from the nucleotides in RNA.

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2. _________________________________________________________________

(2)

(Total 5 marks)

A student investigated the effect of lipase concentration on the hydrolysis of lipids.

He took a beaker containing a suspension of lipids. He placed a pH probe attached to a datalogger into the beaker. After 5 minutes, he added the lipase solution. The data logger recordedthe pH. The apparatus used is shown in the diagram below.

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(a) The student did not add a buffer to the lipase solution.

Explain why.

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(b) Give two variables the student would have controlled in this investigation.

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2. _________________________________________________________________

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(c) Give the suitable control for this investigation.

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The data logger recorded the pH. The graph below shows what happened after he added thelipase solution.

(d) Draw a tangent on the graph and use it to calculate the rate of change at 5 minutes.

Rate of change at 5 minutes = _________________ pH minute−1

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(e) Explain the results shown in the graph.

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(f) The student repeated the experiment with a higher concentration of lipase solution.Describe and explain the results you would expect him to get.

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(Total 11 marks)

(a) Name the monomers from which a maltose molecule is made.

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(b) Name the type of chemical bond that joins the two monomers to form maltose.

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A student wanted to produce a dilution series of a maltose solution so he could plot a calibrationcurve. He had a stock solution of maltose of concentration 0.6 mol dm−3 and distilled water. Hemade a series of dilutions from 0.1 to 0.6 mol dm−3.

(c) Complete the table below by giving all headings, units and the concentration of the maltosesolution produced.

Concentration ofmaltose solution

/ ____________

Volume of 0.6 mol dm−3

maltose solution / cm3 _______________

______ / _______

____________ 5 10

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The student performed the Benedict’s test on six maltose solutions ranging from 0.1 mol dm −3 to0.6 mol dm−3. He placed a sample of each solution in a colorimeter and recorded the lightabsorbance.

His results are shown in the graph below.

(d) Explain how you would use the graph to determine the maltose concentration with a lightabsorbance of 0.45 arbitrary units.

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(2)

(Total 6 marks)


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

The figure below shows his results.

Time after start of reaction / s

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

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

Answer ____________________

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

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(Extra space) _______________________________________________________

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(5)

(Total 8 marks)

Figure 1 shows one base pair of a DNA molecule.

Figure 1

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(a) Name part F of each nucleotide.

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(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

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

Figure 2 represents DNA replication.

Figure 2

(c) Name the enzyme shown in Figure 2.

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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 arrowspoint in opposite directions.

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(Total 8 marks)

(a) Describe how you would test a piece of food for the presence of lipid.

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The figure below shows a phospholipid.

X Y

(b) The part of the phospholipid labelled A is formed from a particular molecule. Name thismolecule.

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(c) Name the type of bond between A and fatty acid X.

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(d) Which of the fatty acids, X or Y, in the figure above is unsaturated? Explain your answer.

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Scientists investigated the percentages of different types of lipid in plasma membranes fromdifferent types of cell. The table shows some of their results.

Type of lipid Percentage of lipid in plasma membrane by mass

Cell lining ileum ofmammal

Red blood cell ofmammal

The bacteriumEscherichia coli

Cholesterol 17 23 0

Glycolipid 7 3 0

Phospholipid 54 60 70

Others 22 14 30

(e) The scientists expressed their results as Percentage of lipid in plasma membrane bymass. Explain how they would find these values.

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Cholesterol increases the stability of plasma membranes. Cholesterol does this by makingmembranes less flexible.

(f) Suggest one advantage of the different percentage of cholesterol in red blood cellscompared with cells lining the ileum.

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(g) E. coli has no cholesterol in its cell-surface membrane. Despite this, the cell maintains aconstant shape. Explain why.

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(2)

(Total 10 marks)

Read the following passage.

Herpes simplex virus (HSV) infects nerve cells in the face, including some near thelips. Like many other viruses, HSV can remain inactive inside the body for years.When HSV becomes active, it causes cold sores around the mouth.

