2.3 Nucleic Acids OCR ExamBuilder

Answer all the questions.

1. A sample of DNA containing only one isotope of nitrogen,

15

N, was incubated with nucleotides containing only

the

14

N isotope along with the enzymes needed for replication.

Which of the following diagrams would represent the resulting DNA after one round of replication?

Your answer

[1]

2. Enzymes are capable of affecting the metabolism and structure of whole organisms. Which of the following

enzymes will have the greatest effect on the development of an organism as a whole?

A Methyltransferase: adds methyl groups to DNA allowing genes to be switched on or off.

B Reverse transcriptase: generates complementary DNA from an RNA template.

C Deoxyribonuclease: digests free DNA molecules outside of the nucleus.

D Telomerase: lengthens ends of chromosomes by adding DNA sequences, preventing them from being

degraded.

Your answer

[1]

3. A length of DNA has the base sequence AATCGCGGTCGCTCA.

Select the row that shows the correct complementary DNA strand and the sequence of mRNA made during

transcription of the DNA sequence above.

Complementary DNA sequence mRNA sequence

A AATCGCGGTCGCTCA UUAGCGCCAGCGAGU

B TTAGCGCCAGCGAGT UUAGCGCCAGCGAGU

C TTAGCGCCAGCGAGT TTAGCGCCAGCGAGT

D TTAGCGCCAGCGAGT AAUCGCGGUCGCUCA

Your answer

[1]

4. An anticodon sequence of five successive tRNA molecules involved in protein synthesis was analysed and found

to have the following percentage base composition.

Adenine 40; Cytosine 27; Guanine 13; Thymine 0; Uracil 20 %

Which row shows the percentage base composition of the template strand of the original DNA molecule?

Adenine Cytosine Guanine Thymine Uracil

A 40 27 13 20 0

B 20 13 27 40 0

C 20 13 27 0 40

D 40 27 13 0 20

Your answer

[1]

5. One of the main functions of the liver cells is the formation of urea by the ornithine cycle, an outline of which is

shown in Fig. 17.2.

(i) Step 1 of the cycle takes place in the organelle represented by D.

Identify organelle D.

[1]

(ii) During the cycle ornithine moves into organelle D and citrulline moves out of the organelle.

Suggest the method by which these molecules move into and out of the organelle during the cycle. Give

reasons for your choice.

[2]

(iii) How has the ammonia that is used in step 1 been formed?

[1]

(iv) Identify the compound labelled X in Fig. 17.2.

[1]

6. The diagram below shows an organic molecule.

What bond is formed when the molecule is polymerised?

A ester

B glycosidic

C peptide

D phosphodiester

Your answer

[1]

7. DNA replication and transcription are two processes that occur in the nucleus of eukaryotic cells.

* Compare DNA replication and transcription by describing the similarities and differences between the two

processes.

[6]

8. Which of the following statements describes an organelle which is not membrane bound?

A contains cristae

B modifies and packages proteins

C contains digestive enzymes

D is made of rRNA and protein

Your answer

[1]

9. The following statements are about the structure of DNA.

Which of the following statement(s) is / are true?

Statement 1: Purine bases pair with pyrimidine bases.

Statement 2: Phosphodiester bonds link adjacent nucleotides.

Statement 3: There are always equal amounts of adenine and guanine.

A 1, 2 and 3

B Only 1 and 2

C Only 2 and 3

D Only 1

Your answer

[1]

10. A gene codes for the production of lactase. This gene is normally switched off after an infant moves to adult

food. Almost all adult mammals are unable to digest lactose. They are said to be lactose intolerant. Humans

are an exception.

Most humans have a genetic mutation that prevents the shutdown of lactase production.

State what structural detail of a polypeptide is altered by gene mutations.

[1]

11(a). DNA is arguably the most important molecule in the whole of biology.

When a cell divides an identical copy of its DNA is made in a process called DNA replication.

Explain how pairing of nitrogenous bases allows identical copies of DNA to be made.

[3]

(b).

(i) Outline how the process of DNA replication is completed, following the pairing of nitrogenous bases.

[3]

(ii) Why is DNA replication described as semi-conservative?

[1]

12. A student tried to extract some DNA from a crushed banana at home. DNA dissolves in water but the student

realised that they needed to add something to break open the nuclear envelope to release the DNA.

Suggest a suitable substance the student could use to release the DNA, and explain why it should work.

[2]

13. Which of the following statements, A to D, about the nature of the genetic code is incorrect?

A It is a degenerate code.

B It is a triplet code.

C It is overlapping.

D It is universal.

Your answer

[1]

14(a). Even the smallest DNA molecules are very long.

A kilobase (Kb) is a unit equivalent to 1000 base pairs of a DNA molecule.

One Kb of double stranded DNA has a length of 0.34 μm.

The DNA in the nucleus of a cell from a fruit fly (Drosophila) is 5.6 cm long.

(i) Calculate the number of Kb in the DNA of the fruit fly.

Show your working. Give your answer to the nearest whole number.

