Author: Philip Colls
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Gene technology booklet revision questions             NAME:
Q1.
Scientists investigated the role of a protein called CENP-W in mitosis. Their method involved cell fractionation and ultracentrifugation.
(a)     The scientists began by lysing (breaking open) cells and organelles using a detergent that dissolves lipids in water.
Suggest how the detergent releases CENP-W from cells.
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(b)     Explain how ultracentrifugation separates CENP-W from other molecules.
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(c)     CENP-W is involved in the formation of spindle fibres in mitosis. Spindle fibres are made of molecules of a protein called tubulin.
The scientists treated cells in a culture with small interfering RNA (siRNA). This siRNA causes RNA interference of expression of the CENP-W gene. The scientists took samples of cells at 0, 48 and 72 hours after adding the siRNA. They then used gel electrophoresis to separate CENP-W and tubulin from these samples.
The diagram shows the results of the electrophoresis. The size of each band is proportional to the amount of CENP-W or tubulin present.

Suggest how the siRNA produced these results.
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(Total 7 marks)

Q2. (a)     Plasmids can be modified by genetic engineering and inserted into bacteria. These bacteria can then make useful substances normally made by another organism. Explain how modified plasmids are made by genetic engineering and how the use of markers enable bacteria containing these plasmids to be detected.
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(b)     In gene therapy, genes are introduced into a person who has defective genes which do not produce an important substance. Three experiments were done to compare techniques for introducing an important substance into a person with defective genes.
1.   The substance was injected directly.
2.   Harmless viruses carrying genes coding for the substance were injected.
3.   The genes were put into a protein capsule which was inserted into the tissues.
The graph shows results of the experiments.
 
Takahiro Ochiya et al, Biomaterials for Gene Delivery: Studies on Metastasis, 
(National Cancer Centre, Research Institute, Tokyo, Japan) 1999
(i)      Describe the results of the three experiments.
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(ii)     Using the information in the graph, suggest one advantage and one disadvantage of the capsule method compared to the others.
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Disadvantage __________________________________________________
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(Total 11 marks)
Q3. (a)     Describe how a gene can be isolated from human DNA.
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(b)     Describe how an isolated gene can be replicated by the polymerase chain reaction (PCR).
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(c)     (i)      Describe how a harmless virus, genetically engineered to contain a CFTR gene, can be used to insert the gene into a cystic fibrosis sufferer.
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(ii)     A virus used in gene therapy has RNA as its genetic material and has an enzyme called reverse transcriptase. Inside a human cell, reverse transcriptase uses viral RNA to make viral DNA.
Explain why the enzyme is called reverse transcriptase.
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(Total 9 marks)
Q4.
The polymerase chain reaction is a process which can be carried out in a laboratory to replicate DNA. The diagram shows the main stages involved in the polymerase chain reaction.
 
(a)     Explain why DNA is heated to 95 °C.
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(b)     What is the role of
(i)      a primer in this process;
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(ii)     DNA polymerase?
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(c)     (i)      How many DNA molecules will have been produced from one molecule of DNA after 6 complete cycles?
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(ii)     Suggest one use of the polymerase chain reaction.
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(d)     Give two ways in which the polymerase chain reaction differs from the process of transcription.
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(Total 7 marks)
Q5. (a)     Plasmids are often used as vectors in genetic engineering.
(i)      What is the role of a vector?
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(ii)     Describe the role of restriction endonucleases in the formation of plasmids that contain donor DNA.
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(iii)     Describe the role of DNA ligase in the production of plasmids containing donor DNA.
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(b)     There are many different restriction endonucleases. Each type cuts the DNA of a plasmid at a specific base sequence called a restriction site. The diagram shows the position of four restriction sites, J, K, L and M, for four different enzymes on a single plasmid. The distances between these sites is measured in kilobases of DNA.
 
1 kb = 1 kilobase
The plasmid was cut using only two restriction endonucleases. The resulting fragments were separated by gel electrophoresis. The positions of the fragments are shown in the chart below.

(i)      Which of the restriction sites were cut?
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(ii)     Explain your answer.
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Q6. A gene was broken into fragments using enzyme Z. The mixture of fragments produced was then separated by electrophoresis.
(a)     What type of enzyme is enzyme Z?
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The table shows the number of base pairs present in the fragments.
 
Fragment | Number of base pairs (× 10^3)
1 | 4.65
2 | 5.72
3 | 10.71
4 | 2.39
5 | 5.35
6 | 7.53
The diagram shows the electrophoresis gel used. The mixture of fragments was placed at the start point marked S and the process started. The boxes indicate the positions reached by the different fragments.
 
