5 - genetic analysis Flashcards

1
Q

what are the benefits of using escherichia coli as a model organism?

A
– small cell size
– 20 min generation time
– easy to culture
– known genetic code
– known mechanisms of DNA replication
– spontaneous nature of mutations

most useful for:
antibiotic resistance, pathogen resistance, basic molecular machinery of a cell

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2
Q

what are the benefits of using saccharomyces cerevisiae (budding yeast) as a model organism?

A

– single celled eukaryote ∴ combines convenience of bacteria with the features of eukaryotes
– easy grown and cultured on plates
– 90 min cell cycle
– used to identify cell cycle genes and mechanisms of recombination
– able to study gene interactions

most useful for:
cell cycle studies, basic eukaryotic cell biochemistry and biology, haploid genetics

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3
Q

what are the benefits of using caenorhabditis elegans (naemotoad) as a model organism?

A
– has a defined number of cells
– transparent ∴ cell fate can be traced 
– can be frozen
– can feed on E.coli and grown on plates
– able to study programmed cell death
– able to study RNAinterferase
– able to study micro RNAs in development
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4
Q

what are the benefits of using drosophila melanogaster (fruit fly) as a model organism?

A
– easy to obtain
– short life cycle (~10 days)
– easy to perform crosses
– large number of mutants
– 70% of cancer genes have drosophila counter parts

most useful for:
developmental biology, balancer chromosomes, maternal effect genes

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5
Q

what are the benefits of using danio rario (zebrafish) as a model organism?

A

VERTEBRAE MODEL ORGANISM

– transparent embryos and external fertilisation ∴ very easy to induce mutation
– easy to breed
– complete generation in 5-6 months
– can model for some human neuronal diseases

most useful for:
vertebrate development, large mapping populations, easy transgenics

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6
Q

what are the benefits of using mus musculus (mouse) as a model organism?

A

MAMMALIAN MODEL ORGANISM

– genetic similarity with humans
– genome sequences available
– able to study genetic basis of skin colour
– able to study gene knockouts
– able to test carcinogens
– model for mammalian development
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7
Q

what are the benefits of using aradiposis thaliana as a model organism?

A

PLANT MODEL ORGANISM

– complete genome sequence
– small and easy to grow, life cycle ~ 5 weeks
– able to identify developmental genes
– able to study plant physiology and environmental effects
– population genetics

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8
Q

what is the difference between forward and reverse genetic analysis?

A

forward:
isolate a mutant based on phenotype → identify the genes which carry the mutant
i.e. mapping

reverse:
identify which genes are present via sequencing → find out function of gene

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9
Q

what is EMS and ENU?

how are they used to study genetics?

A
EMS = induce G:C to A:T transitions
ENU = A:T to T:A transitions + G:C to A:T transitions

feed organism with EMS/ENS

analyse phenotypes of M2 generation
– first generation: recessive mutant exists in heterozygote form and will not be visible in phenotype
– second generation: mutant with separate into homozygous recessive state and can be identified

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10
Q

how do you determine if mutant are located in the same gene?

A

COMPLEMENTATION TEST

mutations are said to be complementary if they are located in different genes

begin by crossing two mutants
if the progeny is heterozygous, the recessive mutations must be located in different genes as the homozygous state is lost

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11
Q

describe how you would test for epistasis between two mutants

A

EXAMPLE: eye colour in drosophila

WT: dark red eyes
cinnabar mutant: bright red eyes
white mutant: white eyes

Begin with a dihybrid cross (ccWW x CCww) to obtain heterozygotes (CcWw = WT phenotype)

Cross heterozygotes and observe phenotype of F2 generation

Expected = 9:3:3:1 = WT : red : white : light red
Observed = 9:3:4 = WT : red : white
(expect double homozygous to have mixed phenotype but it is white-eyed)

White mutant phenotype masks/dominates cinnabar
∴ white is epistatic to cinnabar

This means the white enzyme/protein must be last in the pathway for eye-colour

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12
Q

describe the additive interaction between mutants.

what is the difference between additive and synergistic interactions?

A

additive interaction = the effects of the two mutants combine and are both visible in the double mutant phenotype

EXAMPLE: FLOWERING TIME
• WT = 17 days
• flm mutant = 14 days
• maf2 mutant = 15 days
• flm + maf2 = 12 days

synergy occurs when the contribution of two mutations to the phenotype of a double mutant exceeds the expectations from the additive effects of the individual mutations

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13
Q

what is a homeotic mutations

A

homeotic genes regulate the development of anatomical structures

homeotic mutations can disrupt or displace anatomical parts

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14
Q

how can genetic screening be used to identify the function of a mutation?

A

mutagenise a mutant and look for the effect on the phenotype

enhancer screen: identifies mutations which enhance the phenotype of interest.

suppressor screen: identifies mutations which alleviate/revert the phenotype of interest
new mutants can be:
– second-site (intra-genic, rare)
– interaction (epistatic, synergistic

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15
Q

rewatch leccy on molecular markers and recombination

A

follow along with example problem

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16
Q

what is a mapping population?

A

back-cross or F2 populations

17
Q

how do you map genes when there is no mapping population (e.g. humans) ?

A

ASSOCIATION MAPPING

relies on population history e.g. pedigree analysis

helps identify those marker alleles which are more frequently found in individuals with mutant phenotype

e.g. maker A linked to mutant gene X, marker B of same locus is linked to WT
marker A found more frequently with mutated allele of X & marker B found more frequently with WT allele of X

historical recombination across many generations eventually breaks up linkage —> tighter the linkage = statistically significant association

18
Q

explain the formula:

T = μ + g + e

A

T = value of a qualitative trait

μ = population mean

g = deviation due to genetic factors

e = deviation due to environmental factors

19
Q

what is the formula for the total variance?

A

V(t) = V(g) + V(e)

total variance = genetic variance + environmental variance

20
Q

what is broad sense heritability?

A

broad sense heritability is the proportion of the total phenotypic variance that is due to the GENETIC differences among the population

broad sense heritability = H2

H2 = V(g) / V(t)

H2 ~ 0, little of the variance is due to genetic differences

H2 ~ 1, most of the variance is due to genetic differences

21
Q

what are the components of genetic variance?

A

V(a) = additive genetic variance, variation due to alleles that act additively

V(i) = epistatic genetic variance, variance due to epistatic interactions

V(d) = dominance variance, variation due to dominance

22
Q

what is narrow sense heritability?

A

narrow sense heritability is the proportion of the total phenotypic variance that is due to the ADDITIVE genetic variance

tells how closely the offspring will resemble the parent

narrow sense heritability = h2

h2 = V(a) / V(t)

h2 ~ 1, most of the variance is due to additive genetic variance / the predicted value an the offspring would equal that of the mid-parent value

23
Q

explain the equation:

R = h2 * S

A

R = response to selection
R = T(o) - μ, where T(o) is the offspring mean
measures how much the mean has changed in one generation

h2 = narrow-sense heritability

S = selection differential
S = T(s) - μ, where T(s) is the mean of the selected parents
measures between the mean of the selected parents and the mean of the population

R = h2 * S

24
Q

how is QTL mapping used to identify if two loci are linked

A

start QTL mapping with purebred cross (A and B)

F2 pop. contains either:
A + B
mosaic of A and B
→ three possible markers for particular locus: AA, BB, or AB

for each marker locus, can groom individuals into genotypic classes based on marker (not phenotype)

measure quantitative trait of each marker locus —> test for statistically significant difference in mean trait value

if there is significant difference, allele A and B of marker 1 are linked to a locus which also has allelic variation that contributes to the trait