Recombinant DNA technology

People with the inherited blood disorder haemophilia A (HA) have a reduced ability to clot blood. One variant of HA is due to an inversion mutation in the F8 gene. This mutation causes the production of a non-functional form of the protein, factor VIII. Factor VIII is not an enzyme.

Explain how an inversion mutation can cause the production of non-functional factor VIII.

Currently, the most effective treatment for HA is weekly or monthly injections of factor VIII. Scientists investigated the effectiveness of a gene therapy product, Roctavian, as a treatment for HA. Roctavian contains a viral vector that has the DNA coding for factor VIII.

The scientists:

• selected 112 male volunteers with severe HA, who were at least 18 years of age

• treated each volunteer with a single injection of Roctavian into their blood

• recorded the number of bleeding episodes (internal bleeding) in these volunteers over a two-year period (set A results)

• compared these data with the number of bleeding episodes per year in the volunteers before receiving Roctavian (set B results).

Table 1 shows the scientists’ results.

A value of ±2 × SD from the mean includes over 95% of the data.

Use all the information provided to evaluate the use of Roctavian as a treatment for HA.

Haemophilia C is an autosomal recessive condition.

Two parents are both heterozygous for the haemophilia C allele.

Calculate the probability of these parents producing a son who has haemophilia C.

Bam H1 and Hin dIII are both restriction endonucleases.

Figure 4 shows the positions where these enzymes cut a linear molecule of DNA.

In two experiments, multiple copies of the DNA molecule shown in Figure 4 were all completely cut into fragments using these restriction enzymes. The DNA fragments produced were then separated by electrophoresis.

Experiment 1 – DNA cut into fragments using Bam H1.

Experiment 2 – DNA cut into fragments using Bam H1 and Hin dIII.

Complete Figure 5 to show the relative positions of the bands following

electrophoresis in experiments 1 and 2.

Suggest how you could determine the size of the different DNA fragments produced in these experiments.

How many DNA fragments would be produced in experiment 2 if the original DNA molecule was a plasmid?

Describe how restriction endonuclease and DNA ligase are used to insert a gene into a plasmid.

Short tandem repeats (STRs) are short sequences of DNA, usually 2 to 7 base pairs. STRs are repeated a number of times, one after another. For example, the STR D5S818 is made up of AGAT repeated 7 to 16 times.

STRs are found throughout the whole genome.

The repeated sequences in STRs are common to all humans. Due to variation in the number of repeats, STRs can be used in genetic fingerprinting.

Describe how STRs could be removed from a sample of DNA.

Genetic fingerprinting using STRs requires amplification of the STRs using the polymerase chain reaction (PCR). The short base sequences either side of a specific STR are known.

Explain the importance of knowing these base sequences in PCR.

A single STR molecule consisting of a 12 base pair sequence of AGAT was amplified 50 times using PCR.

Calculate the total number of base pairs in all the STR molecules after 50 cycles of PCR.

Show your working.

During genetic fingerprinting, the different STRs are separated by gel electrophoresis.

Give two features of STRs which enable them to be separated by gel electrophoresis.

Complete the following definitions.

Recombinant DNA technology can involve the transfer of fragments of human DNA into bacteria. The bacteria are then used to produce human proteins.

Give two reasons why bacteria are able to use human DNA to produce human proteins.

Suggest and explain one reason why bacteria might not be able to produce every human protein.

Antithrombin is a protein. Antithrombin prevents blood from clotting too much.

Some people have a deficiency of antithrombin in their blood, so they need to inject the protein.

Genetically modified goats are used to produce this protein. The human antithrombin gene is transferred into goat embryos. The adult goats then make human antithrombin protein.

Figure 7 shows an example of a DNA fragment that can be transferred into the cells of goats.

The enhancer stimulates region M. Name regions M and N shown in Figure 7.

Explain the purpose of the marker gene.

The enhancer only stimulates region M in the milk-producing glands of a goat.

Suggest two explanations for the importance of the enhancer being included in the DNA fragment transferred.

Describe and explain how the polymerase chain reaction (PCR) is used to amplify a DNA fragment.

Figure 9 shows the number of DNA molecules produced using a PCR.

Explain the shape of the curve in Figure 9.

A scientist produced transgenic zebrafish.

She obtained a gene from silverside fish. The gene codes for a growth hormone (GH).

She inserted copies of this GH gene into plasmids. She then microinjected these recombinant plasmids into fertilised egg cells of zebrafish.

