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week 7 Flashcards

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1
Q

Forward Genetics

A

What genes are important for a process? Phenotype to genotype

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2
Q

Reverse genetics

A

What process is the gene important for?
genotype to phenotype

40 years of reverse genetics

if I remove the gene what happens to the phenotype

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3
Q

Testing Forward Genetics

A

Uses mutagenesis to generate a random pool of genome variants

performing a selection or a screen

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4
Q

Selection

A

Identify genes that are important for the effect; only resistant variant survives

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5
Q

Screen

A

Look through all the variants for the phenotype, all the variants survive

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6
Q

Genetic screens

A

genetic analysis requires genetic variants

to dissect the biological process, apply genetic analysis

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7
Q

Saturation screens

A

An attempt to identify as many genes whose products contribute to the process that you are studying as is statistically and technically possible. Only generating in all of these methods a random pool of DNA sequence genetic variants

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8
Q

Genetic screen for leucine auxotrophic yeast

A

looking for genes that are important for synthesis of an important component

mutagenesis of haploid yeast cells, each cells have random DNA sequence changes, we don’t know what the effect of the changes are, look for colonies that reuqires leucine to grow, implying a mutation has inhibited its ability to grow on plates where leucine is absent.

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9
Q

If you have 100 leu auxotrophic yeast mutants does that mean there are 100 genes regulating leu?

A

No. You can have multiple independent alleles in the same gene.

That encodes for components required for the synthesis of leucine. One gene can get his five times at different positions in the gene

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10
Q

Complementation analysis

A

Non-complementation in same complementation group=same gene

Independent mutations in the same gene, if you cross the haploids to create a diploid, this diploid will still lack the ability to synthesis leu, non-complementation

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11
Q

Complementation

A

These two loci have wt alleles and can grow on media without leu; mutations on two different genes

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12
Q

Haploid genetics vs diploid genetics

A

Haploid: only one gene knockout is required for the phenotype

Diploid: need two knockouts to get to the phenotype

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13
Q

Haploid

A

Single generation

Large number of variants to screen or select (10^6-10^8)

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14
Q

Diploid

A

Several generations: we need to perform crosses to get recessive homozygous cells that will show the phenotypes

Small number of variants (10^3-10^4)

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15
Q

Maternal vs Zygotic genomes

A

early phase: information coming from the mother; zygotic genome is not transcribed for hours/days until after fertilization

oocytes doesn’t go onto meiosis until later

the reason for this is that in animals the zygotic genome is not transcribed immediately. So all the early steps have to be run off of the genetic material in the egg which was provided by the mother

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16
Q

Zygotic

A

when we’re thinking about genes expressed from the zygotic genome, to screen for those activities that are important, all we require is a male and a female that carry the mutational change.

that will create a zygote, will be homozygous for that mutational change, and if this gene which has this DNA sequence variant, inactivates it, is important for a process, then you will see a phenotype in the developing zygote.

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17
Q

Maternal affect

A

Transcripts stored in the egg are translated and produced early in the genome.

Now these transcripts are then placed into the egg and after the egg is formed and laid and fertilized, these transcripts that have been stored in the eggs are now translated, producing the product early during development, that allows the egg to develop without transcription of the genetic information of the zygotic genome.

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18
Q

Maternal effect mutants

A

In this case the mother that is homozygous for bicoid loss of fucntion allele lays a normal egg, but every single one of her egg that she ;ays the larvae will develop wihtout a head 100% of the time, so this information that is required to be passed from the mother to her progeny in this case is gone and now because of that missing infromation the ehad doesn’t develop

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19
Q

Zygotic screens

A

3 generation screen

Mutagenize the sperm to generate F1 generation

F1 generation contains one sequence variation change in our gene interest (possibly)

cross with an un-mutagenized individual to generate males and females heterozygous for the seuqence change (F2)

cross the heterozygotes to generate 1/4 of individual homozygous for a sequence change

is the embryo dead or did the mutation not have an effect and we screen for things that are important embryogenesis.

