Name: ________________________

3.2.5 DNA structure and replication Class: ________________________

Date: ________________________

Time: 108 minutes

Marks: 85 marks

Comments:

Page 1 of 29
 Figure 1 shows one base pair of a DNA molecule.
1
Figure 1

(a) Name part F of each nucleotide.

........................................................................................................................
(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)

Page 2 of 29
 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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(4)
(Total 8 marks)

(a) Nucleic acids, such as DNA, are polymers, made up of many repeating monomer units.
2 Name the monomer from which nucleic acids are made.

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

(b) The table shows the percentage of different bases in the DNA of some organisms.

Percentage of each base

Organism
Adenine Guanine Cytosine Thymine

Human 31.2 18.8 18.8 31.2

Cow 27.9 22.1 22.1 27.9

Salmon 29.4 20.6 20.6 29.4

Rat 28.6

Virus 24.7 24.1 18.5 32.7

Page 3 of 29
 (i) Calculate the missing figures for rat DNA and write them into the table.
(2)

(ii) The virus has single-stranded DNA as its genetic material. Explain the evidence from
the table which suggests that the DNA is single-stranded.

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(2)
(Total 5 marks)

The following figure represents part of a DNA molecule.

3

(a) Draw a box around a single nucleotide.

(1)

The table below shows the percentage of bases in each of the strands of a DNA molecule.

DNA Percentage of each base

strand
A C G T

Strand 1 16

Strand 2 21 34

(b) Complete the table by adding the missing values.

(2)

Page 4 of 29
(c) During replication, the two DNA strands separate and each acts as a template for the
production of a new strand. As new DNA strands are produced, nucleotides can only be
added in the 5’ to 3’ direction.

Use the figure in part (a) and your knowledge of enzyme action and DNA replication to
explain why new nucleotides can only be added in a 5’ to 3’ direction.

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(Extra space) ................................................................................................

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(4)
(Total 7 marks)

Page 5 of 29
 The diagram shows a molecule of DNA. It is replicating.
4

(a) Name two substances in the region labelled X.

1 ........................................................................

2 ........................................................................
(1)

(b) Describe how, after the parent DNA strands separated, the second strand of DNA in region
Y was formed.

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(Extra space)................................................................................................

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(3)
(Total 4 marks)

Page 6 of 29
 The bases in DNA nucleotides contain nitrogen.
5
Researchers grew bacteria on a medium containing 15N (‘heavy’ nitrogen) for several
generations. They then transferred the bacteria to a medium containing 14N (‘ordinary’ nitrogen).
They analysed DNA from the bacteria at three stages:

1. whilst the bacteria were growing on the 15N medium

2. after one division of the bacteria on the 14N medium

3. after two divisions of the bacteria on the 14N medium

The diagram shows their results.

Page 7 of 29
(a) Describe how the proportion of DNA that contained 15N changed at each division when
bacteria were grown on the 14N medium.

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(2)

(b) The change in the proportion of DNA containing 15N is due to the way in which DNA
replicates. Explain how.

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(2)
(Total 4 marks)

Page 8 of 29
 The diagram shows the process of DNA replication. The horizontal lines represent the positions
6 of bases.

(i) What is represented by the part of the DNA molecule labelled W?

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

(ii) In the diagram, A represents adenine and C represents cytosine.

Name the base found at

position X; .....................................................................................................

position Y; .....................................................................................................

position Z. .....................................................................................................
(3)
(Total 4 marks)

Page 9 of 29
 (a) Explain why the replication of DNA is described as semi-conservative.
7
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(2)

(b) Bacteria require a source of nitrogen to make the bases needed for DNA replication. In an
investigation of DNA replication some bacteria were grown for many cell divisions in a
medium containing 14N, a light form of nitrogen. Others were grown in a medium containing
15
N, a heavy form of nitrogen. Some of the bacteria grown in a 15N medium were then
transferred to a 14N medium and left to divide once.
DNA was isolated from the bacteria and centrifuged.
The DNA samples formed bands at different levels, as shown in the diagram.

(i) What do tubes A and B show about the density of the DNA formed using the two
different forms of nitrogen?

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(1)

Page 10 of 29
 (ii) Explain the position of the band in tube C.

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(2)

(c) In a further investigation, the DNA of the bacterium was isolated and separated into single
strands. The percentage of each nitrogenous base in each strand was found. The table
shows some of the results.

Percentage of base present

DNA sample Adenine Cytosine Guanine Thymine

Strand 1 26 28 14

Strand 2 14

Use your knowledge of base pairing to complete the table.

(2)
(Total 7 marks)

(a) There are two forms of nitrogen. These different forms are called isotopes. 15N is a heavier
8
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.