Describe how enzymes could be used to insert the GH gene into a plasmid.

Microinjection of DNA into fertilised egg cells is a frequent method of producing transgenic fish. However, the insertion of the transferred gene into nuclear DNA may be delayed. Consequently, the offspring of transgenic fish may not possess the desired characteristic.

Suggest and explain how delayed insertion of the GH gene could produce offspring of transgenic fish without the desired characteristic.

The scientist investigated whether the transferred GH gene increased the growth of transgenic zebrafish. She microinjected 2000 fertilised egg cells with the GH plasmid and left 2000 fertilised egg cells untreated. After 12 months, she determined the mean mass of the transgenic and non-transgenic fish.

The results the scientist obtained are shown in Table 3.

Table 3

A value of ±2 × SD from the mean includes over 95% of the data.

Using Table 3, what can you conclude about the effectiveness of the GH gene on the growth of zebrafish?

Explain how two features of the design of this investigation helped to ensure the validity of any conclusions obtained.

Do not include calculating the mean or SD in your answer.

Sickle cell disease (SCD) is a group of inherited disorders. People with SCD have sickle-shaped red blood cells. A single base substitution mutation can cause one type of SCD. This mutation causes a change in the structure of the beta polypeptide chains in haemoglobin.

Explain how a single base substitution causes a change in the structure of this polypeptide.

Do not include details of transcription and translation in your answer.

Haematopoietic stem cell transplantation (HSCT) is a long-term treatment for SCD. In HSCT, the patient receives stem cells from the bone marrow of a person who does not have SCD. The donor is often the patient’s brother or sister. Before the treatment starts, the patient’s faulty bone marrow cells have to be destroyed.

Use this information to explain how HSCT is an effective long-term treatment for SCD.

A new long-term treatment for SCD involves the use of gene therapy.

Figure 2 shows some of the stages involved in this treatment in a child with SCD.

Some scientists have concluded that this method of gene therapy will be a more effective long-term treatment for SCD than HSCT. Use all the information provided to evaluate this conclusion.

What is a DNA probe?

DNA probes are used to detect specific base sequences of DNA.

The process is shown in Figure 6.

Describe how the DNA is broken down into smaller fragments.

The DNA on the nylon membrane is treated to form single strands. Explain why.

A scientist used DNA probes and electrophoresis to screen four volunteers for five different viral DNA fragments.

Figure 7 shows the results the scientist obtained. The lanes numbered 2 to 5 represent the four volunteers.

Lane 1 of Figure 7 enabled the size of the different viral fragments to be determined.

Suggest and explain how.

The lengths of the viral DNA fragments were:

• 600 base pairs

• 250 base pairs

• 535 base pairs

• 300 base pairs

• 500 base pairs.

Which volunteers had at least one of the viral DNA fragments with 250 base pairs or 535 base pairs?

Write an essay on one of the topics below.

EITHER

The importance of DNA as an information-carrying molecule and its use in gene technologies.

OR

The importance of bonds and bonding in organisms.

What is meant by a genome?

Chromatin immunoprecipitation is one method to determine where a transcription factor binds to DNA. The principle behind this procedure is shown in Figure 8.

Figure 8

Explain why the antibody binds to the transcription factor.

Use Figure 8 to explain what ‘precipitated DNA’ consists of.

Soybeans are used in a number of processed foods. However, soybeans contain a protein known as P34 that causes an allergic response in some people. Scientists have created transgenic soybeans that produce single-stranded cDNA, which prevents transcription of the P34 gene. They used recombinant plasmids as vectors to transform soybean cells. After they had screened these cells for production of the P34 protein, they cultured the transformed cells to form soybean plants.

Suggest how single-stranded cDNA could prevent transcription of the P34 gene.

Describe the roles of two named types of enzymes used to insert DNA fragments into plasmids.

The soybean cells were screened for the presence of the P34 protein. This process involved the use of gel electrophoresis to separate proteins extracted from soybean cells.

Suggest two features of the structure of different proteins that enable them to be separated by gel electrophoresis.

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 8.

Explain the role of reverse transcriptase in RT-PCR.

Explain the role of DNA polymerase in RT-PCR.

Any DNA in the sample is hydrolysed by enzymes before the sample is added to the reaction mixture.

Explain why.

Figure 9 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.

Suggest one reason why DNA replication stops in the polymerase chain reaction.

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.