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20
Q

Maternal effect screens

A

we mutagenize sperm, have an individual

which we cross to have more individuals; males and females that are heterozygous. we crossed them to create a mother F3 that is homozygous, that her maternal genome is homozygous. She will be normal looking but when she lays an egg, which would be the F4 generation, those eggs will not develop.

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21
Q

Mutational tagging

A

DNA sequence variation tgs the gene that you have identified as important

genome sequencing has made this easier.

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22
Q

Reverse genetics requires

A

reintroducing DNA into an organism defines it as a complex refined model organism.

whats the gene important for, make a mutation of the gene and reintroduce an altered copy of the gene into the organism

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23
Q

Attribute of good model organism

A

ability to reintroduce DNA

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24
Q

Four ways to reintroduce DNA

A

Transformation

Injecting DNA

Transposon/viral mediated transformation

Site specific recombination

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25
Transformation
Episome (plasmid; can freely replicate in the organism) Random insertion Homolgous recombination
26
Requirement for transformation
first treat cells to make them competent to take up DNA yeasts; tissue culture cells
27
Episome
dominant selectable marker origin of replication once in the organism they can freely replicate as an epichromosome, one that is not part of the genome
27
Episome
dominant selectable marker origin of replication once in the organism they can freely replicate as an epichromosome, one that is not part of the genome
28
Selectable marker
in bacteria it codes for resistance to a selectable marker in yeast the plasmid will contain a biosynthetic pathway functional marker, Leu2 functional allele, in the chromosome (endogenous locus) the yeast contains a loss of functional allele. This results in the cell requiring leucine for growth
29
Random insertion
occurs in tissue culture cells when they are being transformed with DNA, the DNA is able to be inserted in a eukaryotic chromsome and once its inserted it will be stably maintained generation upon generation. This insertion is random, an not necessarily at the locus of DNA we are introducing, random insertion may lead to lack of gene expression
30
Homologous recombination
Our gene in the chromosome has DNA introduced using transformation, this DNA contains a region homologous to the gene and a dominant selectable marker. This DNA will rarely undergo homologous recombination the marker is incorporated into the endogenous locus and will disrupt the gene targetting your transformation event to a particular locus in the cell
31
Injecting DNA
Very large cells (vertebrate, nematodes and insects) a needle that contains the DNA that you insert into the cell and pump in the DNA generates random insertion
32
Transposon/viral mediated transformation
In some cases, transformation/injection of DNA does not result in reintroduction of DNA. It needs help sometimes
33
Transposon mediated transformation
transposases will bind to the inverted repeats + transposases coding region copy+paste mechanism
33
Transposon mediated transformation
transposases will bind to the inverted repeats + transposases coding region copy+paste mechanism
34
Transposon mediated transformation (two compotents)
transposase: the trans-acting factor inverted repeats: cis acting factor seperate the components; we need to control when transposition occurs inject one plasmid that contains transposase but does contain inverted repeats; and another plasmid that contains the marker and whateber gene we are trying to reintroduce flanked by inverted repeats to which transposase will bind. random insertion
34
Transposon mediated transformation (two compotents)
transposase: the trans-acting factor inverted repeats: cis acting factor seperate the components; we need to control when transposition occurs inject one plasmid that contains transposase but does contain inverted repeats; and another plasmid that contains the marker and whateber gene we are trying to reintroduce flanked by inverted repeats to which transposase will bind. random insertion
35
Viral mediated transformation
incorporate your gene and a marker into a viral genome, take advantage of cirus that integrate their genetic material into the hosts genome. Take the viral genomes and incorporate your gene and marker into the host genome package the genome into a viral particle After viral infection your gene is inserted into the genome of the cell.
