Lecture #1 - Mendelian Genetics

Question 1

Allele relationship of most mutations

Answer 1

Most mutations are recessive (LOF can be complimented by other alleles)

Question 2

Overall Types of mutations

Answer 2

Can have mutations:
1. Inside ORF
2. Outside of the ORF (Mutations outside of the ORF –> affects expression of the gene)

Question 3

Types of mutations inside the ORF

Answer 3

Categories = based on how mutations affects the encoded protein

1. Missense = change in amino acid
2. Nonsense = Early stop codon (Often leads to trunctaed/unstable protein)
3. Indel –> Chnages the reading frame (“frameshift”) if the indel is not a multiple of 3
   - Often leads to a premature stop codon = get unstable protein
4. 4 – Silent = no mutations (Codon encodes the same amino acid)

Question 4

Affect of Mutation (OVERALL)

Answer 4

Mutations = affect the genotype of an organism

Genotype = organism’s genetic makeup (Collection of mutations an organism’s genome)
- Genotype – can refer to a specific gene when talking about if that genes encodes a specific mutation

Question 5

How do we go about understanding gene function (overall)

Answer 5

Explore gene function by examining how a mutation changes a given trait of an organism

Question 6

Allele

Answer 6

Different versions of the same gene

Wild-Type allele – the ‘standard’ genotype/phenotype in a model system

Mutant alleles - when a gene has a mutation

Question 7

2 Broad categories of mutations

Answer 7

1. Recessive - Phenotypes appear in organism with two copies of the mutations (needs to be homozygous)
2. Dominant - Phenotypes appear in organisms with only 1 copy of the mutation (Can be seen when heterozygous or homozygous for the mutant)

Question 8

How does mutation affect a phenotype

Answer 8

Way mutations affects the function of a protein depends on the mutation

Two classes of functional change:
1. Loss of function - protein does not function well or does not function at all OR he protein is not made
2. Gain of function - Protein functions too well OR too much protein is produced OR protein gains new function
- Protein is more active
- Protein can be expressed in different location/tissue

Question 9

Muller’s morphs

Answer 9

Further breaking down affect of mutations on protein
1. Amorph
2. Hypomorph
3. Hypermorph
4. Antimorph
5. Neomorph

Question 10

Amorph

Answer 10

Overall - Null allele –> allele that produces no protein product or a non-functional protein product
- Amorph = complete loss of function
- Allele is functionally equal to completely deleting that gene
- Tends to be recessive (1 WT copy = have WT phenotype)

Question 11

Hypomorph

Answer 11

Overall - Alleles that produces less protein or a form of the protein hat functions less efficiently that the WT allele
- Partial loss of function
- Can be expressed at a lower level
- Tends to be recessive (WT can mask the dysfunctional allele = 1 copy of WT –> get ET phenotype)

Example – Temperature sensitive allele –> less stable protein + promoter down mutation

Question 12

Hypermorph

Answer 12

Overall - Produces too much protein product or a protein product that functions more efficiently than the WT
- Increase in gene function
- Function in the same way and location as WT BUT are more efficient or expressed at a higher level than WT
- Tends to be dominant

Question 13

Antimorph

Answer 13

Overall - Dominant Negative –> Antagonizes/intreferes with the function of the WT allele
- Referred to as poison allele – has reduced function and can interfere with WT function
- Similar to neomorphs BUT the new function that they acquired antagonizes the function of another protein or the WT version of the same protein
- Tends to be dominant

Ex. Truncated receptor
- Can occur with proteins that normally function as multimores or enzymes that function in complexes

Question 14

Neomorph

Answer 14

Overall - Gains a new function or expression pattern (Expressed in new locations)
- Tends to be dominant
- IN selection/screen Neomorphs give the least amount of information

Example – Mutant receptor that responds to new ligand or promoter mutation that leads to misexpression

Question 15

Which mutations are typically Gain of function mutations

Answer 15

Hypermorphs + Antimorphs + Neomorphes = ALL gain of function mutations because adding more function in top of the WT allele OR their new function is independent OR antagonistic of the WT

Question 16

Why do we care about the type of mutation?

