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1
Q

what are the advantages of using zebrafish as a model organism?

A
  • they are easy to maintain in a lab - they only live 3-4 months so low generation time - there are large progeny clutches- foward genetic screens can be carried on on F1 (using techniques) - can survive for 3-4 days as a haploid - can be used to make haploid or gynogenetic offspring- they develop in clear eggs- they are transparent so can see the heart and blood system easily too
2
Q

explain the protocol for a normal Tubingen screen.

A
  1. expose a male fish to a mutagen in the hope the germ line will be effected 2. cross with a female 3. each F1 offspring will represent a different mutation and so will form a single family each 4. each F1 is crossed with a wilf type to produce families whichare 50% het. 5. random F2 crosses result in 25% of the F3 being homozygous mutant
3
Q

give an example of a zebrafish mutant being used to model a human disease

A

anaemia- low blood cell count in fish

4
Q

what two processes can be used to prevent having to screen for mutants in the F3, and in an earlier stage instead?

A

creating haploid offspring or parthenogenetic diploid offspring. They allow screening in the F2.

5
Q

what is the protocol for making a diploid offspring for forward genetic screen?

A

You expose a male to mutagens and cross with a female to produce heterozygous F1s. You then squeeze a female F1 and make her lay her eggs. They you take UV treated sperm which contains no genetic material that can contribute to the offspring but can still activate the egg. Then this egg will continue as a haploid organism for a few days- allowing a mutant from the het chromosome to be seen in 50% of the offspring.

6
Q

how can you use gynogenesis to shorten the length of genetic screening process?

A

-you can apply early pressure to inhibit meiosis ii or you can used heat shock to prveent mitosis i.

7
Q

describe why you would use early pressure in a forward genetic screen.

A

This can be used to allow fish to be screened in the F2 rather than F3. During meiosis I, the sister chromosomes replicate and then the celll divides, creating a cell which only contains the two replicates of one chromosome of the het female. You then squeeze the female and extract the eggs and fertilise with UV treated sperm to activate meiosis II, however you inhibit meiosis II with early pressure. This means that the diplod oocyte doesnt divde and so enters mitosis I as a diploid and becomes a diploid organism which is homozygous for the areas of the chromosome other than those that can recombined. This results in a 50% chance of getting homozygous mutant offspring. However, as the position of the mutation moves further away from the centrosome, recombination is more likely so 50% decreases.

8
Q

describe how you could use heat shock in a genetic screen.

A

This can be used to create a diploid homozygous mutation which is homozygous in 50% of the offspring without the trouble of recombination that early pressure provides. This is carried out by applying heat shock at mitosis I stage after UV treated sperm activate haploid oocytes. This means that the replicated chromosomes do not divide into 2 different cells so they are frozen as a diploid oocyte and therefore embryo.

9
Q

explain why a mosaic screen would be used

A

if you have a mosaic animal then it may not die even though the mutation is lethal because the other non-mutant cells compensate but you may still see a phenotype

10
Q

describe the process of a insertional mutagenesis with a retrovirus forward screen.

A

virsus are injected into 1000-2000 cells at the embryo stage. Hopefully these will affect the germ cells of the P fish which grow from these embryos. - these P fish are then crossed - these F1 fish and then screened via real time PCR or southern blot for multiple insertions and those containing multiple insertions are then intercrossed to produce fish that are heterozygous and wild type for a locus. - you then cross the F2s randomly and you will get the heterozygotes crossing to produce 25% homo mutant. But because at other loci there are mutations, some of the fish which are het or homo dom for the first mutation, may be homo recessive for another mutation and so increase the overall number of mutation phenotypes within the F3- multiple genes homo recessive if not linked

11
Q

give three examples of fluorescence reporters that have been used in genetic screens

A
  • to identify mutants that ar eunable to process phospholipids and cholestrol. a reporter for lipid processing, an engineered quesnched fluorescent moiety was placed in PLA2 cleavge site so that cleaving results in fluos. So mutants whom cant digest lipids properly will not emit flo when fed these phosphplipods . - fluorescent dies injected into the eyes of dat 5 larvae label the entire lengt of the retinal axons, which can be visualised in the skin. you can then screen F3;s for retinal path finding mutants - by linking green fluorescent protein to genes or promoters of interest, it is possible to visualise changes in gene expression in mutants
12
Q

what kind of screens of zebrafish particularly good for and why?

A
  • behaviour screens as they show social preference and hunting behaviour. This can be used to study things such as autism etc - because of their tranparency, optogeneics can also be used to allow the firing of different parts of the brain to be seen in response to visual or other stiimlu- allowing neural curcuitry to be studied.
13
Q

how can pathways been analysed in screens?

A

you can use the concept of enhanced or suppressed phenotupes. This involves using a mutant line and then carrying out a sceond mutant screen for mutation sthat suppress (reverse) the mutation or enhance (amake it more extreme) to find genes that are implicated in the same pathway.

14
Q

describe how allele specificity can assist dissceting a genetic pathway?

A

???

15
Q

list some other techniques that can be used to look at mutant gene function.

A
  1. temperature sensitive alleles2. using morpholinos for knock downs 3. ectopic gene expression by mRNA injection of viral transduction 4. can use chemicals which mimic genetic mutations in order to determine timing that a gene or protein is needed.
16
Q

what disease can be replicated in zebrafish?

A

anaemia

17
Q

what are the disadvantages of performing screens in zebrafish?

A

you have to screen in the F3, although you screen in the F3 for flies (F2 for c.elegans), fish are much harder to maintain in the lab that flies and therefore breeding requires more time and expense.

18
Q

how long can zebrafish live for as a diploid?

A

3 days

19
Q

what are the advantages of producing a haploid F2 fish?

A

they will expose the mutant phenotype and do not have to be crossed to make a homozygous

20
Q

what defects do diploid zebrafish have that could complicate screening?

A

They are shorter, so if you were screening for growth mutants this would be confusing, they contain more and smaller cells, and can have other morphological defects,

21
Q

what genes have experiments using diploid zebrafish uncovered?

A

genes involved in brain development

22
Q

what stage in meiosis does early pressure inhibit and what is created as a result?

A

meiosis II, it prevents the diploid oocytes from doing their last division (they have previously been activated by UV-treated sperm) and so these diploid oocytes develop into diploid embryos

23
Q

what point in meiosis does heat shock inhibit and what is created as a result?

A

after the haploid oocytes have developed normally, they then attempt to undergo mitosis, but heat shock prevents the cells from dividng after the chromosomes have divided so the haploid oocyte then becomes a diploid embryo

24
Q

why are diploid embryos generated from heat shock preferable to early pressure?

A

they are homozygous for all loci so they are more likely to express a mutant phenotype for genes that are positioned away from the centromere.

25
Q

what are the methods that can be used in zebrafish to screen in the F2?

A

gynogenesis (heat shock and early pressure) and haploid embryos

26
Q

what are the drawbacks of gynogenesis

A

there is a low viability of the embryos, only 10-20% form from heat shock.

27
Q

what is the difference between treating fish with ENU and then breeding fairly soon after mutagenesis and treating fish with ENU and then breeding a lot and then starting the screen procedure?

A

If you treat with ENU and then breed straight away, You are using post meiotic sperm cells and therefore, because ENU induces mutations on a single strand of DNA, the mutation only becomes fixed following several divisions after fertilisation. This means the resulting F1 are mosaic for a given mutation .This is good because mosaic fish can withstand a great mutational burden. if you mutagenise and then breed for a while then the mutation becomes fixed in the sperm of the pre meiotic germ cell and so the offspring are not mosaic

28
Q

what are the advantages of breeding the mutagenised fish for a while following mutagenesis and then starting the screen procedure?

A

the fish aren’t mosic so you dont have to worry about the mutant cells not being in the gonads of the F1s.

29
Q

can the efficiency of loci mutation rate vary between genes with ENU?

A

yes

30
Q

what are the downsides of using EMS in zebrafish?

A

it is less potent than in flies

31
Q

what mutagen is potent in flies but not in zebrafish?

A

EMS

32
Q

what is an alternative to chemical or radiation induced mutations? what are the advantages of this approach?

A

insertional mutagenesis with a retrovirus. allows rapid cloning of the gene using inverse PCR

33
Q

what is the name for a screen in zebrafish that uses insertional mutagenesis via a retro virus?

A

a cambridge screen

34
Q

what are the pit falls of using insertional mutagenesis in zebrafish?

A

they are one ninth as effective as ENU.

35
Q

how can you use ectopic genes expression techniques to identify components of a pathway?

A

You can inject mRNA of a gene and see if it can rescue a mutant for another gene, if it can then it could be acting downs stream of the mutant gene or in the same pathway.

36
Q

how can you carry out time specific mutations or spatially controlled expression in zebrafish?

A
  • you can use the Gal4/UAS system in fish too. You can also also use the clarity of the zebrafish: light beams can be shone onto specific cells or at specific time points to release caged RNA molecules or activate transgene expression (cool)
37
Q

why are zebrafish so suitable for chemical screens?

A

the zebrafish chorion is permeable to small molecules and is amenable to high-throughput screening

38
Q

why are chemical screens sometimes preferable to genetic screens?

A

if you want to find components that effect a certain developmental process, you can apply chemicals at a certain point in development. This can allow temporally controlled perturbation. Then you can identify the target protein of the chemical. Chemicals can also potentially interere with one or more targets which circumvents problems with genetic redundancy in a particular pathway.

39
Q

how can you identify components of a pathway of interest in zebrafish?

A

You can use whole mount in situ to see genes that have similar expression to the gene of interest. You can see if ectopic expression of genes can rescue a phenotype (exogenous application of RNA). you can carry out enhancer or suppressor screens.

40
Q

Describe a normal F2 screen

A

Because c.elegans can be hermphrodites, screens can be carried out on the F2. 1. a mutagen such as EMS is applied to the herm 2. one of the germ cells will be heterozygous and so will create offspring which are heterozygous for a mutation. Each F2 can create its own family. 3. these heterozygous F1s will then self-fertilise and produce 25% double mutants.

41
Q

what did Brenners screens aim to isolate?

A

visible mutants such as large small/ fast moving etc

42
Q

what is the simplest way to identify genes that are all involved in one pathway?

A

Look for mutants that all have the same phenotype and then do complementation tests to see if the mutations are occurring at the same loci. If not they you have identified other genes

43
Q

what is a suppressor screen and give an example

A

suppressor screens involve looking for a mutation in a known mutant screen which reverse the mutational phenotype. For examlple a gene called Lin-15 was identified which caused constitutively active RAS pathway signalling (it is a negative regulator). However, another mutation of downstream targets of the pathway reverse the phenotype as it couldnt signal any longer.

44
Q

what types of pathways are supressor screens particularly good for?

A

negative regulatory hierarchies because loss-of -function mutations can cause const activation of a pathway and another mutation can suppress this pathway

45
Q

describe two clever epistatic screen which involved the sex pathway

A

you want to find a mutation which changes a male to a female but how do you spot transformed female among lots of normaly female? You can carry out a screen in a him-5; dpy-1 background. Him-5 causes lots of males to occur and dpy-1 causes any XX organisms to be dumpy. So a mutation which trusn XO males into XO herm which cause a non-dumpy herm to present - XOL-1 is a protein which specifies males and is involved in increasing the doage of X genes to compensate for the XO genotype in males. him-5; XOl-1 produce numerous XO animals because of him-5 but these die because of the xol-1 mutation. Finding a mutation which restores this phneotype to viable males will allow other genes in the pathway to found.

46
Q

give an example of a “ screen from heaven” involving drugs and explain why it is good

A

An ideal screen is one which eradicates all of the organisms not of interest. For example screening for resistance to a drug. For example Aldicarb cause ach to build up in the synaptic cleft and cause death by overexcition. However, some mutants develop resistance as they cant release ach from the synaptic cleft- these are still alive

47
Q

what makes a screen from heaven?

A

a screen in which the relevant mutants can be easily identified- for example, offspring which arent of interest die. Or selection for a suppressor mutant in which a mutation restores viability

48
Q

what are the 3 general advanatges of being able to perform such large scale screens in c.elegans

A
  1. saturation screens: saturation of genes that can be mutated to give a specific phenotype can identify many or all of the components in a particular phenotype. The presence of many alleles in all complementation groups groups shows that a screen is saturated 2. can identify unusual mutations that arise less frequently than null alleles. For example null mutations for most of the components of the RAS signalling pathway but some mutations are partial. Another example is finding allele specific suppressor mutations- this can be found in proteins that interact with each other- a mutation may cause one component to compensate for the loss of function of another and rescue the phenotype- specifically in receptors or cofactors. 3. allows structure and function analysis- different mutations will effect different domains within the gene which can have different functions
49
Q

what are examples of screens from hell?

A
  • phenotypes that cant be screened in the F2- germ line ageing- can take occur over 10 generations and can look for mutations that affect this- microscope screens that require extreme precision - laser ablations
50
Q

why are the standard protocol for lethal screens not applicable?

A

normally you would screen the F2’s, however for lethal homozygotes there are no F2s! SO you need to think of a way of screening so that you can still maintain a source of mutant chromosomes to use for lines

51
Q

when are lethal genes normally implicated temporally?

A

in development normally

52
Q

explain a cool way of screening for mutant effect genes?

A

you can use mutant strains that lin-2 which prevent eggs from being laid. This allows the 25% arrested progeny to be examined easily. You can then keep the F1’s knowing that they are het for a mutant.

53
Q

how do you do a maternal effect screen and why do they pose such as difficulty?

A

The problem with maternal effect is that the F2 mutants are viable because they receive the maternal effects proteins/mRNA from their het mother. As with lethal mutants, you can use a genetic background which has lin-2 mutant preventing egg laying. So you mutagenise worms and then plate each F1 mother and observe her F2 offspring. F2 mothers that are homozygous for a mutation cant give their embryos the right proteins so they die and stay in the mother as a bad of eggs rather than a bag of worms. you can then use the siblings of the F2 mothers which are het so rescue the mutation for lines. (

54
Q

what are the problems with investigating lethal mutations and how can this be circumvented?

A

It is hard to determine the pathways that are involved in the lethal- you could try a suppression screen. or you can anaylse development very closely

55
Q

explain how you carry out an ENU and include the numbers and logic for each step.

A

After mutagenesis, about ten worms (P0 generation) are transferred to large Petri plates and allowed to lay ~100 F1 progeny. The P0s are then removed and the F1s grow to adulthood. After one day of egg laying, the F1 parents and any hatched F2 progeny are washed off. However, the F2 eggs stick on the plate, leaving a 14-h cohort of F2 progeny to survey. So, a fraction (~20–30 individuals) of the 250 potential progeny from any particular F1 are sampled. One-quarter of these 20 progeny will be homozygous for the mutant chromosome, and so about five individuals that are homozygous for any mutation that occurred in the germ line of the P0 are expected. Because each plate of worms descends from 100 F1s, 200 copies of a particular locus can be assayed for mutations. Using typical concentrations of a mutagen (for EMS (ethyl methane sulphate) this is 50 mM), 2,000 copies of a gene need to be screened to find a mutation. So, about ten plates of worms need to be inspected to find a mutation in the gene of interest. See also REF. 105. The F2 progeny are inspected for the phenotype of interest and candidate mutants are removed to a fresh plate to evaluate if the character breeds true.

56
Q

describe a suppressor screen and an example

A

Screens that are suppressor screens as well as selections are particularly powerful. In some cases, the initial genotype can be engineered into the strain by expressing proteins that constitutively acti- vate a pathway. For example, the trimeric G protein Gsα can be locked in the active form by mutating the GTPase domain. The expression of activated Gs under the control of a heat-shock promoter30 or a neuron- specific promoter31 causes neuronal degeneration, which induces death or paralysis in the animal. In the- ory, mutations that suppress this phenotype, by restor- ing viability or movement to these animals, could be found. Specifically, mutations in downstream compo- nents in the pathway — that is, in genes that encode proteins required for neuronal death — were identified as second-site mutations that restored cell viability. Suppressors of neuronal degeneration showed that necrotic cell death acts through cyclic-AMP signals because adenylyl cyclase is required for normal levels of Gs-mediated cell death.

57
Q

there is a problem with lethal mutants in that it is not a very specific phenotype, how can you screen for the specific cause of your mutant lethal gene?