(b)     Explain why base pairs are a suitable way of measuring the length of a piece of DNA.
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(c)     (i)      Write 6 above the appropriate box on the diagram to show the position you would expect fragment 6 to have reached.
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(ii)     Explain how you arrived at your answer.
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(d)     Enzyme Z recognises a particular sequence of bases in the gene. How many times does this sequence appear in the DNA of this gene?
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(Total 6 marks)
Q7.
Read the following passage.
Shark-fin soup is an expensive delicacy. To provide the basic ingredient, fishermen catch the
sharks, hack the fins off and throw the dead bodies back into the ocean. But sharks are slow
to mature and produce only a few offspring at a time, so they are vulnerable to overfishing.
Monitoring the shark-fin trade is difficult, as once a fin has been cut off, it can be extremely
5        difficult to work out precisely from which species it was taken.
The DNA from different species of sharks shows some differences in base sequence. This has
enabled a new genetic fingerprinting technique to be developed. This technique would allow
conservationists and fisheries managers to assess which of the 400 shark species are most
threatened by the trade in shark fins.
10       An identification process has been developed using a range of “primers”. These are short
pieces of single-stranded DNA that are complementary to a particular sequence of DNA.
Each primer is specific to the DNA of one shark species.
The primers are added to DNA taken from a shark’s fin and the polymerase chain reaction is
carried out. Only two primers, one at each end of a certain piece of DNA, will bind. The piece
15      of DNA between the primers is replicated by the polymerase chain reaction. The primers that
bind are specific to a particular species of shark and the length of the DNA fragment
replicated differs for each species. When this DNA is run in an electrophoresis gel it produces
a single band, enabling the researchers to identify which species of shark is involved.
Use information from the passage and your own knowledge to answer the questions.
(a)     (i)      Explain why the DNA for each species of shark shows differences in base sequence (line 6).
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(ii)     Each primer is specific to the DNA of one shark species (line 12).
Explain why a particular primer will only bind to the DNA of one species.
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(iii)     The length of the replicated DNA fragment is different for each species.
Explain why this is important in identifying the shark species involved.
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(b)     In conventional DNA fingerprinting, a series of bands is produced on the electrophoresis gel, resembling the rungs of a ladder. When the DNA in this new genetic fingerprinting technique is run in an electrophoresis gel it produces just one of these ‘rungs’.
Explain the reason for the difference in the number of ‘rungs’ produced.
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(c)     Describe the polymerase chain reaction.
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Q8. Read the following passage.
In 1991, nine skeletons were found in Russia. They were believed to be those of Tsar Nicholas II, his family and staff who were killed in 1917 during the Russian revolution. Very small amounts of DNA were isolated from these skeletons. This DNA was used in the polymerase chain reaction (PCR). Genetic fingerprinting was then carried out on this DNA to identify the skeletons. The chart shows some of the results obtained from the genetic fingerprinting of seven of the skeletons, three children and four adults.
 
Use information from the passage and your own knowledge to answer the following questions.
(a)     Explain why the polymerase chain reaction was used in this investigation.
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(b)     In the polymerase chain reaction, DNA is heated to 95 °C and nucleotides, enzymes and DNA primers are added to the mixture.
(i)      Explain why the DNA is heated to 95 °C.
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(ii)     What are DNA primers?
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(iii)     Why are DNA primers added during the polymerase chain reaction?
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(iv)    What is the advantage of the enzyme used in the polymerase chain reaction being thermostable?
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(c)     Describe how genetic fingerprinting is carried out.
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(d)     All three children on the chart had the same parents. One of the parents was Adult 1.
Which of the other three adults on the chart was the other parent? Give the reason for your answer.
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Q9. ‘Take-all’ is a disease of wheat caused by a fungus. It can cause serious damage to the crop.
There is no gene for resistance to this fungus in wheat. There is, however, a gene for resistance to this fungus present in oats.
The diagram shows how this gene might be transferred to wheat.
 
(a)     (i)      The wheat plant with the resistance gene contains recombinant DNA. What is recombinant DNA?
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(ii)     The plasmids act as vectors for the resistance gene. What is a vector?
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(iii)     Suggest how cells with the resistance gene might be selected.
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(b)     A laboratory has oat plants containing the resistance gene and a supply of plasmids. Describe how bacteria may be produced which have the resistance gene in their plasmids.
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Q10. Plasmids can be used as vectors to insert lengths of foreign DNA into bacteria. The diagram shows how this is achieved.
 
(a)     Name enzyme E.
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(b)     Cut plasmids and lengths of foreign DNA can join. What features of their ends allows them to join?
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(c)     Draw three different structures that could be formed by incubating cut plasmids and lengths of foreign DNA with ligase. Use the spaces provided on the diagram.

 
 
(3)
(Total 6 marks)
Q11.
Read the following passage.

DNA tests were used to confirm the identity of deposed Iraqi leader Saddam Hussein, after
his capture in December 2003. DNA tests were carried out to prove the suspect was not one
of the many alleged “look alikes” of the former leader.

5

Firstly, the DNA was extracted from the mouth of the captured man using a swab. Great care
was taken to check that the swab did not become contaminated with any other DNA. DNA
extracted from the swab was then subjected to a standard technique called the polymerase
chain reaction (PCR), which takes a couple of hours. Lastly, the sample was “typed” to give
the genetic fingerprint. This was produced within 24 hours of Saddam Hussein’s capture.
Tests for use in criminal cases often take much longer because samples are very small or

10 | contaminated.

It appears that Hussein’s genetic fingerprint was already stored away for comparison. This was
obtained from personal items such as his toothbrush. DNA from the toothbrush would have
been subjected to PCR before a DNA fingerprint could have been obtained.

Source: adapted from SHAONI BHATTACHARYA, New Scientist 15 December, 2003
          Use information from the passage and your own knowledge to answer the questions.