Human cells infected with a virus may undergo programmed cell death. While HSVis inactive inside the body, only one of its genes is transcribed. This gene is thelatency-associated transcript (LAT) gene that prevents programmed cell death of aninfected nerve cell.

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Scientists have found that transcription of the LAT gene produces a microRNA.This microRNA binds to some of the nerve cell’s own mRNA molecules. ThesemRNA molecules are involved in programmed cell death of nerve cells. Thescientists concluded that production of this microRNA allows HSV to remain in thebody for years.

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

(a) HSV infects nerve cells in the face (line 1). Explain why it infects only nerve cells.

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(b) HSV can remain inactive inside the body for years (lines 2–3). Explain why this virus canbe described as inactive.

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(c) Suggest one advantage of programmed cell death (line 4).

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(d) The scientists concluded that production of this microRNA allows HSV to remain in thebody for years (lines 10–12).

Explain how this microRNA allows HSV to remain in the body for years.

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(4)

(Total 10 marks)


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

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(a) Name the process shown.

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(b) Identify the molecule labelled Q.

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

the complementary DNA base sequence __________________________________

the missing anticodon _________________________________________________

(2)

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

Amino acid Encoding base triplet

Aspartic acid GAC, GAU

Proline CCA, CCG, CCC, CCU


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

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

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(Extra space) _______________________________________________________

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(3)

(Total 8 marks)


Newborn babies can be fed with breast milk or with formula milk. Both types of milk containcarbohydrates, lipids and proteins.

• Human breast milk also contains a bile-activated lipase. This enzyme is thought to beinactive in milk but activated by bile in the small intestine of the newborn baby.

• Formula milk does not contain a bile-activated lipase.

Scientists investigated the benefits of breast milk compared with formula milk.

(a) The scientists used kittens (newborn cats) as model organisms in their laboratoryinvestigation.

Other than ethical reasons, suggest two reasons why they chose to use cats as modelorganisms.

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2. _________________________________________________________________

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(b) Before starting their experiments, the scientists confirmed that, like human breast milk,cat’s milk also contained bile-activated lipase.

To do this, they added bile to cat’s milk and monitored the pH of the mixture.

Explain why monitoring the pH of the mixture could show whether the cat’s milk containedlipase.

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The scientists then took 18 kittens. Each kitten had been breastfed by its mother for the previous48 hours.

The scientists divided the kittens randomly into three groups of six.

• The kittens in group 1 were fed formula milk.• The kittens in group 2 were fed formula milk plus a supplement containing bile-activated

lipase.• The kittens in group 3 were fed breast milk taken from their mothers.

Each kitten was fed 2 cm3 of milk each hour for 5 days.

The scientists weighed the kittens at the start of the investigation and on each day for 5 days.

The figure below shows the scientists’ results.

Type of milk given to kittens


(c) What can you conclude from the figure about the importance of bile-activated lipase inbreast milk?

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(3)

(Total 7 marks)

Apple farmers want to harvest their fruit when it is ripe enough for eating but also when it can bestored to sell later.

One method apple farmers use to decide when to harvest their fruit is to determine the starchcontent. As apples ripen, starch in the apple is converted into soluble sugars that make themtaste sweet.

Scientists investigated the best time to harvest apples for storage before being sold.

To determine the starch content, they picked samples of apples. They cut each apple in half andcovered the cut surface with iodine solution. They left it for 1 minute and then compared it withthe diagram below to give it a starch index score between 1 and 10.

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They collected samples of apples at 5 different days during the ripening period and tested themfor starch content. These results are shown in the table below.

When apples were collected / dayduring ripening period

Mean starch index

117 3.7

124 4.4

131 6.3

138 7.7

145 8.2

The scientists stored samples of apples from each collection day for 180 days. They thendetermined the percentage of apples that were rotten. These results are shown in the graphbelow.