36
Site specific recombination
recombination sites are recognized by recombinase, recombinase mediates recombination between the two sites. place a recombination site in the genome, and add DNA in the plasmid that also contains a recombination site + GOI+ marker, express recombinase in these cells as well
37
Use of transformation/transgenesis
Functional complementation Misexpression and gain-of-function Knocking out genes
38
Functional complementation
In a forward genetic screen we have identified mutations in genes important for a process what are the genes associated with the mutations phenotype to gene What are the genes associated with the mutation, what is the normal function of the gene
39
Functional complementation steps
put random pieces of DNA in the plasmid and replicate them in yeast 1-extract DNA from wt yeast; all loci have functional alleles 2-fragment the genome with restriction enzyme 3-take random fragments from the genome and clone them into yeast plasmids 4-create a library of random fragments form the WT yeast genome 5-yeast with non functional cdc24 gene 6-transform each one of the episomes into the yeast and look for phenotype (functional cdc24) 7-extract the plasmid and make mutations in cdc24 gene 8-What part of the cdc42 gene are important for function, do this by making mutation, ask which mutation do and do not complement lf Map out functional important regions
40
Misexpression and gf
What happens when we mis-express the gene, do we see an alteration the phenotype. Do a designed approach Antennapedia is expressed in cells that will give rise to the second leg. What if we misexpressed antennapedia in the antenna cells of the organism
41
Transposon mediated transformation of antennapedia
marker, hsp and antp cds heat shock promoter results in heat shock inducible ANTP protein expression, results in ANTP expressed in all cells make a very specific construct Mis-expression results in antenna to leg transformation
42
Knocking out genes
Knock out genes via homologous recombination great way to test the function of genes
43
Homologous recombination
DNA molecule that contains a marker is introduced into the cell, introduce that into the cell and this DNA will search through the genome looking for homologous DNA sequences and it it finds it the homologous recombination event will occur.
44
Dominant selectable marker
the marker selects for homologous recombination the insertion of the marker into the gene is the mutational event that disrupts the gene the marker is inserted into the gene via hr we need a very strong selection in order to identify cells where this event has occured. the marker is the mutaitonal event that disrupts the gene
45
Systematic gene knockouts in yeast
functional genomic, all yeast genes have been knocked out and you can screen strains for which gene is required for your process you can screen these strains for which gene is required for your process Strain one is missing a gene. Screen each strain for a gene required for your process.
46
RNAi
Double stranded RNA complementary to mRNA Inhibits mRNA by degrading it, the gene is unaffected no mRNA no protein
47
Mechanism of RNAi
Double stranded RNA is digested by an RNAse called dicer fragments are recognized by ago and with other factors creates a RISC. It is target to our specific mRNA because of the ability of the RNA fragment to bp with mRNA
48
Major Points of RNAi
Specific mRNAs are targeted for degredation because of the complementarity of the RNA bound to AGO and the mRNA The RNAi mechanism is thought to have evolved as a mechanism that supresses parasitic genetic elements: viruses, transposable elements
49
miRNA and soil worm development
speical future of development is its invariant lineage it goes through a stereotypic series of cell divisions: We know at specific times what cell will be present in the developing embryo or larvae. each one these divisions can be followed by using microscopy because it is a transparent organism
50
lin-14
Timing is affected, allows us to identify mutants. Lf no complicated first lineage, l2 lineage is occurring during L1. Gf leads to a complicated L1 lineage which is repeated again. lin-14 encodes a nuclear protein
51
lin-4
This gene is important for supressing the L1 lineage. lin-4 encodes a non-coding RNA a micro-RNA-the first found
52
Lin-14 gf
lin-4 rna base pairs with the lin-14 3' utr lin-14/lin-4 duplex 2 (bulged) in lin-14gf mutants the lin-4 binding sites are deleted
53
lin-14/lin-4 mRNA/Protein expression