Answer 16

1 - Want to know the type of mutation to understand how disease is passed on
2 - In yeast –> Need to know if the mutant has the LOF or GOF mutation
- Example - Might have a GOF enzyme that diverts precursors of Uracil to a different pathway and prevent making Uracil (not making uracil BUT this mutation is not in the uracil pathway itself)
- To know WT version you need to know if mutant increases or decreases activity

Question 17

How do you know what category the allele is in?

Answer 17

To know what category an allele is you do a phenotypic analysis

Question 18

Phenotypic analysis

Answer 18

Phenotypic analysis - Compare phenotypes of organisms with known genotypes to various combinations with the allele of interest

If we know the phenotypes of the WT and the null THEN we can compare them to various combinations of a WT, a deletion mutant (del), or a mutant (m)

Question 19

Phenotypic Analysis - Amorphic alleles

Answer 19

Amorphic alleles often gave a strong loss of function phenotype –> THEREFORE the phenotype of a homozygous mutant should be equivalent to a heterozygous m/del phenotype

m/m = m/del = del/del –> ALL 100% lethality
- m = NULL alelle (null/null)

ALSO WT/WT = WT/m - because amoprhic alleles are recessive
- WT/m and WT/Wt phenotypes should be ‘better off’ than m/del or m/m

ALSO – adding one extra copy of WT allele through duplication [dup(WT)] to a WT/m genotype should not change the phenotype
- WT/m = WT/WT = WT/m;dup(WT)

Question 20

Phenotypic Analysis - Amorphic alleles ALL together

Answer 20

Combining the relationships gives a spectrum of phenotypes from worse to better off

Worse off = m/del = m/m < WT/m = WT/WT = WT/m;dup(WT)

Question 21

Hypomorphs vs. Amorphs

Answer 21

Hypomorphs are similar to amorphs but have some expression or function at a reduced level (amorphs would have no expression or function)

Have very similar phenotypic analysis (Ex. Hypomorphs is still compensated for by a WT allele because the hypomorph allele is recessive)

Question 22

Phenotypic analysis - Hypomorphs

Answer 22

Worse off - del/del < m/del < m/m < WT/m = WT/WT = WT/m;dup(WT)

m/del is worse off the m/m BECAUSE would have some activity out of the additional hylomorphic allele

Know that the hypomorph has some residual function because hypomorph/hypomoprh is less severe than hypomorph/deletion or null/null

Question 23

Phenotypic analysis - Hypermorphs

Answer 23

Overall - As more copies of the hypermoprhic allele is added in the presence of the WT the phenotypes because progressively more severe

Worse off - WT/m;dup(WT) </= m/m < WT/m < del/m </= WT/WT (better off)

Depending on how much stronger the hypermorph is than the WT THEN del/m can be equivalent or worse off than the homozygous WT phenotype BUT the WT/m will almost always be worse off than WT (because there is too much gene function)
- Adding extra function is not always good

Hypermorph = dominant = see phenotype when only 1 copy of the mutant is present

Question 24

Which is worse of?

Answer 24

C = worst because have too much of a good thing
- Worse – C –> B –> A

As you add more of the hypermorph in the context of the WT = get progressively worse

Question 25

Phenotypic analysis – Antimorphs

Answer 25

Overall - Antimorphs negatively affect WT allele function –> MEANS the more copies of WT allele can improve the phenotype (WT allele can dilute out the antimorph)

Worse off - WT/m < WT/m;dup(WT) </= WT/WT (Better off)

Question 26

Phenotypic analysis of Amorphic vs. Antimoprhic alleles

Answer 26

• Because antimorphs function by impairing WT allele function they will behave as if they were independent LOF allele when the WT allele is absent –> Del/m or m/m phenotypes can mimic amorphic allele

ONLY when include WT allele in analysis can you see the difference between the amorphic and the antimorphic

Question 27

Which is better off?