(a) The cut surface of the apple covered with iodine solution is left for 1 minute before beingcompared to Figure 1.Explain why each apple must be left for the same length of time.

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(b) Describe and explain the change in appearance of the cut surface of the apple whentreated with iodine solution from underripe (starch index 1) to overripe (starch index 10).

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(3)

(Total 5 marks)

Glucose is a monosaccharide. Two glucose molecules join together to form a disaccharide.

(i) Name the products of this reaction.

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(ii) Name the type of reaction that joins the glucose molecules together.

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(Total 3 marks)


A student investigated the glucose concentration in five different drinks.His results are shown below.

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(a) Using the data, calculate how many grams of glucose would be in 220 cm3 of drink F.

Answer = ____________________________ g

(1)

(b) Calculate how much more glucose is in drink C than in drink F. Show your answer as apercentage.

Answer = ____________________________ %

(1)

(Total 2 marks)


(a) Most blood glucose comes from starch and disaccharides in the diet.Describe a test you could use to check if food in the diet contained starch.

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(b) Explain how digestion of starch in the gut (small intestine) leads to an increase in theconcentration of glucose in the blood. Details of co-transport are not required.

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(c) Suggest a method you could use to estimate the concentration of glucose in severaldifferent solutions that all turned brick red with Benedict’s reagent in 3 minutes.

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(Total 6 marks)

(a) Describe the difference between the structure of a triglyceride molecule and the structure ofa phospholipid molecule.

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

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(c) Animal fats contain triglycerides with a high proportion of saturated fatty acids. If peoplehave too much fat in their diet, absorption of the products of fat digestion can increase therisk of obesity. To help people lose weight, fat substitutes can be used to replacetriglycerides in food.

Describe how a saturated fatty acid is different from an unsaturated fatty acid.

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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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(2)


(e) This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surfacemembranes of cells lining the gut.

Suggest why it cannot cross cell-surface membranes.

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(1)

(Total 7 marks)


Mark schemes

(a) 8;

Accept eight1

1

(b) Phosphodiester (bond);

Accept phonetic spellings1

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

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

1. Accept H bonds.

1. Accept hydrolyses for breaks

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

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

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

be different / varies;

Both parts of each MP needed

3. Reject Uracil / U

3. Accept C, T or G for different bases

Accept annotated diagram for any of the three marks2 max

[6]

(a) Biuret;

Ignore any other detail

Accept

• Copper sulfate and sodium hydroxide• CuSO4 + NaOH• Alkaline copper sulfate• Copper sulphate and sodium hydroxide• Alkaline copper sulphate• Biurette• Buiret• Biruet• BieuretReject burette or Beirut

1

2


(b) Draw around

1

(c) Nitrogen;

Ignore N1

(d) Choice: (Student’s) t-test;Reason for choice: Looking for differences between two means;

Reason: Allow comparing contrasting two means

Explanation: Difference is significant / not due to chance because the P value is 0.04/ is less than 0.05;

Explanation: Assume ‘it’ means difference

Explanation: Reject result / data is significant / not due to chance

Explanation: do not accept P value is less than 0.043

[6]

(a) 1. Cellulose is made up of β-glucose (monomers) and glycogen is made up ofα-glucose (monomers);

2. Cellulose molecule has straight chain and glycogen is branched;3. Cellulose molecule has straight chain and glycogen is coiled;4. glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has only 1,4-

glycosidic bonds;

Ignore ref. to H bonds / microfibrils2 max

3

(b) Any two from:1. Insoluble (in water), so doesn’t affect water potential;

2. Branched / coiled / (α-)helix, so makes molecule compact;ORBranched / coiled / (α-)helix so can fit many (molecules) in small area;

3. Polymer of (α-)glucose so provides glucose for respiration;4. Branched / more ends for fast breakdown / enzyme action;5. Large (molecule), so can’t cross the cell membrane