Answer = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Kb [2]

(ii) The DNA of the fruit fly was analysed and 22% of the bases were adenine.

What % of the bases were guanine? Show your working.

Answer = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ % [2]

(b). A DNA molecule contains polynucleotide strands.

(i) Individual nucleotides are joined together to make a polynucleotide strand.

What type of chemical reaction takes place when two nucleotides in a single polynucleotide strand are joined

together?

[1]

(ii) Name the chemical released when the bond is formed between the two nucleotides.

[1]

(iii) A DNA molecule contains two polynucleotide chains.

Describe how these two chains are held together.

[3]

15. This figure represents an influenza virus. Various protein antigens are attached to the outer surface of the virus.

When a virus infects a human host, it causes the host's cells to produce many new copies of the virus.

(i) The influenza vaccination must be given each year because there are frequent mutations in the RNA of the

virus.

The antigens on the surface of the virus are made of protein.

The virus uses the organelles and enzymes in the host's cells to produce new copies of itself.

Suggest the role of the viral RNA in the production of viral proteins.

[2]

(ii) Explain why a mutation in the viral RNA leads to a change in the 3-D shape of the protein antigens.

[3]

(iii) The head teacher of a school decided to offer teachers free influenza vaccinations every year.

Suggest why the head teacher thought this would be a good use of the school's money.

[1]

16.

ATP is described as the ‘universal energy currency’.

Describe how the structure of ATP is similar to and differs from the structure of a DNA nucleotide.

In your answer, you should use appropriate technical terms, spelled correctly.

[5]

17. This question is about genetic control and selective breeding.

Fill the gaps in the following passage using the most appropriate term: Much of the _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

in cells contains sequences called genes.

Many genes code for _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ that fold to make enzymes. Often, enzymes are kept in an

inactive form until needed. These enzymes may then be activated by cAMP, which involves changes in their _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .

[3]

18(a). Fig. 5.1 shows part of a DNA strand.

Fig 5.1

(i) Name the base represented by the letter T.

[1]

(ii) On Fig. 5.1, draw a section of the mRNA strand that is complementary to the section of the DNA strand

shown.

This should be answered on Fig. 5.1

[2]

(b). Table 5.1 contains a list of statements about DNA replication. Some of these statements are incorrect.

Put a cross (X) in the box next to each incorrect statement.

DNA replication Incorrect

statement

The DNA molecule unwinds

Hydrogen bonds between the base pairs break

Free RNA nucleotides join to bases on the exposed DNA strands

Both polypeptide strands act as a template

Hydrogen bonds form between complementary bases

Three hydrogen bonds form between bases A and T

DNA polymerase links the new nucleotides

Covalent bonds form between the phosphate of one nucleotide and the

pentose sugar of the next nucleotide

Table 5. 1

[3]

19. Genes are important in the process of natural selection. Genes are made of DNA.

(i) State the role of a gene.

[1]

(ii) Explain how the structure of DNA allows replication.

[5]

20(a). Gene sequencing is an important technique in molecular biology.

Fig. 3.1 shows part of a computerised graph obtained from an automated gene sequencing machine.

The section of the DNA molecule represented in Fig. 3.1 is from base position 117 (on the left of the graph)

to base position 137 (on the right of the graph).

The bases in the DNA sequence are labelled with four different coloured fluorescent dyes.

The identities of some of the bases (117 to 119 inclusive and 129 to 137 inclusive) are indicated below the

graph.

Use Fig. 3.1 to identify the order of bases from positions 120 to 128.

[1]

(b). To produce the type of graph shown in Fig. 3.1, the automated gene sequencing machine needs to be loaded

with the following:

the DNA to be sequenced

short primer sequences specific to the DNA to be sequenced

many normal DNA nucleotides

some chain-terminating DNA nucleotides labelled with coloured dyes

the enzyme Taq polymerase.

A regular cycle of temperature changes allows many DNA fragments of different lengths to be built up by the

polymerase chain reaction (PCR).

Fig. 3.2 shows the end parts of the sequences of seven of these different length fragments, labelled 1 to 7. The

end parts of the sequences for fragments 1 to 4 are complete but those for fragments 5 to 7 are not.

These seven fragments correspond to the last seven peaks on the right hand side of the graph in Fig. 3.1.

The letters in boxes represent labelled chain-terminating DNA nucleotides.

The letters not in boxes represent normal DNA nucleotides.

(i) Use the information in Fig. 3.1 to fill in the missing nucleotide bases on fragments 5 to 7 on Fig. 3.2.

You should distinguish between the normal and labelled nucleotides in the sequence for each fragment.

Fig. 3.2

[2]

(ii) Explain how the automated sequencing machine orders the DNA fragments from the PCR reaction into the

size order shown in Fig. 3.2.

[3]

(c). Asthma in children may be treated with drugs. One of the most commonly used drugs is salmeterol.

Salmeterol acts by binding to protein receptors in the lining of the bronchioles. However, in approximately 14% of

children with asthma, salmeterol is not very effective. This is thought to be the result of a genetic mutation in

these children.