Answer 27

Better off = C because WT dilutes the antimorph

Question 28

What do you need when assigning morphs

Answer 28

Need to look at entire spectrum when assigning morphs

Question 29

Assigning phenotypic Neomorphs

Answer 29

Because neomorphs gain a new function - function can vary between genes or different mutations of the same gene –> makes it hard to place them on a spectrum with WT and deletion alleles

All mutant genotypic combinations have the phenotype because mutations causes the gene to act in a new way (equally mutant in all situations)

Hard to infer Wt function because the mutant is adopting an extra function that is not related to WT

Question 30

How to characterize a suspected neomorph

Answer 30

The best way to characterize a suspected neomorph is to experiment with the allele to determine its GOF phenotype

Exmple – Antenopedia in flies
- Normally – Flies grow antena on their head
- Neomorph – grows legs in place of antenei

Question 31

Loss of function vs. Gain of Function Mullers Morphs

Answer 31

Loss of function - Amoprhs and Hypomorphs (recssive)

Gain of function - Hypermoprh + Antimoprh + Neomorph (dominant)

Question 32

Haploinsufficney

Answer 32

Exception to the difference between hypomorphic and amoprhic

There are some enzymes where knocking out the allele in hypomoprhic means there wil not be enough protein from the dcerased effcicney/remaining WT = the null and the hypomorph have the same phenotype BUT when add extra WT copies then the phenotype can go away

REALLY.- 1 copy if hypomorph or morph vel dies = have phenotype of the ahypomorph or the amorph even though they are usually recssive

Haploinsufficincey = have phenotype present in 1 copy even though it is a LOF mutation

CHECK THIS

Question 33

What can we do if our mutation is lethal or the gene we are intersted in is essential for life

Answer 33

Answer – Use a conditional Allele

In cases where we are interested in essential genes conditional alleles of the gene might provide a way to study it

Question 34

Conditional Allele

Answer 34

Alleles whose protein products function as WT under permissive conditions but become unstable under restrictive conditions
- Alleles have additional utility because allow us to control gene expression or function in other types of experiments

Example of conditions used – Temperature + Light + prescence of chemical compounds + induction of other genetic elements

Question 35

2 broad classes of experiments

Answer 35

Foward vs. Reverse genetics

Tell you about what a gene does+ what genes are responsible for a phenotype

Question 36

Reverse Genetics

Answer 36

Start with the gene and observe he phenotypes that result from manipulating it
- NOT identifying new genes from phenotypes BUT instead taking information about what we know about the gene and working backwards to learn more about that gene

Might look at what happens when a gene is deleted or might want to characterize a disease causing mutation

Question 37

Tools used in Reverse Genetics

Answer 37

1. CRISPR
2. RNAi
3. TALENs/Zinc Finger nucleases

ALL make specific point mutations or deletion or insertions into a gene of interest

Question 38

Forward Genetics

Answer 38

Take a phenotype and work to identify the genes responsible for it
- Used before full genomes were sequenced
- Usually trying to find new parts of a known pathway or pin down the cause of a phenotype

Tools used = Chemical mutagens is EMS or MMS

Question 39

Mutogensis in forward genetics

Answer 39

Mutations occur in population BUT at a impractically slow rate

ALSO mutations occur randomly across the genome = if we intersted in a particular gene the odds that a mutaton lands in 1 of them is slim

Solution - Mutagens are used in forward genetics because they increase the probability that a mutation will occur in an organism

Question 40

Applying mutagens in foward genetics

Answer 40

Mutagens = increase the rate of mutations –> applying mutagen to system will reduce the amount of time required to find mutation that we are interested in

Issue - High concentration of a mutations will reduce survival

Question 41

Concentraion of mutagen that should be used

Answer 41

Need to balance the rate of mutation against the percent survival –> get enough survivor organisms that carry a limited number of mutations (ideally want 1 mutation per organism)