Require feature and explanation for 1 mark

1. Accept Ψ or WP1. Accept Insoluble so doesn’t affect osmosis

1. Do not allow ref to ‘doesn’t affect water leaving cells

4. Ignore ‘surface area’

4. Accept ‘branched so glucose readily released’2 max

(c) Iodine/potassium iodide;1


(d) For correct answer of 40 (μm) award 2 marks;Evidence of division by 500: award 1 mark

Allow tolerance of 0.5mm i.e. 20±0.5mm2

(e) 1. Scanning electron (microscope);2. 3D (image);

Accept SE(M)

2. Ignore any other correct features2

[9]

(a) 1. Low starch, fewer copies;2. Ranges overlap almost completely;

ORRanges overlap from 2 − 13 copies;

3. (surprisingly) very few / 2 or 3% have only 2 copies / are diploid;4. the mode / highest percentage for low starch is 4 copies and for high starch is

6;5. the range / spread is greater with high starch;

4. “most people” is not equivalent to mode3 max

4

(b) 1. More mRNA / more transcription;2. More translation / enzyme;3. So reaction faster;

The idea of “more” must be stated at least once.

2. Accept ‘amylase’ for enzyme

3. “More starch digested” is insufficient3

(c) 1. Mutation(s) produce extra copies of (AMY1) gene;2. Those with more copies / this adaptation/mutation reproduce / survive better on

high starch diet;2. And pass on multiple copies / this adaptation/mutation (to offspring);

Ignore ref. to single allele/gene3

[9]

(a) 6 (g dm−3);1

5

(b) Correct answer of (−)0.14;1 mark for correct difference in concentration (5) divided by 35 / (69 − 64) ÷ 35 / 1 ÷ 7

Ignore +/- sign

Ignore additional d.p.

Accept 0.31(4) for 1 mark if female data used2


(c) 1. Protein content decreases with age and decreases more in females;2. Difference (between sexes) only significant at 95 years because SDs do not

overlap;ORDifferences not significant because 2 × SD would overlap;

2

(d) 1. Produce known concentrations of protein;2. Measure absorbance of each concentration

ORMeasure each concentration with colorimeter;

3. Plot a graph of absorbance on y-axis against concentration (on x-axis) anddraw curve;

4. Use absorbance of sample to find protein concentration from curve;

1. Idea of known concentrations required.

Accept % transmission / absorption for absorbance3 max

(e) 1. (Lower plasma protein concentration suggests) fewer antibodies;

Ignore ref. to other proteins.

Reject answers which refer to white blood cells as proteins.1

[9]

(a) 1. Degenerate: more than one (base) triplet for each amino acid;

2. Non-overlapping: each base is part of only one triplet.

Accept codon (as would be applicable to mRNA code)2

(b) A = adenine

C = cytosine

G = guanine

U = uracil

All four correct = 2One error = 1Two or more errors = 0

2 max

(c) AGT;1

[5]

6


(a) 1. Sodium ions actively transported from ileum cell to blood;

2. Maintains / forms diffusion gradient for sodium to enter cells from gut (and with it,glucose);

3. Glucose enters by facilitated diffusion with sodium ions;3

(b)

Biochemical test Liquid from beakerLiquid inside

Visking tubing

Biuret reagent ✔

I2/KI ✔ or blank

Benedict’s ✔ ✔

1 mark for each correct row3

7

(c) 1. Biuret: protein molecules too large to pass through tubing;

Neutral: enzyme molecules

2. Iodine in potassium iodide solution: starch molecules too large to pass throughtubing;

If no tick in 04.2, allow no starch hydrolysed

3. Benedict’s: starch hydrolysed to maltose, which is able to pass through tubing.

Reject: glucose3

[9]

(a) 1. Polar molecule;

2. Acts as a (universal) solvent;

OR

3. (Universal) solvent;

4. (Metabolic) reactions occur faster in solution;

OR

5. Reactive;

6. Takes place in hydrolysis / condensation / named reaction;

Polar molecule so acts as (universal) solvent so (metabolicreactions are faster = 3 marks

4

8


(b) Name of ion;

Correct function within cell;

Ions other than sodium in specification are H+, Fe2+ and PO43– butaccept any correct ion (other than sodium) plus relevant function= 2.

