Other

Question 1

How can DNA be damaged.

Answer 1

Oxidation- h is removed or o is added. It acts on double bonds and breaks them.

Hydrolysis- water splits a chemical bond. Removes NH2 or separates the base from the sugar so the sequence is no longer coding.

Methylation- adding a methyl group to a base which affects their ability to pair. Happens at an N atom.

Question 2

Depurination

Answer 2

Most frequent and spontaneous form of DNA damage.

Water is used to spilt the base from the sugar.

The sequence has a base missing and this can be replicated into new strands.

Question 3

Deamination

Answer 3

Most frequent and spontaneous form of DNA damage

Water is used to change cytosine into uracil.

NH2 on the cytosine is swapped for a double bond O.

The new U will bind to an A.
But originally it should have been a C and G

Making a mutated sequence

Question 4

Base excision repair

Answer 4

Deaminated and depurinated bases are repaired.

The enzyme uracil DNA glycosylase recognises uracil in the DNA sequence and will cleave it off because it’s abnormal.

The enzyme apurinic endonuclease removes the sugar.
Phosphodiesterase removes the phosphate
DNA polymerase adds a new nucleotide and DNA ligase seals the nick.

Question 5

Pyrimidine dimer formation

Answer 5

UV radiation causes it.

Thymine and cytosine are pyrimidine bases.

Covalent bonds form between the carbon atoms in adjacent pyrimidine bases on the same strand.

They distort the structure so the bases can’t make the correct paring.

Question 6

Nucleotide excision repair

Answer 6

Repairs pyrimidine dimers.

The enzyme excision nuclease removes a whole section of the sequence that contains the pyrimidine dimer and the backbone is cut.

DNA helicase cuts the hydrogen bonds

DNA polymerase used a primer to fill the gap and DNA ligase seals it.

Question 7

Defective excision repair

Answer 7

XP genes code for the enzymes used in excision
There are seven.

Xeroderma pigmentosum XP. A skin cancer that renders patients extremely sensitive to sunlight and causes many small skin tumours.

The lesions caused by UV light cannot be respires to sir causes cancer.

XP homologues in E. coli are the UVr proteins.

Question 8

Base vs nucleotide excision repair

Answer 8

Nucleotide EP is an emergency response.
Base EP is constantly surveying the DNA and finding background mutations.

Nucleotide EP is triggered by changes in UV and located to areas that need it most such as protein coding DNA.

Nucleotide EP Is physically coupled to RNA polymerase because it checks the RNA of protein coding genes before they are transcribed. (Transcription coupled).

Question 9

Ionising radiation and how it can be repaired basic

Answer 9

Produces double strand breaks and large sections of chromosomes can be lost.

Non homologous end joining- non specific and random sticking back together of fragments. Can be done wrong and lead to mutations.

Homologous recombination- causes cross over in cell division. It’s a last line of defence because EP produces a single strand break that can become a double strand break.

Question 10

Non homologous end joining

Answer 10

Staggered double strand break

Degradation to even strands.

Two strands are ligated

Some sequence is lost and can cause mutations.

Question 11

Homologous recombination

Answer 11

One of the two sister chromatids has a double break.

Degredation by exonuclease at the 5’ end to make a 3’ overhang.

RecA promotes strand invasion of undamaged template sister chromatid which acts as a primer.

Where the strands cross over is the invasion/branch point.

DNA polymerase extends the 3’ tail and the branch point migrates left.

The tail is joined to the 5’ end by DNA ligase.

The new strand is used as a template for the other broken strand

Question 12

Homologous recombination genes mutations.

Answer 12

Cause cancer

BRCA2 - breast ovarian prostate cancer.

ATM - ataxia telangiectasia, leukaemia and lymphoma

Fanconi anaemia - 13 different fanc genes Leukaemia

Once HR isn’t working in BRCA2 then the cells become dependant on base excision.
To treat this cancer we can inhibit base excision and cause them to be unable to repair their DNA so they die.
Synthetic lethality is where the DNA is too damaged for survival.

Inhibiting PARP treats prostate cancer.

Question 13

Homologous recombination in meiosis

Answer 13

For crossing over.

Spoll does a double strand break in one of the sister chromatids using endonuclease.