Suggest why this mutation reduces the effectiveness of salmeterol.

[3]

(d). In a recent medical trial, 62 children with this genetic mutation were studied.

Their asthma was not controlled well by salmeterol.

31 children continued using salmeterol and the remaining 31 were given an alternative drug, montelukast.

Montelukast is not routinely prescribed because salmeterol is far more effective for most children with

asthma.

(i) After one year, the children taking montelukast had better control of their asthma and were able to reduce

their use of montelukast.

Suggest why these children responded better to montelukast than to salmeterol.

[2]

(ii) Comment on the reliability of the results of this medical trial.

[1]

(iii) It is proposed that a simple saliva test could identify those children who have the mutation.

What would be the source of the genetic material used in this test?

[1]

21. In a colony of bees, about 5% of the workers are more adventurous than other workers. These bees are known

as scout bees. They actively seek out new food sources and, if necessary, new nest sites.

Researchers investigated how gene expression differed in the brains of the scout bees compared to the normal

worker bees.

The researchers extracted mRNA from the brain cells of normal worker bees.

This mRNA was used to produce lengths of single-stranded DNA, which were then attached to a fluorescent

dye.

These lengths of single-stranded DNA were used as gene probes fixed onto a device known as a ‘microarray

DNA chip’.

mRNA extracted from the brain cells of scout bees would only bind to the gene probes that matched it,

causing these probes to fluoresce.

The locations of the brightest fluorescent spots on the DNA chip revealed which genes were most active.

(i) Name the enzyme that can be used to convert mRNA to single-stranded DNA.

[1]

(ii) Explain how the locations of the fluorescent spots on the DNA chip reveal which genes are most active.

[3]

(iii) The researchers found many differences in gene activity in the scout bees compared to the normal worker

bees. One of these differences in activity was in a gene used to make the neurotransmitter, dopamine.

In a follow-up experiment, scout bees became less adventurous if dopamine signalling was prevented.

Use your knowledge of the DRD4 dopamine receptor in humans to comment on the findings of this research

into scout bee behaviour.

[3]

22(a). The genetic code carries instructions for the synthesis of polypeptides.

(i) State the number of DNA nucleotide bases that code for a single amino acid.

[1]

(ii) There is a maximum of 64 different base combinations in DNA that could each code for an amino acid.

How is this number of combinations calculated?

[1]

(iii) Twenty different amino acids are commonly used for protein synthesis. In theory, this would need only 20

different base combinations.

Explain the uses of the remaining 44 combinations.

[2]

(iv) Which nucleotide bases are common to DNA and RNA?

[1]

(b). Describe how a nucleotide base sequence in a gene is used to synthesise a polypeptide.

In your answer you should describe the steps of the process in the correct order.

[7]

23. Which of the statements, A to D, shows that the genetic code is degenerate?

A CCA and CCT code for proline

B rRNA is manufactured in the nucleolus

C tRNA is not complementary to DNA

D uracil is not found in DNA

Your answer

[1]

24. Fig. 22 shows four nucleotides.

Fig. 22

On Fig. 22, draw and label the bonds holding the nucleotides together as part of a DNA molecule.

[2]

25. Potatoes often suffer bruising, which reduces their value as a food crop.

A variety of crop potato that does not bruise has been developed using a technique called gene silencing.

Scientists carry out gene silencing by inserting small sequences of RNA into potato cells. These RNA sequences

are complementary to mRNA from genes responsible for bruising.

Use this information to suggest why the technique is called ‘gene silencing’.

[2]

END OF QUESTION PAPER

Mark Scheme

1 A 1

Total 1

2 A 1

Total 1

3 B 1

Total 1

4 A 1

Total 1

5 i mitochondrion 1 ALLOW mitochondria.

ii either

facilitated diffusion (1)

conversion of ornithine into citrulline

creates concentration gradients

or

(molecules are not lipid soluble so) require

protein channels to cross membrane (1)

or

active transport (1)

ornithine and citrulline need to be moved

into and out of D

more quickly than would be met by

diffusion (1)

2

iii deamination / removal of NH

2

group from

amino acid (1)

1

iv ATP (1) 1

Total 5

6 D 1

Total 1

Question Answer/Indicative content Marks

Guidance

Mark Scheme

7 Please refer to the marking instructions

point 10 for guidance on how to mark

this question.

In summary:

Read through the whole answer. (Be

prepared to recognise and credit

unexpected approaches where they show

relevance.)

Using a ‘best-fit’ approach based on the

science content of the answer, first decide

which of the level descriptors, Level 1,

Level 2 or Level 3, best describes the

overall quality of the answer.

Then, award the higher or lower mark

within the level, according to the

Communication Statement (shown in

italics):

award the higher mark where the

Communication Statement has been

met.

award the lower mark where aspects of

the Communication Statement have

been missed.

The science content determines the

level.

The Communication Statement

determines the mark within a level.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

Level 3 (5–6 marks)

A comparison of all or most aspects of the

two processes is included, with no

significant errors.