Page 11 of 29
 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)

Page 12 of 29
 (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)

(a) DNA helicase is important in DNA replication. Explain why.

9
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(2)

Page 13 of 29
Scientists investigating DNA replication grew bacteria for several generations in a nutrient
solution containing a heavy form of nitrogen (15N). They obtained DNA from a sample of
these bacteria.

The scientists then transferred the bacteria to a nutrient solution containing a light form of
nitrogen (14N). The bacteria were allowed to grow and divide twice. After each division,
DNA was obtained from a sample of bacteria.

The DNA from each sample of bacteria was suspended in a solution in separate tubes.
These were spun in a centrifuge at the same speed and for the same time. The diagram
shows the scientists’ results.

Page 14 of 29
(b) The table shows the types of DNA molecule that could be present in samples 1 to 3.
Use your knowledge of semi-conservative replication to complete the table with a tick if the
DNA molecule is present in the sample.

(3)

(c) Cytarabine is a drug used to treat certain cancers. It prevents DNA replication. The diagram
shows the structures of cytarabine and the DNA base cytosine.

(i) Use information in the diagram to suggest how cytarabine prevents DNA replication.

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(2)

Page 15 of 29
 (ii) Cytarabine has a greater effect on cancer cells than on healthy cells. Explain why.

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(1)
(Total 8 marks)

Figure 1 shows a short section of a DNA molecule.

10
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)

Page 16 of 29
(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)

Page 17 of 29
 (a) (i) Describe the role of DNA polymerase in DNA replication.
11
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(1)

(ii) Other than being smaller, give two ways in which prokaryotic DNA is different from
eukaryotic DNA.

1 ................................................................................................................

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

(b) The table shows the percentage of each base in the DNA from three different organisms.

Percentage of each base in DNA

Organism
Adenine Guanine Thymine Cytosine

Human 30.9 19.9 29.4 19.8

Grasshopper 29.4 20.5 29.4 20.7

Virus 24.0 23.3 21.5 31.2

(i) Humans and grasshoppers have very similar percentages of each base in their DNA
but they are very different organisms.

Use your knowledge of DNA structure and function to explain how this is possible.

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(2)

Page 18 of 29
 (ii) The DNA of the virus is different from that of other organisms. Use the table above
and your knowledge of DNA to suggest what this difference is. Explain your answer.

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[Extra space] ...........................................................................................

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(2)
(Total 7 marks)

Cells constantly hydrolyse ATP to provide energy.

12
(a) Describe how ATP is resynthesised in cells.

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(2)

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

1 ..........................................................................................................................

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2 ..........................................................................................................................

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(2)

Page 19 of 29
(c) This is a photograph (micrograph) of a mitochondrion taken using a scanning electron
microscope.

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

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(1)

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

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(1)

Page 20 of 29
 (e) The actual length of the mitochondrion between points A and B in the photograph is 4 μm.

What is the magnification of the mitochondrion in the photograph?

Show your working.

Magnification ............................
(2)
(Total 8 marks)

When one mole of glucose is burned, 2800 kJ of energy are released. However, when one mole
13 of glucose is respired aerobically, only 40% of the energy released is incorporated into ATP. Each
mole of glucose respired aerobically produces 38 moles of ATP.

(a) (i) Calculate how much energy is incorporated into each mole of ATP. Show your
working.

Answer ................................................. kJ
(2)

(ii) When glucose is respired what happens to the energy which is not incorporated into
ATP?

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

(b) (i) When one mole of glucose is respired anaerobically, only 2 moles of ATP are
produced. Explain why less energy is released in anaerobic respiration.

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(1)

Page 21 of 29
(ii) At the end of a sprint race, a runner continues to breathe rapidly for some time.
Explain the advantage of this.

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(2)
(Total 6 marks)

Page 22 of 29
Mark schemes
(a) Deoxyribose.
1 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]

(a) nucleotide;
2 1

(b) (i) 21.4, 21.4; 28.6;

2

(ii) amounts of A and T / C and G / complementary bases different;

therefore no base-pairing;
2 max
[5]

(a) Box around single nucleotide.