A

Pierre Gönczy and colleagues screened chromo- some III for all maternal-effect-lethal mutations, by using an inverted chromosome to balance and main- tain the induced mutations. By re-screening these strains under high magnification for defects in cell division in the one-cell-stage embryo, these researchers identified 34 loci that are required for mitosis and cytokinesis in the first cell division. Another successful strategy for identifying interest- ing lethal-mutant phenotypes has been to manipulate worms physically and thereby predict the relevant phe- notype for which to screen. For example, because it has an invariant lineage, C. elegans was originally character- ized as a simple developmental system that depended on the segregation of determinants into specific lineages — quite different from the development of higher organisms. However, Jim Priess removed a single cell in the four-cell-stage embryo and discovered that the development of a neighbouring cell, called ABa, was dis- rupted. ABa made epidermis and neurons, as it did usually, but failed to produce pharyngeal muscles. By screening for mutants in which ABa failed to produce pharyngeal cells, Priess and colleagues identified three genes that are involved in the Notch signalling path- way. These studies and others revealed that the stereoptypical lineage of C. elegans relies on invariant cell–cell interactions more than it does on the segrega- tion of determinants into specific lineages.

58
Q

what is genetic redundancy and how does it occur, and how can you identify redundant genes?

A

In the simplest situation, functional redundancy arises because a gene has been duplicated and two or more closely related genes exist, any of which can carry out a particular task. This phenomenon is called homologous redundancy. The synergism between two redundant genes can be discovered because a deletion exists that removes both genes. Interestingly, the C. elegans genome contains many small duplications that carry pairs of potentially redundant genes. Mutagens that generate deletions might therefore be useful for the design of screens that circumvent redundancy- can also do sensitised screens - or double RNAi

59
Q

how long is the generation time in flies?

A

10 days

60
Q

what are the advantages of using flies as a model organism?

A
  • easy and cheap tp rear- no recombination in males so tracking mutations is easy - only 4 chromosomes which are easily visualised- the exoskeleton of flies is good to look at patterning development of the embryo - has many homologues in humans- twice as many than c.elegans
61
Q

what is the problem with screening for mutations in genes that have many different effects at different times?

A
  • if you produce a mutant, only its first role will be disrupted- you cant get past this stage without using a conditional mutation. At example is Wg in flies. This is involved in development and in the adult
62
Q

what is the problem with screening for pehnotypes of offspring that express lethal maternal effect genes?

A

the phenotype wil lonly be expressed in the offspring once the maternal source has run out- so you can see the role that it has before this point.

63
Q

how did scientists try to address the problem of onlybeing able to screen for the role of the first role of a gene that is expressed at different time sin development?

A

They looked at mutations which were homozygous from large deletions of the genome. They then let the embryo develop until it died. They then noted that the gene that was deleted from the space which was the earliest acting from that which were deleted, would show its phenotype. They then repeated this with different regions there were abe to identify all the regions and genes which held the early acting genes. they found that there were only 7 early acting zygotic (non maternal genes)

64
Q

give an example of using a dominant enhancer screen and why they are good

A

Loss-of-function mutations in almost all genes are recessive, which indicates that 50% of the wild-type level of a protein is sufficient for normal development. When a particular process is already par- tially disrupted by another mutation, however, this amount might no longer suffice, and mutations in the genes that are involved in the pathway can therefore be identified as dominant enhancers or suppressors in this sensitized genetic background.- because this relies on a heterozygote mutation (dominant) you can screen for these in the F1.

65
Q

what are three good things about using sensitise backgrounds for screens: enhancers?

A
  • you can screen in the F1 for dominant enhancers - if the sensitized background only effects one part of the body (such as the eye) which isnt required to live then you can see the effect of mutations to otherwise lethal genes that function as many stages in development as the sensiitzed background only affects the eye - Third, the whole genome can be screened at once, because there is no need to use balancers to make particular chromosomes homozygous
66
Q

what are 2 reasons why sometimes looking for dominant enhancers or suppressors isnt good?

A
  • not all components of a pathway will necessarily be dosage sensitive- the levels are not crucial to functioning sometimes- many mutations may turn out to be homozygous viable and have no phenotype when removed from the sensitive background- no guarantee that a gene has an essential role in the process.
67
Q

why is the eye a good target for enhancer and suppressor screens?

A
  • it is not essential for viability or fertility ad is easy to score. It is particularly good for modifier screens because it is composed fo 800 regulalry packed ommatidia and numerous defects in cell-fate determination and differentiation give a rough eye phenotype, the severity of which reflects the number of ommatidia affected. It is even possible to carry out screens for modifiers of genes the functions of which have not been characterized in the eye, if the eye-spe- cific expression of a wild-type or mutant construct of the gene produces a rough-eye phenotype
68
Q

explain how the Flp/FRT can be used to generate mitotic clones

A

In mitosis one paternal and one maternal copy is inherited by each daughter cell. Flp is a recombinase which causes recombination between sites with a FRT site. Therefore, when activated and present with a transgenic line that has FRT sites on the same site on the matenral and paternal chromosomes (inserted via homologs recombination), then flp will cause recombination between the two and one paternal chromosome will the mutant so the daughter cell will get two copies and be homozygous.

69
Q

when can the Flp/Frt be used?

A
  • it can be used to study the difference in growth rates between cancerous cells and normal cells. This can be done by also putting a marker that is linked to the mutation- e.g red vs white eye s
70
Q

explain how the Flp/Frt system can be used to screen for maternal effect genes in the F2.

A
  1. cross a male which expresses the dominant ovoD sterility gene in its germ line and a balancer, with a female that expresses a mutant and a balancer. These both have an FRT below the mutant position. The mother will also express a heat sensitive Flp allele so that after fertilisation it can be triggered2. The F1 female is produced that has a genoptype of the maternal expressing ovoD and the paternal expressing the mutation. It also expressed Flp which is heat sensitive. 3. Flp can be activated and the germ line cells can undergo recombination when they divide. This will result in either no recombination which will not be layed due to the ovoD, a double ovoD which also wont be laid and a double mutant which will be laid 4. the F2 can be screened and the male siblings used to keep the mutation.
71
Q

describe how tissue specific Flp/FRT can be designed in such a way to distinguish between each cellular outcome of the FRT/Flp

A
  1. You can introduce a “Minute” muttation (M) onto the non mutation atm that cause cell death when homozygous. Therefore, twin spot cells die and the non recombined hets have inhibited growth.This means that the twin-spots die and the non recombined are out competed by the recombined homo mut which has normal growth. By having a mutant white eye allele next tot the minute mutation, the different cell types can be viewed. This is good because you can basically erradicate all of the non non-clonal cells in order to look at the effects of your homo mutation in one part of the organism 2. another way is to put a mutated White eye allele in the arm that has a mutation on the other chromsome. This means that the twin spots are dark red, the hets (non-recombined) are red, and the homo mutatnts are white. However, this doesnt get rid of the other cell types. This can be used to look at cancerous mutation’s growth relative to a wild type cell.
72
Q

why is using the FRT/Flp system useful? (2)

A
  • it allows you circumvent the problems that arise with looking at the effects of mutations which are affective at different times of development. If you knock out a gene, you will only see the phneotype for the first time that it acts. So you can express genes specifically in different tissues at different times - it is good for comparing cancerous mutation as you can compare the growth of the different cell types
73
Q

why are drosophila good model organisms in terms of compromise for human studies?

A

they are very easy to work with and easy to do genetics in but they also share many conserved genes with humans

74
Q

name 2 famous examples of pathways in flies being transferable to vertebrates?

A

patterning of the D/V axis: Dpp; BMP4 and sig;chordin

75
Q

are flies good for forward genetic screens?

A

yes

76
Q

what is the easiest way to mutagenise a fly?

A

feeding them EMS which induces point mutations

77
Q

why are flies particularly good for developmental investigation?

A

their exoskeleton (cuticles) provide an exquisite read out of the patterning of the embryo. Secondly, the mother provides her eggs with most of RNA required for embryogenesis so very few mutantionf or embryonic lethal and very few block embryonic development at early stages- this means screens were ver efficient at identifying the transcription factors and signalling molecules that generate positional information in the embryo.

78
Q

what does a saturated screen mean?

A

Saturation screens are used to uncover all genes involved in a particular phenotype of an organism or species. The screen is carried out by mapping mutants of a biological process until no new genes/gene mutations can be found.

79
Q

what is a general problem with using screens for genes that are recessive lethal?

A

you can only observe their first essential role, as after this point the embryo is dead so can not longer be screened for future roles of the gene

80
Q

how did Wieschaus identify the first early acting genes in drosophila?

A

they made flies that were homozygous for large deletions. They then were able to work out the regions which contained the essential early acting genes- they found that 7 were required

81
Q

what is the general premise of enhancer and suppressor screens?

A

Loss-of-function mutations in almost all genes are recessive, which indicates that 50% of the wild-type level of a protein is sufficient for normal development. When a particular process is already par- tially disrupted by another mutation, however, this amount might no longer suffice, and mutations in the genes that are involved in the pathway can therefore be identified as dominant enhancers or suppressors (mutations) in this sensitized genetic background.

82
Q

what are the advantages of performing an enhancer of suppressor screen?

A

First, the progeny of mutagenized flies can be screened directly (an F1 screen), because the mutations do not need to be made homozygous, which means that an order of magnitude more flies can be screened than in an F3 screen. Second, lethal mutations in essen- tial genes that function at many stages of development can be isolated, because the sensitized background only affects the eye, which is not required for viability ( so basically the identified gene may be involved in early processes so you can’t make a homozygous mutant but you can have a sensitised background in an area that will not cause lethality when perturbed). Third, the whole genome can be screened at once, because there is no need to use balancers to make particular chromosomes homozygous.

83
Q

what do you do if you have used a sensitised background to carryout a screen but you can only identify the pathway so far before the components become less sensitive to the sensitised background, the more downstream they are. and what if the most downstream gene you have identified is an early essential gene so you can’t produce a sensitised background using it? (fly)

A

you can express the component which you want to know the downstream components of in the eye, if this gives a particular phenotype, you can then look for mutations which perturb this phenotype- having the ey in the drosophila which you can perturb without causing lethality is a very useful trick

84
Q

what are the drawbacks of sensitised screens?

A

not all pathways are dosage sensitive. many of the mutations turned out to be homozygous viable and have no phenotype when removed from the sensitized background. Suppression or enhancement of the phenotype of interest is therefore no guarantee that a gene has an essential role in the process. It may also be necessary to carry out screens in a variety of different sensitised backgrounds to uncover all of the components of the pathway.

85
Q

where is the bets place to carry out sensitised screens in the fly and fly?

A

In the fly eye, Furthermore, it is particularly suited to modifier screens because it is composed of ~800 regularly packed OMMATIDIA, and numerous defects in cell-fate determination and differentiation give a rough-eye phenotype, the severity of which reflects the number of ommatidia affected. It is even possible to carry out screens for modifiers of genes the functions of which have not been characterized in the eye, if the eye-spe- cific expression of a wild-type or mutant construct of the gene produces a rough-eye phenotype

86
Q

what are clonal screens in flies?

A

screens in which only the area of interest expresses homozygous alleles for the mutation.

87
Q

why are fly clonal screens useful?

A

they can be used to study pleiotropic lethal mutant genes

88
Q

what are redundant signalling molecules?

A

signalling molecules which when knocked out or down, can have their signalling compensated for by other genes.

89
Q

Why can’t redundant signalling molecules be discovered in normal loss of function screens?

A

because when they lose their expression other genes will compensate

90
Q

how can redundant signalling molecules be uncovered in the fly?

A

by inducing the over expression of genes. This was done by Rorth et al who generated a p-element vector called the EP element which carries a UAS site at one end, so that any gene it inserts next to can be activated by gal 4. this makes it possible to carry out gain- of function screen by simply crossing a half 4 driver that is expressed in the appropriate tissue ( to which the process being investigated is relevant) to a large number of EP insertions and looking for a phenotype.

91
Q

what did the Rorth group us EP element insertion to investigate? how many lines did they have to test? what is the down side of this approach?

A

border-cell migration when overexpressed in either the border cells or the germ cells through which the border cells migrate. 8500. requires LOTS of crossing and EP element strain formation

92
Q

What the main model organisms?

A

Yeast, plant (arabidopsis), c.elegans, drosphila melanogaster, zenopus, zebrafish and mice (musculus musculus)

93
Q

List of the order of genetic similarity to humans of the main model organisms.

A

Mice, xenopus, fish, fly, c.elegan, yeast, e.coli

94
Q

what percentage of coding DNA is different in mice compared to humans?

A

10%

95
Q

what percentage of coding DNA is different in fish compared to humans?

A

25%

96
Q

what should be kept in mind when studying the role of regulatory non-coding DNA in your model organism in relation to human application?

A

differences in non-coding DNA greatly, a lot more than protein DNA, for example there is an average change of 50% of non-coding DNA when compared to primates. It is better to use mice for such experiments as closely related.

97
Q

what makes a good model organism?

A
  • short-breeding cycle - large brood size- high relating freqeuncy - robustness- simple feeding and other maintenance requirements- ability to store special strains in an inactive state- history (linkage maps, mutant collections, genome sequence) - molecular techqniues (transgenesis, targetted mutations) - you can do gene targeting and targeted mutagenesis
98
Q

what type of study is normally done in zebrafish?

A

development, disease, behaviour

99
Q

what type of study is normally done in mice?

A

behaviour, optogenetics, brain imaging, genetics, obesity

100
Q

what type of study is normally done in mice?

A
  • cell signalling, nervous system study
101
Q

give two examples of model organisms not mirroring the behaviour of genuinely wild type animals?

A
  • wild mice are more aggressive and less maternal instincts- c.elegans in the wild form clumps in response to high oxygen concentrations
102
Q

Why are c.elegans good for Knock out experiments?

A

there are libraries containing a strain expressing a KO for every c.elegan gene (save this)

103
Q

why list 9 reasons why c.elegans are a good model organism

A
  • easy to maintain - short life cycle (around 3 days)- small number of cells- around 1000 somatic cells - transparent - fully cell-lineaged- detailed anatomy and full connective - genome fully sequenced (first animal)- ease of transgenesis - forward and reverse RNAi genetics
104
Q

why are c.elegans good to study development in?

A
  • there re conserved genes in worms and in other organisms (transcription factors, homeodomains, bHLH, miRNAs, signalling pathways such as Notch, Wnt and TGF-beta)-There are conserved mechanisms: cell autonomous dell fate specification (cascades of TFs and miRNAs), cell non-autonomous signalling (tissue patterning, lateral inhibition)- cell lineage is invariable in the worm so you can ablate cells and look at inductive vs autonomous development .- because worms are transparent so can track lineage the entire time translational and transcriptional reporters can be used.
105
Q

how can c.elegns be stored for a long time?

A

in the dater state- dont need to be fed

106
Q

how many neurons do c.elegans have?

A

302

107
Q

how many glial cells to c..elegans have?

A

56

108
Q

how many morphological classes of neurons are there?

A

118

109
Q

what is considered the brain of the worm?

A

the nerve ring

110
Q

why is having a connectome not hugely important?

A

you want to know the precise interactions between the components, are the signals positive or negative long range or short range?

111
Q

why is the c.elegan being transparent a good thing?

A
  • because worms are transparent so can track lineage the entire time translational and transcriptional reporters can be used.
112
Q

when are transcriptional reporters used? what are their down falls?

A
  • to monitor gene expression or for cis-regulatory analysis (identify enhancer regions) - not all mRNA is translated- can’t be used to look at protein localisation
113
Q

when are translational reporters used?

A
  • monitor gene expression - monitor protein localisation - cis regulatory analysis - over-expression or mutant rescue
114
Q

how do you make translational reporters in the c.elegans and how are they put into c.elegans in particular?

A

you produce a DNA construct containing the gene of interest and its regulatory regions upstream of it. Then you produce an overhanging sticky end which is complimentary to another construct’s end that contains the a gene for a reporter such as GFP with a 3’ unc- 54 UTR on the end of it. SO that they can anneal to each other and act as primers to each other during PCR. These can then be injected into the gonad for form an array (mini chromosomes) that is inherited in a mosaic fashion (can be generated in a week). They can be integrated into the genome for mendelian segregation using gamma/ uv-irradiation (how?)- cam also bombard using gold particles into the genome (how?)

115
Q

when can a mini chromosome be used in c.elegans and for what?

A

They can be used to study cell lineages- if you have a construct that has a fluorescence reporter downstream of a universal promoter and inject, it will be inherited mosaically so one cell will inherit fluorescence and the other won’t after division.

116
Q

what is a fosmid construct and why is it used?

A
  • In order to ensure that the gene of interest that is being inserted into the worm genome is being authentically regulated, a fosmid construct. This fosmid can contain a bicistronic region. A GFP can be inserted after a gene within a fomid but due to the intercistronic region, the primary transcript will undergo trans-splicing. This will result in the protein localising to the normal site while the GFP can undergo nuclear localisation (if has an NLS) and allows for cell identification.
117
Q

what can be carried out in c.elegans?