There is a well-developed line of

reasoning, which is clear and logically-

structured and uses scientific terminology

at an appropriate level. All the information

presented is relevant and forms a

continuous narrative.

Level 2 (3–4 marks)

A description of some similarities and

differences between the two processes is

included, with only minor errors.

There is a line of reasoning presented with

some structure and use of appropriate

scientific language. The information

presented is mostly relevant.

Level 1 (1–2 marks)

A description of similarities or differences

between the two processes is included, but

with significant omissions or errors.

There is a logical structure to the answer.

The explanation and use of scientific

language, though basic, is clear.

0 marks

No response or no response worthy of

credit.

6 Indicative scientific points may include:

Similarities

DNA unwinds and unzips

Helicase enzymes

Template DNA

Complementary base pairing

Hydrogen bonds

Free, activated nucleotides

Polymerase enzymes

Differences

Only a small section of DNA (where the

gene is located) unzips during

transcription

Both strands act as templates in

replication

RNA vs DNA free nucleotides

RNA vs DNA polymerase

Different helicase enzymes

Products are two new daughter strands

of DNA in replication and one mRNA

strand in transcription

mRNA leaves nucleus whereas the

new DNA strand remains bound to the

template strand

Total 6

8 D 1

Total 1

9 B 1

Total 1

10 sequence / order, of amino acids 1 ALLOW primary structure.

Total 1

Question Answer/Indicative content Marks

Guidance

Mark Scheme

11 a three from

adenine / A pairs with thymine / T and

cytosine / C pairs with guanine / G (1)

(because of) hydrogen bonding (1)

idea that purine can only bind with

pyrimidine because they are different sizes

(1)

idea that if one base is known it can pair

with only one other base (1)

3

ALLOW 2 H bonds between A and T and 3

H bonds between C and G.

b i (involves) DNA polymerase (1)

sugar-phosphate backbone (re)forms /

condensation reaction between phosphate

and sugar (1)

3

i DNA winds into double helix (1) ALLOW higher level answers

e.g. role of DNA ligase in joining sugar-

phosphate backbone lagging strand filled

in with Okazaki fragments.

ii (new molecule consists of) one old strand

and one new strand (1)

1

Total 7

12 detergent (1)

works as an emulsifier / attracts

phospholipid molecules and water

molecules (1)

it will break up the plasma / nuclear

membranes (1)

2

Total 2

Question Answer/Indicative content Marks

Guidance

Mark Scheme

13 C 1

Examiner's Comments

Candidates found this difficult, many

suggesting A or D. It is possible that they

had misread the question and gave an

option that was a correct statement about

the genetic code rather than an incorrect

one. Candidates should be encouraged to

take care when reading questions rather

than rushing into answering the question

that they thought had been asked.

Total 1

Question Answer/Indicative content Marks

Guidance

Mark Scheme

14 a i 164 706 ✔✔ 2 Correct answer with no working = 2 marks

If the answer is incorrect, look for a

working mark:

either

(incorrect rounding) ALLOW 1 mark for

seeing 164 705 or 164 705.88 or 164

705.9 anywhere

or

ALLOW 1 mark for any ref to

56 ÷ 34

(e.g. 5.6 ÷ 0.34 or 5600 ÷ 34)

Examiner's Comments

Most candidates recognised that the

number of kB would be obtained by

dividing the length of DNA by the length of

a kB to arrive at the number of kB in the

length of DNA. However, they were not

confident converting units of cm and

micrometres to standard form, and also

failed to state the answer to the nearest

whole number.

ii 28 ✔✔ 2 Correct answer with no working = 2 marks

If answer incorrect, ALLOW 1 mark for

seeing 100 – 44 or 50 – 22

Examiner's Comments

The majority of candidates correctly scored

maximum marks for this calculation, but

the most common mistake that was

presented was through poor arithmetic,

e.g. 100 – 44 = 66.

b i

condensation ✔

1 If additional incorrect answer given, then 0

marks

ACCEPT esterification

Examiner's Comments

Most candidates identified the correct

reaction involved and stated that the

chemical released was water. Esterification

also gained credit for some candidates. A

minority of candidates wrongly answered

hydrolysis, with hydrogen given off.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

ii

water ✔

1 If additional incorrect answer given, then 0

marks

ACCEPT H

2

O (correct formula only)

Examiner's Comments

Most candidates identified the correct

reaction involved and stated that the

chemical released was water. Esterification

also gained credit for some candidates. A

minority of candidates wrongly answered

hydrolysis, with hydrogen given off.

iii

1 phosphodiester bonds in, backbone /

described ✔

2 hydrogen / H, bonds / bonding (between

chains / bases) ✔

3 purine to pyrimidine / A to T and C to G

✔

4 ref to correct number of bonds between

base pairs (A-T & C-G) ✔

max 3 IGNORE antiparallel

1 ACCEPT covalent bond in backbone

2 DO NOT CREDIT if other bond

mentioned to connect between the two

chains

DO NOT CREDIT H

+

bonds

IGNORE strength of bond

3 DO NOT CREDIT thiamine / cysteine /

adenosine

Note:

‘Two bonds between A and T and three

bonds between C and G’ = 2 marks (mp 3

and mp 4)

‘Two hydrogen bonds between A and T

and three hydrogen bonds between C and

G’ = 3 marks (mp 2, mp 3 and mp 4)

Examiner's Comments

Generally this was a well answered

question with candidates recalling correctly

the base pairs and the relevant number of

hydrogen bonds between the pairs. Fewer

candidates were able to describe the

correct location of the phosphodiester bond

in the sugar‐phosphate backbone. A few

candidates were unsure of DNA structure,

incorrectly identifying them as polypeptides

and then going on to list the bonds found in

protein structure.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

Total 9

Question Answer/Indicative content Marks

Guidance

Mark Scheme

15 i viral RNA, acts as, host cell / m, RNA;

RNA, carries, code / sequence (for viral

protein);

(to) ribosomes;

2 max ACCEPT RNA / DNA, produced from viral

RNA

DO NOT CREDIT tRNA

ACCEPT RNA is, translated into / used as

a template to produce, (viral) protein (or

description)

ACCEPT RNA codes for (viral) protein

DO NOT CREDIT tRNA

ACCEPT as a standalone mark

Examiner's Comments

This question presented a challenge both

to the candidates and examiners.

Candidates often could not express the

difference between viral RNA and host

mRNA and many candidates thought that,

contrary to the diagram provided, the virus

contained DNA. Thus both host DNA and

the supposed viral DNA became

entangled. Examiners then had to unravel

which RNA and DNA was being referred to

by the candidates. A little less than half of

candidates described RNA as carrying the

code for protein, often viral protein, a

slightly larger number identified ribosomes

as the ultimate destination of RNA. A

smaller number correctly suggested a

specific role for the viral RNA.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

ii altered base sequence (of viral RNA)

means, altered, primary structure /

(sequence of) amino acids;

R-groups / disulphide bonds / hydrogen

bonds / ionic bonds, interact differently;

tertiary structure is determined by, bonds /

R-groups / secondary structure / primary

structure / sequence of amino acids;

3-D shape is tertiary structure;

3 max ACCEPT if a nucleotide (in RNA) is

different the amino acid (in the protein) is

different

ACCEPT changed as AW for interact

differently

ACCEPT implication that 3D is tertiary

structure

Examiner's Comments

The first marking point was not awarded

often because most candidates failed to

mention the link between base sequences

and amino acid sequences. Close to half

the candidates realised that an alteration in

primary or secondary structure would lead

to an altered tertiary structure and a similar

number linked this to 3D shape. Less than

a quarter of candidates gained the second

marking point – usually for reference to

bonds rather than R-groups.

iii money would be saved / education

improved / fewer sick days / reduced

spread (of virus) / good example of health

practice / few teachers will have immunity

(to current strain);

1 IGNORE so they don't get the flu without

further qualification

IGNORE because they are at risk of

infection

Examiner's Comments

was fairly easy achieved by most

candidates.

Total 6

Question Answer/Indicative content Marks

Guidance

Mark Scheme

16

similarities

S1 (both) have, pentose / 5C, sugar;

S2 (both) have phosphate;

S3 (both) have, adenine / (nitrogenous)

base;

differences

D1 DNA (nucleotide) has deoxyribose

and

ATP has ribose;

D2 DNA (nucleotide) has 1 phosphate

and

ATP has 3 phosphates;

D3 DNA (nucleotide), has (4 possible)

different bases / can have thymine / can

have cytosine / can have guanine

and

ATP has, only 1 (possible) base / adenine;

4 max Accept all mark points from diagrams

that are clearly labelled IGNORE

letters for bases

IGNORE bonds

S2 ACCEPT P

1

/PO

4

(3-)

/

DO NOT CREDIT phosphate heads /

phosphoric acid / phosphorus / P

S3 DO NOT CREDIT adenosine

D2 ACCEPT P

i

/ PO

4

(3-)

/ / ‘a

phosphate group’ for DNA

DO NOT CREDIT phosphate heads /

phosphoric acid / phosphorus / P

D3 DO NOT CREDIT thiamine / cysteine /

adenosine

Note: ‘ATP has adenine but DNA has

adenine or thymine’ = 2 marks (S3 and

D3)

Question Answer/Indicative content Marks

Guidance

Mark Scheme

QWC; 1 Award if 3 of the following terms have been

used in a correct context with correct

spelling:

adenine thymine / cytosine /

guanine

deoxyribose ribose

pentose phosphate

Examiner's Comments

Some excellent answers were seen to this

question with points being made in quick

succession and awarding of the QWC

mark. Common mistakes were to state that

ribose is present in both ATP and DNA

(seeming to treat ribose as synonymous

with pentose), ribose described as a

hexose or not recognising that adenine is a

base and stating that DNA doesn't contain

adenine but does have a nitrogenous base.

Some had not read the question carefully

enough and attempted to distinguish

between ATP and a DNA molecule rather

than a nucleotide.