(a)     Describe how the technique of genetic fingerprinting is carried out and explain how 
it can be used to identify a person, such as Saddam Hussein.
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(b)     Explain how DNA could be present on a toothbrush (line 12).
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(c)     (i)      Explain why the polymerase chain reaction was used on the sample of DNA from the toothbrush (lines 12-13).
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(ii)     Explain one way in which the polymerase chain reaction differs from DNA replication in a cell.
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(d)     Tests for use in criminal cases often take much longer because samples are very small or contaminated (lines 8-10). Explain why it takes longer to obtain a genetic fingerprint if the sample is
(i)      very small;
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(ii)     contaminated.
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(Total 15 marks)
Q12.
A husband and wife wanted to know whether they were carriers of the mutated form of a gene. This mutation is a deletion that causes a serious inherited genetic disorder in people who are homozygous.
A geneticist took samples of DNA from the husband and the wife. He used a DNA probe to look for the deletion mutation. The DNA probe was specific to a particular base sequence in an exon in the gene. Exons are the coding sequences in a gene.
The geneticist compared the couple’s DNA with that of a person known not to carry this mutation.
The chart shows the geneticist’s results.
 
(a)     The geneticist told the couple they were both carriers of the mutated gene.
Explain how he reached this conclusion.
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(b)     The DNA probe the geneticist used was for an exon in the DNA, not an intron. Explain why.
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(c)     To make the DNA probe, the geneticist had to find the base sequence of the normal gene. Once he had copies of the gene, what methods would he use to find the base sequence of the gene?
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(Total 8 marks)

Q13. Read the following passage.

The giant panda is one of the rarest animals in the world and is considered to be on the brink of extinction in the wild. Giant pandas have been kept and bred in zoos with the hope that they could be released into the wild. One worry is that small populations, like those in zoos, reduce the genetic variation needed to allow the species to adapt to changing conditions. Unfortunately, pandas find it difficult to reproduce in captivity. Fertilisation of the females is guaranteed only by insemination with semen from several males. With so many potential fathers, the true paternity of the cubs is not clear. It is important to identify the fathers to maintain genetic variation.

Panda faeces can be collected in the wild. The faeces contain DNA from the panda, from the bamboo on which they feed and from bacteria. The DNA is subjected to the polymerase chain reaction (PCR). The primers used attach only to the panda DNA. The resulting DNA is subjected to genetic fingerprinting. This can help us to count the number of individuals in the wild because it allows us to identify individual pandas.

5

 

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Use information in the passage and your own biological knowledge to answer the questions.
(a)     Describe how genetic fingerprinting may be carried out on a sample of panda DNA.
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(b)     (i)      Explain how genetic fingerprinting allows scientists to identify the father of a particular panda cub.
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(ii)     When pandas are bred in zoos, it is important to ensure only unrelated pandas breed. Suggest how genetic fingerprints might be used to do this.
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(c)     (i)      Suggest why panda DNA is found in faeces. (line 10)
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(ii)     Explain why the PCR is carried out on the DNA from the faeces. (line 12)
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(iii)     Explain why the primers used in the PCR will bind to panda DNA, but not to DNA from bacteria or bamboo. (line 12)
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(d)     DNA from wild pandas could also be obtained from blood samples. Suggest two advantages of using faeces, rather than blood samples, to obtain DNA from pandas.
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(2)  (Total 15 marks)
Q14. There are wolves in many European countries. Scientists investigated the genetic diversity of these wolves. They collected samples of DNA from the mitochondria of wolves from different countries. For each sample they identified which haplotypes were present in the DNA. A haplotype is a particular sequence of bases on DNA. Mutations can produce new haplotypes.

Country

Number of wolves
sampled

Number of different
haplotypes in
mitochondrial DNA

Spain | 84 | 3
Portugal | 19 | 2
Italy | 101 | 1
France | 7 | 1
Bulgaria | 29 | 6
Sweden | 93 | 1
The scientists wanted to find out whether one of the haplotypes in the Portuguese wolves was the same as one of those in the Spanish wolves. They used a restriction endonuclease, electrophoresis and a labelled DNA probe.
(a)     For what purpose did they use
(i)      the restriction endonuclease
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(ii)     electrophoresis?
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(b)     Explain why the labelled DNA probe could be used to find out whether the haplotypes were the same.
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(c)     The scientists analysed the DNA on the Y chromosome and the DNA in the mitochondria of the Swedish wolves. They concluded that the Swedish wolf population descended from one male wolf from Finland and one female wolf from Russia.
(i)      Explain why DNA on the Y chromosome helped them to reach this conclusion.
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(ii)     Suggest why DNA in the mitochondria helped them to reach this conclusion.
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Wolves eat different mammals. An ecologist investigated factors that affect wolf numbers in North America. He collected data from different field studies carried out in different places. The graph shows his results.
 