IF concentration of mutagen is too high = most organism die and survivors are riddled with mutations
- If the survivors have too many mutations = hard to single out 1 mutation that is cause of the phenotype

IF concentration is too low = requires looking at many more organisms to find a phenotype

Question 42

Issues with mutagenesis

Answer 42

1. Even with ideal concentration mutagen it is possible that a single organsim displaying a phenotype will be harboring multiple mutations that could be the causal mutation
2. There are some classes of genes that are infrequently recovered from mutagenesis experiments

Question 43

How do you ensure that a phenotype is caused by 1 mutation when using a mutagen

Answer 43

To ensure that a phenotype is caused by 1 mutation – need to backcross the mutant organism and cross it with the WT strain for several generations) –> eventually you can isolate a single causative mutation in an otherwise WT background

Question 44

What classes of genes are infrequently recovered from mutagenesis experiments

Answer 44

1. Essential genes are frequently missed because mutations in them produce infertile or dead organisms
2. Short genes are less likely to be hit by a mutations
3. Redundant genes (similar genes that preform the same function)
   - To identify them in mutagenesis = need simultaneous mutations in both paralogs in a single organism (otherwise organism looks WT)

Question 45

Screening

Answer 45

Screening = sifting through organisms and picking those out that have a phenotype

Question 46

Mutagenesis Screen (overall)

Answer 46

Overall - producing mutation and looking for phenotypes

Question 47

How long do we need to screen for before we have decided that we have hit all of the genes that there are to hit?

Answer 47

To address the question - can plot the number of mutant alleles that we found in the given gene against the number of genes containing that many allele

Start of screen (found few alleles) - expect that distribution leans toward the right skew (most genes we identify are represented by 1 mutant allele (grey bars)

As screen progresses (reach saturation) –> As we find more alleles and their associated genes –> the number of alleles per gene will increase and the distrubution will be more normal curve
- Every time you do a screen you get more alleles (number of alleles is infinite) BUT eventually you start getting alleles in the same gene

Question 48

Saturation

Answer 48

Process of diminishing returns (identifying fewer new genes as we find more new alleles as the screen progresses)

Once hit a saturation point – tells us that we have reached a good stopping point (can stop screen/selection)

Question 49

Selection Experiment

Answer 49

Second type of mutogensis experiment that helps identfy genes responsible for a phenotype

Requirements:
1. Need to know what the phenotype is that we want to select for beforehand
2. Need to impose conditions on the organisms such that only organisms that display the phenotype are able to survive

Issue - NOT all phenotypes can be selected for
- Example of selectable phenotype – Drug resistance + Metabolic phenotype

Question 50

Selection vs. Screen

Answer 50

Selection = faster because only have organisms with the phenotype you are intrested in = fewer organisms to look through
- BUT limited to certain phenotypes and need prior knowledge of phenotype of interest
- Selection = only thing survives the treatment is the mutant you are intersted in
- Look sat more specific mutants in a subste of genes in a pathway
- Selection = good if you already know the gene of interest then it is faster
- Can screen more EMS mutants (not looking indvidual)
- Less work intensive
- Selection = Fast way to get many alleles of specific pools of genes of a specific genotypes

Screens = more unbiased (target all the genes involoved in a pathway) + don’t resuire a starting phenotype BUT require more time + requires more attention to detail and resources
- Screen = look at al EMS colony and ask if it it grows on a second plate
- Screen = more open ended and can recover more different types of genes

Question 51

Yeast (overall)

Answer 51

Unicellular Eukaryote

Many mammalian processes + their components +their regulation are very conserved in yeast

Question 52

Yeast Life cycle

Answer 52

Yeast = grow by dividing (yeast divide by budding)
- One divsion takes 90 minutes

Haploids = divide to make a new haploid cell that is genetically idetical to the mother cells
- Each haploid has a mating type (a or alpha)

