Recombinant DNA technology Flashcards
What does recombinant DNA technology involve?
- transfer of fragments of DNA from one organism to another. The organisms may be the same species or different species. The DNA produced is called recombinant DNA, as it
contains genes from at least 2 combined sources.
Why can transferred DNA be translated within cells of the recipient organism?
- genetic code is universal (the same triplet codes for the same amino acid in all species), as are transcription and translation mechanisms (the same in all species)
What is the recipient organism called and what does it do?
- transgenic organism or GMO (genetically modified organisms), it will manufacture proteins using the DNA as if it was its own.
Describe the process of bacterial production of human protein
- Isolation of DNA from cell (isolation)
- Cutting out the required gene (producing DNA fragments) (restriction)
- Insertion of gene into a vector (ligation/insertion)
- Transformation of host cells (transfer of DNA into suitable host) (transformation)
- Identification of host cells that have taken up the gene (using gene markers) (selection)
- Growth/cloning of host cells (culturing)
How can fragments of DNA be produced?
- conversion of mRNA to complementary DNA
(cDNA), using reverse transcriptase - using restriction enzymes to cut out a fragment containing the desired gene from DNA
- creating the gene in a ‘gene machine’
Describe restriction method 1: using reverse transcriptase
- Select cell that produces known protein eg. Insulin
- Extract (intron-free) mRNA, which is purified and mixed with reverse transcriptase + free DNA nucleotides. mRNA acts as a template on which
cDNA (complementary DNA) is formed - Hydrolysis of the mRNA with an enzyme allows cDNA to be removed. DNA polymerase
allows base pairing on the single DNA strand to make a double strand - Now have DNA coding for the required gene
Describe restriction method 2: using restriction endonuclease
- Many types of restriction endonuclease exist - each one cuts DNA at a specific sequence of bases called recognition (restriction) sites/sequences
- Some enzymes perform straight cuts, which gives DNA blunt ends.
- Others cut in staggered fashion producing restriction fragments of DNA with ‘sticky ends’ – ends of exposed bases
Often research scientists wish to use bacteria to
produce eukaryotic proteins. This is only possible if bacteria are given an intron-free version of the eukaryotic gene.
Explain why.
- Prokaryotes such as bacteria do not have introns;
- Therefore they lack the ability to splice them out of pre-mRNA;
- Introns need to be spliced out from any pre-mRNA before the mRNA can be translated into a polypeptide
Describe restriction method 3: gene machine
- The bases of a gene can be sequenced by working backwards, by determining the amino acid sequence and then the mRNA codons for each amino acid, then the DNA triplets
- The desired sequence is fed into a computer & checked for biosafety and biosecurity. (important ethically since this method, in theory, be used to make an entirely artificial gene, coding for a protein that may not exist naturally)
- The computer constructs oligonucleotides
(small, overlapping single strands of DNA). These are then joined together to make the gene (with no introns) - The gene can be converted into double stranded DNA in the Polymerase Chain
Reaction (This gene can ultimately be inserted into vectors such as bacterial plasmids - The whole process is quick, gives great accuracy and removes all non-coding sections
Scientists used restriction mapping to investigate
some aspects of the base sequence of an unknown piece of DNA. This piece of DNA was 3000 base pairs (bp) long. The scientists took plasmids that had one restriction site for the enzyme Kpn1 and one restriction site for the enzyme BamH1. They inserted copies of the unknown piece of DNA into the plasmids. This produced recombinant plasmids. The diagram shows a recombinant plasmid. When the scientists digested one of the recombinant plasmids with Kpn1, they obtained two fragments. One fragment was measured as 1000 bp. The other fragment was described as “very large“. What does this show about the base sequence of the unknown piece of DNA?