(d)     (i)      The wolf numbers are given per unit area. Explain why.
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(ii)     The ecologist calculated the total prey index for each of the places that had been studied. In order to do this, he gave each prey species a value based on how much food was available to wolves from the prey animal concerned. He called this value the prey index.
The ecologist considered that the prey index gave a better idea of the food available than the prey biomass in kg. Suggest why the prey index gives a better idea of food available.
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(e)      The ecologist calculated the total prey index by combining the prey indices and the total number of animals of each species present in 1000 km^2. He plotted this information on the graph. What does the graph suggest about the factors that determine wolf numbers in North America? Explain your answer.
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(Total 12 marks)
Q15. Scientists wanted to measure how much mRNA was transcribed from allele A of a gene in a sample of cells. This gene exists in two forms, A and a.
The scientists isolated mRNA from the cells. They added an enzyme to mRNA to produce cDNA.
(a)     Name the type of enzyme used to produce the cDNA.
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The scientists used the polymerase chain reaction (PCR) to produce copies of the cDNA. They added a DNA probe for allele A to the cDNA copies. This DNA probe had a dye attached to it. This dye glows with a green light only when the DNA probe is attached to its target cDNA.
(b)     Explain why this DNA probe will only detect allele A.
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(c)     The scientists used this method with cells from two people, H and G.
One person was homozygous, AA, and the other was heterozygous, Aa.
The scientists used the PCR and the DNA probe specific for allele A on the cDNA from both people.
The figure shows the scientists’ results.
 
(i)      Explain the curve for person H.
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(ii)     Which person, H or G, was heterozygous, Aa? Explain your answer.
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(Total 8 marks)
Q16. One way to detect and measure accurately the amount of RNA in a tissue sample is by RT-PCR (reverse transcriptase-polymerase chain reaction).
RT-PCR uses a reaction mixture containing:
•   the sample for testing
•   reverse transcriptase
•   DNA nucleotides
•   primers
•   DNA polymerase
•   fluorescent dye.
The principle behind this method is shown in Figure 1.
 
(a)     Explain the role of reverse transcriptase in RT-PCR.
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(b)     Explain the role of DNA polymerase in RT-PCR.
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(c)     Any DNA in the sample is hydrolysed by enzymes before the sample is added to the reaction mixture. Explain why.
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(d)     Figure 2 shows the results from using RT-PCR to detect RNA in two different samples, A and B.
 
A quantitative comparison can be made of the amount of RNA in samples A and B. This involves determining the number of cycles required to reach 50% maximum concentration of DNA (C). The amount of RNA in a sample can be measured as: 
Use this information to calculate the ratio for RNA content in sample A : RNA content in sample B.

Answer ___________
(2)
(e)     Suggest one reason why DNA replication stops in the polymerase chain reaction.
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(f)     Scientists have used the RT-PCR method to detect the presence of different RNA viruses in patients suffering from respiratory diseases.
The scientists produced a variety of primers for this procedure.
Explain why.
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(Total 9 marks)
Q17.
People suffering from pituitary dwarfism do not make enough human growth hormone (HGH). They can be treated using injections of HGH.
A geneticist wants to transform the bacterium, Escherichia coli, to make HGH by adding the gene coding for HGH.
The geneticist could obtain the HGH gene using any one of three methods.
1.      Use restriction enzymes to cut out a fragment of DNA containing the HGH gene from a human genome.
2.      Convert mRNA for HGH into cDNA using reverse transcriptase.
3.      Create the HGH gene using a ‘gene machine’.
(a)     The geneticist decided not to use restriction enzymes to cut out a fragment of DNA containing the HGH gene from a human genome. She made this decision because only methods 2 and 3 would produce DNA that E. coli could use to make HGH.
Explain why only methods 2 and 3 would produce DNA that E. coli could use to make HGH.
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(b)     The geneticist concluded it would be faster to create the HGH gene using a gene machine than by using reverse transcriptase to convert mRNA for HGH into cDNA.
Suggest why the geneticist reached this conclusion.
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(c)     After obtaining copies of the HGH gene, the geneticist will attempt to insert them into plasmid vectors.
Describe how the geneticist would attempt to insert copies of the HGH gene into these plasmids.
(3)
(d)     The geneticist plans to use the plasmids containing the HGH gene to try to transform cells of E. coli. She knows that some E. coli might not take up the plasmid.
To enable her to identify which bacteria have taken up the plasmid with the HGH gene, the plasmids she intends to use contain a gene that codes for a green fluorescent protein (GFP). Bacteria that contain this plasmid glow green under UV light.
Suggest one advantage of using this gene for GFP to identify bacteria that have taken up plasmids.
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The diagram below shows part of the plasmid containing the gene that codes for GFP. It also shows the roles of two genes that control the GFP gene.
 
(e)      Arabinose is a sugar that can bind to the araC protein.
Use information in the diagram to suggest why the geneticist must include arabinose in the agar on which she hopes to grow E. coli containing the transgenic plasmids.
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(Total 9 marks)
Q18.
Some populations of flies are becoming resistant to insecticides intended to kill them.
Scientists developed a method for finding out whether a fly was carrying a recessive allele, r, that gives resistance to an insecticide. The dominant allele, R, of this gene does not give resistance.
The scientists:
•        crossed flies with genotype RR with flies with genotype rr
•        obtained DNA samples from the parents and offspring
•        used the same restriction endonuclease enzymes on each sample, to obtain DNA fragments.
(a)     Explain why the scientists used the same restriction endonuclease enzymes on each DNA sample.
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The scientists added two different primers to each sample of DNA fragments for the polymerase chain reaction (PCR).
•        Primer A3 only binds to a 195 base-pair fragment from allele r.
•        Primer A4 only binds to a 135 base-pair fragment from allele R.
The scientists separated the DNA fragments produced by the PCR on a gel where shorter fragments move further in a given time.
Their results are shown in Figure 1.