Total 5

Question Answer/Indicative content Marks

Guidance

Mark Scheme

17

DNA;

polypeptide(s);

tertiary, structure / shape;

3 Mark the first answer. If that answer is

correct and an additional answer is given

that is incorrect or contradicts the correct

answer then = 0 marks

IGNORE chromosomes

IGNORE protein

ACCEPT 3D, shape / structure

IGNORE active site

Examiner's Comments

Examiners were surprised by the number

of candidates who failed to score well on

this question. Often chromosomes and

genetic material were given as the

incorrect answers to DNA for the first ‘fill in

the blank’. Many answers named proteins

instead of polypeptides for the second,

which gained no credit. Lack of detail such

as just mentioning shape and not

extending their answer to tertiary

structure/shape or 3D structure/shape

meant that many candidates did not gain

this mark for the third blank.

Total 3

Question Answer/Indicative content Marks

Guidance

Mark Scheme

18 a i thymine; 1 Examiner's Comments

The vast majority of candidates achieved

the mark.

ii correct complementary sequence;

bases joined by a backbone drawn below

the letters;

2 IGNORE bonds between bases

=2 marks

Examiner's Comments

Around half of responses gained 2 marks.

Most candidates drew the backbone

correctly. A large minority wrote T, where

the U should have been.

b 3 Four ‘X’s – max 2

Five ‘X’s – max 1

Six or more ‘X’s – DO NOT CREDIT any

marks

If candidate does not use ‘X’, ACCEPT

unambiguous statements system of

indicating correct answers.

Examiner's Comments

Most candidates scored 2 marks on this

question. The most commonly awarded

mark was for there not being 3 hydrogen

bonds between A and T. Candidates

usually scored one other mark for noticing

either that RNA nucleotides are not

involved in DNA replication or that DNA is

not a polypeptide. Crosses were seen in

every box.

Total 6

Question Answer/Indicative content Marks

Guidance

Mark Scheme

19 i code for (one or more) polypeptide(s); 1 ACCEPT protein

IGNORE amino acid sequence

Examiner's Comments

Around half of candidates mentioned

protein or polypeptide and gained the

mark.

ii

1 double stranded;

2 each / both (strands) act as template;

3 hydrogen bonds, easily, break / form,

between bases;

4 complementary (specified) base,

pairing / AW;

5 purine (only able to) bind to

pyrimidine;

6 (due to) different sizes of purines and

pyrimidines;

7 hydrogen bonding different between A

& T and C & G

or

3 H bonds between C & G and 2 H

bonds between A & T;

5 max AWARD marks from clearly annotated

diagram

1 ACCEPT double helix or two,

polynucleotides / strands / chains or

antiparallel strands

1 IGNORE one old and one new strand

2 IGNORE either

NOTE ‘there are 2 strands which act as

templates’ = 2 marks (mp 1 and 2)

3 ACCEPT weak H bonds between bases

break

3 IGNORE refs to H bonds, breaking /

forming, without qualification that the

bonds are weak or, form / break, easily

4 IGNORE complementary nucleotides

unless qualified with examples of base-

pairing

7 ACCEPT names of bases with phonetic

spellings

7 DO NOT CREDIT thyamine

7 ACCEPT A=T and C≡G without

reference to hydrogen bonds

Examiner's Comments

Question Answer/Indicative content Marks

Guidance

Mark Scheme

This question required candidates to apply

their knowledge of DNA structure and

replication, which was, in most cases, very

thorough, and identify those aspects of

structure which facilitated accurate

replication. Many candidates clearly found

this challenging and it was rare to award

full marks. Most responses merely

described DNA replication and scored 2 or

3. Lower scoring responses often veered

into discussions of transcription or

translation, or described DNA as a

polypeptide made of amino acids.

Total 6

Question Answer/Indicative content Marks

Guidance

Mark Scheme

20 a AAA TCT GGT; 1

Examiner's Comments

The vast majority of candidates were able

to identify the bases correctly from the

automated sequencing graph, with base 6

labelled as A being the most common

error.

b i the correct bases inserted in all 3 rows

before box;

correctly identifying the last base in each

sequence as the labelled base;

2

Examiner's Comments

This was also well answered by the

majority of candidates, though some

candidates failed to attempt the question.

The most common error was a failure to

inset the correct sequence of bases in all

three rows, before putting in the labelled

base.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

ii electrophoresis;

(negatively-charged DNA) moves towards,

positive electrode / anode;

smallest/smaller (fragments) move, fastest

/ faster; ora

resolution on gel sufficient to register 1,

nucleotide / base;

3 max

ACCEPT positive, end /terminal

IGNORE ref to distance

ACCEPT lightest / shortest

ACCEPT description ‘machine detects

fragments to one base in length’

IGNORE pair

Examiner's Comments

This question proved to be a good

discriminator, with few candidates gaining

full marks. Most candidates realised the

question was asking them to explain the

process of electrophoresis, though some

explained the PCR process instead, or the

process gene sequencing in general.