A

forward genetic screens ,genetic dissection of developmental pathways (epistasis), mapping and cloning, reverse genetics (RNAi)

118
Q

how many chromosomes does the fly have?

A

4

119
Q

what percentage of human disease genes have genes in the fly?

A

60%

120
Q

how many of the fly genes have homologues in mammals?

A

more than half

121
Q

why are flies good for tracking recessive genes?

A

there is no recombination in male flies so you can associate a recessive gene to a visible marker (such as curly wings or white eyes) and you know there will not be recombination between this (but does it not matter about the females?- maybe if passed on from father to and only males used then it doesn’t|)

122
Q

what are two methods of tracking recessive alleles un flies?

A
  • no recombination in males- using balancers
123
Q

explain how balancers can be used to track recessive mutations

A

you can use a balancer which has many inversions on in which prvent recombination. Then have a recessive lethal allele on the balancer and a dominant visual trait so that you can locate the balancers. SO you can cross a mut/bal with a mut/bal, the offspring will be straight winged double mutants or curly winged het mutants. If there are none without curly wings then you know that the mutation is recessive lethal

124
Q

how can you use balancers to determine if mutations are recessive lethal?

A

have a recessive lethal allele on the balancer and a dominant visual trait so that you can locate the balancers. SO you can cross a mut/bal with a mut/bal, the offspring will be straight winged double mutants or curly winged het mutants. If there are none without curly wings then you know that the mutation is recessive lethal

125
Q

are balancers specific to flies?

A

no but they are most widley used in flies

126
Q

describe how p elements can be used for fly transformation

A

1.Clone the P element into a plasmid and transform and grow this in bacteria.2.Eliminate the P transposase and replace it with your gene of interest.3.Microinject the posterior end of an early-stage (pre-cellularization) embryo with DNA coding for transposase and a plasmid with the reporter gene, gene of interest and transposase recognition sequences.4.Random transposition occurs, inserting the gene of interest and reporter gene.5.Once the gene of interest has been inserted it is no longer mobile because it cannot produce its own P transposase.6.Grow flies and cross to remove genetic variation between the cells of the organism. (Only some of the cells of the organism will have been transformed. Hopefully, some of these transformed cells end up in the germ line. A transformed gamete will give rise to an organism with no variation between its cells).7.Look for flies expressing the reporter gene. These carry the inserted gene of interest, so can be investigated to determine the phenotype due to the gene of interest.

127
Q

describe how p elements can be used for mutagenesis

A

Insertional mutagenesis[edit]1.Microinject the embryo with DNA coding for transposase and a plasmid with the reporter gene and transposase recognition sequences (and often the E. coli reporter gene and origin of replication, etc.).2.Random transposition occurs, inserting the reporter gene randomly. The insertion tends to occur near actively transcribed genes, as this is where the chromatin structure is loosest, so the DNA is most accessible.3.Grow flies and cross to remove genetic variation between the cells of the organism (see above).4.Look for flies expressing the reporter gene. These have experienced a successful transposition, so can be investigated to determine the phenotype due to mutation of existing genes.

128
Q

list the types of mutations that can arise from using p-elements

A

Possible mutations:1.Insertion in a translated region => hybrid protein/truncated protein. Usually causes loss of protein function, although more complex effects are seen.2.Insertion in an intron => altered splicing pattern/splicing failure. Usually results in protein truncation or the production of inactive mis-spliced products, although more complex effects are common.3.Insertion in 5’ (the sequence that will become the mRNA 5’ UTR) untranslated region => truncation of transcript. Usually results in failure of the mRNA to contain a 5’ cap, leading to less efficient translation.4.Insertion in promoter => reduction/complete loss of expression. Always results in greatly reduced protein production levels. The most useful type of insertion for analysis due to the simplicity of the situation.5.Insertion between promoter and upstream enhancers => loss of enhancer function/hijack of enhancer function for reporter gene

129
Q

once you have used a p-elemnt to induce a mutation, how can you map where in the genome the p-element has inserted?

A

Using a process called reverse PCR. This involves you inserting a primer site into your p-element which will allow you to PCR amplify. These known sequences of DNA will be either side of a restriction site that you have put into your p-element. So once, you p-element has inserted, you use restriction enzymes to cut up all of the genome and make linear fragments. Then you ligate up using the sticky ends to make circular DNA. Then you use your known restriction site inside you p-element and apply the restriction enzyme so that you circle of DNA of interest is linear with your known DNA seuqneces of the p-element at each end then yu can make primers of these ends and PCR amplify and then sequence the PCR product to find the gene or area of DNA in which the p-lement has been inserted.

130
Q

why are zebrafish a good model organism?

A
  • vertebrate so closely related to humans - rapid development-available in the lab all year round- optically transparentexcellent genetics- forward, reverser and transgenic- excellent embryology - easily injected- easy to make mosaicsconservation of gene function
131
Q

what are the downfalls of flies?

A

??

132
Q

what are the downfalls zebrafish??

A
  • Don’t have good inbred lines- because seem to get bad fecundity when inbreed but just because hasn’t been used a model organism for a long time. This is bad because there is always going to be some genetic background which means that sisters and brothers are going to be different because the parents aren’t isogenic - have gene duplications because of genome duplication events
133
Q

what are the advantages of using zebrafish for genetics?

A
  • unlimited supply of embryos- accessibility - optic clarity- imaging - in vivo cell manipulation techniques from early to late development - genetic and genomic techniques- unlimited mutants- cost - more genes?
134
Q

what is the gene knock-down technique in zebrafish?

A

Morpholinos

135
Q

How do morphosinos work?

A

they bind to the parts of the mRNA and stop it from being translated. SO you can target an oligo to the AUG site of the mRNA

136
Q

why would you want to use a moprholino in zebrafish?

A

If you wanted to do a knock down test to look for gene function in processes. Or if you could see if the knock down of a candidate gene function phenocopies the genetic mutation. You can carry out ubiquitous gene specific knock-down or mosaic knock-down by only injecting into specific areas.

137
Q

what is a draw back of morpholinos

A

they have to be injected into the embryo so they are mosaic- not every cell affected.

138
Q

what is a gain of function techqniue in the zebrafish?

A

you can inject RNA or DNA in a ubiquitous manner or in a mosaic manner.

139
Q

once you have a candidate gene for a function, how can you consolidate the role of these gene (2)?

A

you can use a knock down technque to see if it can phenocopy the mutant or you can use gain of function technique to see if DNA or RNA expression of the candidate gene can rescue the mutant

140
Q

what are the drawbacks of using the injection of DNA or RNA into the zebrafish to cause gain or loss of function effects?

A

they are not integrated into the DNA. This means that you can’t make stable lines from these animals

141
Q

how can you make stable transgenic lines with zebrafish?

A

You inject transposes and a plasmid containing DNA with a transposon construct (told sites either side of gene of interest- such as a reporter) then you inject this into the fertilised egg and the transposes will insert it into the genome of the cells of the embryo will all incorporate the plasmid hopefully. However, if you dont get this at the start of development then you will get a mosaic and can’t guarentee a mendelian ratio of inheritence if not all of the germ cells have inherited the construct

142
Q

if you wanted to make a reporter transgenic line and you wanted to ensure that the reporter represented the expression of the gene as interest as closely as possible, how would you do this?

A

you would use a BAC (bacterial artificial chromosome), these are large amount of DNA that should contain all of the regulatory regions of your gene. these can then be inserted into your genome ??

143
Q

what are the use of dual BACS?

A

??

144
Q

what are the drawbacks of BACs?

A

the larger piece of DNA the harder it is for them to be integrated into the genome?

145
Q

how can you make chimeras in zebrafish? (2)

A
  1. you can make a transgenic universal GFP fish. this can act as the donor. Cells can be taken from this fish and transplanted into an acceptor fish and then the fate of these cells can be viewed relative to the area in which they were derived in development. 2. you can also move cells from one side of the fish and place it into the other side of the embryo- such as when looking at the organiser cells etc- can induce two headed fish
146
Q

how can you study the effect of recessive maternal effect genes, that cause the mother when mutant to die?

A

if you cant have a homo mum because the gene is lethal then you never have access to a homo mum (so the homos die even though the mum contributes early, because of later things) but you still want to see the developmental effects of the gene, you inject the donor embryo (which is homo mutant) with a marker that goes into the germ line. Then you can suck out these cells very early out of an inject into a wildtype fish that doesnt have any germ line cells (killed with morpholinos) so then you are putting these mutant germ cells into a surrgate that can actually survive, unlike the mutant recessive mother,

147
Q

give an example of when the fish having 2 gene duplications has been a problem in research

A

autism (find jason’s paper)

148
Q

what is an advantage of using mice as a model organism for human studies

A

99% of human genes have a clear orthologue in the mouse

149
Q

give an example of the difference between mice and human genetics

A

estimated >20% of human essential genes are non-critical in the mouse

150
Q

what are the three approaches to genetically modifying mice and what are each of these approaches used for?

A
  • microinjection of DNA into fertilised ouse eggs (transgenic mice): gene reg studies to characterise promoters, over expression or miss express studies on mutated or native proteins to induce or rescue a phenotype, expression markers (GFP, LacZ) - genetic manipulation of mouse embryonic stem cells and blastocyst/morular injections: knock-out and knock-in mice, gene trap strategies- genome editing
151
Q

what is the process of DNA microinjection into the mouse egg?

A

the transgene is injected into the pronucleus of fertilised mouse eggs (BAC DNA or plasmid DNA containing regulatory elements and a gene of interest). The insertion into the genome is at random. it is injected into the male pronucleus because it is large. then the eggs transferred into the recipient mother and she produces a transgenic offspring- which is het?

152
Q

what is the process of the knock in and knock outs?

A

using homologous recombination to and make a transcript with homologous arms and inject this into ES cells which have a maker (normally neo for resistance) and then inject into a blastocyst and transfer into a mother to form a chimeric mouse. then if incorporated into the gonad then fill produce a het knockout or knock in which you can use to make a homo mutant

153
Q

what is the ROSA26-stop-reporter mouse?

A

you can use the area called the ROSA26 region which is redundant to insert genes. the ROAS26 promoter is also ubiquitously expressed so you can put your gene downstream but with a stop codon infront of it . you can use this region to insert reporter genes which can report the expression of a gene by using a cre loxp system. You can have a construct that has a SA and a loxp surrounding a neo, tpA (stop codon) upstream of a reporter gene, so that when cre is driven by gene of interest promoter, there is recombination and the reporter is transcribed

154
Q

what are the advantages of using a ROSA26-stop-reporter mouse rather than a nuclear injected reporter construct?

A

you know where it will be inserted so dont risk being inserted into unwanted harmful area or unexpressed hetereochromatin area

155
Q

how can the tamoxifen, oestrogen receptor in mice?

A

you produce a mouse that expresses a cre attached to a oestrogen receptor protein with one that expresses a floxed ROSA26 reporter construct. then when you add tamoxifenit will enter the nucleus.

156
Q

what can you use the ROSA26 area for?

A
  • for reporters for normal cre- for reporters or gene exogenous gene expression driven by cre of tamoxifen - for cell death driven by floxed DTA
157
Q

why is being able to do homologues recombination in mice really important for gene expression analysis?

A

if you can do it right you can get fully physiological control of the reporter- all of the promoter regions are already there

158
Q

give an example which demonstrates the importance of being able to see where your gene of interest is being expressed by using full regulatory influence?

A

Sloan and Barres 2014 wanted to stop gliotransmission by driving a gene that did this in glial cells by using a glia “specific promoter” but they didn’t realise that it is expressed in neurons too

159
Q

what is in utero electroporation used for?

A

you can inject DNA in certain sites by using electroporation - certain areas of the brain- dont have to do genetics in the one cell stage- also works in the fish.

160
Q

what are retroviruses?

A

the genetic material of retroviruses is in the form of RNA molecules. When a retroviruses infects a cell it will introduces its RNA together with some enzymes, namely reverse transcriptase and integrease into the cell. The reverse transcriptase can produce a cDNA copy of itself and then the integrate will insert the DNA molecules into the cell

161
Q

what are the problems with retroviuses?

A

they can insert randomly into the genome which can disrupt gene function

162
Q

how can the insertion site of retroviruses be controlled?

A

they can be used in combination with a double stand break inducer.

163
Q

what is an adenovirus?

A

viruses that carry their genetic matieral as double stranded DNA- the genetic material is not incorporated and is instead free in the cell.

164
Q

what is an adeno-associated virus?

A

it is a small virus that can infect cells and will induce a negligible immune response. it can infect non-dividing cells and will integrate into the host genome- it normally integrates into the human chromosome 19- cool!

165
Q

how could you make a translational reporter than didn’t interfere with the proteins functioning and instead just labelled which cell was affected?

A

you add a IRES spacer and then a NLS attached to the GFP protein

166
Q

go through the entire process of making a translational reporter in c.elegans

A
  • you has a fosmid which you have ordered which you know contains the gene on interest - you know the sequence of the gene so you can make homologous arms on your GFP construct that will recombine din - your GFP construct contains 4 GFP genes, with a selection maker in the middle Galk surrounded by FRT sites. - you insert the fosmid and your recombining structure into a a yeast strain and grow on the medium galactose. - you then select those colonies that survive- you introduce FLP arabinose - then you will have strains elect dfor by growing on deoxy-galatcose cntaing your fosmid which you can purify out and inject!
167
Q

how can transcription reporters be implemented in c.elegans?

A

via a plasmid which doesn’t integrate and is inherited in a mosaic fashion or by actually inserting into the genome for a trasngenic line

168
Q

how would you carry out a transcriptional reporter study using a transcription reporter that is in a plasmid?

A

Plasmids containing marker and target DNA are injected into the syncytial gonad arm. Injected DNA then forms an extrachromosomal array or minichromosome. Progeny that inherit the array display the marker phenotype in a mosaic fashion (?)

169
Q

why are balancers used?

A

Balancers are most often used in Drosophila melanogaster (fruit fly) genetics to allow populations of flies carrying heterozygous mutations to be maintained without constantly screening for the mutations but can also be used in mice.[1] Balancer chromosomes have three important properties: they suppress recombination with their homologs, carry dominant markers, and negatively affect reproductive fitness when carried homozygously.

170
Q

how would you use Homolgous recombination to find out where your gene of interest is expressed- you would want to do this to make sure your Cre is going to affect the right cells.

A

you can HR you GFP into a part of the protein you know won’t interrupt the functioning into the ES cell and then inject into th blastocyst then the mouse mother

171
Q

what should an organiser be able to do?

A
  • form the dorsal anterior mesoderm - induce both the AP and DV axis- induce a secondary axis when transplanted - change in its inducing abilities over time
172
Q

what is the difference between a BL transplanted from an early gastrula as opposed to from a later gastrula?

A
  • it will form an entire axis early on but later it will only be able to for more posterior structures
173
Q

what do the cells from the blastopore lip go on to form in the xenopus?

A

anterior endoderm, prechordal late, head mesoderm, notochord

174
Q

where is head inducing tissue found in the early blastula?

A

in the dorsal vegetal region- the presumptive anterior endoderm- acts as the AVE

175
Q

what do the AVE cells express in the mouse?

A

crescent, cerberus, frzb, otx2, hex- they are involved in head patterning

176
Q

where is cerberus expressed in xenopus?

A

anterior endoderm

177
Q

how does cerberus block BMP WNT and Nodal?

A

binds to them outside the cell

178
Q

how is neural induction different in the chick than in the xenopus?

A

BMP inhibition can not replicate the actions of the node to induce neural tissue- they can stabilise sox 3 expression after 5 hours but that is it- there is a different mechanism for neural induction- involving FGF from the node and from the underlying hypoblast

179
Q

how has the organiser beens how to not be required for neural tissue in xenopus and what has been proposed as a result?

A

a neural plate will still develop in frogs, birds and mice the organiser or node is excised during gastrulation, although chordin and noggin encode for neural inducing signals in zebrafish, double knock out does not prevent a neural plate from developing - therefore it has been suggested that neural fate is given before gastrulation- maybe involving the WNT signalling in the early dorsal patterning of the xenopus embryo

180
Q

what pieces of evidence suggests that gene expression in the mesoderm may be influencing gene expression in the in the ectoderm?

A

several box genes int he notochord and in the pre-somatic mesoderm and ectoderm are at the same position along the axis.

181
Q

where can hox gene expression not be found in the mouse and what is thought to act in their place?

