Genetics part 3 Flashcards

Question

What is Jjj1?

Answer 1

1. Ribosome maturation | 2. Involved in ribosome biogenesis

Answer 2

* The rDNA locus in the genome – is the DNA that encodes for the ribosomal proteins * It is repeated massively in vast tracks as we make a lot of ribosomes * We transcribe the locus and process it and process it down again and get 2 subunits forming * The large subunit will try to get out of the nucleolus to the cytoplasm * To do that, it needs accessory proteins, there are a number of enormous accessory proteins involved in getting a ribosome made e.g. 200 proteins involved in ribosome assembly * They get onto the ribosomes to get it out into the cytoplasm where it is joined to two other proteins which all stick together and cause reaction and kick the red protein back into the cytoplasm

Answer 3

Disturbs stem cell function Also leukaemia promotion

Answer 4

1. BMF 2. pancreatic insufficiency 3. Predisposition to leukaemia

Answer 5

The hypo-proliferative to hyper-proliferative paradox

Answer 6

a gene in a different species that evolved from a common ancestral gene

Answer 7

Ancestral gene in yeast

Answer 8

Causes a difference in methylation Regulates gene expression

Answer 9

* Differences in severity of clinical phenotype e.g. BMD and DMD * Mutations in which you have changes in the frame of MRNA that will cause catastrophic changes to protein (DMD) * If the protein stays inframe but becomes either larger or smaller – (BMD)

Answer 10

Targeted and specific modification of a genome of living organism

Answer 11

1. Zinc finger nucleases 2. Talens 3. Crispr-Cas9

Answer 12

1. once double strand break is generated - cell attempts to repair it as quickly as possible 2. Via non-homologous end joining 3. Error-prone technique 4. Introduce or delete a few nucleotides

Answer 13

You cannot influence the type of mutation generated

Answer 14

1. Provide the cells with a donor template 2. Cell uses donor template to manufacture missing sequence to fill gap 3. o When the donor template is designed the 5’ and 3’ regions need to be exactly the same sequence as sequence you attempt to target – introduce mutation within the middle

Answer 15

Use zinc finger nucleases

Answer 16

Combined DNA and are the most common DNA binding in eukaroytes ``` Different versions - Cys (2) his (2) - Cys (2) hiscys Hold a zinc atom Zinc fingers bind 3-4 BP of DNA ```

Answer 17

Libraries with zinc fingers – known to bind to triplets of nucleotides If the right zinc fingers are put together molecularly – bind to particular sequence of choice The zinc finger itself wont make a double-stranded break – an endonuclease is required FOK1 molecules are added FOK1 is endonuclease but it needs to dimerise – 2 fok1 molecules required Fok1 dimerizes and makes a double stranded break Problem with zinc finger is that they don’t always bind to their target Assay: 3 zinc finger together = determines if it binds and a reporter gene

Answer 18

Pathogens Xanthomonas

Answer 19

Transcription factors that bind DNA and drive host genes which are beneficial to Xanthomonas

Answer 20

* Contains repeat domains – essential for binding * Transcription factors need to be in the nucleus and contain activation domains * Contains 34 amino acids – repeated many times * Amino acids between 12 and 13 which nucleotides of DNA it binds to – called repeat variable di-residues

Answer 21

Host promoters

Answer 22

1. DNA binding domain | 2. FokI DNA-cleaving domain

Answer 23

Editing of the genome in specific locations

Answer 24

* E.g. histidine and aspartic acid – mainly binds to C * E.g. asparagine and glycine mainly bind to T * Make molecule and clone the right repeat domain in sequence * If you get the right amino acid together in 12 and 13 – target any sequence you would like to bind to * In Xanthomonas these are transcription factors – so they don’t make a double stranded break

Answer 25

combining (Targeted/specific) DNA binding capability with FOKI endonuclease activity

Answer 26

* Edit parts of the genome – removing, adding or altering sections of the DNA sequence * Comes from single cellular organism – Achaea and bacteria * Used as an adaptive immunity * Invasion of single cellular organism by phage