Figure 1
 
(b)     Explain why primer A3 and primer A4 only bind to specific DNA fragments.
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(c)     Use all the information given to explain the results in Figure 1.
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(d)     The scientists wanted to know on which chromosome the gene with alleles R and r was located. From the flies with genotype RR, they obtained cells that were in mitosis and added a labelled DNA probe specific for allele R. They then looked at the cells under an optical microscope.
Explain why they used cells that were in mitosis.
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(2)
(e)     Another group of scientists thought that pesticide resistance in some flies was related to increased activity of an enzyme called P450 monooxygenase (PM).
This enzyme breaks down insecticides.
The scientists obtained large numbers of resistant and non-resistant flies. They then set up the following experiments.
•        Non-resistant flies exposed to insecticide.
•        Resistant flies exposed to insecticide.
•        Resistant flies treated with an inhibitor of PM and then exposed to insecticide.
They then determined the percentage of flies that were dead at different times after being exposed to insecticide.
Figure 2 shows their results.
Figure 2
 
(i)      Explain why the scientists carried out the control experiment with the non-resistant flies.
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(2)

(ii)     The scientists concluded that the resistance of the flies to the insecticide is partly due to increased activity of PM but other factors are also involved.
Explain how these data support this conclusion.
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_____________________________________________ (4) (Total 15 marks)
Q19. Agrobacterium tumefaciens is a bacterium that is often used in recombinant DNA technology to produce transformed plants that benefit humans.
A. tumefaciens contains a plasmid which can be used as a vector to transfer a desired gene into plant cells. These plant cells may then develop into plants which produce the protein coded for by the desired gene.
The diagram outlines this process.
 
(a)     (i)      In stage 1, an enzyme is used to cut open the plasmid.
Name the type of enzyme used to cut open the plasmid.
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(1)
(ii)      In stage 1, another enzyme is used to insert the desired gene into the plasmid DNA. 
Name the type of enzyme used to insert the gene into the plasmid.
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(1)
(b)     In stage 4, some plant cells had plasmid DNA only in their cytoplasm. In other plant cells, the plasmid DNA had become inserted into plant DNA in the nucleus.
In stage 5, only cells with plasmid DNA inserted into the plant DNA in the nucleus grew into plants where all the cells contained the desired gene.
Explain why some of the plants in stage 5 contained the desired gene in all of their cells and others did not.
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(3)
(c)     The desired gene in the diagram was from an insect. In stage 6, the plant containing this gene was able to use it to synthesise an insect protein.
The plant is able to synthesise the insect protein. Explain why this is possible.
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(3)
(Total 8 marks)