Allow ion to be named in words but not as element, e.g, iron ion butnot iron.

2

(c) 1. Comparison: both move down concentration gradient;

2. Comparison: both move through (protein) channels in membrane;

Accept aquaporins (for water) and ion channels

3. Contrast: ions can move against a concentration gradient by active transport3

[9]

(a) 1. IV on x axis and DV on y axis and both axes on linear scales;

2. Axes labelled clearly and with correct units separated from variable by solidus or inbrackets;

3. All rates calculated correctly;

4. Points plotted correctly and joined by ruled lines and no extrapolation;4

9

(b) Yes:

1. Expect optimum temperature of enzyme to be same

OR

Similar to temperature where bacterium lives;

2. Optimum temperature for enzyme (appears to be around) 15 °C;

No:

3. Need data from more temperatures (between 10 °C and 20 °C);

4. Data for only isolated enzyme

OR

Isolation may affect activity;4


(c) 1. Initial / starting substrate concentration

2. Enzyme concentration

3. pH.

Any 2 for 1 mark1 max

[9]

(a) 1. Contents of phloem vessel pushed into insect’s mouth by high pressure;

2. (High pressure in phloem vessel) caused by loading of sugars into phloem in leaf;

3. And (resulting) entry of water by osmosis.3

10

(b) 1. Polysaccharides are insoluble;

2. So do not affect water potential of gut.2

(c) 1. (Only few bacteria passed from parent, so) only a few (copies of) genes passed on (inbacteria);

2. May not / does not include all alleles (of genes, so diversity reduced)ORSmall number of bacteria transmitted means unrepresentative sample.

2

(d) 1. Number / mass / density of insects per plant;

2. Stage of development / size of plants / insects;

Ignore any abiotic factor2

(e) Draw around leaf on graph paper and count squares;1

[10]

(a) Lactase hydrolyses lactose in to glucose (and galactose);111

(b) No lactase in the milkOREnzyme can be reused.

1

(c) 100 cm3 minute–1 is too fast to bind to active site / converse for 50 cm3 minute–1;1

(d) 14.1(4);1


(e) 1. Galactose is a competitive inhibitor / attaches to the active site (of lactase);

2. Fewer enzyme substrate complexes formed.2

[6]

(a) × 20 000

Accept range from 18 000 to 22 0001

12

(b)

✔

✔

✔

1 mark for each correct column2

(c) 1. DNA contains thymine and RNA contains uracil;

2. DNA contains deoxyribose and RNA contains ribose.2

[5]

(a) Student was measuring change in pHORBuffer would maintain a constant pH.

1 max

13

(b) 1. Volume of suspension of lipids;

2. Concentration of suspension of lipids;

3. Volume of lipase solution;

4. Temperature;2 max

(c) Boiled lipase solution;1

(d) –0.34 = 2 marks

0.34 = 1 mark2


(e) 1. Fatty acids produced;

2. Curve levels off as all substrate used up.

accept the lower pH inactivates / denatures the enzyme2

(f) 1. Faster fall in pH and levels off at same point;

2. More enzyme = substrate complexes formed;

3. Same amount of fatty acids produced / product3

[11]

(a) Glucose (and glucose);1

(b) (α1,4) Glycosidic;1

(c) 1. Headings correct – mol dm–3 and volume of water / cm3;

2. Concentration correct. ie 0.2;2

(d) Line of best fit drawn;

Read off value at 0.45.2

[6]

14

(a) Concentration of substrate solution / of enzyme solution / pH.115

(b) 1. 2.5 / 0.04;

1 mark for correct value

2. g dm–3 minute–1 / g dm–3 s–1;

1 mark for related unit2

(c) 1. Initial rate of reaction faster at 37 °C;2. Because more kinetic energy;3. So more E–S collisions / more E–S complexes formed;4. Graph reaches plateau at 37 °C;5. Because all substrate used up.