3 1

(b)
DNA Percentage of each base
strand
A C G T

Strand 1 (16) 34 21 29

Strand 2 29 (21) (34) 16

2 rows correct = 2 marks;

1 row correct = 1 mark.
2

Page 23 of 29
 (c) 1. Reference to DNA polymerase;
2. (Which is) specific;
3. Only complementary with / binds to 5’ end (of strand);
Reject hydrogen bonds / base pairing

4. Shapes of 5’ end and 3’ end are different / description of how different.

4
[7]

(a) (Pentose) sugar/deoxyribose and phosphate;

4
Reject ribose and phosphorus
1

(b) Semi-conservative replication;

Complementary pairing;
Hydrogen bonding (of bases/nucleotides);
Condensation/described of nucleotides;
DNA polymerase involved;
Accept example (A, T and C, G)
3 max
[4]

(a) Decreases by 50%;

5
Per generation / per division;
Only accessible if linked to first marking point

OR

15
N makes up ½ after 1 division;

Makes up ¼ after 2nd division;

2

(b) In DNA replication strands separate;

Each acts as template (for formation of new strand);
One strand in each new molecule / semi-conservative replication;
New strands made using 14N.
2 max
[4]

Page 24 of 29
 (i) sugar or phosphate / S-P / nucleotide chain / backbone /
6 original / parent DNA;
1

(ii) X thymine; Y guanine; Z adenine;

(Allow T, G and A) Reject: thiamine
3
[4]

(a) each strand copied / acts as a template;

7 (daughter) DNA one new strand and one original / parent strand;
2

(b) (i) 15N / tube B (DNA), more / greater density;

(reject heavier)
1

(ii) DNA with one heavy and one light strand;

new / synthesised strand, made with 14N / light strand;
2

(c) 32;
28 32 26;
2
[7]

(a) (i) base / named bases;

8
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]

Page 25 of 29
 (a) 1. Separates / unwinds / unzips strands / helix / breaks H-bonds;
9
1. Q Neutral: strands / helix split
1. Accept: unzips bases

2. (So) nucleotides can attach / are attracted / strands can act as templates;
2. Q Neutral: bases can attach
2. Neutral: helix can act as a template
2

(b)

One mark for each correct row

3

(c) (i) 1. Similar shape / structure (to cytosine) / added instead of cytosine / binds
to guanine;
1. Accept: idea that only one group is different
1. Reject: same shape

2. Prevents (complementary) base pairing / prevents H-bonds forming /

prevents formation of new strand / prevents strand elongation / inhibits /
binds to (DNA) polymerase;
2. Accept: prevents cytosine binding
Neutral: ’prevents DNA replication‘ as given in the question stem
Neutral: ’competitive inhibitor‘ unqualified
Neutral: inhibits DNA helicase
2

(ii) (Cancer cells / DNA) divide / replicate fast(er) / uncontrollably;

Accept: converse argument for healthy cells
1
[8]

(a) (i) Deoxyribose;

10
pentose / 5C sugar = neutral
1

(ii) Phosphate / Phosphoric acid;

phosphorus / P = neutral
1

Page 26 of 29
 (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]

(a) (i) Joins nucleotides (to form new strand).

11
Accept: joins sugar and phosphate / forms sugar-phosphate
backbone
Reject: (DNA polymerase) forms base pairs / hydrogen bonds
1

(ii) (Prokaryotic DNA)

1. Circular / non-linear (DNA);
Accept converse for eukaryotic DNA
Ignore: references to nucleus, binary fission, strands and plasmids

2. Not (associated) with proteins / histones;

Accept does not form chromosomes / chromatin

3. No introns / no non-coding DNA.

Accept only exons
Q Neutral: no ‘junk’ DNA
2 max

(b) (i) 1. Have different genes;

Reject: different alleles

2. (Sobases / triplets) are in a different sequence / order;

Accept: base sequence that matters, not percentage

3. (So) different amino acid (sequence / coded for) / different protein /

different polypeptide / different enzyme.
Unqualified ‘different amino acids’ does not gain a mark
Reject: references to different amino acids formed
Ignore: references to mutations / exons / non-coding / introns
2 max

Page 27 of 29
 (ii) (Virus DNA)
1. A does not equal T / G does not equal C;
Accept: similar for equal
Accept: virus has more C than G / has more A than T

2. (So) no base pairing;

3. (So) DNA is not double stranded / is single stranded.
2 max
[7]

(a) 1. From ADP and phosphate;

12
Accept

Reject P/Phosphorus
Reject use of water in the reaction

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

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

Reject ‘produce’ energy
2. To add phosphate to other substances and make them more reactive/change their
shape;
2

(c) (Can see) 3D image;

1

(d) Crista/cristae;
Ignore matrix
1

(e) Value between 20,750 (83mm) and 21,250 (85mm) two marks;;
Formula given/used but calculation wrong, award 1 mark
Magnification = image size
Object size
(Large number divided by 4)
2
[8]

(a) (i) 29.47(29.5);

13
(2 marks for correct answer)

40% / 0.4 of 2800 / 38;

2

(ii) released as heat;

1

Page 28 of 29
(b) (i) glucose only partly broken down / only broken down to lactate;
1

(ii) lactate / lactic acid has built up / been produced;

oxygen used to break down lactate / convert it back to
pyruvate / glucose / glycogen;
2
[6]

Page 29 of 29