Mark schemes
Q1.
(a)     1.      Cell membranes made from phospholipid;
Note: candidates may make both mark points in once statement
2.      (Detergent) dissolves membranes / phospholipid (bilayer);
Ignore breaks down
Reject hydrolysis
2
(b)     1.      Spin (liquid / supernatant) at (very) high speed / high g;
Need context of high, not just ‘faster’, ‘higher’ in context of use of bench centrifuge
Accept high centrifugal force
Award 1 max if ref. to organelle separation
2.      Molecules / CENP-W separates depending on (molecular) mass / size / density;
Accept weight
2
(c)     1.      siRNA binds to / destroys mRNA for CENP-W;
Reject if siRNA binds to gene / DNA
Context is important, siRNA acts on mRNA for CENP-W, not tubulin.
Ref. to CENP-W required once for MP1 and MP2
2.      Prevents translation of CENP-W;
Context is important, siRNA acts on mRNA for CENP-W, not tubulin.
Ref. to CENP-W required once for MP1 and MP2
Accept reduces translation of CENP-W
3.      (After / as) CENP-W reduces so does tubulin production;
3
[7]
Q2.
(a)     isolate wanted gene / DNA from another organism / mRNA from cell / organism;
using restriction endonuclease / restriction enzyme / reverse transcriptase to get DNA and produce sticky ends;
use ligase to join wanted gene to plasmid;
also include marker gene e.g. antibiotic resistance;
add plasmid to bacteria to grow (colonies)then (replica) plate onto medium where the marker gene is expressed;
bacteria / colonies not killed have antibiotic resistance gene and (probably) the wanted gene;
6
(b)     (i)      injection, rapid rise and fall;
virus, slower rise and longer in effective / harmful range;
capsule slowest rise, longest in effective / harmful range;
injection and virus give harmful concentrations but capsule does
not;
3 max
(ii)     advantage e.g.:
substance never reaches harmful levels / no side effect / less likely to harm the organism, longer relief from symptoms / less frequent treatment needed / longer effective range / longer but without harmful side effects;
1 max
disadvantage e.g.:
takes longer to take effect;
1
[11]
Q3.
(a)     use restriction enzyme / endonuclease / named, e.g. Bam / Eco;
to cut DNA in specific place / base sequence;
2
(b)     heat DNA to 90 – 95 °C;
strands separate;
add primers;
and nucleotides;
cool so that primers bind to DNA;
(DNA) polymerase forms new strands / joins nucleotides;
4 max
(c)     (i)      virus is inhaled / sprayed into the lungs;
gets into cells, inserting the healthy gene;
2
(ii)     makes DNA from RNA
rather than other way round
1
[9]
Q4.
(a)     to separate the two strands / break hydrogen bonds;
1
(b)     (i)      enables replication / sequencing to start (allow keeps strands
separate);
1
(ii)     joins DNA nucleotides (not complementary bases);
1
(c)     (i)      64;
1
(ii)     replication of DNA from crime scene / tissue sample / 
for DNA sequencing / gene cloning;
1
(d)     (transcription uses) RNA polymerase;
RNA nucleotides / uracil;
one (template) strand / PCR both strands;
start / stop codons;
(accept enzyme separates strands)
2 max
[7]
Q5.
(a)     (i)      transfer / carry genes from one organism to another / into
bacteria / cells;
1
(ii)     cut open plasmid;
cut donor DNA, to remove gene / length of DNA;
cut donor DNA and plasmid with the same enzyme / enzyme
that cuts at the same base sequence;
sticky ends / (overhanging) ends with, single strand / bases exposed;
association / attachment / pairing of complementary strand;
2 max
(iii)     annealing / splicing / backbones joined / phosphodiester bonds;
1
(b)     (i)      L and M;
1
(ii)     fragments 64 and 36(kilobases obtained)
1
[6]
Q6.
(a)     Endonuclease / restriction enzyme;
1
(b)     DNA made of base pairs;
Each base pair is same length / occupies same distance
along backbone;
2
(c)     (i)      Second blank box from left labelled 6;
1
(ii)     Distance moved depends on length / number of base pairs / 
second longest fragment / second shortest distance identified;
1
(d)     5;
1
[6]
Q7.
(a)     (i)      Different genes / characteristics / features;
Reference to mutations;
Or
Base sequence determines protein;
Different species have different protein sequences;
max 2
(ii)     Primer has different DNA sequence;
DNA specific / complementary base-pairing;
2
(iii)     Electrophoresis separates DNA;
(So they can be) identified by position on gel;
Smaller / shortest fragments travel furthest / quicker / or
reverse argument;
3
(b)     (conventional) Many lengths / all DNA / (new) one length;
Each rung is DNA of one / specific length;
2
(c)     1 Heat DNA;
2 Breaks hydrogen bonds / separates strands;
3 Add primers;
4 Add nucleotides;
5 Cool;
6 (to allow) binding of nucleotides / primers;
7 DNA polymerase;
8 Role of (DNA) polymerase;
9 Repeat cycle many times;
max 6
[15]
Q8.
(a)     only small amounts obtained / PCR increases the amount / mass of DNA;
so enough DNA available for genetic fingerprinting;
2
(b)     (i)      to separate the two strands of the DNA / 
to break the hydrogen bonds;
(Reject “unzip”)
1
(ii)     short lengths / fragments of DNA / nucleotides / 
single stranded DNA;
1
(iii)     to mark beginning and / or ends of the part of DNA needed / 
for attachment of enzymes or nucleotides / initiator / 
keeps strands apart;
1
(iv)    would not be denatured;
must be heated to 95 °C / must withstand high temps;
2
(c)     1 DNA extracted from sample;
2 DNA cut / hydrolysed into segments using restriction endonucleases;
3 must leave minisatellites / required core sequences intact;
4 DNA fragments separated using electrophoresis;
5 detail of process e.g. mixture put into wells on gel and electric 
   current passed through;
6 immerse gel in alkaline solution / two strands of DNA separated;
7 Southern blotting / cover with nylon / absorbent paper (to absorb DNA);
8 DNA fixed to nylon / membrane using uv light
9 radioactive marker / probe added (which is picked up by required