A

it can’t be found in the anterior most neural tissues such otx and em are expressed anterior to the hindbrain and these gene encode homeodomain TFs and specify th pattern of the neural plate

182
Q

what is the simple two step process by which neural patterning is thought to occur along the neural tube?

A

mesoderm induces the antire overlying ectoderm to become anterior and then posteriorising signals transform the posterior parts of tissues after.

183
Q

what are the posteriorising tissues thought to be?

A

FGf ants WNT3a

184
Q

what are the two signals implicated in AP neural tube patterning ?

A

the first are the anterior signals thought to be noggin ad chordin, the second are the posteriorising such as FGF and WN3a

185
Q

how do the node and the hypoblast act together to induce the AP axis along the neural tube in the chick?

A

the node emits FGF and posteriorisig signals, the anterior neural tissue must be protected from these signals- the anterior mesoendoderm is though tot achieve this as it moves anterior to the node after ingression and releases factors such as cerberus, frzb, dkk etc

186
Q

what are the two organiser within the neural plate, what are their roles?

A

One of them is the anterior neural ridge, which lies at the junction between the prosencephalon and the anterior ectoderm, and is necessary for the maintenance of forebrain identity. The second one is the isthmic organizer (IsO), which lies at the junction between the midbrain and hindbrain, and is necessary and sufficient, for the development of mesencephalic and metencephalic structures.

187
Q

what does the anterior neural ridge express?

A

FGF8, otx2, chordin, shh,

188
Q

what is the Anterior neural ridge essential for? how has this been shown?

A

patterning of the telencephalon, removal of it in fish or mice will result in lack of forebrain expression of the telencephalon marker and gratifying an ectopic one into the diencephalon will induce telen marker expression

189
Q

what is the role of the isthmus organiser? how was this shown?

A

found at the midbrain hindbrain junction, if you transplant it, it will induce the formation of an ectopic midbrain in the hindbrain

190
Q

what is found in both the anterior neural ridge and the isthmic organiser?

A

FGF8

191
Q

what is different about the neural induction in the chick and in the xenopus?

A

it doesn’t occur by default in the chick- BMP inhibition is not sufficient- instead it is a multiple step process

192
Q

what is the FGf signalling pathway?

A

FGF TK receptor -grb, SOS, RAS, RAF, MAPKK, MAPK, acts with TF

193
Q

what does the prechordal plate underly?

A

the anterior forebrain

194
Q

what does the notochord underly?

A

the ventral head mesoderm

195
Q

what is the neural default model?

A

this proposes that the defualt model of neural induction in xenopus. this proposed that the default stae of the dorsal ectoderm was to develop as a neural tissue but this pathway is blocked by the presence of BMPS which promote epidermal fate. The role of the organiser so to lift this block by producing proteins that inhibit BMP actvity, the region of ectoderm that comes under the influence of the antagonist form the organiser act on ectoderm and cause it to form neural tissue. according to this model the signals from the organiser act on the ectoerm that lies adjacent to the organiser at the bedginning of gastrulaion. as gastrulation process, internalises cells continue to act on the ectoderm that now overlies them.

196
Q

what does the elimination of one BMP antagonist do compared to all 3?

A

when you remove 1 no effect but when you MO against them all- ventralise

197
Q

explain how FGF has been implicated in neural induction the xenopus?

A
  • neural development in xenopus also required FGF when even BMP is lifted by noggin or chordin. FGF is produce by cells in the blastula that are thought to act as early neural inducers. - it has also been shown that apparently spontaneous differentiaiton of explanted newt ectoderm into neural tissue is caused by activation to the MAPK pathway, this occurs in the FGF signalling pathway and treatment of explanted newt ectoderm with small molecule inhibitors of the pathway prvents neural differentiation. the ras-mapk pathway is also activated when xenopus extodermal cells are dissociated which might explain as noted ealrier hwy this dissociation forms neural fate so easily. induction of the neural plate therefore needs not only the antagonism of BMP signalling but also FGF siganlling (like in chick!)
198
Q

how is FGF signalling though tot act in the neural patterning of the chick?

A

the effects of FGF neural induction may result from a direct inducting effect, activated MAPK can interefere with BMP signalling by causing an inhibitory phoshprylation of smad 1 which is part of the intracellualr BMP sgnalling pathway.

199
Q

describe the ‘qualitative’ and ‘quantitative’ components of neural plate AP patterning

A

qualitative= the mesoderm secrets different antagonists at different points along the AP axis- it secretes BMP, Nodal and WNT inhibitors at the very anterior. But then only BMP antagonists are secreted more posteriorly, thought to induce spinal cord. quantitative: WNt/b-cat signalling along the AP axis is the neurula acts int a posterior to anterior graded manner, injected animal cap explants with different levels of NWT produces different posterior markers. FGF is also a posteriorising factor in a gradient.

200
Q

what is a forward genetic screen

A

this involves using a screen to identify a gene involved in a process in an unbias and from an uninformed background.

201
Q

what is a reverse genetic screen?

A

a bias approach in which candidate genes are screened for their involvement in a pathway or process

202
Q

what is the general approach to performing a screen in drosophila?

A

You use a DTS (temperature sensitive dominant lethal mutation) and a balancer female. You have a wild type male which you expose to a mutagen. You then cross these to produce the F1. You are given DTS/balancer gets and mutagenised chromosome/ balancer. You then grow these at 29 degrees so you kill the DTS/balancer. then you are left with only must/bal. You cross these and get a ball/bal which is lethal, must and ball (which will have the balancer marker) and a homo mutagenised . You can observe these for a phenotype

203
Q

how is a screen generally carried out in c.elegans?

A

A population of wild-type hermaphrodites is exposed to a mutagen and genes are randomly mutated in the germ cells (mutated germ cells are indicated in red). For example, one sperm could be mutated for the gene unc-32, which is required for the correct functioning of the nervous system. Fertilization of an egg by this sperm will result in a heterozygous F1 individual. Because this animal is a self-fertilizing hermaphrodite, it will produce eggs and sperm that bear this mutated gene; one-quarter of its F2 progeny will be homozygous for the mutation and result in a coiled phenotype (shown in b). Such an animal can be transferred to a plate, and in three days, its F3 progeny can be inspected to determine whether the mutant phenotype breeds true.

204
Q

how do you carry out a screen in zebrafish?

A

mutagenise a malecross with a femaleeach F1 offpsring will make a family So cross with a wild type to make a family which contains heterozygous F2 fish when these two hets cross you get 25% homozygous in the F3

205
Q

what type of mutations so UV and x-ray normally induce?

A

large deletions

206
Q

what type of mutations do chemical mutagens normally make?

A

single point mutations

207
Q

describe how EMS works

A

it is an alkylating agent, it induces G-c pairings to change to A-T. it does this because it methylates guanine which makes it pair with T so when the DNA replicates, an A-T pairing is formed

208
Q

describe how ENU works?

A
  • most toxic but most well used mutagen- it can change G-C to A-T and vice versa.
209
Q

how do optogenetics mutations work and in what organism are they used in?

A

Caenorhabditis elegans expressing His-mSOG in the germline behave and reproduce normally, without photoinduction. Following exposure to blue light, the His-mSOG animals produce progeny with a wide range of heritable phenotypes. We show that optogenetic mutagenesis by His-mSOG induces a broad spectrum of mutations including single-nucleotide variants (SNVs), chromosomal deletions, as well as integration of extrachromosomal transgenes, which complements those derived from traditional chemical or radiation mutagenesis.

210
Q

what are the 4 type of mutation phenotypes that can be generated following a genetic screen?

A
  • no signalling (null)- less signalling (loss of function)- increased signalling (gain of function) - increased signalling (over expression)
211
Q

can some loss of function mutations be temperature sensitive?

A

yes

212
Q

how can the order of a genetic pathway be achieved by using epistasis? what are the conditions when making double mutants and how can this issue be circumvented?

A

You have candidates for a pathway that you wish to order. You make single mutation mutants of each the candidate genes and take note of them. You then make double mutants of all of the possible pairings. The downstream gene will have the same mutant in the single and in the double mutants. The double mutants must have opposite phenotypes - you can use gain of function mutations for one if they both have the same phenotype

213
Q

what is good about ordering genetic pathways in organisms such as the c.elegans? give an example

A

they are very easy to carry out genetics in, once the pathway has been order it can be investigated in other more complex organisms. For example, the TGF0beta pathway homologues (?) are found in c,elegant, drosophila and verterbates and are all involved growth patterning

214
Q

describe a GFP screen carried out in c.elegans and cite it.

A

flowers and poole et al., 2010: They sought to find gene involved in the patterning os the ASEL neurons in c.elegans. they used GFP labelling by driving the expression of ASER and ASEL specific markers in the worms. They then did an RNAi screen and looked form probes which called defects in the left/right patterning. They identified lsy-22 and lsy-5. When they were both knocked down there were no ASER reporter expression so they are involved in ASER patterning. They did epistasis analysis by using lsy-6 mutants which express the opposite phenotype (2 ASERs) and found they were both epistatic to lsy-6 but act upstream of die-1 which is after lsy-6. they found that they were both alleles of grouch which is a transcriptional repressor and work in conjunction with cog-1 to help repress die-1 which stimulates right fate. They did this by looking at binding domains.

215
Q

when was RNAi as a mechanism first used?

A

2003

216
Q

what organism does RNAi not work in?

A

zebrafish

217
Q

do people know why RNAi doesn’t work in zebrafish?

A

no

218
Q

how was RNAi first discovered?

A

sense RNA was injected but didn’t work and antisense RNA didn’t work very well either, but they tried double stranded RNA this worked! first done in the c.elegan

219
Q

what is the mechanism of RNAi gene silencing?

A

double stranded RNA is injected into he organism or introduced in some way. You have a protein called dicer which chops the double stranded RNA into short interfering RNA (siRNA). Then you have a protein belonging to he argonauts or piwi proteins which form a RISC complex (RNA-induced silencing complex) which binds to the siRNA and converts them to a single strand which is used as a guide. If the match is really good then the RISC complex will “slice” the mRNA that is complimentary (degradation) or it will induce translational repression

220
Q

what is the human RISC called?

A

argonaute

221
Q

how many sub units does the argonauts protein have?

A

4

222
Q

what is the function of the RISC complex?

A

it stretches the guide RNA strand and the active site which has 2 Mg atoms needed for the slicing function. The other 3 subunits help to modify the structure

223
Q

how does RNAi cause transcriptional silencing?

A

siRNA is formed from dicer and then it binds to RITs, which act in the nucleus, they make sure the single stranded RNA binds to nascent mRNA while it is being transcribed. the consequence of this is the RITS inducing histone methylation, DNA methylation and transcriptional repression via chromatin condensation

224
Q

why did the RNAi system evolve?

A

it is a defence mechanism from dsRNA viruses

225
Q

what type of viruses incorporate into the genome? and how does this link to the RNAi mechanism

A

retroviruses - their transposons are often transcribed from both sides and then forms dsRNA which is then targeted by the RNAI

226
Q

who can RNAI be supplied to worms?

A

They can be fed with e.clo expressing dsRNA against the gene because they have a dsRNA transport channel called SID-1 which allows the RNAi to spread across the body of the worm. It can also be injected. into the gut.

227
Q

how can genome wide screens be carried out in c.elegans?

A

can fill 96 well plates with c.eleganas and each well contains e.coli expressing a different RNAi probe almost all of the genes in the worm’s genome. You then inspect either the P0 worms or the F1 worms of these- the RNAi is passed onto the offspring. You can use reporters to then look for affected phenotypes

228
Q

how can you carry out RNAi in drosophila?

A

use DNA directed RNAi- create a DNA plasmids from which you transcribe an RNA (double stranded) which contains a spacer in between the two homology arms so that you get a hairpin loop of RNA. then this triggers the RNA response. To introduce the this construct into the fly. You can induce the expression of the RNAi using the GAL4/ UAS system. You can have gal4 downstream of a universal promoter or in a specific tissue. If you keep flies at a cold temperature, Gal4 will be less active that at 29 degrees- just by changing temp you can modulate the level f gal4 expression

229
Q

can the gal4/uas system be used in other organisms other than flies?

A

yes- zebrafish

230
Q

how can Gal4 is repressed?

A
  • via temp- via a reposer called gal80. there is a temp sensitive version of gal 80- at 18 degrees it will bind to gal4 and there is no transcription, at 29 degrees, gal80 changes its conformation and can no longer bind- this can allow you to temporally control your genes expression. so you could cross a fly that expresses. you normally have GAL80 expressed ubiquitously, then Gal expressed in a tissue specific manner and then cross this fly with a fly that expresses the UAS construct.
231
Q

what must gal4 have attached to it in order for it to work in the UAS system?

A

a DNA binding domain

232
Q

what is the split gal 4 system? and when would it be used

A

you have two different enhancers- one expressing the gal4 DNA binding domain and the other expresses the gal4 activation domain. If you know that there are two genes that are expressed in different places in the organism but you want to see where the expression overlaps then you can use this system by using the enhancers for the two genes with the two protein subdomains. Then you have a UAS driving GFP expression

233
Q

how can you use the split gal4 system to subtract cells that express both of 2 genes of interest?

A

you have enhancer-1 driving the expression of gal4, then you have enhancer-2 driving the expression of gal80. then you have a UAS driving GFP expression. so that the tissues which express both enhancer-1 and enhancer-1 will be turned off.

234
Q

how can you turn on gene transcription using gal4 via drugs? why is this a good method?

A

have a gal4 that is drug inducible. SO you have a tissue specific transcription of gal4 but only fed the flies are put on foo containing this drug- then you can temporally and spatially control the expression of a gene in a way that isn’t temperature sensitive. This is a goo method because you have a good control- one group on the food and others not on the food.

235
Q

what must always be considered when you are using mechanisms such as the gal4 UAS system?

A

you need to make a control for every gene that you introduce to ensure there are no indirect artefacts

236
Q

describe two examples of RNAi being used in flies to look at clocks .

A

1.they were looking at circadian rhythms. The used a GAl4 system so with a looper and RNAi gene sites. they knocked down the gene period which is needed for the clock to fun. They made two constructs targeting different regions of the per gene. One was against the Ct domain and the other against the PAS. They timeless- gal4 (a gene expressed in the same cells and is a clock gene). They then looked at the behavioural rhythms. These flies were rhythmic but they have a slow running clock. 2. They wanted to knock down pigment dispersing factor in neuronal subsets. only 16 of the clock neurons express pigment dispersing factor. They want to dissect which neurons are important for the circadian function of PDF (PDF loss of function: they become active in the afternoon earlier than wild type flies, flies do not become active before the lights come on, when you put the flies in darkness they lose their close.) so which of these neurons are important for these diff phenotypes. They used 5 different gal4 drive which are expressed in a diff subset of PDF expressing neurons (they used the gal4 driven by expression of each of the to drive the expression of UAS galactosidase expression as a marker to find out which ones each is expressed in). First they investigated how well the RNAi was working. SO they stained the nervous system with a PDF antibody to show tis expression. Then they did this in tim-gal4/UAS-pdfpRNAi (tim is subset promoter) and they saw it was knocking out PDF. (did verification for all drivers) Then they analysed the behaviour of these flies.

237
Q

what does a RNAi produce?

A

a knock down not a knock out- you reduce expression depending on the quality of the RNAi construct.

238
Q

what are the limitations of RNAi and what are the remedies for these limitations?

A
  1. off-target effects: RNAi targets different of additional mRNAs with similar sequence 2. partial gene function normally remains 1. you can use different RNAi constructs, targeting different regions of the mRNA- if you get the same phenotype by knocking down two different regions of the same gene, you are sure it is due to the KD of that gene and not another gene. You can rescue the phenotype by creating RNAi resistant rescue. This will show that the phenotype was only caused by the knock down of specific gene 2. You can boost the expression of the dicer machinery by using a UAS construct to drive the expression of another copy (but you have to add another construct then). You can also add more than one transgene of the RNAi construct. or you can add a copy of gal4 driver. You could increase the number of UAS-repeats.
239
Q

how can you rescue RNAi induced phenotypes?

A

introduce a transgene of the target gene that cannot be KD by the RNAi, then the phenotype must be due to the RNAi of that gene. One way of doing this is to use a gene from a sufficiently divergent different species that still have the same function but is different enough to not be targeted by the RNAi. Or you can use the degenerate genetic code and then make a different DNA for the same protein using different triplets.

240
Q

how are transgenes normally made in flies?

A

by P-element- mediated germline transformation- injected vector with gene of interest and with p-element surrounding and then with transposes.