Expressed at all stages of larvae development but is not active during L2 protein expression falls during L2 lin4 miRNA expression starts at the L2 stage and remains present at all stages, as it accumulates and binds to the mRNA it suppresses the translation of lin14 mRNA
54
miRNA mechanism
miRNA is apart of your genome, they are transcribed genes (RNAPII) produces a primary miRNA transcript that is inactive, it is processed to a pre-miRNA and exported from the nucelus where it is processed by dicer dicer slices it into small pieces and AGO binds to the miRNA recognized by cofactors to form the RISC
55
Argonaute protein family
Multiple proteins encoded in the genome AGO1 miRNA silencing AGO2 RNAi mRNA cleavage 8 AGO proteins have different biochemical properties
56
miRNA vs RNAi
miRNA surpresses translation of mRNA RNAi degrades the mRNA No protein is produced by distinct mechanism
57
miRNA role in RISC
it guides the RISC to specific mRNA it does not activate the RISC proteins If you could get AGO onto the mRNA idenpendently of the miRNA, the RISC proteins should still work tethering ago to lambda n RNA binding protein still allows for RNA silencing
58
Functional genomics with RNAi
Ecoli expresses a double stranded RNA, nematodes eat ecoli and absorb the double stranded RNA Double stranded RNA is getting to all the cells and inhibting gene products
59
Ecoli nematode plasmids
cDNA is inserted between two promoters; to get double stranded RNA 20,000 strains with a cDNA for every gene, into each ecoli. transfect 20,000 nematode with one ecoli each to knock out every single gene and look at the phenotype nematode eats ecoli and absorbs dsRNA, that silences the gene of interest
60
Defenses against genomes
1-RNAi/miRNA 2-Restriction modification 3-CRISPR natural defenses against genomes, selfish elements, viral DNA
61
Bacteriophage
Bacteriophage infects the bacteria viral DNA takes over the host machinery replicates its RNA and is packaged into viral shells The bacteria lysis open and the phage viruses are released
62
Restriction modification immunity
Source of restriction enzymes Origin of recombinant DNA technology - RE sees phage DNA, binds to recognition site and chops it up into little pieces - methylase methylates all the genomic DNA restriction sites so that the genomic DNA is protected from the action of the restriction enzymes
63
CRISPR mediated immunity
1-Acquire 2-Immunity 3-Reeingineered acquire information from invading DNA store it in its genome and use it to defend itself
64
CRISPR Components
Clustered regularly interspaced short palindromic repeats cas: CRISPR associated PAM: protospacer adjacent motif
65
Bacterial CRISPR Steps
1-Cas enzymes are able to cut off a section of the invadgin DNA next PAM, cas binds to PAM and then cuts the DNA 2-Cas binds the piece of viral RNA and inserts it into the genomic DNA at a CRISPR array 3-When a phage infects again, pre-cr RNA is transcribed 4-pre-cr RNA is processed to release each spacer which is homologous to past invading viruses 5-Tracer RNA binds to a motif adjacent to the cr-RNA spacer 6-Tracer RNA allows for Cas9 to bind 7-Complex formation, Cas9 protein with guide RNA will find the PAM in the phage genome and then unwind the DNA using guide RNA as a template and then induce a double stranded break in phage genome inactivating it
66
Why doesn't cas9 cleave the genome
no PAM in genome, a sequence called PAM is just NGG, that is next door to the DNA that is going to cut out highly unlikely that the genome is cleaved, 44 bits of information is very high specificity
67
Reeingeenred CRISPR
Make RNA which contains the recognition seqeunce and the tracer RNA into a chimeric RNA Two component system
68
Double stranded breaks
Needs to be repaired: End joining homologous recombination DSB are very dangerous, they initiate mitotic recombination
69
End-end Joining
NHEJ, enzymes reseal the break, they are not very precise and in the process of repairing the break introduce indel mutations that can affect gene functions.
70
Homologous recombination
DNA contains a selectable marker that has homologous arms to the gene you want to insert your selectable marker in to create a mutation/bring in a new DNA element Enzymes are brought to the spot to induce recombination
71
Mutation of Cas9
Introduce mutation in the areas that make the double stranded break, they cannot make cuts because of the high information content of its binding you can locate unique sequences in the genome Inhibitor of transcription Activator of transcription Florescence protein

Genetics (16 decks)

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