   fragments) / complementary to minisatellites;
10 (areas with probe) identified using X-ray film / autoradiography;
max 6
(d)     adult 3;
this is only one which, (with number 1), can provide (all) the DNA
fragments which children have / all bars match;
(Reject ‘genes’)
2
[15]
Q9.
(a)     (i)      contains genes / nucleotides / sections of DNA / artificial
DNA from two species / 2 types of organisms;
1
(ii)     carries gene / DNA (into the other organism / gene carrier);
1
(iii)     expose cells to the fungus;
non-resistant ones die, resistant ones survive;
OR identify by adding marker gene / gene probe / (qualified)
marker probe; description of positive result
e.g. radioactivity / fluorescence / complementary base pairing;
2
(b)     EITHER      1 cut desired gene (from DNA) of oat plant;
                   2 using restriction endonuclease / restriction enzyme;
OR             1 use mRNA from oat which will code for resistance;
                   2 and use reverse transcriptase to form desired DNA;
OR             1 make artificial DNA with correct sequence of bases;
                   2 using DNA polymerase;
                   3 cut plasmid open;
                   4 with (same) restriction endonuclease / restriction enzyme;
                   5 ref. sticky ends / unpaired bases attached;
                   6 use (DNA) ligase to join / ref. ligation;
                   7 return plasmid to (bacterial) cells;
                   8 use of Ca^2+ / calcium salts / electric shock;
                   (if ref. to ‘insulin’ allow 5 max.)
max 6
[10]
Q10.
(a)     restriction (enzyme) / endonuclease / named example;
1
(b)     unpaired bases / sticky ends / staggered;
complementary / explained;
2
(c)     1 mark for each correct outcome
plasmid with foreign DNA joined in ring;
ring with plasmid only; ring of foreign DNA only;
ignore linear structures
3
[6]
Q11.
(a)     1.      DNA is cut;
2.      using restriction enzyme;
3.      electrophoresis;
4.      separates according to length / mass / size;
5.      DNA made single-stranded;
6.      transfer to membrane / Southern blotting;
7.      apply probe;
8.      radioactive / single stranded / detected on film / fluorescent;
9.      reference to tandem repeats / VNTRs / minisatellites;
10.    pattern unique to every individual;
6 max
(b)     cells on toothbrush;
DNA present in cell;
2
(c)     (i)      toothbrush gives small sample of DNA / need more DNA
for analysis;
PCR gives many copies;
2
(ii)     uses heat;
to separate strands;
OR
PCR replicates pieces of DNA;
because DNA has been cut;
OR
primer added in PCR;
to initiate replication
2 max
(d)     (i)      PCR / amplification needed;
1
(ii)     other DNA present; need to identify ‘required’ DNA from rest;
2
[15]
Q12.
(a)     1.      Carriers are heterozygous / have one normal copy and one mutant copy of gene / have one recessive allele / don't have the condition;
2.      Both have DNA that binds (about) half / 50% amount of probe (that non-carrier does);
3.      Probe binds to dominant / healthy allele so only one copy of exon in their DNA / have one copy of gene without exon / base sequence for probe to bind to;
3. Accept normal and gene
3. Accept have a deletion mutation
3
(b)     1.      Introns not translated / not in mRNA / (exons) code for amino acids / introns do not code for amino acids;
1. Accept not expressed
1. Accept polypeptide / protein for amino acids
2.      Mutations of these (exons) affect amino acid sequences (that produce) faulty protein / change tertiary structure of protein;
2. Accept deletion leads to frameshift
2. In this context, accept affects protein made
3.      So important to know if parents’ exons affected, rather than any other part of DNA / introns;
Accept converse arguments involving - eg introns do not code for amino acids / proteins
Reject references to making amino acids, once
3
(c)     1.      Restriction mapping / described;
2.      DNA / base sequencing (of fragments) / description / name of method;
2
[8]
Q13.
(a)     1       DNA is cut;
2       Using restriction enzyme;
3       Use electrophoresis;
4       Separates according to length / mass;
5       Southern blotting / transfer to (nylon) membrane;
6       Make single-stranded;
7       Apply probe;
8       Radioactive / fluorescent;
9       Reference to tandem repeats / VNTRs / minisatellites;
10     Autoradiography / eq; 
8 and 10 should be consistent
max 6
(b)     (i)      All bands in cub which don’t come from mother;
Must be in father’s DNA fingerprint;
Principle that all bands in cub must come from mother and father = 1
2
(ii)     Select pairs with dissimilar DNA fingerprints;
1
(c)     (i)      Cells (from panda) in faeces / gut cells / blood cells;
1
(ii)     To increase amount of DNA / only small amount present;
1
(iii)     DNA / primer has specific base-sequence;
Reference to specific / complementary base-pairing;
2
(d)     Taking samples from animals causes stress / injury to animal;
Difficult to find animals;
Pandas are dangerous / threat to human;
max 2
[15]
Q14.
(a)     (i)      To cut the DNA;
Reject breakdown, cutting out
1
(ii)     To separate the (pieces of) DNA;
1
(b)     Complimentary base sequence / complementary DNA; binds to both (haplotypes);
Label would show up in both;
Idea of complimentarity required
2
(c)     (i)      Y chromosome inherited / comes from male parents / only found in males;
1
(ii)     Mitochondria in egg / female gamete / no mitochondria come from sperm / male gamete;
1
(d)     (i)      Allows comparison;
Different (sized) areas covered;
2
(ii)     Wolves do not eat all of prey animal / do not eat (large) bones / skin;
Inedible parts make up different proportions / wolf eats different proportions;
2
(e)      Limited by food / prey; as prey increases so do wolf numbers / positive correlation;
Large range so other factors involved;
2
[12]
Q15.
(a)     Reverse transcriptase;
1