241
Q

when generating a transgenic fly, what must you include in the construct and what is the egenral genetic background that should be used? (CHANGE ANSWER DEPENDING ON RALPH’S REPLY)

A

You should use a genetic background that has a known phenotype that can be rescued by thee expression of another allele- for example use a strain of white eyed flies. You then included W+ gene (encoding red eye) in the construct next to your gene that you want to express so that you know your transfection has been successful when the fly expresses red eyes. Therefore, you would include the regulatory elements of w+ gene.

242
Q

how many genes for the VDRC cover and what percentage of the genome does this cover? what is a bonus of this library? (2)

A

12,671 genes vs 15000??, (91%) of all dros protein coding genes. 8267 genes more than one RNAi line available and they have the KK line which inserts into known regions

243
Q

why are c.elegans better for genome wide screens than flies?

A

the RNAi libraries available for c.leegans are more complete than those for the flies

244
Q

what can you order for fly RNAi libraries?

A

you can order libraries for flies which contain DNA fragments containing all genes that are flanked then with a T7 RNA polymerase primers which you can then use T7 RNA polymerase to transcribe dsRNAs and transfect either in cells or into a fly strain by inserting it into a construct

245
Q

how do the libraries differ?

A

they vary in how the construct is introduced into the fly, one has p-elements which means that the RNAi is randomly located into the genome. There is a KK line which uses phi integrase which allows you to precisely insert theh transgene in a known location in the genome- there are landing sites which are defined and we know that they are viable when inserted here and know they will be expressed- not in a highly chomatinised region.

246
Q

what do you need to keep in mind when you are designing RNAi probes? how can this same concept be useful?

A

genes have different transcripts due to different alternative splicing. So many of the libraries sign probes that will knock down all of the splice variants by identifying gene regions that are common to all transcripts. But if you want to find out the different functions of the different transcripts, you can design probes which will target specific transcripts.

247
Q

what is the role of a ftz intron at the end of the RNAI cDNA construct in dros RNAi?

A

its increases the efficiency of the RNAi

248
Q

what are three drosophila libraries that are available?

A

vienna drosophila resource center, national institute of genetics, japan, transgenic RNAi project, harvard.

249
Q

what are the the two libraries that the vienna drosophila resource center used and what is the difference between them?

A

GD library and the KK library they differ in the transposons that they use the GD has “p-element” which gives you a random insertion and the kk library used a phiC31 which means all are inserted into a known location within the fly genome in a safe, efficient “landing site”

250
Q

which library accounts for multiple transcripts?

A

VDRC

251
Q

how many genes does the NIG-Fly library cover?

A

11909

252
Q

what is the construct of the VDRC?

A

p-element, 10 UAS sites, hsp70, homologues arms of RNAi probe with linker in the middle, ftz intron, poly A tail and mini white gene, p-elemt.

253
Q

what is the construct of the NIG-Fly?

A

UAS-hsp70, clone two inserted repeats with a heterologous oncogene (exon-intron-exon) which improves RNAi efficiency.. Guessing there are known primer sites in the construct too?

254
Q

how many genes does the harvard medical school library cover?

A

10108

255
Q

what kind of transposon does the harvard library use?

A

phiC31 so it integrates into known region

256
Q

what is a cool bonus about using the harvard library?

A

their constructs allow for alteration of the number of UAS repeated to generate a series of different RNAi expression levels.

257
Q

what is a quirk about the harvard medical school RNAi library?

A

instead of using long dsRNA they use short dsRNA which work during oogenesis unlike normal long dsRNA (valium 10 based vectors) - short hairpins.

258
Q

how can you boost RNAi?

A

boost dicer, increase temperature and boost UAS

259
Q

what is a valium 20 based vector?

A

a short strand of dsRNA used by the harvard RNAi library which produce short hairpin RNAs not long ones.

260
Q

describe a screen that looked at wound healing using RNAI in flies?

A

they made wounds into fly larvea that normally takes 24 hours to repair in wild types. They screened for gene which are responsible for this wound healing. They used a red reporter driven by gal4/uas to stain nuclei and stained membranes with anti fasciclinIII to visualise the wound well. They were able to observe thephenotye of wound closure without dissecting ( saves a lot of time). They used 190 different RNAi strains from NIG-fly library. they found lots of genes in involved in this

261
Q

describe a genome-wide behavioural screen using the NIG fly RNAi library..

A

They used genes from the NIG fly RNAi library using 6000 strains and crossed with a strain expressing a neuronal sub-type specific GAL4 driver (in PDF neurons) and then they looked for weird circadian rhythms- found strip.

262
Q

what is miRNA?

A

the naturally occurring double stranded RNA that is present, but people aren to 100% sure what they do

263
Q

how did people try to wok out what the role of miRNAs were?

A

they wanted to find what the targets of the miRNA are and what happens when they dont find these targets. they create miRNA sponges. this is a construct that contain many perfect binding sites for the miRNA and when expressed (by the UAS system) they will sponge up the miRNA and will titrate these miRNAs away from the nucleus where they normally work.an miRNA sponge was made that soaked up 141 miRNAs, . the construct has a reporter (cherry) so you would know where it was being expressed. they increased the number of sponge copies (2) to increase efficacy. They then checked for expression of miRNAs so see if they are down regulated when comapred to a scrambled sponge construct (control). some were and some weren’t. they showed that the expression of the sponge was able to rescue the expression of the targets (see paper). The found that there were eye development defects etc.

264
Q

how many miRNAS are there in flies?

A

200

265
Q

how can you use the Gal4/UAS system to over express genes? why would you want to do this

A

instead of having reporter or RNAi, you can have an extra gene copy to see if you can rescue a mutant to prove the role of this gene in the process of interest. you can also look at the effect of dosage. you can express extra copies in the native tissue or in different tissues- induce eyes in fly legs etc.

266
Q

how can you use UAS for over expression?

A

can use a library and over express many genes to see the effect and look at function. and you can drive over expression in different areas of the embryo and in different points in development

267
Q

describe the entire process of making a transgenic fly.

A
  1. You have two parent flies that are mutant for the white eye gene 2. You cross these and make a fertilised egg. 3. You insert your gene of interest with its promoter (Gal4 with tissue specific promoter or UAS site with hsp downstream), and a reporter gene with its promoter, into a transposable p-element. This p-element is then injected into the fly egg into the area of the germ cells and helper plasmic containing transposes (under standard promoter?) that isn’t integrated is also injected. 4. the transposase then interrogates the P-element into the germ cells of the F1. F1. You can then cross the F1 which will have het of the p-element in the germ line and the F2 will have a red eyed fly that expresses the gene of interest too (Gal4 or UAS)
268
Q

how would you actually increase UAS or DICER expression with your RNAi?

A

increase the expression of UAS in the plasmid which is going into one fly and gal4 in the crossing fly

269
Q

what are 7 advantages to using a mouse as a model organism?

A
  • easy to breed and house ( but expensive) - easy to control the environment that the mice are in - physiological similarities with humans - well developed physiological techniques for mousegenetically homogenous (inbred) strains available - deposition of the vaginal plug - strains can be used across labs across the world
270
Q

when do mice reach sexual maturity?

A

around week 4 post birth

271
Q

what is the litter size of mice?

A

up to 10 pups

272
Q

is the physiology of the mouse strain dependent?

A

yes

273
Q

what about the genetics of the mouse makes them suitable for human studies?

A

all human genes have counterparts in the mouse genome. thus cloning of a human gene leads directly to cloning of a mouse homologue which can be used for genetic, molecular and biochemical studies that can then be extrapolated back to an understanding of the function of the human gene. In only a subset of cases are mammalian genes conserved within a genome of the flies of worms

274
Q

what is the most commonly used mouse strain?

A

C57BL/6

275
Q

why is it important to keep buying new mice from your original source when keeping a mouse strain?

A

it is important to keep repopulating your strain with founders because genetic drift can occur within the populations causing th strain to diverge from the original, meaning that there is less similarity between strains used by other labs and less reliable and comparable phenotypes therefore

276
Q

why is it important to always get the same strain from the same supplier even though they are the same strain?

A

because there can always be divergence between two sub strains due to genetic drift

277
Q

why is homogeneity so important in mouse strains?

A

normal development and physiology can vary significantly from one strain of mice to the next and in the analysis of, mutants, it is often not possible to distinguish subtle effects due to the mutation itself, from effects dye to other genes within the background of the mutant strain. to make this distinction, it is essential to be able to compare animals with the same genetic background. Phenotypic differences that persist between these a mutant and a wild type from a strain must be a consequence of the mutant allele.

278
Q

what species of mouse are normally used in mouse experiments?

A

musculus musculus domesticus

279
Q

how many generations of inbreeding do you need to ensure the strain is 98.7% isogenic?

A

20

280
Q

how many generations of inbreeding do you need to do to ensure your strain is 99.9999% isogenic?

A

150

281
Q

what is important to remember about different strains (inbred lines) of mice that are used. give 2 examples of this.

A

they all have different characteristics and are different genetically. For example the C57BL/6 is seizure resistant so epilepsy analysis in this mouse would not be good. The C3H has poor spatial learning so experiments relating to this would not be suitable

282
Q

what is the C3H inbred line of mouse bad at?

A

spatial learning

283
Q

what is the C57BL/6 mouse resistant to?

A

seizures

284
Q

in an ideal world, how would mutations occur in mouse strains and why?

A

they would occur spontaneously in strains, the way you would know that the phenotype observed from the mutation would only be due to it, and not from any other genetic background

285
Q

Due to the fact it is unlikely for interesting mutations to occur spontaneoulsy, how can mutations be induced and what is the protocol for ensuring the mice are coisogenic?

A

You can induce mutations and then either view them in heterozygous animals if they are dominant, or in homozygous animals if they are recessive. They can then be continuously inbreed to create its own strain that is homozygous at all loci. Or if it cannot be maintained as a homozygous then it can be maintained as a heterozygous by continuous backcrossing into the original strain. In both cases they are coisogenic meaning that they are genetically identical to the sister strain expect at the mutant loci

286
Q

what is genetic drift?

A

the constant tendency for genes to evolve. This is the introduction of new alleles into the strain. This can result in one strain being developed into many different strains.

287
Q

after how many generations of a colony being separated from its parent colony should it be considered a sub colony and what is the mechanism behind this?

A

20 generations. This is due to genetic drift

288
Q

how common are spontaneous mutations?

A

5 X 10^-6 per gene per generations

289
Q

how many nucleotides are there in the mouse genome?

A

3 x 10^9 nucleotides

290
Q

after how many generations is there a 90% chance that two substring differ at one or more locus?

A

16

291
Q

after 16 generations, what is the probability that two substring differa t one or more loci?

A

0.9

292
Q

how can substring arise from bad strain maintenance? (2)

A

substring can arise fi there is residual heterzygosity within a strain caused by incomplete inbreeding at the time of separation from progenitor strain. This means that not all loci were homozygous for an allele and so the diffferent allele could replace the normally and cause a sublime. this is why it is important to ensure inbreeding is carried out extensively. 2. substring can also arise from genetic drift if the strain is not constantly replenished by the progenitor strain

293
Q

how do you maintain in bred lines?

A

you cross brother and sisters

294
Q

how can you get rid of spontaneous mutations which could eventually be incorporated into genetic drift and cause the formation of a substrain?

A

Once you have identified a mouse that has a phenotypic mutation, you then “inheritance test” them. This involves backcrossing the mice and scoring the ratios of the offspring. if the homo mouse is crossed with a het then the ratio will be different than if it was crossed with a homo non mutant (all wild type if recessive). Once identified, you then get rid of the het rogue mice.

295
Q

what is another word for inheritance testing?

A

test crossing

296
Q

although spontaneous mutations can be damaging, how can they also be useful? give 3 examples

A

They can be useful if they give rise to interesting phenotypes. reduced insulin secretion, impaired memory and retinal degeneration have manifested themselves in spontaneous mutations

297
Q

how can you reduce genetic drift?

A

obtain mice from a reputable source, maintain good records to enable mutations to be bred out, cryropreserve a large number of embryos from your colony and return them every 10 generations to replenish, replenish from the original source

298
Q

what is quantitative genetics?

A

it is the study of phenotypes that are continuous rather than discrete and the genes associated with them

299
Q

what are the 3 approaches to quantitative genetics in rodents (and their subtypes is relevant)?

A
  • artificial selection - inbreeding - crosses (recombinant inbred strains, consomics, outbred (heterogeneous stock).
300
Q

what is the process of artificial selection when carrying out quantitative genetic study?

A
  • you choose a trait of interest and repeatedly mate animals with that trait of interest (for example, long and short tail length). You then end up getting a segregated population. You can then use sequence the animals (how) and identify differences in alleles or point mutations when you compare. This should allow you to identify regions of the genome that contain genes for the trait of interest
301
Q

how can recombinant inbred strains be used to identify genes for traits of interest?

A

recombinant strains are generated by crossing two or more strains that are very polymorphic. due to recombination events, F2s are inbred repeatedly for about 20 or more generations and produce a strain that is a random mixture of the different strains. different F2 crosses are then used to produced different strains. Traits that are restricted to certain IRS and not others are of interest. The genome of the strains are compared and SNPs that are exclusive to the trait in interesting strains are considered to be within areas of the genome where the gene for the trait is. The sequence of SNPs will belong to one of therapist strains and can be identified via sequencing. The area of interest can then be looked at the candidate genes can be proposed. These can then be subject to further testing

302
Q

what are the downsides of using recombinant inbred strains?

A

it is very hard to refine the candidate site down to a sufficiently small size that candidate genes can be extrapolated very well- specific genes being derived from these studies are very unlikely

303
Q

what is a recombinant inbred strain?

A

an organism with chromosomes that incorporate an essentially permanent set of recombination events between chromosomes inherited from two or more inbred strains. F1 and F2 generations are produced by intercrossing the inbred strains; pairs of the F2 progeny are then mated to establish inbred strains through long-term inbreeding.

304
Q

how do you produce inbred recombinant strains

A

you cross two strains that produce a heterozygous F1 that has no recombination between the two strains. You then cross these F1s together and produce an F2 which is recombinant. You then take two F2s and inbred for generations. eventually, the strain will be homogenous at all loci. after around 20 generations.

305
Q

what is a consomic strain?

A

chromosome substitution strain- a strain that is homogenous to its sister strain bar one chromosome.

306
Q

what are consomic strains used for?

A

they are used pinpoint which chromosomes are important for certain phenotype- whether a trait is encoded predominately by a single chromosome.

307
Q

how are consomic strains formed?

A

you identify many SNPs that are present on a single chromosome. around 233 micro satellite markers have been used before. You can backcross f1 females arising from a interstrain cross, into one of the strains, and continuoulsy check for the SNPs or microsatellites and select this for the backcross, and so on and so forth for 7-16 generations.

308
Q

what is a congenic strain?

A

a strain that is isogenic to its sister strain bar a region containing the gene which encodes the phenotype of interest

309
Q

how are congenic strains formed and how can it be used to identify genomic regions of interest?

A

generated in the laboratory by mating two inbred strains (usually rats or mice), and backcrossing the descendants 5-10 generations with one of the original strains, known as the recipient strain. Typically selection for either phenotype or genotype is performed prior to each backcross generation. So you us this to map by backcrossing until you have the entire genome of one strain bar the region which contains the gene of interest- you can then look for SNPs that are not associated with the majority strain and are instead related to the other strain and you will know this is where the gene is interest is. you then input this into a database of the mouse genome and identify which genes lie in this region

310
Q

after how many generations of back crossing is a strain considered congenic?

A

10 generations

311
Q

what are the requirements in terms of the types of strains that can be used when forming a recombinant inbred strain for quantitative genetic analysis?

A

the strains used must be polymorphic and significantly different to each other

312
Q

describe how you would make a congenic strain for a mutation that was recessive or one that was dominant.

A

following the crossing of two strains, the differing phenotype may be able to be viewed in the F2. If this is the case, then you can carry out SNP PCR analysis and find out which regions of chromosomes is linked to the phenotype. You then carry out progressive back crosses but select for the genotype of phenotype of interest each time. It may be easier to select for the genotype as then you dont have to expose the mice to whatever it is that you are looking for. you then repeatedly backcross the heterzygotes. Once this has occurred you can eventually keep the allele as a homozygous by sister brother mating after 10 back crosses. 2. if the trait is homozygous then you can also hopefully view in the F2 in 25% of offspring. You can then use SNP PCR to identify the region and then continue to backcross as heterozygous if you can detect the QTL every time.

313
Q

Is it preferable to maintain congenic strains in het or homo and why?