Diploids can divide to make new diploid cells OR can go through meiosis
- Meoisis = makes 2 mating type a and 2 mating type alpha haploids (process = called sporulation)

Question 53

Yeast states

Answer 53

Yeast can exist in both haploid and diploid stages (goes between 2n and n haploid state)
- Haploid = have a single set of 16 chromosomes
- Diploids = have two sets of 16 –> have 32 total chromosomes

Goes to diploid and haploid by mating
- 2 haploid can mate to form a diploid (ONLY different types of halpoids can mate –> Only alpha and a can mate)
- Diploid goes back to haploid by meiosis

Question 54

Sequencing of the yeast genome

Answer 54

Yeast = fully sequenced in 1996 (First Eukryotic genome to be fully sequenced)

Yeast = Have 16 chromsomes (15 mb) ; Have 1 gene every 2 kb

Have 6,617 predicted ORFs (80% of ORFs have been verified ; 20% of known ORFs but don’t know if they are expressed)

20% of yeast genes are required for growth (majority of yeast genes are doing something else)

Question 55

Biological Processes Yeast genes are implicated in

Answer 55

Lots of genes are required for basic Eukaryotic maintainance (Ex. Translation has many genes)

Have many genes for response to chemicals –> because respond to chemicals is how yeast responds to environment

Question 56

Yeast genetic nomenclature

Answer 56

WT gene names = italicized and uppercase (Ex. ROG1)

Mutant alleles = italicized and lower case (often followed by an alleles number to differentiate mutant allele types) (ex. rog1-3)

Gene deletion = italicized + lowercase + followed by a delta (Ex.rog1d)

WT proteins = only have teh first letter capital and are often followed by a p (p = protein)

Mutant proteins = all lowercase BUT are also followed by a p

Question 57

Types of selectable markers

Answer 57

1. Auxotrophic
2. Antibiotic Marker

Question 58

Antibiotic marker

Answer 58

2 – Antibiotic marker –> conferes resustence to Antibiotic

Any yeast strain can use these markers (no endogenous genes offers resistance)

Question 59

Auxotrophic marker

Answer 59

Auxotrophic marker –> Lack gene or genes that are needed for producing an essential metabolite
- Mutants can’t make a specific molecule (yeast would only survive if that molecule is added to the media)

Question 60

Prototroph

Answer 60

Prototroph = WT yeast –> MEANS the yeast generates its own food from salts + minerals + sugar

Phototrophs can grow on minimal media –> yeast makes all Amino acids and bases that it needs to make proteins + DNA + RNA

Question 61

Types of media Yeast can Grow on

Answer 61

1. Minmal media
   
   • Has no amino acids or bases (more expensive)
   • WT prototraphs an grow on minimal media
2. YPD Yeast extract + epptone + dextose
   - Made up of ground up yeast cells + additional amino acids + sugar sources
   - Less expensive
   
   • More standard way to grow yeast

Question 62

Transforming selectable markers into cells

Answer 62

Selectable markers are often assembled into constructs with a gene of interest –> construct is introduced into yeast through transformation

In the construct the marker is placed close to the gene of intrest (ensures that the marker and gene remain linked)
- Constructs ALSO have 50-500 bp homology arms upstream and downstream of the gene and the marker

Question 63

Homology arms in constructs

Answer 63

Homology arms – Help direct insertion of the construct into a target locus by taking advantage of endogenous machinery that is usually used for homologous recombination
- Use diploid cells when transfroming constructs because have fewer off traget insertions compared to halpoid cells

Question 64

Recombination in yeast vs. other mammals

Answer 64

Mammalian cells = occurs in 1 in every 20,000 cells (recombination is more rare)

Yeast = recombination is more common (1 in 100 cell divisions)

Question 65

Selection after construct is added to cells

Answer 65

Once we recombined the gene of interest and selectable marker into diploid cell –> CAN select for cells that are carrying the marker and therefore the gene of interest by plating in selective media