Many explained that the DNA was

negatively charged so moved towards the

positive anode, but many then failed to get

the third marking point as they stated that

the smallest fragment moved further and

failed to say that they moved fastest,

possibly because this had been the correct

marking point in a previous question. In

this case the question was related to the

use of electrophoresis in sequencing,rather

than as a method of just separating

fragments. A number of candidates lost a

marking point for saying 'the fragments

moved towards the positive cathode'.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

c (mutation) change in (DNA) nucleotide/

base, sequence; (mutation causes) change

in, amino acid sequence / primary structure

(of protein);

change in, tertiary structure/ 3D shape /

binding site, of receptor;

salmeterol unable to bind;

idea that no response triggered in cell / no

second messenger system activated;

3 max IGNORE triplet / codon / gene / frameshift

DO NOT CREDIT active site

ACCEPT salmetorol not complementary

shape to receptor

ACCEPT salmeterol cannot bind as easily

e.g. adenyl cyclase not activated

IGNORE ‘has no effect’

Examiner's Comments

This question discriminated well, with many

candidates gaining 3 marks, for correctly

linking a change in nucleotide sequence to

a change in primary and tertiary structure

of the receptor, so that salmeterol was

unable to bind. A common error was to talk

about proteins in general, rather than

receptors, or to not mention that a mutation

leads to a change in base sequence.

Weaker candidates related their answers

to enzymes, and discussed changes to the

shape of active sites which gained no

credit.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

d i (mutation resulted in) receptor having

complementary shape to montelukast;

montelukast able to bind;

(whereas) salmeterol cannot;

montelukast may have a different receptor;

2 max DO NOT CREDIT active site

IGNORE fit

ACCEPT attach

ACCEPT cannot bind as easily

ACCEPT montelukast receptors not

damaged

Examiner's Comments

Many candidates scored both marks on

this question by applying their knowledge

about receptors to this new situation, and

suggesting that montelukast could bind to

the mutated receptor due to it being

complementary in shape, whilst salmeterol

could not. Other acceptable suggestions

included that montelukast had different

receptors which were unaffected by the

mutation or that the drug worked in a

different way which did not involve the

mutated receptors. A minority of

candidates talked about building up a

resistance to salmeterol or having no

resistance to montelukast, which did not

gain credit.

ii not reliable because, sample size too

small / only 62 children in study;

or

could be reliable because 31 is quite a

large sample;

1

Note

31 is a suitable number for a phase 1 trial

Examiner's Comments

A surprisingly large number of candidates

failed to get this mark, which asked about

the reliability of this trial. Some candidates

talked about the lack of a control group,

length of study or age range of the group,

rather than the fact that the sample size

was too small to be reliable. Generalised

statements such as 'if you increase the

sample size then reliability will be

increased' were not credited.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

iii (epithelial) cells lining cheek; 1 ACCEPT (named) white blood cells in

saliva / salivary gland cells

Examiner's Comments

It was surprising to many examiners that

most candidates struggled to answer this

question. Many candidates believed that

genetic material can be found in proteins

and enzymes, or just in saliva in general.

Few candidates realised that cells from the

cheek, white blood cells and salivary gland

cells can be found in saliva, and that this

can be used as a source of DNA.

Total 13

Question Answer/Indicative content Marks

Guidance

Mark Scheme

21 i

reverse transcriptase;

1 Mark the first answer. If the answer is

correct and an additional answer is given

that is incorrect or contradicts the correct

answer then = 0 marks

DO NOT CREDIT DNA (reverse)

transcriptase

Examiner's Comments

Surprisingly few candidates could correctly

identify the enzyme. A variety of incorrect

answers were given, including DNA / RNA

polymerase, restriction endonuclease and

ligase.

ii 1 mRNA binds to, (gene) probes / cDNA /

ssDNA, by complementary base pairing;

2 idea that the more active the gene the

more mRNA produced;

3 during transcription;

4 more fluorescence indicates more

mRNA (bound);

3 max 1 DO NOT CREDIT in the context of the

gene probe binding to DNA

3 IGNORE translation

Examiner's Comments

This was a challenging question, which few

candidates scored full marks on. Many

missed the point of what an active gene

means in terms of mRNA transcription, and

so referred to the gene binding to the gene

probe, rather than mRNA. Some

candidates did not understand the use of a

gene probe in this context, and referred to

gel electrophoresis and automated gene

sequencing, or just restated the question

by saying the most active gene fluoresced

the most.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

iii 1 dopamine linked to, ADHD / addiction /

risk-taking / adventurous behaviour /

hyperactivity / erratic behaviour (in

humans);

2 idea of common mechanism in bees and

humans (for adventurous behaviour);

3 idea that as they are different organisms

the mechanisms may not be comparable

(even though apparently similar);

4 AVP;

3 max 1 IGNORE ref to schizophrenia /

Parkinson's

This mark is for the effect of the

chemical dopamine, not the dopamine

receptors alone.