A

in het because then ever time you backcross the heterozygosity decreases by 50%. also this may not be a choice if the allele can’t be maintained as a homozygous if it is lethal or causes sterility etc.

314
Q

what is the mendelian principle behind congenic backcrosses?

A

for every backcross, the heterozygosity should decrease by 50% at all loci that aren’t linked. This means that after 10 generations, the strain will be 99.8% isogenic for the recipient strain. But, due to linkage, after 10 generations of back crossing there will be 20cM of linked donor strain.

315
Q

in terms of making a congenic strain, after 10 generations of back crossing, what percentage will be isogenic to the recipient strain and what will the length of linked donor strain DNA?

A

99.8% and 20 cM of linked

316
Q

what 2 things should always be considered when comparing congenic to coisogenic strains?

A

First, congenic strains, especially those that have undergone only a minimum number of backcrosses, will have small random remnants of the donor strain — so-called passenger loci — scattered throughout the genome. In congenic strains maintained by inbreeding, the same passenger genes will be present in all members of the strain. In rare instances, traits attributed to the selected donor allele may actually result from one of these cryptic passenger genes. Such effects can be sorted out by breeding the congenic strain back to its original inbred partner. If a trait is due to a passenger gene, it will assort independently of the donor locus in subsequent backcrosses.The second difference between a congenic strain and a coisogenic strain is in the chromosomal vicinity of the differential locus. Congenic strains will always differ from their inbred partner along a significant length of chromosome flanking the differential locus; coisogenic strains will only differ at the differential locus itself and nowhere else. Thus, there is always the possibility that phenotypic differences between the two members of a congenic pair are actually caused by a closely linked gene rather than the selected differential locus. This potential problem is much more difficult to resolve by simple breeding protocols.

317
Q

what is the process of genotyping SNPs using PCR?

A

the process relies on allele specific primers. You use for primers, two outer primers (one 5’ up from the SNP site on the 5’ to 3’ strand and one 5’ up from the SNP site in the 3’ to 5’ strand) You then have two inner primers, one that matches one SNP allele on the 5’ to 3’ strand and another which matches the other SNP allele on the 3’ to 5’ strand. When you PCR with all of these primers and with the two alleles, you will get a non-speific allele product which is very long and connects the two outer primers. You will also get two products, one for each allele, they are designed in such a way that the length is easily distinguishable

318
Q

draw a picture of how SNP genotyping works when using PCR primers

A

http://bmcbioinformatics.biomedcentral.com/articles/10.1186/1471-2105-9-253

319
Q

how do you genotype SNPs?

A

using PCR primer technique

320
Q

what are heterogenous stocks?

A

you can create recombinant inbred strains from 8 different strains

321
Q

what is the advantage of using more strains when creating recombinant inbred strains?

A

in enables high resolution link mapping. 2cM between each recombinant

322
Q

when you use 8 strains for RIS, what is the distance between the recombinants?

A

2cM

323
Q

If you you want the highest resolution linkage mapping using inbred strains, what strains can you use?

A

outbred stock. These are colonies descended from a small number of inbreds. This means that 9%% sequence variants.

324
Q

what percentage sequence variance do you have when using outbred stock?

A

95%

325
Q

what is the mendelian principles behind the resulting homozygous it that results from sister brother mating of F2s during the production of recombinant inbred strains?

A

there is a 12.5% chance that both F2 progenitors are identically homozygous at any one locus, then approximately 12.5% of all loci in the genome will fall into this state at random. The consequence for these loci is dramatic: all offspring in the following F3 generation, and all offspring in all subsequent filial generations will also be homozygous for the same alleles at these particular loci. Another way of looking at this process is to consider the fact that once a starting allele at any locus has been lost from a strain of mice, it can never come back, so long as only brother-sister matings are performed to maintain the strain.At each filial generation subsequent to F3, the class of loci fixed for one parental allele will continue to expand beyond 12.5%. This is because all fixed loci will remain unchanged through the process of incrossing, while all unfixed loci will have a certain chance of reaching fixation at each generation. At each locus which has not been fixed, matings can be viewed as backcrosses, outcrosses, or intercrosses, which are all inherently unstable since they can all yield offspring with heterozygous genotypes- after 20 generations, 98.7% of the loci in the genome of each animl should be homozygous, at 40 generations, 99.98% will be homozygous. However, chromosomes do not assort randomly and some are linked- inherited in junks. e, there is an ever-increasing chance of complete homozygosity as mice pass from the 30th to 60th generation of inbreeding (Bailey, 1978). In fact, by 60 generations, one would be virtually assured of a homogeneous homozygous genome if it were not for the continual appearance of new spontaneous mutations (most of which will have no visible effect on phenotype). However, every new mutation that occurs will soon be fixed or eliminated from the strain through further rounds of inbreeding. Thus, for all practical purposes, mice at the F60 generation or higher can be considered 100% homozygous and genetically indistinguishable from all siblings and close relatives

326
Q

what are the downsides of congenic strains?

A

infertility and bad health can occur

327
Q

why do some people use outbred stocks for quantitative trait mapping

A

Outbred mice are used for the same reasons as F1 hybrids — they exhibit hybrid vigor with long life spans, high disease resistance, early fertility, large and frequent litters, low neonatal mortality, rapid growth, and large size. However, unlike F1 hybrids, outbred mice are genetically undefined. Nevertheless, outbred mice are bought and used in large numbers simply because they are less expensive than any of the genetically-defined strains.Outbred mice are useful in experiments where the precise genotype of animals is not important and when they will not contribute their genome toward the establishment of new strains. They are often ideal as a source of material for biochemical purification and as stud males for the stimulation of pseudo-pregnancy in females to be used as foster mothers for transgenic or chimeric embryos. It is unwise to use outbred males as progenitors for any strain of mice that will be maintained and studied over multiple generations; the random-bred parent will contribute genetic uncertainty which could result in unexpected results down-the-road.

328
Q

when creating transgenic mice, why is it important to do the same experiment in 3 or more founder lines?

A

potential problem can result from the insertion of the transgene into a normally-functioning endogenous locus with unanticipated consequences. In approximately 5-10% of all cases studied to date, homozygosity for a particular transgene locus has been found to cause lethality or some other phenotypic anomaly. (Palmiter and Brinster, 1986). These recessive phenotypes are most likely due to the disruption of some normal vital gene. In less frequent cases, a transgene may land at a site that is flanked by an endogenous enhancer which can stimulate gene activity at inappropriate stages or tissues. This can lead to the expression of dominant phenotypes that are not strictly a result of the transgene itself. 39 For all of these reasons, it is critical to analyze data from three or more founder lines with the same transgene construct before reaching conclusions concerning the effect, or lack thereof, on the mouse phenotype.

329
Q

what is the process of calculating linkage using markers?

A

330
Q

what is a quantitative trait locus?

A

an area of the genome that’s been implicated in containing a gene that of interest for a particular trait

331
Q

what are the most commonly used two strains for recombinant inbred mice strains?

A

C57BL/6J and DBA/2J

332
Q

describe the drug and alcohol sensitivity experiment using recombinant mice

A

..

333
Q

how can reduce the number of backcrossing generations needed for congenic production from 10 to 5?

A

by not only mapping for the SNP of interest but also mapping for the greatest loss of the donor strain

334
Q

what does ENU do?

A

The chemical is an alkylating agent, and acts by transferring the ethyl group of ENU to nucleobases (usually thymine) in nucleic acids. It changed GC pairings to AT pairings and AT to GC pairings

335
Q

who determined the mutation rate of ENU?

A

Bill russel

336
Q

how did bill Brussel work out the mutation rate of different mutagens?

A

he used a mouse from the T-stock which have 7 mutations across different chromosomes. They are all recessive mutations. He then applied the mutagen to wild type mice and then crossed with these T -stock. If the mutagen mutagenised one oft the loci then then the phenotype would be exposed in the cross. He could then work out the rate of mutation rate per locus.

337
Q

what strain did bill russel use to estimate the mutation rate per nucleus for different mutagens?

A

t-stock- 7 recessive mutant alleles loci with phenotypes

338
Q

what is the most powerful mutagen? (rate)

A

ENU fractionated dose (1/750)

339
Q

list the success of mutagens in order

A

ENU fractionated dose, ENU, procarbazine, x-irradiation, sponatenous

340
Q

does the mutation rate vary from locus to locus?

A

yes

341
Q

how did bill Russel use his t-stock experiment to work out the types of mutation that ENU and x-irradaition induced?

A

Two of the 7 loci in the strain were present on the same chromosome (9), he noticed that when irradiaition they were always both present and they were separate when used ENU. This was because irradiation caused deletions but ENU cause base pair mutations

342
Q

what are the 5 classes of genetic mutation?

A

amorph, hypomorph, hypermorph, antimorph, neomorph

343
Q

what is an amoprh?

A

complete loss of function (null), generally recessive unless haploinsufficient

344
Q

what is a hypomorph?

A

partial loss of function, usually recessive,

345
Q

what is a hypermorph?

A

increase a normal gene function, normally dominant or semi dominant,

346
Q

what is an antimorph?

A

dominant negatie opposes normal function

347
Q

what is a neo morph?

A

dominant and gives a completely new function

348
Q

what is an allelic series?

A

Allelic series describe different mutant alleles of a gene that cause a range of phenotypes, whereby each one carries a single point mutation within different regions of the same gene. They uncover new functional domains

349
Q

give an example of ENU uncovering an allelic series

A

the trembler (PMP22) locus,a point mutation in the transmembrane section caused a very severe loss of myelin phenotype whereas the intracellular membrane mutation cause a less severe phenotype

350
Q

what is a nonsense mutation?

A

a mutation which results in a the insertion of a premature stop codon

351
Q

what is a missense mutation?

A

a mutation which results in a different amino acid

352
Q

how can ENS mutations induce the same phenotypes as knockouts?

A
  • missense- nonsense - splice site mutatnt - mutations in promoters and other control elements - dominant negative mutations
353
Q

when can ENU phenotypes be preferential to knock out phenotypes and why?

A

when the knockout produces embryonic lethality but the ENU doesn’t because it only targets a specific functional domain rather than the entire gene.

354
Q

what is the process of ENU mutagenesis?

A

• Inject male mice with ENU• Fractionated dose more efficient (3 x 100 mg/kg)• ENUinducestemporaryperiodofsterility• New germ cells re-populate testis• Sperm carry independent mutations• Generation of approximately 50 progeny per injected male

355
Q

what is the process of a dominant screen strategy?

A

there is a single founder individual (male of female) that is mutagenised and then crossed with a wild type, you then screen the F1. Conduct whole-genome scan of affected individuals, Successive out crossing of affected individuals for fine mapping (more recombination events), Candidate gene(s)/ sequencing

356
Q

what are the two options for carrying out a screen for recessive mutant alleles?

A

cross a male mouse that has been mutagenised with a female normal mouse. This will result in each F1 mouse being heterozygous for a different mutation. So each F1 male mouse is then crossed with a wild type and produces a variety of offspring, some will be heterozygous. The F2s can then either be crossed back with the het F1 or they can be intercrossed (inbred or backcrossed) to produce F3s, some of which will be homozygous mutant. If you have identifies G3 mutants, you can then cross the G2 or G2/G1s again and get more mutants

357
Q

how have dominant screen strategies been revolutionised?

A

Once you have performed a few backcrossed of your identified mutant, you dont have to do lots of linkage mapping, you can simply send it off for next generation sequencing and they will identify potential missions mutations.

358
Q

what are the differences between the procedure for a dominant screen and for a recessive screen?

A
  • you have to screen more animals with recessive screens- recessive screens require an F3- more expensive and time consuming
359
Q

what is a modifier screen and give an example

A

when you carry out a screen on a certain genetic background. for example you could have a mouse that is a carrier for a huntington disease and then you would cross with a mutagenise mouse and then look for F1s that have a phenotype that modifies the carrier phenotype- so for example fives a later or earlier insert than normal

360
Q

are modifier screens difficult to carry out in mice?

A

yes

361
Q

what are challenger screens?

A

this involves challenging the mutant with environmental stresses, drugs, disease or ageing etc and then carrying out a screen to see if there is a progression of phenotypes with age or under the other stresses- then do congenics and mapping etc

362
Q

what is a sensitised screen?

A

where you carry out a screen with animals that have a predisposition for a certain disease- you would do this to see what other genes may be acting to produce diseases

363
Q

what are region specific saturation screens?

A

where you look for all potential mutations within a small region of genome. This is done using various mutations that have

364
Q

how can you carry out a saturation screen for a certain loci?

A

You use a mouse that you already have a strain containing a deletion for that gene such as the Del36H deleted region. you expose a mouse to a mutagen and then cross it with another inbred strain. This gives you a het for the mutation and one normal section from the other strain. You then cross this het with a mouse that is het for a deletion at this section. This will give rise to some of the animals in the G2 expressing a recessive phenotype for mutant section as there is a deletion in the other chromosome. This is easier than producing G3 homozygotes and guessing that it is at the same loci because of a similar phenotype- this way you know it is at the deleted loci and it is in the G2.

365
Q

what is the aim of a saturation screen in mice?

A

Saturation screens aim to identify every possible mutant phenotype within a particular deletion region

366
Q

what are gene driven screen ?

A

identifying mutations in a specific gene

367
Q

how would you carry out a gene driven screen?

A

There are sperm banks containing sperm of 10,000 mutagenise mice. if you were interested in a paritcular gene then you design primers for the gene of interest and then amplify the DNA from this particular gene for 10,0000 mutagenised mice and then look for mutations. The number of DNAs screen (mutagenised animals) the more likely you are to find a mutation of the allele- provided that the mutation rate is 1/1000 per locus and the mutation detection rate is 90% then you will find 4 or more alleles with a porbability of 0.9% if you screen 5000 animals:

368
Q

Why would you want to do a gene driven screen? give an example?

A

If you know that the knock out is lethal you may want to screen for alleles that have mutations in certain domains of the gene. This was used to look at the alleles for the L-type voltage-dependent calcium channel which normally produced cardiac deficits. They wanted to look for hypomoprhic mutations. The Ko was embryonic lethal- so wanted to look for hypomorphic mutations- screened for mutations in 2 regions of the gene- it is an intracellular protein- so they screened for certain exons which they thought would give hypomorphic phneotype- screened for exon 6 which is an rea important for the interaction with the beta subunit of the complex and another region which is inolved in the activation of the channel- using this ENU gene driven screen they found 1 missense and one nonsense (stop) mutation they then resurected the two mouse lines which contained these mutations (using the sperm) and then screened them for phenotypes

369
Q

what is an additive transgenesis?

A

increase gene dosage or adding a reporter gene

370
Q

how can you produce a loss of function mutation using transgenics? (4)

A

use homologus recombination to replace DNA with a selection marker or a mutated form of the gene.- gene trap - conditional KO

371
Q

how do you produce transgenic mice?

A

via pronuclear injection or by homologues recombination in an ES cell

372
Q

how do you carry out transgenics using pronuclear injection?

A

you inject into a single fertilised cell into the male pronucleus. You can construct any piece of DNA and put it into a vector, then linearise your vector and then inject this into the nucleus and it will randomly insert into the genome. You then implant this egg into a surrogate mouse and one of the offspring will be het for the transgene. You can do DNA analysis to identify which of the offspring contains the construct by using primers within the transgene which will produce a known length of DNA (should normally span the promoter gene junction)

373
Q

what are the downsides of random insertion of the pronuclear injection approach?

A
  • if there is a promoter in the construct then it can affect other gene expression which could cause a phenotype that isn’t ue t the expression of your transgene - it could be inserted into a low expression region - it could be inserted into a gene and disrupt the expression of that gene and cause a phenotype unrelated to the injection of your construct- for small DNA constructs you can get tandem insertions so you need to analyse how many genes have inserted and where
374
Q

what is a BAC?

A

bacterial artificial chromosome- - large pieces of DNA. BAC libraries exist for the human, mice and other species.

375
Q

what are BACs used for?

A
  • they are used to create large constructs inside e.coli. This involves recombining a construct such as one being used for homologues recombination inside an ES cell . This is done inside an e.coli and then the BAC can be removed, the construct removed from the BAC and then injected into the ES cell. - BACS can be used to inject into pronucleus of mice embryos. There are BAC libraries available for the mouse, they can be engineered to put reporters in to report gene expression well. They are linearised and then injected
376
Q

what is the good thing about using BACs?

A

large and faithful expression of gene or reporter

377
Q

what are the downsides of BACs?

A

very large and may contain other genes so you also get expression of this gene too which can interfere

378
Q

how can you produce knock out mice?