- It must have one restriction site present,
1000bp away from the Kpn1 on site of plasmid
Scientists used restriction mapping to investigate
some aspects of the base sequence of an unknown piece of DNA. This piece of DNA was 3000 base pairs (bp) long. The scientists took plasmids that had one restriction site for the enzyme Kpn1 and one restriction site for the enzyme BamH1. They inserted copies of the unknown piece of DNA into the plasmids. This produced recombinant plasmids. The diagram shows a recombinant plasmid. One of the fragments that the scientists obtained was described as “very large”. What is represented by
this very large fragment?
- Most of the plasmid and the rest of the unknown DNA fragment.
Scientists used restriction mapping to investigate
some aspects of the base sequence of an unknown piece of DNA. This piece of DNA was 3000 base pairs (bp) long. The scientists took plasmids that had one restriction site for the enzyme Kpn1 and one restriction site for the enzyme BamH1. They inserted copies of the unknown piece of DNA into the plasmids. This produced recombinant plasmids. The diagram shows a recombinant plasmid. When the scientists digested another of the recombinant plasmids with BamH1, they obtained three
fragments. How many BamH1 restriction sites are there in the unknown piece of DNA?
- 2
Describe how fragments of DNA are amplified by in vivo techniques
- Gene of interest it cut as special recognition sites (specific base sequences of DNA) using a specific restriction endonuclease
- Bacteria have rings of DNA called plasmids, which are cut with the same restriction endonuclease as target DNA. Both form sticky ends which complement one another
- Target DNA is inserted into the plasmid; the ‘vector’. This is called in vivo gene cloning (involves living organisms). DNA ligase enzyme
seals their sugar phosphate backbones together, catalysing a condensation reaction - Recombinant plasmid is inserted back into the bacterial cell - we then look to see if the cell has
been successfully transformed
What are two requirements of the DNA fragment before we can insert the plasmid (vector) back into the host bacterial cell and transform it?
- Addition of extra length of DNA (promoter) required at the start of the gene, to allow RNA polymerase to attach, for subsequent transcription’ of the gene and production of messenger RNA
- Addition of extra length of DNA (terminator) required at the end of the gene, to allow RNA polymerase to detach from the messenger RNA, to mark the completion of transcription
Suggest the simplest way to add a promoter and terminator to a DNA fragment?
- Use a plasmid that already has the correct promoter and terminator base sequences. The gene can then be inserted between them
What do vectors have which ensure the inserted gene is transcribed in the host cell?
- regulatory DNA sequences
What are two steps needed to insert the plasmid into the host cell?
- Calcium ions to increase membrane permeability
- Heat shocks- temperature lowered to freezing then raised to 40 degrees
What is the problem with in vivo method?
- Only 1% success rate: some bacteria don’t take the plasmid up and others simply close up without the DNA fragment in place
Cut plasmids and lengths of foreign DNA can join. What features of their ends allows them to join?
- unpaired bases / sticky ends / staggered; complementary
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.
- plasmid with foreign DNA joined in ring; ring with plasmid only; ring of foreign DNA only
How can we check to see if a plasmid has been successfully transformed?
- inserting target gene into already existing gene segments of a plasmid (act as gene markers)
What would happen if we were to cut one of the genes?
- no longer function
In an experiment using the R-Plasmid, a scientist used restriction enzymes to cut the tetracycline resistance gene, and insert a section of DNA coding for insulin protein. The recombinant plasmid was then inserted into a bacterial cell which was allowed to multiply. Some bacterial cells were transferred to a petri dish containing ampicillin antibiotic, and some were transferred to a dish containing tetracycline antibiotic. Explain what would happen to the bacterial cells after 24 hours
- The bacterial plasmid contains a functional gene for ampicillin resistance, but the tetracycline resistance gene is no longer functioning, as it has been cut. This means that the bacteria growing on the dish of ampicillin will survive, and multiply, but the bacteria growing on a dish of tetracycline will be killed by the antibiotic
After 24 hours, the scientist has noticed that all bacterial cells have died. Suggest an explanation for this observation?
- Either the plasmid was not originally taken up by the bacterial cells, or it was but the gene for tetracycline resistance was not successfully transformed (i.e., cut open)