Answer 27

- Enzyme: Cas 9 – acts as molecular scissors that cut 2 strands of DNA at a specific location in the genome – bits of DNA can then be added or removed - A piece of RNA called guide RNA (gRNA) – consist of small piece of pre-designed RNA sequence located within a longer RNA scaffold - Scaffold part bind to DNA - Pre-designed sequence guides cas9 to the right part of the genome - The guide RNA is designed to find and bind to a specific sequence in the DNA - The cas9 follows the guide RNA to the same location in the DNA sequence and makes a cut across both strands of DNA - At this stage the cell recognises that DNA is damaged and tries to repair it

Answer 28

- CRISPR RNA - TracrRNA - Cas9

Answer 29

* CRISP RNA is complementary to the sequence you want to target * NGG – protospacers – the sequence you are trying to target in the genome * Protospacers adjacent motif – important sequence for the complex * For Cas9 to bind to the complex – need particular sequence called NGG * Any nucleotide followed by two G * To target sequence by CRISP RNA – NGG has to be adjacent

Answer 30

• Viral method • Lipid transfection - Lipid binds to the DNA - By endocytosis you get all the RNA into the cell

Answer 31

* Mix the cells with DNA or RNA – use current to get the macromolecules in * Optimization is required because high current is used – destroy the cell whereas if you don’t use enough – nothing can happen * After the molecules are introduced into the cell – make clonal line – dilute the cells

Answer 32

* Assay used * Introduce CRISPR-cas9 or talen into cells- by non-homologous end joining you will get introduction of mutations (small deletions/small insertions) * Pool of different mutations in this dish * T7E1 detects little mismatches in DNA and it can cleave * See cleavage products on agarose gels when you separate these * Some cells will have mutations but they may be different mutations (small BP insertion 1,3) and some will be wild type * After a few days you isolate the genomic DNA from the cells, so you have a mixture * And you do PCR amplification with primers which flank the region your targeting * When you do PCR you have some molecules which have a mutation at a location ur tryna target and some will be wild type

Answer 33

the better your reagent works

Answer 34

* Clonal lines of cells * Non-homologous end joining – PCR – flank region you are trying to target * Larger deletions are detected on an agarose gel * Do PCR run on agarose gel – smaller products * Resolution of gels aren’t great – wont see a difference * DNA sequencing done – determine if mutation is present or not

Answer 35

o Mutation that occurs when the reagent doesn’t bind to regions, bind elsewhere in the genome and cause mutation there o For talen – 1% mutations somewhere o CRISPR-CAS9 – up to 77% o The differences in percentage is because for talens you need 2 fok1 molecules to dimerise – chance of this happening elsewhere in the genome is low

Answer 36

- Adv.: eliminates risk of random genome integration of expression plasmid DNA - Lowers cytotoxicity - Higher efficiency - Expanded range of cell types - Eliminated the need to use different promoters for ZNF expression - Dis: off-target effect

Answer 37

- Adv.: all DNA sequences can be targeted - The cloning of construction of DNA molecules with TALEN repeats is more cumbersome - Highly repetitive nature of TALEN- coding sequence creates barrier to their delivery – viral vectors - Dis: only single site at one time - Designing construct is time consuming and are not cost-effective - Sensitive to target DNA methylation

Answer 38

- Adv.: highly specific - Highly efficient - Simultaneous targeting of multiple sites - Dis: target selection limited by requirement of PAM sequence - Off target effects

Answer 39

• Technique to remove the disease causing mitochondria by producing 3 parent babies • Genomic engineering techniques - Correct pathogenic mtDNA mutations in vivo in the mouse • Mutation: TRNA is affected • Mouse model: leads to cardiomyopathy • Mitochondrial zinc finger nucleases are used to target this mutation • Zinc fingers that specifically target these sequences • Types of zinc fingers - Wtm1: will always bind - Mtm25: won’t bind if its wild type • Ta – mutation – bind and caused double stranded break • Degradation of mtDNA – double stranded break • Change the ratio from wild type to mutated mtDNA • >60% mutated mtDNA – see a phenotype • Push the percentage down below threshold – stop the disease from happening • Injected 2 zinc finger nucleases construct into the tail vein of mice • Then used adeno-associated virus – specifically targets the heart – change the ratio