Example – use uracil synthesis deficient strain (Uracil auxotroph and URA3 marker) –> plate cells on media lacking uracil
- No URA3 marker the cells can’t make their own uracil = only the ones with the URA3 marker (and therefore the gene of interest) survive

Question 66

Cells used for selection/screens

Answer 66

Need haploid cells

WHY - Need haploid cells because most mutatiion will be reccessive –> chance of having both mutations in the diploid is low (likley will only hit 1 copy BUT to see the phenotype to see the phenotype you want to use haploid because only need 1 mutation)
- Need to force yeast to sporulate to obtain haploid cells

Question 67

Each dot on a plate after yeast are plated

Answer 67

Each dot on the plate is a colony derived from 1 yeast cell

Colony = clones that are genetically identical to the first cell

Question 68

Sporulation

Answer 68

Sporulation - Yeast undergo meiosis and put each meiotic product into an ascus membrane (the product in the ascus membrane in now haploid)
- Whole structure (haploid + ascuse) = tetrad

Sporulation = occurs when yeast are placed in nutrient poor condition

Question 69

Tetrad Dissection

Answer 69

Using tetrad dissection - tetrads can be separated and moved onto non-selective plate

AFTER a few days the haploid cells will form clonal colonies –> NOW have haploid yeast where each has the gene of interest

Issue - because plated in a non-selective media = don’t know which colonies ave the gene of interest or if meitic segregation is occurring as expected –> TO ANSWER you can use replica plating

Question 70

Replica plating

Answer 70

Purpose - replica plating = genotypes haploids

Process – Use a specialized pad to transfer some of each colony form the master plate onto a new plate of selective media and a copy plate of non-selective media
- Stamping the colonies onto media then stamping onto a second plate

END - Compare the two plates to observe if the expected 2:2 ratio of the tetrad is produced

Question 71

Can a mutant in DHP dehydrogenase grown in miminal media

Answer 71

Answer – NO

Image – shows the Uracil pathway to make UTP –> IF have a mutation in pathway then the cell would need uracil added to the media to be able to grow

Question 72

Example function of screen

Answer 72

Goal – screen for genes that are required to build UTP

Question 73

EMS

Answer 73

EMS = used as a random mutagen

EMS – chemcial that he yeast absorbs –> EMS modifiees the DNA so that when DNA is replicated the polymerase will introduce a mutation
- Depending on the dose of EMS = introduce a certain number of mutations randomly in the genome

Question 74

Example URA screen procces

Answer 74

Mutogensis with EMS –> Grow haploid yeast on YPD to spread out single cells (yeast can grow because all of the nutrients are provided) –> let the cells grow into colonies–> replica plate onto minimal media + all AA and bases – URA –> look for colonies that grow on master plate (YPD plate) but not on replica plate (minmal plate that lacks URA)
- Auxotrophs can survive on YPD (YPD has everything that is needed to grow) = grow all haploids including mutants and WT after mutogensis on YPD because YPD has the uracil that they needed
- Grow replica on minimal media + all AA and bases – URA –> because want to mutants that can’t make uracil to die but need other amino acids and bases because could have mutations in other pathways and wouldn’t know which amino acid impacts growth

Image – the red arrpw colony failed to grow on replica plate because the minmal media is not proding te nutrinet that the colony needs

Question 75

What happens to ura-3 mutants on plates containing with FOA

Answer 75

FOA – 5-fluroorotic acid is converted to a toxic compound by URA3 gene product

Answer – Mutants LIVE
- FOA only becomes toxic when have active uracil pathway
- WT = die because they use URA3 to turn it to a toxic compound but ones that are defective in the pathway will survive

Question 76

Example Selection Processes

Answer 76

Mutagenizie yeast culture with EMS –> Spread out on minimal media + all amino acids/bases + FOA –> Colonies that grow are FOA resistant (things that grow are mutants in URA3 gene)

Selection = does NOT use replica plating
- Select for yeast colonies that survive (Only survivors are the mutants you are interested in)

Can use a master plate as a control to make sure FOA selection is working

Question 77

How do you characterize mutations without sequencing mutants?