Allow converse for correct descriptions and explanations for curveat 25 °C

5

[8]

(a) Deoxyribose.116


(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]

(a) 1. Dissolve in alcohol, then add water;2. White emulsion shows presence of lipid.

2

17

(b) Glycerol.1

(c) Ester.1

(d) Y (no mark)Contains double bond between (adjacent) carbon atoms in hydrocarbon chain.

1

(e) 1. Divide mass of each lipid by total mass of all lipids (in that type of cell);2. Multiply answer by 100.

2

(f) Red blood cells free in blood / not supported by other cells so cholesterol helps tomaintain shape;

Allow converse for cell from ileum – cell supported by others inendothelium so cholesterol has less effect on maintaining shape.

1

(g) 1. Cell unable to change shape;2. (Because) cell has a cell wall;3. (Wall is) rigid / made of peptidoglycan / murein.

2 max

[10]

(a) 1. Outside of virus has antigens / proteins;2. With complementary shape to receptor / protein in membrane of cells;3. (Receptor / protein) found only on membrane of nerve cells.

Accept converse argument3

18

(b) 1. No more (nerve) cells infected / no more cold sores form;2. (Because) virus is not replicating.

2


(c) Prevents replication of virus.1

(d) MicroRNA binds to cell’s mRNA (no mark)1. (Binds) by specific base pairing;2. (So) prevents mRNA being read by ribosomes;3. (So) prevents translation / production of proteins;4. (Proteins) that cause cell death.

4

[10]

(a) Translation.119

(b) Transfer RNA / tRNA.1

(c) TAC;

UAC.2

(d) Have different R group.

Accept in diagram1

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

codon from UAC to ACC.3

[8]

(a) Two suitable suggestions;E.g.1. (Are mammals so) likely to have same physiology / reactions as humans;2. Small enough to keep in laboratory / produce enough milk to extract;3. (Can use a) large number.

Ignore references to ethical issues2 max

20

(b) 1. Hydrolysis of lipids produces fatty acids;2. Which lower pH of mixture.

2

(c) 1. (Bile-activated lipase / it) increases growth rate (of kittens);2. Results for formula with lipase not (significantly) different from breast milk / are

(significantly) different from formula milk alone;3. Showing addition of (bile-activated) lipase is the likely cause (of increased

growth);4. Lipase increases rate of digestion of lipids / absorption of fatty acids.

3 max

[7]


(a) 1. Allow equal (time for) diffusion of iodine into apple cells;

2. For comparison between apples / between harvest dates;

1. Accept equal time for reaction / colour change to occur

2. For comparison alone is insufficient.

Ignore unqualified references to fair test, controlling a variable,standardising the method.

2

(b) 1. Starch lost from the centre first / area with no starch getsbigger as it ripens;

2. (Less starch / blue / black as the) starch is converted tosugars / maltose;

3. (Less starch) as it is hydrolysed;

4. By amylase;

Less starch as it is hydrolysed into sugars scores MP2 and MP3.

3. For ‘ hydrolysed’ accept ‘as a result of hydrolysis’ or ‘broken downby hydrolysis’.

3 max[5]

21

(i) 1. Maltose;2. Water;

Accept H2O2

22

(ii) Condensation;1

[3]

(a) 1.1 (g);1

(b) 300(%);1

[2]

23


(a) 1. Add iodine / potassium iodide solution;

Reject if heated

2. Blue-black colour (with starch);

Accept black

Ignore purple2

(b) 1. Hydrolysed by enzymes / hydrolysed by amylase / maltase;

If named enzyme given, it must relate to the correct substrate

2. Produces glucose (in the gut);

3. Small enough to cross the gut wall (into the blood) / monomers / monosaccharides(can) cross the gut wall (into the blood);