2 e.g. both have, DRD4 / dopamine

receptors e.g. dopamine has the same

effect in both

4 e.g. other genes also involved in, bee /

human, behaviour

Note:

‘both have dopamine receptors which are

linked to adventurous behaviour’ = 1 mark

(mp 2 only)

‘both have dopamine receptors and

dopamine is linked to adventurous

behaviour’ = 2 marks (mps 2 & 1)

Examiner's Comments

Many candidates did not link dopamine to

its effect on humans, but rather stated it

was the receptor or gene alone causing the

effect. However, most did gain credit for

identifying a common mechanism in both

humans and bees.

Total 7

Question Answer/Indicative content Marks

Guidance

Mark Scheme

22 a i 3; 1 IGNORE triplet

Examiner's Comments

Almost all candidates achieved this mark

by correctly stating ‘3’.

ii 4

3

or

4 × 4 × 4

or

4 × 4

2

;

1

Examiner's Comments

The majority of students correctly stated

either 4

3

or 4×4×4. Students who did not

achieve this mark usually tried to explain

methods to identify the number of

combinations of bases.

iii Several, triplet(s) / codon(s), code for one

amino acid;

(some are used as) start / stop /

termination;

idea that mutation may,

not result in change in amino acid / have a

neutral effect / result in silent mutation;

2 max Must be clear that base combination is

a group of 3 bases

IGNORE degenerate

DO NOT CREDIT makes / produces /

creates, amino acids

DO NOT CREDIT deletion / insertion (as

would create frame shift)

Examiner's Comments

Most candidates understood what the

question was asking, but lost marks

because they repeated the wording of the

question, using the term ‘base

combinations’ rather than triplet or codon

for mp1. The most common marking point

awarded was for stop codons, and more

able candidates also achieved mp3,

though this was less common.

Question Answer/Indicative content Marks

Guidance

Mark Scheme

iv

adenine / A

and

cytosine / C

and

guanine / G;

1 Mark the first 3 answers.

If the answer is correct and an additional

answer is given that is incorrect or

contradicts the correct answer then = 0

marks

DO NOT CREDIT adenosine

DO NOT CREDIT cysteine

DO NOT CREDIT glycine

Examiner's Comments

It was clear that some candidates did not

know what the phase ‘common to’ meant

and answered which ones were common in

DNA and RNA (AGCTU). Many wrote

‘AGC’ correctly, but those who also wrote

the full names sometimes negated the

mark by writing argenine / adensine /

adenosine or cysteine.

b

transcription

1 DNA / gene, copied / transcribed, into

mRNA;

2 free / activated, (RNA) nucleotides /

(RNA) nucleoside triphosphates;

3 (line up by) complementary base-pairing

/ described;

4 (to) one / template / reference / sense,

(DNA) strand;

5 (catalysed by) RNA polymerase;

translation

6 (mRNA moves to) ribosomes;

7 tRNA (molecules) bind to mRNA;

8 anticodon(s), match / pair with / bind to,

codons;

9 specific / correct, amino acid attached to

tRNA;

10 formation of peptide bond between

amino acids;

6 max Marks may be awarded from an

annotated diagram

1 IGNORE ‘used to make’

2 DO NOT CREDIT DNA nucleotides

3 CREDIT ‘A-T, C-G and A-U’

4 ACCEPT ‘non-coding’ for ‘template’

5 DO NOT CREDIT in context of breaking

H bonds‘

6 CREDIT translation occurs at ribosomes

Note: tRNA anticodons bind to mRNA

codons = 2 marks (mps 7 & 8)

10 DO NOT CREDIT dipeptide /

polypeptide, bond

Question Answer/Indicative content Marks

Guidance

Mark Scheme

QWC; 1 Award QWC if two mps from 1 – 5 have

been awarded before two mps from 6 – 10

Place a tick or a cross alongside the

pencil icon to indicate whether or not

the QWC mark has been awarded.

Examiner's Comments

Candidates had obviously revised this topic

and demonstrated a clear understanding of

transcription and translation, often scoring

in excess of the 6 factual mp's required

and very few did not also gain the QWC

mark as well . Some candidates began

their answer by talking about the ribosome

and giving a detailed account of

translation. Such candidates failed to

realise that the question required an

explanation of both transcription and

translation, and thus did not access several

marking points. Other candidates gave

superficial explanations of transcription and

translation that lacked sufficient detail.

Sequencing activities using statements

written on cards that need to be arranged

in the correct order can help students

develop a logical and detailed

understanding of protein synthesis as a

whole.

Total 12

23 A ✓ 1

Total 1

Question Answer/Indicative content Marks

Guidance

Mark Scheme

24 bond drawn between phosphate and

carbon 3 of sugar

and

labelled phosphodiester bond ✓

two bonds drawn between bases T & A

and

three bonds between C & G

and

labelled hydrogen bonds ✓

2

ACCEPT just one phosphodiester bond

drawn

Total 2

25 base sequence in genes is unchanged ✓

idea that mRNA is inhibited, therefore

translation does not occur ✓

gene is not expressed ✓

2 max

Total 2

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Question Answer/Indicative content Marks

Guidance