A

you can produce a construct with homology arms to the area around the gene you want to kick out, then you replace the gene of interest with a resistance gene that allows the successfully transfected ES cells to be identified. Then you inject these ES cells into a blatsocyst.

379
Q

how do you create a knock in mouse?

A

same as with the knock out mice but you put an altered allele into the gene area instead but you have also have a resistance gene with its own promoter too. you could then potential surround this resistance gene with a loop site so that it can be flexed out via gene trapping mechanism.

380
Q

how can you test that your gene has inserted? and where it is inserted?

A

checking for for resistance given by the resistance gene. You can check it is inserted by using primer for the SNP within the allele and a primer just outside within the homology arms?

381
Q

if a knock out is embryonic lethal, what can you do?

A

produce a conditional knock out

382
Q

how do you produce a conditional knock out?

A

using the cre inducible system and inserting a gene via homologues recombination that replaces the gene of interest with one that has loop sites around it so that it can be knocked out when there is Cre expression which causes it to recombine out.

383
Q

when do knock outs normally cause lethality?

A

when the gene is pleiotropic and has an essential function during development but others later

384
Q

What can cre be used for in terms of gene reporting?

A

You can have a reporter downstream of a flowed stop codon insetting into the ROSA26 region. Then Cre can be driven by a gene of interest and this will result in the reporter being expressed whether this gene is expressed or has been expressed- can be used for fate mapping

385
Q

what is the tamoxifen method of cre induction?

A

rather than having genes stimulating Cre expression, you can express cre everywhere but have it as a cre- oestrogen receptor fusion protein, that when activated ny tamoxifen, will enter the nucleus of the cell and drive recombination

386
Q

what is Cre more commonly used that flp/FRT?

A

Cre historically was much more efficient than FLP recombinase (FLP thermo-unstable).• Consequently, many deletor strains generated in past have been based on cre.• Cre more adaptable due to mutant lox sites.• However, Flp now improved my Flpe and Flpo versions

387
Q

what is the cre multiplex system? and what is it used for?

A

It is a modified Cre/lox system that relies upon transgenes that contain three pairs of self-compatible lox sites (loxN, lox2272 and loxP). Recombination has three different possible outcomes, each with the same probability. The use of tamoxifen inducible Cre recombinase allows for recombination to be regulated temporally and enables one to limit recombination to few cells. this can be used in the brainbow for example to generate different coloured cells. You can use this in lineage tracing to map the fate of individual cells over time??

388
Q

what should you always do when you are wanting to use a CRe drive? give an example of when this is needed to be done.

A

check the pattern of expression by using it to drive a reporter. it could look like it was only expressed in a precise area in adulthood when in reality it was expressed everywhere in development.

389
Q

what are the three things you should consider before going ahead with your Cre experiment?

A
  • measure cre activity - understand quirks of cre driver ?? (checking that the BAC you are using doesn’t contain other genes that would complicate results) - inducer delivery and side effects
390
Q

what controls should you do before using a Cre experiment?

A

Check for phenotype in driver lineCheck pattern of cre activity (may be substrate dependent)Check for variability of activity (leakiness may be in the germ line!)

391
Q

how many matings must you do when you are performing a cre loop experiment and why?

A

you can’t just cross a Cre with a homo loxp because you will only get a het loxp chromosome. You must cross and then cross this one with a homo loxp. This will give you a homo loxp cre

392
Q

is it important that both the Cre and floxed mice are from the same line?

A

yes

393
Q

what 3 controls should you produce when carrying out a cre/loxp experiment? And how do you make these?

A

homo loxp, cre and het loxp, just cre and no loxp. first you cross a cre with no loxp with a homo loxp. This gives you a het loxp and cre. Then cross this with a homo loxp and you get your desired mouse and your three controls.

394
Q

how do you orchestrate directed gene trapping?

A

by producing your cassette with homologues arms for the exon either side

395
Q

why would you want to carry out directed gene trapping into certain genes.

A

To cause a truncated protein and to determine where it is that the gene is being perturbed via the reporter. The reporter shows that the gene trap has worked. Depending on what cassette you use you can induce a mutation via the insertion and then flp is back out to look at the control to show that the cassette is causing the mutation and then you can flp it back into again for another control. you want to report where the mutation is occurring because CRe is probably tissue specifically driven

396
Q

what is an example of a gene trap construct that could be inserted?

A

FRT site- loxp- SA- B-geo-pA-loxp- FRT

397
Q

what does the inserted poly adenylate site of a cassette do?

A

prematurely stops transcriptions

398
Q

how can you use a gene trap in conditional knock outs?

A

you can have a FRT’d resistance gene (neo) and a floxed essential exon. This way you can remove the resistance gene as a control to show there is no artefact. Then you can remove the exon for the knock out

399
Q

Describe the knock out conditional first cassette and the 4 genotypes available from using this, including why you would want each of the genotypes.

A

the cassette in inserted around an exon that is required for the functioning of a gene. it would be as follows: exon1- FRT-lacZ-loxp-promoter driven Neo-FRT-loxP-exon 2-loxp-exon 4. This would be tm1a. tm1b would occur when crossed with Cre and the neo and exon could be removed (this would result in a labeled truncated proteinto show successful cassette and where the gene is being mutated (Cre will be tissue specific). tm1c would be when crossed with Flp. this would remove marker and neo and leave normal gene expression. (this could be done after you know cassette is successfuly inserted as a control). Then tm1d would occur when crossed again with Cre to cause a mutant without the exon and because you have done the controls you know it can only be down to the removal of the exon and nothing else.

400
Q

what are the three types of gene traps? what is the general thought behind these traps?

A

enhancer traps, promoter traps and gene traps. The cassettes contain a reporter which will report expression when inserted into the genome.

401
Q

describe the cassette used for an enhancer trap and how it would work.

A

hsppromoter-lacz-pA-HSV promoter, Neo, pA. When inserted into near a regulatory element, the promoter will be activated and the reporter gene transcribed. and the neo will be transcribed as a separate protein (this is needed for the insertion into the ES cell). This will therefore report an enhancer which can then be identified using 5’ RACE

402
Q

describe the cassette used ofr a gene trap and how it works.

A

SA (splice acceptor)-lacZ-pA-actin promotoer-neo-pA. This will insert into an intron and will induce a truncated protein which will be labelled by the b-gal. The splice acceptor ensures it is attached to the truncated protein. Then the gene that has been perturbed can be identified by 5’ RACE

403
Q

describe the cassette used for a promoter trap and how it works.

A

no-promoter- lacZ pA promoter neo pA. This will insert in frame with a protein and has no promoter so needs to land very close to a promoter. The lacZ can be used to demonstrate protein expression levels. 5’ RACE

404
Q

once your gene trap has randomly inserted into a site of interest, how can you then determine where it has inserted?

A

first you extract the RNA from your organism. Then you put it through electrophoresis, then you apply a probe for lacZ via northern blot. Then you cut out the RNA and perform RNA race : you have the RNA containing the galactosidase gene you use reverse transcriptase to create one strand of cDNA from the RNA and the transcriptse adds 3-5 C’s on the 3’ end of the cDNA> then you add an oligonucleotide which has 3-5 G’s on the 3’ end which will anneal to the Cs of the cDNA with an added known artificial primer site you then use reverse transcriptase to extrend from the C’s and on the primer site you then use primers for the known primer site and for a known sequence within the cDNA (your reporter gene) to amplify the cDNA via PCRthe PCR product can then be sequences and the 5” region of the gene that has been trapped can be identified and compared to the sequenced genome in order to find which gene has been trapped.

405
Q

draw out the process of 5’ RACE.

A

..

406
Q

how can you carry out a knock in using Cre loxp recombination?

A

A gene of interest (GoI) can be inserted into the ATG position of a targeted gene (Fig. 4), and the GoI is then driven by the endoge- nous promoter of the targeted gene. Design of such a replacement vector is shown in Fig. 4a. Importantly, the selection marker in this vector should not contain a pA signal to provide further selection of the recombination event, as described below. The replacement and Cre expression vectors are coelectroporated into the targeted ES clone in their circular forms. The cre gene is transiently expressed and mediates recombination. Since the replacement plasmid and the targeting vector in the ES genome both carry two lox sites with the same spacer region (RE or LE mutant lox and loxP), it is expected that intramolecular recombination between the two lox sites should occur after coelectroporation, resulting in the production of two intermediate molecules, as shown in Fig. 4b. Integrative recombi- nation then occurs between LE mutant lox sites in the genome and RE mutant lox sites in the intermediate molecule. ES cells in which (RE is the white arrow head, black arrow bod) (Le is the opposite)- SO first the RE or LE site recombine with normal loxP site. then the LE and RE recombine with each other.

407
Q

describe the process of 5’ Race

A

You extract tissue that is expressing the gene trap- you then extract the RNAelectrophores the RNA to seperate by size then transfer RNA to membrane for northern blotting, fix RNA to the membrane by using UV or heat hybiridze membrane with labeled probes for the galactosidase gene which you know is in the gene trap cut out the RNAyou have the RNA containing the galactosidase gene you use reverse transcriptase to create one strand of cDNA from the RNA and the transcriptse adds 3-5 C’s on the 3’ end of the cDNA> then you add an oligonucleotide which has 3-5 G’s on the 3’ end which will anneal to the Cs of the cDNA with an added known artificial primer site you then use reverse transcriptase to extrend from the C’s and on the primer site you then use primers for the known primer site and for a known sequence within the cDNA (your reporter gene) to amplify the cDNA via PCRthe PCR product can then be sequences and the 5” region of the gene that has been trapped can be identified and compared to the sequenced genome in order to find which gene has been trapped.

408
Q

what are the general aims of someone wanting to edit a genome?

A

people want to be able to target a a specific area of the genome and induce double stranded breaks with no off target effects with very efficient cut rates

409
Q

what are the two response to double strand breaks?

A

non-homologue end going or homologous recombination

410
Q

what normally happens when double strand breaks are repaired by no homologues end joining?

A

mistakes are made by the repair machinery and point mutations, deletions, insertions which causee frameshift and maybe truncated protein (KO)

411
Q

what are the three main genome editing tools used?

A

zinc finger nucleases, CRISPR/Cas9, TALENS

412
Q

what is the general structure of a zinc finger motif?

A

a 2 strand beta sheet and an alpha helix which interact via a zinc molecule.

413
Q

what is the general structure of a zinc finger domain?

A

3 zinc finger motifs that each bind to 3 nucleotides.

414
Q

how is the zinc finger domain transformed into a zinc finger nuclease?

A

you attach a fok1 which is a restriction endonuclease

415
Q

what is the best way to improve binding specificity of gene editing tools?

A

have two different binding sites either side of the target site which then interact to cause cutting

416
Q

what is the general structure of a zinc finger endonuclease tool?

A

two zinc finger domains that are attached to a split fok1. each interacting with 9 nucleotides either side of the target site on opposite strands.

417
Q

what, in theory, is the process by which zinc fingers can be constructed in relation to the base pairs they interact with and why is this not true?

A

you have different zinc finger motifs for each triplet and you can put three motifs together depending on the three triplets you want it to bind to. This is not the case as each finger can alter the binding specificity of the other . This is because they can reach across and interact with the major and minor grooves to influence binding to the DNA and interactions with each other

418
Q

how do the three zinc finger motifs interact to complicate binding specificity?

A

they can reach across and interact with the major and minor grooves to influence binding to the DNA and interactions with each other

419
Q

Because of the complicated interactions that the zinc fingers have with each other and the complications that this can have on binding specificity, how can you ensure that your zinc finger is right/

A

you have to undertake a lot of screening and selection. methods include OPEN, CODA and alternate zinc finger selection strategy.

420
Q

what are the 3 screening strategies for zinc fingers?

A

OPEN, CODA and 2 other random alternate zinc finger selection strategy.

421
Q

what is the Oligomerized pool engineering selection mechanism for selecting the right zinc finger domain?

A

oligomerized pool engineering (open) selection method: this involves having finger pools- contains vectors containing genes for sifferen zinc finger motifs- you can then use primers to amply a random finger from each pool- they have complimentray ends and begininngs so that the fingers can anneal and act as primers for each other. Then you have a random triplet ZFD, then this is cloned into a B2H phagemid and expressed as a fusion protein with Gal11p. This is then put into a B2H selection strain(e.coli). This selection strain contains a ZFN “half-site” positioned upstream of a B2H promoter which drives expresion of selectable markere genes and a hybrid alpha gal4 which acts as an RNA polymerase. To make this strain the target site is cloned into a plasmid and use homologous recomb to clone into the site and mating of the e.coli to pass along the target site. When the phagemid infects the e.coli, the expression of a zinc finger domain that binds to the target site will bring the Gal11p next to the hybrid alpha protein Gal4 and drive the expression of the marker genes- the e.coli colonies expressing the reporters can then be selected and teh zinf fingers extractaed and anylised.

422
Q

how could you slightly alter the OPEN selection protocol?

A

you could use the ZFD to drive transcription of a resistance gene instead of a reporter and grow the cells on something that causes cell death (neomycin?)

423
Q

what is the CODA selection process for zinc finger domains?

A

This uses known zinc fingers for targeting known triplets. For example, if you have a ZFD which targets successful the first 6 nucleotides that you are interested in and you know another ZFD that successfully targets the last 6 nucleotides of your site of interest, and they both share the second finger, then it is likely that the F1 from your first and the F3 from your second will interact well together with the shared F2 in the middle. You can produce this using Primers and PCR for each finger with overlapping ends which act as primers for each other- construct the three finger domain and it is done

424
Q

what are the draw backs of the CODA selection process?

A
  • Only works for about 2.55% of all possible 9-pair sequences (only this amount of genes have known common F2s that will ensure complimentary ZFD)- Mostly works with triplets that have the code 5’-GNNGNNGNN-3’. Thus, can target only about 81% of genes (not sequences) in fish 63% of all genes in Arabidopsis
425
Q

what is the success rate of using CODA to target zebrafish genes?

A

there is a 50% success rate

426
Q

describe the 2 other random inc finger selection strategies

A
  1. have a zinc finger library and randomly sticking zinc fingers together and then you have a target site that you want to target and you have an out of frame GFP on the right, a stop codon at the end of target site and an GFP on the left. If the zinc finger binds and allows for cutting- you will break this stop and it will cause the cell to express GFP. This construct would be injected into cells previously. 2. take a library of zinc fingers and randomly put triplets together to create zinc finger domains and put each triplet into a construct with a Fok1+ and then another with a Fok1- (so the two halves of the endonuclease). These two constructs also both express a specific marker gene. Then you construct a plasmid which expresses two marker genes: one is between the two target site of interest and one is not. You then infect a yeast with these three plasmids (that all have promoters upstream of the ZFD and upstream of trapped reporter. If the zinc fingers both target the trapped gene then the trapped reporter will be cut. You then select for yeast colonies that express the three reporters but not the trapped one. You put combinations different plates and then keep track of which combinations in which plate so you can refernce which ones are successful.
427
Q

what are the drawbacks of zinc fingers?

A

want to be able to target any sequence but you have to screen for the correct sequences etc and certain zinc fingers can hinder each other etc so have to check all this before and might not work for the sequence that you want cut rate is fairly low

428
Q

what needs to be considered when testing zinc finger domains in vitro?

A

this may be no representation of how they may work in vivo- how sensitive they will be heterochromatin

429
Q

what were TALE proteins naturaly used for?

A

bacteria secrete these into plants (repeating structures that bind the DNA- and activate transcription of genes- nutrient genes that make it easier form the bacteria to get in and infect it)

430
Q

what is the general structure of TALENS?

A

They are repeating units, each modular unit is identical bar a diamino acid structure which binds specifically to a nucleotide. They have increased specificity due to their ability to bind to around 14 base pairs either side of the target site.

431
Q

what needs to be considered and are potential down sides to using TALENS?

A
  • they dont bind as tightly as ZFN- they are very laborious to make: the repeating structures mean that they can recombine within bacteria when they are being cloned. so you have to use bacteria that are recombination deficient but these have their problems too.- They are generally very tedious to put together
432
Q

what are the three method of TALEN construction?

A

FLASH assembly, golden gate cloning or standard restriction digest.

433
Q

what are the areas of the TALEN modules which are repeated, called?

A

repeat variable diresidues.

434
Q

what is the flash assembly of TALENs?

A
  • a dna fragment encoding a single TALE repeat is labelled on its 5’ end with biotin and is initially ligated to a second DNA fragment encoding 4 specific TALE repeats and then attached to a streptavidin-coated magnetic bead. Additional DNA fragments encoding pre-assembled ATLE repeates are ligated until an array of the desired length is assmbled. The DNA fragment is then removed from the bead by restriction digestion which removes the biotin bead.
435
Q

what is the golden gate cloning process of TALEN construction?