Accept cell membranes / epithelium / cells for ‘gut wall’3

(c) 1. Time how long it takes to go brick red;

2. Weigh precipitate;

3. Dilute glucose samples / use smaller volume of glucose samples / use greatervolume of Benedict’s reagent;

Ignore references to colorimeter1 max

[6]

24

(a) 1. In phospholipid, one fatty acid replaced by a phosphate;

Ignore references to saturated and unsaturated

Accept

Reject P/Phosphorus

Accept annotated diagrams1

25

(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 precipitate2


(c) Saturated single/no double bonds (between carbons)ORUnsaturated 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 saturated1

(d) 1. (Fat substitute) is a different/wrong shape/not complementary;ORBond 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 penaliseonce

2

(e) It is hydrophilic/is polar/is too large/is too big;

Ignore ‘Is not lipid soluble’1

[7]


Examiner reports

Students found questions (a) and (b) accessible, with the majority scoring the marks.1Answering question (c) should have involved only simple recall, however only 28% of studentsscored two marks. When describing the role of DNA helicase, many used very simple language,such as ‘unwinds’ or ‘unzips’. Such terms were not accepted at GCE level. When describing therole of DNA polymerase, the most common mistake was stating that it forms hydrogen bondsbetween bases. There was also confusion with protein synthesis; many students stated that DNApolymerase forms pre-mRNA.

For question (d), 25% of students failed to score any marks, usually because they did not followthe command word and thus failed to give a contrasting statement; for example, ‘ATP has ribose,DNA has deoxyribose.’ Many thought adenosine is a base.

Other than with spelling, there were no issues with question (a); 76% of students answeredcorrectly.2

Question (b) proved to be harder than expected; again, this tested simple understanding but only55% of students answered correctly.

68% of students scored the mark on question (c), which was fewer than expected for simplerecall.

Question (d) showed that AS students lack an understanding of when to use each of the threestatistical tests stated in the specification. Many thought that “a value for P of 0.04” meant thatthe difference between them was only 0.04. There was also a large number of students referringto the ‘sign test’; this is not in the AQA specification. 50% of students scored zero marks and only4.6% scored three marks.

More than 40% of the students gained both marks on question (a), usually for correctly

contrasting the types of glucose (α or β), or by referring to the branched or straight chain of themolecule. Those who failed to score often did not give differences, referring only to one of themolecules.

3

Question (b) discriminated well. Many students wrote about starch being insoluble and thereforenot affecting the water potential, and the branched nature of the molecule making it compact.However, a large number of students did not follow the instruction to both describe and explainfeatures of starch that make it a good storage molecule.

In answering question (c), 87% of students correctly identified the correct test.

In answering question (d) most students were able to attempt the calculation but many wereunable to convert millimetres to micrometres and therefore gained only 1 mark.

Whilst half of the students scored both marks for question (e), there were many who were unableto identify the correct type of microscope and often their evidence was that the photograph wasblack and white rather than in 3-D.


Students clearly have difficulty in describing data with almost a quarter gaining no marks at all forquestion (a), and many of these confusing data for low-starch and high-starch diets. However,most could state that a low-starch diet was linked to fewer copies of the gene and many correctlyidentified the modal values. There were few references to the overlapping ranges and the higherrange in populations with a high-starch diet. Students would benefit from learning to use termssuch as mode and range in their answers to similar questions.

4

In answers to question (b), students seemed less confident in their knowledge of proteinsynthesis than in their knowledge of enzyme action. Consequently, very few students couldexplain that, with more copies of the gene, there would be more transcription or more mRNAformed. However, most realised that there would be more enzyme present and then went on toexplain the advantage in terms of faster starch digestion.

Often students failed to use the information given to them in question (c) and wrote genericanswers about natural selection which failed to gain any marks. Examiners were looking toaward marks for points made in the context of multiple copies of the AMY1 gene. Those studentswho did this often gave excellent answers referring to mutation resulting in extra copies; theythen went on to explain that this mutation resulted in a higher chance of survival andreproduction in populations with a high-starch diet, with the beneficial mutation being passed onto their offspring.