Mre11 is an exonuclease that degrades the 5’ end to make 3’ tails.

RecA promotes strand invasion

The 3’ tails are extended by DNA polymerase using the sister as a template.

The tail is ligated to the 5’ end and this forms a second branch point called a double holiday junction. Creating two pathways.

1- only the internal strands are broken and rejoined. Outer strands don’t cross.

2- external and internal strands are broken and rejoined.

Question 14

What can enzymes do

Answer 14

Lower the activation energy

Physically being substrate into close proximity

Bend the substrate

Provide electron donors or acceptors

Question 15

Hydrolase

Nuclease

Proteases

Synthases

Isomerase

Answer 15

Catalyse hydrolitic cleavage reaction

Break down nucleic acids by hydrolysing bonds between nucleotides.

Break down proteins by hydrolysing bonds between amino acids.

Condense and synthesise a new molecule

Catalyse rearrangement of bonds

Question 16

Polymerase

Kinases

Phosphatase

Oxidoreductase

ATPase

Answer 16

Catalyse polymerisation

Addition of a phosphate group

Hydrolytic removal of a phosphate group

One molecule is oxidised and the other is reduced

Hydrolyse ATP

Question 17

Dissociation equations

Association equations

Answer 17

AB=A+B
Dissociation rate= rate constant x con of AB
Or = k off means dissociation

This is because the more AB there is the more it will have dissociated so there will be a higher dissociation rate.

A+B= AB
Association rate = rate constant x (conc of A) x (con of B)
Or k on

Question 18

What happens at equilibrium

What kind of reactions is this for

Answer 18

The association rate equals the dissociation rate.

K on [A] [B]. = k off [AB]

Non covalent reactions

Question 19

What happens if the surfaces of two proteins match well

And three binding combos

Answer 19

They can form enough weak bonds to withstand thermal jolting and can remain bound for longer.

Surface string

Helix helix

Surface surface

Question 20

EFTu

Answer 20

Binds to GTP and becomes activated.

GTP is hydrolysed to GDP and the EFTu is inactivated

The GTP holds one of the helix structures a certain way allowing it to interact with domain 2 of the EFTu complex.

Question 21

Conserved domains

Answer 21

SH2- binds phosphorylated tyrosine

SH3- binds proline rich motifs

PH- binds phospholipids

EF hands- binds Ca and Mg in structural or signalling roles

Zinc finger- binds zinc in a structural mode

Leucine zipper- protein to protein or protein to DNA binding.

Question 22

SH2

Answer 22

Src homology 2 domain involved in signalling

Kinases and phosphatases phosphorylate the tyrosines so they can bind to SH2
These enzymes control the binding rate of tyrosine to SH2

The binding is done by ionic interactions between the negative phosphate group and the positive amino acids.

Question 23

SH3

Answer 23

Src homology 3 domain

A poly proline binding domain acting as a adaptor to link proteins.

Used in many signalling pathways.
The minimum consensus sequence is PaaaaP

They contain many aromatic residues that interdigit between the prolines of the PaaaaP motif.
This is stabilised by aromatic stacking.
There are electrostatic interactions due to aromatic stacking of proline and tyrosine.

Question 24

PH domain

Answer 24

Pleckstrin homology domain involved in lipid binding, signalling and anchoring proteins to the membrane.

Kinases modify phospholipids in the membrane to create binding sites for proteins that contain PH domains.

Cytoplasmic lipids can also be phosphorylated and dragged up to the membrane by kinases to create more binding sites for proteins with PH domains.

A combination of hydrophobic and charged interactions bind the phospholipid and drive association with the membrane.

Question 25

Different sizes and charges of metal ions means

Answer 25

They need to be liganded by different amino acids.

Question 26

EF hands

Answer 26

Bind to Ca for a regulatory function. Or bind to Ca and Mg for a structural function. Ca binds to EF hands induces a structural change. The EF hand motif is ocatdentate so it has eight atoms in the binding site that interact with Ca. There is an invariant glycine residue to accommodate the tight turn of O2 around Ca. There are two alpha helixes in the EF hand. The Ca is surrounded by seven oxygens. They have to have a very tight geometry for the oxygen and glycine to make a tight turn.