A

You use PCR to amply each unit, you may have for example 6 in each pool. You then add appropriate ligation adaptors. so that each of the units will ligate to the next one along although the 1st and the 6th will not ligate together instead they will have the same restriction site, but there will be a spacer complimentary site annealing site. Then, using the intermeidate space between the 1st and 9th, the hexamers are amplified and purified. The next step involves cutting the hidden sites at after the 1st and 6th and then 6th restriction site is then revealed which will then attach to the 7th start. and the same for the 12 to the 13th. Then they are put into a cloning backbone which may be a viral vector which then fuses it to a functional domain. this is done by ligating the backbone with the same restriciton enzyme that is at either end of the TALEN units.

436
Q

what is the standard cloning based method?

A

making two modules at a time and fusing them using standard restriction enzyme digestions and ligations. Then these two modules are fused to each to form 4 modules. then the 4 modules are fused together etc etc

437
Q

are zinc fingers good with methylation?

A

no

438
Q

does the efficiency of the TALEN cutting decrease when there are more methylation sites?

A

yes

439
Q

when there are around 4 methylation sites, what is the percentage cutting of the TALEN?

A

40-50%

440
Q

how can you use TALE to modulate the methylation of a DNA site?

A

you can construct the same DNA targeting mechanism but then instead of attaching a for 1 you can instead attach a DNA methyltransferases (DNMT). This then methylates the DNA and turns it off. This would mean that you would follow the same procedure to produce the TALE part but when you ligate it into the cloning backbone you would fuse it to a DMT rather than a Fok.

441
Q

how can you use TALE to modulate the acetylation of a histone at a DNA site?

A

You you can construct the same DNA targeting mechanism but then instead of attaching a for 1 you can instead attach “histone acetyltransferase” (HAT) or “histone deacetylase” (HDAC). This will either unwind the DNA or cause it wind by adding an acetyl group to the histones associated to the DNA

442
Q

what are the pitfalls of TALENS?

A

modular build is very tedious and long winded, they are very big constructs, not all sites are equally well targeted.

443
Q

what was the CRISPR/Cas9 originally used for?

A

bacteria used it as a mechanism to capture viral DNA and store it as a memory so that they could then chop up the DNA if it entered again.

444
Q

what is the structure of the DNA target sequence of the guide strand within the Cas9?

A

The content of the guiding sequence should have a GC content of 40%-80%, the length and should be 17-24 (smaller = less of target effects)

445
Q

how does the CRISPR/Cas9 sequence work within bacteria?

A

The area of the genome encoding the CRISPR looks as follows: - the tracrRNA followed by a CAS operon which encodes different CAS enzymes, followed by a cRISP repeat spacer array which contains genes encoding the viral DNA with spacers in between. When this gets transcribed, the tracrRNA binds with the Cas9 and allows the CAs9 in conjunction with RNAaseIII cleve at each spacer and produce mature crRNA. This results in an RNA duplex consisting of the tracrRNA and the mature guide RNA which is then used by Cas9 to cleave the target DNA in two places on each strand using two domains of enzymatic activity

446
Q

how did people take the matures structure of the guide tracrRNA and crRNA duplex and use it to create a more efficient guide?

A

They looked at the tracrRNA crRNA duplex and realised that if you add a linker between the two then you can create a single guide strand that can be provided to the Cas9 enzyme. this is called a single guide RNA The tracrRNA has different hairpin loops

447
Q

what is the structure of the tracrRNA?

A

has hair pin loops

448
Q

what is the difference between the specificity of the Cas9 system compared to the ZFN and TALENS? what does mean for the application?

A

there is only one guide and so you lose the specificity with Cas9

449
Q

describe the experiment which looked at the specificity of the CAs9 system

A

create cell that that expressed short lived GFP gene which, when broken by a double stranded break, will result in the cell losing its GFp expressiosn They designed missmatches in the site and guide and asked how efficient was the Cas9 and guide to cut at the site of the GFP. They showed there could be up to 2-3 mismatches in the bases between the target and the guide and you still get efficient cutting.

450
Q

why does the Cas9 system naturally not want to be too specific?

A

because bacteria want to be able to recognised evolved viral DNA

451
Q

what is a Cas9 nickase? how is it formed?

A

it only cleaves one strand of the DNA. It is formed by mutating the Cas9 protein

452
Q

how can Cas9 nickases be used to produce increase Cas9 specificity?

A

you can use two nickases which two guides that target either end of the target site, one cleaving one strand and one the other.

453
Q

how is the spacing important with nickases?

A

if you have the two pairs too far apart then it doesn’t work

454
Q

what is the guide called for Cas9?

A

sgRNA (single guide RNA)

455
Q

what is the PAM site?

A

it is a specific sequence that has to be targeted at the end of the guide (3’ end of the guide but on opposite target strand) and for the classical cas9 this is NGG. this allows for the proper alignment between the guide and the double strand and the loop structure. You need a 5’ to 3’ sequence that ends in NGG

456
Q

how are people making increased specificity Cas9?

A

by mutating the Cas9 protein

457
Q

because Cas9 guides can have off target effects, what is an experiment that you can do to ensure that the mutation that is produced following a Cas9 approach is targeting your desired sequence?

A

you use one guide that has known off-target sites A,B or C. Then you have another guide which has the off-target sites D,E,F. you use both on different organisms then you get a phenotype. Then you can cross these and see if you still get a phenotype (compliment testing ). if the phenotype is being cause by cuts at two different sites then a het will be formed when crossed and you will lose the phenotype- but if you dont lose phenotype then it is at the same site (fail to complement) 2. do the two guides give the same phenotype- if they do then likely they are not targeting the same gene 3. you can outcross you animals to clean them up

458
Q

what is the PAM site for classical Cas9s?

A

5’–> 3’ NGG

459
Q

how much of the human reference genome contains GG? what does this means about how many target sites there are per how many bases?

A

5.21%, 1 target site every 42 bases

460
Q

describe a selection experiment that can be used to produce mutant Cas9 proteins that require a different PAM sequence.

A

mutate Cas9, put into selection screen by using a plasmic containing Cas9 under the control of a T7 promoter then a sgRNA under teh control of a T7 promoter followed by a resistance gene for selection. Then you also have a plasmid which has your target site that doesn’t have a GG downstream of a toxic gene target site that when cleaved will allow growth. An example of a toxic gene would be arabinose-inducible toxic gene. Then the bacteria can be plated in the arabinose and only those bacteria colonies containing Cas9 mutants that can cleave the mutant PAM will allow the colony to grow. Then you can make sure that the Cas9 is specific to that Pam sequence and not others but selecting those mutants which will cut a certain PAM sequence but then use the positive selection screen to test whether, when you mutate the PAM back to NGG, does it still work. If not then that is good! The positive selection screen is done by using target site containing a resistance gene is upstream of a NGG. If the PAM is refractory to the Cas9 then the colony will be allowed to grow when grown of a medium with the compound requiring resistance against. As long as you keep track of which Cas9 mutants you are using you have work out which work with specific pam sites and not with others.

461
Q

describe an experiment which will allow you determine what makes a good gRNA for CRISPR

A

CRISPR scan- they have taken different gRNAs for different sites within the same gene and asked what was the cut rate was for each gRNA target. and then they looked for rules within the guide which were determining what makes a good guide.

462
Q

how can you use the Cas9 system to silence transcription?

A

once you can target a certain area you can add enzymes which activate transcription. You can mutate the nickase sites of the Cas9 so that it isnt actually cutting, instead it will just bring a fused protein. You can use this to silence ranscription by using a enzyamtically dead Cas9 that will bind to a DNA region and will interact with the PAM sequence so that ranscription is haulted when ti reaches this point.

463
Q

how could you produce a light sensitive cas9 and why would you do this?

A

You can put a light sensing domain on the Cas9 which will allow it to be taken into the nucleus and then out again. This could be used to spatially or temporally control transcription. you can do this by using a light sensitive nucleus localising signal which is a protein that could be fused to a protein and is based on the LOV protein domain- this could be used to activate or suppress transcription when you wanted.

464
Q

how could you use crispr cas9 to report the transcription of a mutant gene?

A

inject into the early embryo, a cas9 which a guide RNA for an area downstream of a enhancer r promoter. Then also inject a plasmid which contains homologous arms and a reporter gene with a stop codon. When the cas9 cuts, this will be inserted and stop transcription of the protein but will express the reporter. this can be used in a gene trap way

465
Q

are crisprs efficient ?

A

yes

466
Q

how can you perform a Cas9 genome wide-esque screen.

A

inject a pool of guides targeting different genes into the embryo with the cas9 or into an embryo which already expressed the cas9. And then look for a certain phenotype. Then you can identify which guides were injected into this pool then do individual screen to find which gene was being targeted

467
Q

in what organism is a cas9 screen should e carried out in?

A

zebrafish because can be injected early into the embryo and will start cutting early

468
Q

describe how the gene drive works.

A

you target a gene with a construct that encodes both the cas9 and the guide RNA, for itself this is inserted into the gene on one of the chromosomes this is then transcribed and the crispr cas9 forms a KO in the other chromosome because of the homology on either side of the CAs9 on te other chomosome, homolgous recombination results in the cas9 construct recombining into the other chromosomethis means that any cell that expresses this construct will be homozygous this means that when you cross a male whci expresses this and you get a “het” offspring, all cells will then become homo because the cas 9 construct will attack the other in most casesthis means that instead of getting a mednialin ratio you can get all offspring eventually expressing as there is an over 50% chance of inheriitng it. if you target a gene involves in sterility you can knock out an entire species.

469
Q

how can you detect mutant events in zebrafish?

A

you inject crispr/cas9 stuff into the embryos. You can’t screen embryos and then put them back. SO you screen half of the clutch you injected and then use DNA melt of Primer PCR to identify a mutant event. You then infer that the other 50% has been equally successful.

470
Q

what is the high resolution melt curve and what is it used for?

A

it is an efficient way to ID mutants, you melt the DNA and differentt DNA sequences melt at different times do you can identify changes within a certain region of DNA by comparing to wild type

471
Q

what are the problems with using th melt curve in zebrafish?

A

you dont have homogenous lines in zebrafish so need to watch out for SNPs.

472
Q

how can you sequence mutants and ID the type of mutation is a way that is better than melting? why is this better?

A

You can PCR region of interest and then deep sequence many different fish. This will allow you identify different mutant loci and identify which one you want. then you keep the fish that contains these mutations and breed this one as a strain

473
Q

what type of errors can non homologous end joining result in?

A

insertions, deletions, substitutions at the break site

474
Q

when can non homologous end going mediate translocation?

A

when there are two double stranded breaks then the ends can be swapped when repaire .

475
Q

what is the process following breakage of DNA that stimulates DNA repair?

A

Following the detection of a DSB, the histone variant H2AX is targeted for phosphorylation in chromatin close to the break site. The phosphorylated form of H2AX, known as gH2AX, serves as a molecular beacon, signaling the presence of damage by marking nucleosomes in one or more megabases of DNA surrounding the DSB (Rogakou et al., 1998). gH2AX is central in linking damaged chromatin to the DNA repair machinery, directing the recruitment of multiple DNA repair and signaling proteins into repair centers, microscopically visible nuclear aggregates known as ‘‘foci.’’

476
Q

do the last two lectures!!

A

477
Q

what are the enzymes that clear DNA?

A

nucleases or endonucleases

478
Q

in what organisms have TALENs been used?

A

flies, yeast, zebrafish, rat, tips cells and human somatic cells.

479
Q

why are TALENS good to use in relation of the construction?

A

they rely on a repeated domains that only change for the two amino acids that give them their base pairs specificity- these are two residues at position 12 and 13

480
Q

has the Binding efficient of ZFN and TALENS been directly compared?

A

no but studies have shown that they can cleavage specific genes with similar efficiency.

481
Q

where are the 2 varying residues in the TALEN modules?

A

usually position 12 and 13

482
Q

what do the the two hyper variable resides of the TALEN modules interact with upon DNA binding?

A

positioned in the DNA major groove

483
Q

what are the names of the modules of the TALENs and what is each module’s base pair?

A

NN= guanine NI= adenine HD= cytosine NG= thymine

484
Q

once you have made a double strand break, what can you do?

A

mediate homologous end joining or allow the DNA repair machinery to make mistakes and introduce mutations.

485
Q

Do TALENS have a high success rate?

A

one study said that they did have a high success rate in any DNA sequence of interest that was performed in human cells.

486
Q

how can you check if you mutation has been successful?

A

All of the mentioned techniques require that you amplify up using PCR your target region, you can do this because you know the site around the mutation and you can use primers. 1. you can insert your PCR product into a plasmid vector that contains a reporter gene. impairement go recovery of the reading frame of the acZ gene occurs via insertions or deletions- you can then see how many colonies exrpress the reporter from each PCR product 2. you can use Sanger sequencing your PCR product 3.use heteroduplex formation and then electrophoresisform heteroduplexes then use high resolution melt curves

487
Q

what is the heteroduplex process of analysis if a mutation has been induced?

A

because you obviously know the surrounding region of your gene due to targetting it, you can make primer amplifying the target locus once you have extracted the DNA and used restriction enzymes. Once you have the PCR product you can mix this mutant DNA with wild type DNA and then denature and induce hybridisation- this will form a heterodulpex which is retarded in electrophoresis- can determine the presence of mutations

488
Q

what is sanger sequencing?

A

During Sanger sequencing, DNA polymerases copy single-stranded DNA templates by adding nucleotides to a growing chain (extension product). Chain elongation occurs at the 3’ end of a primer, an oligonucleotide that anneals to the template. The deoxynucleotides added to the extension product are selected by base-pair matching to the template.The extension product grows by the formation of a phosphodiester bridge between the 3’-hydroxyl group on the primer and the 5’-phosphate group of the incoming deoxynucleotide (Watson et al. 1987). Growth occurs in the 5’ -> 3’ direction (Figure 1).DNA polymerases can also incorporate analogues of nucleotide bases. The dideoxy method of DNA sequencing developed by Sanger et al. 1977 takes advantage of this characteristic by using 2’,3’-dideoxynucleotides as substrates. When dideoxynucleotides are incorporated at the 3’ end of the growing chain, chain elongation is terminated selectively at A, C, G, or T. This is because once the dideoxynucleotide is incorporated, the chain lacks a 3’-hydroxyl group so further elongation of the chain is prevented. you then do electrophoreses and the labelled terminating bases will read out downwards giving the sequence

489
Q

how can you induce large deletions using gene editing tools?

A

target two sites surrounding a region

490
Q

how can genome editing be used in disease study?

A

you can use these techqniues to target genes involved in diseases- such as dystrophin and then create a disease model that you can work with

491
Q

which of the TALEN making mechanisms has the highest throughput?

A

solid phase

492
Q

give some precise data demonstrating the superiority of TALEN cutting to ZFD. what organism?

A

85% of the 32 TALEN pairs used induced somatic indeed stations but only 25% of the 84 ZFD used induced indeed mutations (ZFD were made from the CODA technique) in zebrafish

493
Q

which gene editing tool has been shown to be sensitive to CpG sites?

A

TALENs - more than 2 or site sites decrease mutagenicity

494
Q

how can the sensitivity of talens to CPG sites be circumvented?

A

can use a N* module rather HD module

495
Q

when can TALEN mutations be inherited?

A

if injected into the embryo early then they can enter the germ line

496
Q

how can TALENs, CRISPR and ZFD be induced into the entire embryo and what embryos can this method be used in?

A

If you inject the mRNA of the TALEN, ZFD, or CAS9 sgRNA into a single cell embryo. You can also inject lentiviruses expressing the construct. You can inject the TALEN mRNA into the mouse male pronucleus.

497
Q

what is the problem with the fok1 dimerisation theory? and how has it been circumvented?

A

you can separate the uncle domain must homodimerise to function. sometimes the foc1 domain can homodimerise when only one monomer binds to DNA and can produce off target effects. Instead you can create a fok1 that must heterodimerise- this reduced off target effects

498
Q

what is a big downside of ZFN and an issue that comes with this?

A

The binding of the 3 motifs are not always complimenty to each other so it requires a lot of screening to check the zinc ginger will work. it is hard to produce high activity zinc finger domains that are not cytotoxic

499
Q

what is the problem with he resources available for ZFN?

A

there isn’t a library containing 64 inc fingers all combinations of the triplet sites

500
Q

what are the criteria for a good zinc finger target site? compared to talens?

A

guanine rich and have the sequence GNN, talent can target any site