(b)     1.      Probe (base sequence) complementary (to DNA of allele A / where A is (and) binds by forming base pairs / hydrogen bonds;
Accept gene A
2.      So (only) this DNA labelled / has green dye / gives out (green) light;
Accept glows for green light
2
(c)     (i)      1.      More probe binding / more cDNA / mRNA / more allele / gene A means more light;
2.      DNA (with A) doubles each (PCR) cycle;
3.      So light (approximately) doubles / curve steepens more and more (each cycle) / curve goes up exponentially / increases even faster;
3
(ii)     (G because)
1.      (Heterozygous) only has half the amount of probe for A attaching / only half the amount of DNA / allele A (to bind to);
Accept only one A to bind to
2.      (So,) only produced (about) half the light / glow / intensity (of H) (per cycle of PCR);
If reference to ‘half’ for point 1, allow ‘less light’ in 2.
2
[8]
Q16.
(a)     Produces (c)DNA using (m)RNA;
Accept: ‘converts’ (m)RNA to (c)DNA.
Reject: tRNA
1
(b)     Joins nucleotides to produce (complementary strand/s of) DNA;
Accept: ‘joins DNA nucleotides’.
1
(c)     1.      To remove any DNA present;
2.      As this DNA would be amplified / replicated;
1.      Must be idea of removal / destruction.
2.      Accept: idea of DNA not being used as template.
2
(d)     1.      Ratio in range of 1.4 :1 to 1.5 :1 = 2 marks;
2.      One mark for answers which shows incorrect ratio but
Shows 0.24 as a number or line on the graph
OR
Ratio in correct range, but the wrong way round
OR
Ratio in correct range but not expressed to 1
OR
Ratio shown the other way round in range
1: 0.67 to 1:0.71;
Note: ratio not expressed to 1 in correct range may be shown in different ways, for example as:
3:2 or simply as 1.5 for one mark.
2
(e)     Limited number of primers / nucleotides;
Accept: DNA polymerase (eventually) denatures
Accept: primers / nucleotides ‘used up’.
1
(f)      1.      Base sequences differ;
2.      (Different) complementary primers required;
1.      Accept: reference to either RNA or DNA base sequences but reject reference to DNA base sequence in viruses.
2
[9]
Q17.
(a)     1.      Human DNA / human gene / HGH gene contains introns
OR
Methods 2 and 3 produce DNA / HGH without introns;
2.      E. coli cannot remove introns / cannot splice mRNA / cannot splice pre-mRNA;
2
(b)     Faster to use gene machine than all the enzyme-catalysed reactions (involving reverse transcriptase);
Accept extra step / more steps involved in isolating mRNA
1
(c)     1.      Cut the plasmid with a restriction endonuclease;
Allow ‘add base sequences to blunt ends of plasmid and HGH gene’
2.      (So that) both have complementary / sticky ends;
3.      (Mix together) and add ligase to join the complementary / sticky ends;
3
(d)     Can quickly identify transformed bacteria using UV light;
1
(e)     1.      Arabinose alters structure of araC protein / reduces effect of araC protein;
2.      So stops / reduces inhibition of promoter gene and GFP gene is transcribed;
OR
So stops / reduces inhibition of promoter gene and GFP is produced;
2
[9]
Q18.
(a)     1.      Cut (DNA) at same (base) sequence / (recognition) sequence;
Accept: cut DNA at same place
2.      (So) get (fragments with gene) R / required gene.
Accept: ‘allele’ for ‘gene’ / same gene
2
(b)     1.      Each has / they have a specific base sequence;
2.      That is complementary (to allele r or R).
Accept description of ‘complementary’
2
(c)     1.      Fragments L from parent rr, because all longer fragments / 195
         base pair fragments;
Ignore: references to fragments that move further / less, require identification of longer / shorter or 195 / 135
Accept: (homozygous) recessive
2.      Fragments N from parent RR, because all shorter fragments / 135 base pair fragments;
1 and 2 Accept: A3 for 195 and A4 for 135
2. Accept: (homozygous) dominant
3.      (M from) offspring heterozygous / Rr / have both 195 and 135 base pair fragments.
Accept: have both bands / strips
Reject: primer longer / shorter
3
(d)     1.      (Cells in mitosis) chromosomes visible;
2.      (So) can see which chromosome DNA probe attached to.
2
(e)     (i)      1.      For comparison with resistant flies / other (two) experiments
        / groups;
Ignore: compare results / data / no other factors
2.      To see death rate (in non-resistant) / to see effect of insecticide in non-resistant / normal flies.
Accept: ‘pesticide’ as ‘insecticide’
Accept to see that insecticide worked / to see effect of enzyme
2
(ii)     (PM must be involved because)
1.      Few resistant flies die (without inhibitor);
2.      More inhibited flies die than resistant flies;
3.      (PM) inhibited flies die faster (than resistant flies);
(Other factors must be involved because)
4.      Some resistant flies die;
5.      But (with inhibitor) still have greater resistance / die slower than non-resistant flies.
Accept: (with inhibitor) die slower than non-resistant flies
4 max
[15]
Q19.
(a)     (i)      Restriction endonuclease;
1
(ii)     (DNA) ligase;
1
(b)     (For those plants that contained the desired gene in the nucleus/plant DNA)
1.      (DNA of desired gene) copied/replicated with host DNA/inside nucleus;
2.      Passed on by mitosis/plant grows by mitosis;
3.      Produces genetically identical cells/clones;
Ignore references to protein synthesis or plasmids not taking up the gene
1. Accept DNA replication during mitosis
1. and 2. Accept converse for plants with the gene in the cytoplasm
3. Neutral ‘identical unqualified’
3. Accept description, e.g., DNA is the same
3
(c)     1.      Genetic code is universal/triplets in DNA always code for same amino acid;
2.      It/insect DNA can be transcribed;
3.      Can be translated (process/mechanism same in all organisms/cells);
2. Accept (basic) transcription (process/mechanism) same in all organisms/cells;
2. Accept descriptions of process
3. Accept descriptions of process
3
[8]