Enzyme and restriction mapping Flashcards
What is a recombinant protein?
Protein produced from the transcription and translation of recombinant DNA
What is recombinant DNA?
DNA produced via bringing together DNA from multiple sources to create sequences not found in the genome
Name some examples of recombinant proteins
- Insulin
- Interferon - Group of signaling proteins made and released by host cells in response to the presence of several viruses
- G-CSF (Granulocyte-colony stimulating factor) - Glycoprotein that stimulates bone marrow to produce and then release granulocytes, a type of white blood cell, into the bloodstream.
What is a transgenic organism?
An organism with an altered genome
What are some of the uses of transgenic organisms?
- Produce disease models
- Improve agricultural yields - genetically modify crops to be resistant to certain pests
What is the function of a nuclease?
An enzyme that cleaves nucleic acids by the phosphodiester bonds between the nucleotides
What are the two different types of nuclease enzyme?
- Ribonuclease (RNase): Cleaves RNA
- Deoxyribonuclease (DNase): Cleaves DNA
What are the 2 types of deoxyribonuclease?
- Exonuclease: Cleaves the DNA from the end of the molecule
- Endonuclease: Cleaves the DNA from within the nucleotide chain
Why are restriction endonucleases (restriction enzymes) called that?
Because they are expressed in bacteria and their function in bacteria is to limit (restrict) the transfer of nucleic acids from the bacteria to phages.
Explain the basic mechanism for how a restriction endonuclease works?
- They recognise a specific sequence
- Then cleave that specific sequence at a recognition site
Name an example of a restriction enzyme. What is the specific recognition sequence that this restriction enzyme recognises?
- EcoRI
- Specific recognition sequence: G/AATTC
- Palindromic complementary sequence: CTTAA/G
What does it mean when it is said that recognition sequences are “palindromic?”
- It means that they can be read the same way in both directions - from 5’ to 3’ or from 3’ to 5’
- It also means that every recogntion sequence also has a complementary palindromic sequence which is the same sequence as the recognition sequence but is read in the opposite direction.
Where specifically does EcoRI cleave within the recognition sequence?
It cleaves between the Guanine and adenine nucleotides of the recogniton sequence
What are the effects of the cleavage from EcoRI on the recognition sequence?
- It creates a 5’ overhang on both the recognition sequence and the palindromic sequence
- Due to the hydrolysis rection that occured on the 3’ end of the guanine nucleotides there are OH groups attached while at the 5’ end of the adenine nucleotides there are phosphate groups attached
What is an overhang?
- A stretch of unpaired nucleotides within a DNA molecule
- Can be 3’ or 5’ overhang
Apart from overhang what other effect can a restriction endonuclease produce within the recognition sequence?
- Can also produce blunt ends which is when both strands of a DNA molecule end in a base pair.
What does the result of the EcoRI lane for gel electrophoresis tell you about how it digests the pDTL3 plasmid?
- EcoRI only has one band at 13kb which is the same size as the plasmid
- This means that it didn’t digest the particular sample or if it did it only cleaved the sample at one position
What does the result for the BamHI lane for gel electrophoresis tell you about how it digets the pDTL3 plasmid?
- BamHI produced 2 fragments, one of which is 6kb and the other 7kb
- This means that it cleaved the pDTL3 plasmid twice
How can the results from the BamHI lane for gel electrophoresis be used to locate the restriction sites for BamHI on a pDTL3 restriction map?
- You know that BamHI produced a 6kb fragment and a 7kb fragment
- You also know that the entire pDTL3 is 13kb in size
- With this information on a restriction map you simply place one of the restriction sites on the top of the map.
- You can then work out where on the map the other restriction site is because you know the length of the two fragments produced.
Can the results from the EcoRI lane be used to locate where its restriction site/s are on the pDTL3 restriction map?
No because the EcoRI only produced one fragment which was the same length as the plasmid so you have no indication of where exactly EcoRI cleaved the plasmid
In this example what can be used to locate the restriction site for EcoRI on the pDTL3 restriction map? Explain how it can be used to do this
- You have to look at the size of the fragments produced when both restriction endonucleases were used.
- In this example the 6Kb fragment is still produced meaning that within that fragment there is no EcoRI restriction site present.
- However, the 7Kb band produced by BamHI is now split into a 3 and a 4Kb fragment which means the EcoRI restriction site is located within the 7kb fragment produced by BamHI
How does the sickle cell anaemia mutation affect the activity of the restriction enzyme Dde1?
- The normal 𝛃-globin gene (𝛃A) contains the recognition sequence for Dde1: CTGAG
- However, the sickle cell anaemia mutation causes this sequnce to change to CTGTG in the mutated 𝛃-globin gene (𝛃S) which Dde1 is unable to recognise
- This means that in the mutated 𝛃-globin gene Dde1 isn’t able to cleave the gene at that position
What size fragments would you expect to see on the gel electrophoresis gel if a person was homozygous for the normal 𝛃-globin gene? why is this?
- You would expect to see 3 fragments:
- One 175 bp fragment
- One 201 bp fragment
- One 67 bp fragment
- This is because on both copies of the gene Dde1 is able to recognise and cleave the recognition site 175 bps away from the start of the sequence
- Dde1 also cleaves second recognition site 67bp away from the end of the sequence which also creates a 201bp fragment
What size fragments would you expect to see on the gel electrophoresis gel if a person was heterozygous for the 𝛃-globin gene? why is this?
- You would see 4 fragments:
- One 376 bp fragment
- One 201 bp fragment
- One 175 bp fragment
- One 67 bp fragment
- This is because on one of the copies of the 𝛃-globin gene Dde1 isn’t able to recognise the first recognition sequence and so doesn’t produce a cut at 175 bp so instead you have a 376 bp fragment
- On the other copy of the gene the recognistion site is able to be recognised by Dde1 so it does cleave at 175 bp producing a 175bp fragment and a 201 bp fragment
- Dde1 able to recognise second recognition sequence on both copies of the gene so 67 bp fragment produced.
What size fragments would you expect to see on the gel electrophoresis gel if a person was homozygous for the mutated 𝛃-globin gene? why is this?
- You would see 2 fragments:
- One 376 bp fragment
- One 67 bp fragement
- This is because on both copies of the gene Dde1 is unable to recognise the first recognition sequence so instead of a 175 bp fragment being produced you just have a 376 bp fragment
- On both copies of the gene the Dde1 able to recognise the second recognition site so a 67 bp fragment is produced.
What is DNA ligase?
An enzyme that forms phosphodiester bonds in the sugar-phosphate backbone at areas within a DNA strand where there is an overhang or a sticky end
What are some of the uses of DNA polymerase in the lab?
- Used in PCR
- Used to generate DNA probes
- Used to turn DNA overhangs into blunt-ends on a DNA sample so they can then be ligated together.