Question 27

Calmodulin

Answer 27

Has two calciums bound on either side and four in total. The structure has a helix then loop then helix. A glycine makes sure the loop is a very tight bend. Ca binding exposes a hydrophobic aa patch on the protein which allows the protein to interact with amphipathetic alpha helixes. The calmodulin will wrap around the helix Amphipathetic means having both hydrophobic and hydrophilic amino acids.

Question 28

Protein DNA binding domains

Answer 28

Always have an overall basic charge with basic amino acids because they have to interact with acidic DNA. Interact with the major groove Zinc fingers, leucine zippers, helix loop helix, Beta sheet.

Question 29

Restriction enzymes

Answer 29

Cut DNA into manageable sizes Act as dimers and recognise short palindromic sequences. Can leave overhang sticky ends or cut the DNA flush in the middle and are called blunt restriction enzymes. One subunit of the dimer recognises one strand and and the other does the other. Each subunit cuts one of the backbones and the H bonds remain but are not strong enough to hold the strand together to they break.

Question 30

Main steps of recombinant technology.

Answer 30

1) DNA is isolated and digested into fragments. 2) fragments are separated by gel electrophoresis. And Ethidium bromide is used to stain the DNA. 3) purify the DNA of interest. 4) ligate the fragments using ATP to create recombinant DNA. It works best if the sticky ends are complementary.

Question 31

Cloning DNA using plasmids

Answer 31

Plasmids are small circular DNA in bacteria. They have their own origin of replication that results in 50 copies of the plasmid being made in each bacteria. They usually carry antibiotic resistance genes. A piece of DNA can be inserted into the plasmid. This is put into bacteria and can be replicated many times and now you have a permanent supply of the DNA of interest.

Question 32

What is the multiple cloning site.

Answer 32

Where many restriction enzymes are added to the plasmid to cut it.

Question 33

Plasmid vectors BAC Yeast

Answer 33

Plasmid vectors only hold less than 30 kilo bases of DNA. Any more and the bacteria wouldn’t have the energy to cope. Bacterial artificial chromosomes can hold up to 300 kilo bases. It can mimic a full size chromosome to do recombination with larger DNA. Yeast artificial chromosomes can hold up to three megabases. The bigger the DNA the more difficult because they are fragile so plasmids are best.

Question 34

Transformation of bacteria

Answer 34

After making the recombinant plasmid it is put back into the bacteria. The bacteria has temporary holes in its cell membrane. Electroporation or chemical treatment. Competent bacteria are ready to take up new DNA. It is not very efficient and there is 1 in a million success rate. The bacteria are put into a Petri dish and an antibiotic is added. All those that did not take up the plasmid will die because the plasmid had an antibiotic resistance gene. The surviving bacteria are incubated and will reproduce to make thousands of clones of the DNA.

Question 35

Where does the DNA of interest for recombination come from.

Answer 35

By making a library of genomic clones we can see all of the regulatory sequences and allow the study of transcriptional regulation. By making a library of cDNA from mRNA we can see all the genes that will have been expressed by that mRNA. This is good for looking at which genes are expressed in diseased tissue.

Question 36

Cloning cDNA

Answer 36

Extract mRNA from cancer tissue Reverse transcriptase will create cDNA from the mRNA. The DNA is ligated into a plasmid vector and then put into bacteria. They are grown and the DNA is purified. The cDNA will express genes that are found in the original tissues transcriptome. The RNA extracted will give a population of 1000s if different RNAs. So many RNAs will make many cDNAs in the bacteria.

Question 37

To isolate a single clone from the mixed population of bacteria

Answer 37

You need one insert in each plasmid One plasmid in each bacteria One bacteria starting each colony.

Question 38

ESTs

Answer 38

Sequencing the ends of all the clones in the library to find the regulatory genes is called expressed sequence tags. Many of the clones will contain the cancers house keeping genes and some genes will be tissue specific. We need ESTs to find the sequences we desire.

Question 39

Making a genomic library

Answer 39

A human cell is used to purify and digest chromosomal DNA with restriction enzymes. The DNA is put into plasmids and then into bacteria and colonies are produced. Every piece of DNA in the chromosome is represented in the bacteria. This is useful for sequencing genomes.