Answer 77

Answer – Cross haploid yeast (to make diploid offspring) and look at the phenotype
- Mate URA- with WT yeast and look at the phenotype of the offspring

Question 78

Is URA- recessive or dominant?

Answer 78

Answer – mutation is recessive

When URA- is mated with URA+ –> get diploid –> IF the diploid is URA+ THEN the URA- mutation is recessive

Question 79

How can we determine how many genes are actually represented in the data

Answer 79

Use a complementation test –> determines if two mutants are in the same gene or diffferent genes

Example - Generate 20 independent lines that cause small wings BUT do not know if this represents 20 different genes or 20 alleles of a single gene

Question 80

What types of alleles can be used for a complementation analysis

Answer 80

Can only use recessive alles (should exclude any alleles that we know are dominant)

Question 81

How does a complementation test work?

Answer 81

Complementation implies that 2 alleles when crossed will restore the WT phenotype in the F1 generation

Example – want to determine if allele a and allele b are in the same gene controlling wing size

Steps - Cross a/a X b/b (cross homozygous flies from each line together and scrore the F1 generation for wing size)
- IF the mutations is in the same gene for both alleles –> then allele a and allele b would expect to see progeny with short wing (because progeny would have 2 non-functional copies of the gene and still be homozygous for the defective copy of that gene) –> HERE IS NO COMPLENTATION between these alleles because the WT phenotype is not rescued
- IF the mutations are in seperate genes –> Expect to only have flies with WT wing length in the F1 generation bceause each allele is complementaed/compensated for by a WT copy received from the opposite parent

If we make crosses for more of the mutants identified we can organize the outcomes into a complenetartion table

Question 82

Complementation table

Answer 82

Each intersection of complementation table represents the phenotype of the F1 generation

+ = means complementation (have rescue of the WT phenotype)
- = no complementation (Still see small wings)
0 = self cross (Already know what the phenotype should be)

Purpose - Table lets us visualize complementation group –> gives an estimate for how many genes are represented among the alleles

Example (Red box) - aa and bb –> have neither allele complement the other = can assign a and b to the same complementation group

Question 83

Complementation groups

Answer 83

Each complementation group should correspond to a singe gene

Example - Look at a, b, c –> can see that c complements both a and b
- Tells us that c must be in its own gene = belongs in its own compmentation group WHILE a and b are still in the same gene
- IF we anlyzye the whole table = get 4 different complementation groups

In image - Notice that in e is part of every complementation group –> suggests that e is not a recessive allele because it can’t be a mutation in 4 seperate genes

Question 84

Exception to complementation groups

Answer 84

1. Intragenic complementation
2. Non- allelelic non- complementation (aka secon site non-complenentation)

Question 85

Intragenic complementation

Answer 85

Overall - Some mutations within the same gene can complement one another

Need to consider how a partiuclar mutation will affect the gene procuct

Can occur through different mechasnims:
1. Different domains
2. Reduced dosage
3. Stabilization

Question 86

Intragenic complementation - Different domains

Answer 86

Example – protein with a protease domain + dimerization domain + transmembrane domian

If have a separate mutation affect the protease domain and the transmembrane domain but the dimer can still form it is possible that this protein can function close to normal

Question 87

Intragenic complementation - Reduced dosage

Answer 87

Reduces the occurrence of improper interaction caused by a mutation

Example – protein that functions as a dimer
- Mutations that disrupts the dimerization domain may not be complemented by a WT allele because the not enough dimer is formed BUT if the same mutant allele can dimerize with itself and the dimer is still functional THEN the mutant may be able to complement a null allele because ow a enough functional dimer can form