Two-thirds of the students gave the correct answer of 6 g dm−3 for question (a).5Approximately 40% of the students were able to gain 2 marks for question (b). A significantnumber did not read the key on the graph correctly and so gave figures for females instead ofmales. Some students left the calculation as a fraction, perhaps through not bringing a scientificcalculator to the examination – these students scored 1 mark.

In question (c), many students realised that the protein concentration decreased more in femalesthan males, but very few used the given information about the standard deviation to comment onthe significance of this difference. Students should take more care in interpreting graphicalinformation since, again, many confused the male and female data.

Question (d) was answered poorly, with 10% of students giving no response and over halfscoring no marks. This was disappointing as the idea of preparing a dilution series to produce acalibration curve is in Required Practical 3. There were some good answers in which studentswrote about measuring the absorbance of known concentrations of protein, but many suggestedproducing different concentrations of blood plasma rather than protein. These students were stillable to gain some marks. Very few went on to describe how to draw the calibration curve insufficient detail to gain a mark. Many students suggested plotting age as part of their calibrationcurve, and few were able to give both axes. Those who did were usually able to continue theiranswer to gain one mark for the use of the curve to find the protein concentration from theabsorbance value.

In question (e), almost 40% of students gained 1 mark for suggesting a decrease in proteincontent could be linked to fewer antibodies. Examiners were surprised by how many studentsthought that cells were proteins.


(a) Many partial responses were seen here that did not fulfil the mark points. “For comparison”needs further qualification for credit at this level. It was surprising how few linked theinvestigation they had carried out into diffusion of iodine to include this when scoring markpoint 1, many achieved mark point 1 with a statement more similar to that in the commentscolumn.

(b) The command words used in this question were “describe” and “explain”. Most responsesstarted with the explanation and those that attempted the description often did not givesufficient detail to achieve mark point 1. Many responses showed clear understanding ofthe starch being broken down by amylase although “hydrolysed” was less common.

21

In part (i), a significant minority of students omitted water and therefore gained only 1 mark, butmost correctly identified condensation as the reaction with very few giving hydrolysis in error.22

Both parts of this question proved to be good discriminators as very few students were able tocarry out the calculations correctly.23

(a) Most students were able to identify the correct test for starch but a significant number didnot refer to iodide solution and so did not gain the first mark point. A significant numberincorrectly referred to the colour produced by starch as being purple or blue instead ofblue/black or black.

(b) The correct enzymes involved in the hydrolysis of starch were known by many students butthe small size of the molecules produced was rarely mentioned. When it was, the mark wasoften not gained due to vague references to how they were absorbed.

(c) Many students referred to the use of a colorimeter. This would not work due to the densenature of the red precipitate. Better answers clearly described a dilution technique whichwould work (not serial dilution), or mentioned timing the appearance of the colour orweighing the precipitate.

24

About two thirds obtained the mark in (a) for a correct description of the difference between atriglyceride and a phospholipid. Those who failed to score either did not know about the structureof these molecules or just described the structure of a phospholipid (or triglyceride).

In (b), about 40% of students could fully describe how to test for a lipid and obtained both marks.A minority described tests for other biological molecules. Many made errors in their descriptionsof the emulsion test or of a positive result. These errors included: adding water before ethanol,heating the mixture, the presence of a precipitate and failing to note that the colour of theemulsion would be white.

Most students scored the mark in (c). Those who didn’t got saturated and unsaturated the wrongway round in terms of carbon-carbon double bonds.

In (d), 28% obtained both marks for stating that the fat substitute would not bind to the active siteof lipase because it has a different shape to a triglyceride. A similar percentage obtained onemark. Those who failed to score often ignored the question’s reference to lipase and wrote aboutbile salts, micelles and methods of absorption.

25


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