Question 40

Deoxy terminator sequencing of DNA

Answer 40

Denature the DNA so it’s single stranded. Then add the primer and allow DNA synthesis to begin. The synthesis needs to fail. A deoxy terminator will be added to the strand instead of a normal nucleotide. And the strand can’t be extended further. The OH on the deoxy allows more dNTPs to attach. But the terminator deoxy only has a H.

Question 41

A mix of deoxy terminators and normal ones. All terminators.

Answer 41

The strands being made will all be terminated at the base complementary to the terminator. So a C terminator will have all the strands ending at G. This means you can calculate where all of the G bases are in the strand. You can make a nucleotide ladder. You can work out the full sequence because the strands will terminate at every single base. This sequence will be complementary to the strand of interest. The primer that was used is a radioactive oligonucleotide and x rays can be used to find it. The gel bands will show the primer and you can Start at the top and read off each base going down.

Question 42

Automated sequencing

Answer 42

Rather than labelling the primer. The dNTPs are labelled with a fluorescent dye. There is a different colour for each base. This allows for it all to be done in a single reaction and there is no need for radioactive substances. An electronic camera takes pictures of each band and measures its intensity. These readings are called traces and show colour intensity over time. There will be four coloured lines showing the peaks of each base. It only works for up to 1000 base pairs and got its name from the robot doing everything.

Question 43

Progressive sequencing

Answer 43

Used when you want to sequence 1000+ bases. You need a genomic library to start with. We already know the vectors DNA so we can design primers on either end of the vector that will go in the direction of the unknown insert. DNA synthesis occurs along from the primer up to 1000 bases. And now we know the first section of the unknown insert. Another primer is designed to go on the end of the new sequence. And another 1000 bases can be sequenced onto that This continues until the strands meet in the middle.

Question 44

How to sequence a genome using progressive sequencing

Answer 44

Sequence many BAC clones They can be localised to particular regions of the genome and chosen for sequencing

Question 45

Shotgun sequencing

Answer 45

Can be used for 1000+ bases. Same as progressive and needs a genomic plasmid library to start with. Primers are designed for the vector and the end 1000 bases are sequenced using the primers. The same primer is the used again. A new one is not designed for the next bit of the sequence like in progressive. This makes this method a lot easier. Although. It will results in lots of random sequences and we don’t know where they are in the genome. All the sequences are put into a computer and are assembled based on how much they overlap and together they show the contig. They show which sequence carries on from the one before. However you need to sequence more than 6x the size of the genome to get large contigs so it is inefficient and there will always be gaps.

Question 46

What was used to sequence the human genome What can be done today instead.

Answer 46

Progressive and shot gun sequencing. 1990-2003. One machine can sequence a human genome in 56 hours

Question 47

How to find genes from a string of bases

Answer 47

Gene prediction software. Scans the sequence for promotors, start and stop sequences and intron splice sites to predict what the gene will be like. This may not be accurate and it could be a non coding sequence. Or you could assume that all of the strand is coding DNA and translate it in all 6 reading frames. There are three on each strand. BLAST software shows similarities to known proteins and sequences. Or if they have a similar charge to known proteins. If the proteins are similar it suggests that they evolved from the same ancestor and will have a similar function.

Question 48

Microarray

Answer 48

Compares the transcriptomes of different tissues. Normal liver vs cancer liver cell. It is high through put so is small as a cover slip fast and can be automated. A very precise robot manufactures an array and each spot in the grid contains one of every gene in the genome as mRNA. The fluorescent cDNA is added from the tissue And it will bind to the mRNA in the grids. The genes that it binds to will be those expressed in the liver. It is rinsed and a camera takes pictures. Some will be red and some will be green. Yellow areas are where the gene is found in both groups of cells.

Question 49

Microarray findings in the liver

Answer 49

Any genes that the normal liver has and the cancer liver doesn’t could be lost tumour surpressors. The genes that the two tissues have in common will be the liver housekeeping genes Extra genes found in the cancer that are not in the normal liver could be oncogenes that have been upregulated.

Question 50

Evaluate microarrays

Answer 50

Very powerful as you can test the whole genome at once. Need a lot of cells to do it.