Question 88

Intragenic complementation – Stabilization

Answer 88

Two distinct mutations that are capable of interacting with one another and neither of the mutations on their own can interact with the WT protein
- Complementary changes in structure allow complementation between the two mutant alleles

Question 89

Non- allelelic non- complementation (aka secon site non-complenentation)

Answer 89

Overall – Exception to complementation that comes when mutations in seperate genes can’t rescue one another

Example – Halpoinsuddiciency –> occurs when having 1 WT copy of a gene is not enough to produce the WT phenotype
- Can affect complementation where 2 genes with only 1 WT copy preforms poorly
- Another example of dosage BUT in this case we are below the required threshold to produce the WT phenotype

Question 90

Non- allelelic non- complementation (aka secon site non-complenentation) example #2

Answer 90

Occurs with Dominant negative alleles (aka antimorphs/poison alleles)

Dominant negative alleles – can sequester WT protein and therefore reduce its function even when only 1 copy of the dominant negative alleles is present

Question 91

Complementation Test Process

Answer 91

Take haploid mutants –> cross haploid mutants to make a diploid –> IF the two different mutants are in two different genes then the WT copy of each gene should compensate and the offspring should display the WT phenotype
- Easier and faster than sequencing

Example - Cross URA mutants –> make diploid cells and get WT alleles for each mutations = diploid can grow on media without any uracil

Question 92

Why is it hard to determine if two mutants are in the same gene from sequencing?

Answer 92

Issue – EMS generates more than 1 mutation per yeast = when you do sequencing you don’t know which mutation is actually causing the phenotype

TO potentially resolve - Sequence 100s of independent mutants and find 1 gene where mutations arise in more than 1 strain then you can do “in silico complementation” –> guess that when that gene is mutanted it causes URA-)
- Hard to know if you have few mutants

Overall – sequencing is not economical because ned you sequence many strains

Question 93

Why can’t you use dominant alleles in Complementation Analysis

Answer 93

WATCH class video

Question 94

How many genes?

Answer 94

ANSWER - don’t know how many genes

Know 1 and 2 = + –> they are on different genes (means they complement each other)
- Two alleles complement that each other are on different genes

Know 1 and 1 = - –> fails to complement to self
- Logical - Mutations that fail to complement = on the same gene

3 is likley dominant because get - with 1 and 2 BUT can’t be on 1 and 2 because 1 and 2 have to be different genes
- Mutant 3 = dominant because over WT it shows the phenotype in 1 and 2
- Could be teh same gene as 1 or 2 buyt wouldn’t be able to know from this

Question 95

Tests you have to do alongside a screen

Answer 95

When you do a screen = do complementation tests to ask if the alles that you are getting are in new genes or if you are getting the same gene

Question 96

Example of reaching Saturation

Answer 96

Black bars = show the number of single alleles in a single complementation group

White bars = idealized distribution when reach saturation (hit all of the genes in uracil biosynthesis pathway once
- Have a normal distrubution across genes

Need to compare the observed allele frequencies and the idealizes frequencies and conclude if you need to keep going to get more genes or if you can stop

Image - Have 45 complnementations defined by 1 alelle ; 15 comeplementaions defined by 2 alleles ; 8 groups defined by 3 alelles

Question 97

Do you need to keep going?

Answer 97

Answer – YES keep going (because you are not at saturation yet – not at a normal distribution)

The distribution in black does not match the distribution in the white –> means you need more screening to get more alleles of loci in a allele and push to 2 aleles and get new mutants in genes with 1 alelles
- Need to keep going to idetofy more genes

Question 98

Question – What classes of genes makes up classes with 0 alleles vs. 8 alleles

Answer 98

1. Non-essnetial vs. Essential genes
   
   • 0 = essential genes (because null would die ; hypomorphs would survive)
   • 8 = non-essnetial (can get a mutation in gene and yeast would survive)
2. Short genes = less alleles Vs. longer genes have 8 alleles because EMS is more likely to hit a longer gene
3. Duplicated or redundant genes