Answer 40

determined % of wild type mtDNA with mutated mtDNA in heart and earo Since the virus only targets the heart – expected to change the ratio only at the heart o Virus will bind to mutated DNA in heart – double stranded break – degradation of mutated DNA o First control: use without zinc finger nuclease – compare heart and ear - % of mutated DNA is around 0 o Second control: one zinc finger nuclease – similar quantities between heart and ear o Put both in increasing concentration of viral vector – see an effect

Answer 41

o Technique that uses cells o Lady who has mitochondrial disease o Aim: prevent transmitting the mutated mitochondria o Use oocyte of a donor female without disease o Fertilize both oocytes with sperm from the father o Remove the pronuclei from donor o Insert pronuclei from parents into donor egg o You have genomic DNA from the parent but mtDNA from donor o By IVF – get a baby without mitochondrial disease

Answer 42

o Before the oocytes are fertilized o Remove genetic material from the donor o Insert the genetic material from lady who can transmit the mitochondrial disease o Fertilize it with sperm from the father o IVF: get lady pregnant and deliver baby * Too early to determine if the methods are completely safe * Some carry over of mutated mitochondria (2% reported) – affect future generations – effect female to the donor * Technique approved by house of common * UK license granted last year * Law has been passed in London to create baby via pro nuclear transfer technique and not spindle transfer technique

Answer 43

• Mutation in factor 8 can give rise to haemophilia A • Factor 8 has region in intron 1 which is identical to a sequence which is 140,000 BP over to this side • Recombination between these two • Net effect - Exon 1 of the gene with all the regulatory element pointing in one direction - Exon 2 with non-regulatory elements and the rest of gene pointing in other direction • Severe phenotype – no factor 8 produced • Inversion – refer it back to wild type • Number of different talen reagents - which one is best to use for inversion • T7E1 assay • In control – don’t see smaller products – cleavage with T7E1 • Talen 1 – good – 33% mutations incorporated • Talen 3 is very bad – don’t see affect any of these smaller products

Answer 44

o Wild type line + some controls o Control from patient o Had double strand break o Talen is identical – regions are identical in sequence o To make double strand break in both – make inversion

Answer 45

o Wild type intron 1 – have homolog 1, consist of primers 1F and 1R - 140,000 bp you have 2F and 2R in homolog 2 o Inversion: primer 2R sits opposite 1R and 1F is opposite 2F o By using PCR reaction – you can tell if you’ve managed to get this inversion o First line of PCR → looking with primer 1F and 1R o Wild type situation: amplify band o In patient where 1F and 1R are both pointing in the same direction and are 140,000 bp apart – never going to amplify the PCR product here – the line is empty o 1R-2R – flipped over when we have this inversion o Patient in junction 1: primer 1R and 2R are used – amplify the PCR product

Answer 46

o Take one of the inversion clones and reverted it back to a normal gene o Used same talen o Can flip it back to normal o Homolog 1 – 1F and 1R – wild type give you a PCR product o The patient and inversion clone’s don’t o Cells are hematopoietic lineages and can be isolated – returned relatively easily o All processes conducted at a clinical grade o Off-target effects have to be controlled for

Answer 47

o Mutations in dystrophin can lead to BMD and DMD o Mutation in exon 44 o Goes out of frame and you get a stop codon – lead to DMD - REMOVE splice acceptor from exon 45, get splicing from 43-46 – the protein goes back into frame – shorter protein - Creating reading frame shifts – add extra nucleotide to exon 45 – back into frame – it can be 1 bp or multiple - Knocking exon 44 in front of exon 45 – restore the whole of the protein donor template required - - This is intron-exon boundary

Genetics part 3 Flashcards

(75 cards)