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Evolutionary Genetics and Genomics > Lecture 11 > Flashcards

Lecture 11 Flashcards

1
Q

RFLPs (Restriction length polymorphisms) can be used to calculate genetic distance between two alleles. They are good because they are.
(6)

A
  • co-dominant
  • the genetic map can be directly related to DNA-sequence
  • Can be found new ANY gene
  • Many markers per cross (they can even be anonymous)
  • Can crease a genetic map with high marker density spanning the two genome
  • Useful for many species
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2
Q

Menedelizing Quantitative Traits

A
  • Different segments of the genome can be marked with molecular variants that can be analysed individually with respect to the phenotype of interest.
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3
Q

Monkey Flowers:

A
  • A type of flower with two species (of interest), with a region of sympatry, but no hybrids. One (lewisii) is designed for bees, the other (cardinalis) for hummingbirds.
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4
Q

Mimulus lewisii

A
  • Designed for bees
  • Pink
  • Yellow nectar guides
  • Vide corolla
  • Small volume of concentrated nectar
  • Short anther and stigmas
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5
Q

Mimulus cardinalis:

A
  • Designed for humming birds
  • Red
  • No nectar guides
  • Tubular corolla
  • High volume of nectar
  • Long anther and stigmas
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6
Q

Are the hybrids viable and fertile when artificially bred?!

A

Yes!

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7
Q

Step 1 in QTL mapping

A
  • Start with two lines that differ for the trait of interest
  • Best if the genetic variation between the lines is maximised
  • Best if the genegtic variation within lines is minimized
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8
Q

Step 2 in QTL mapping

A
  • Cross the two lines
  • Allow recombination
  • The more progeny, the more recombination, the greater the mapping resolution, can use F2 (3 progeny types, 2 homozygotes and 1 heterozygotes) or backcross (2 progeny types, 2 homozygotes)
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9
Q

Step 3 in QTL mapping:

A
  • Choose molecular markers (RFLPs, AFLPs, SNPs)
  • Marker density must e informative to the cross
  • Consider cost, labour, marker density and co-dominance
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10
Q

Step 4 in QTL mapping:

A
  • Score the F2 or backcross progeny and parents for molecular markers and create a linkage map
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11
Q

Step 5 in QTL mapping:

A
  • Score the F2 or backcross progeny and parents for trait of interest
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12
Q

Step 6 in QTL mapping:

A
  • For each marker (or interval) perform a statistical test for association with phenotype
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13
Q

Input (dataset) for a QTL experiment:

3

A
  • Trait values of individuals related by a known cross (eg. backcross of F2)
  • Marker states for each individual
  • Markers arranged in a map
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14
Q

T-tests:

A
  • Compares the mean between two populations, high and low, given their variance
  • Generally the L marker will have a lower trait value than the H marker
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15
Q

When the P value approaches 1:

A
  • The means are about the same
  • There is no significant difference
  • There is no gene of interest there contributing to the effect
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16
Q

P < 0.05:

A

There is a significant difference

17
Q

When there is a QTL above the threshold line:

A
  • There is a significant association between the marker and the trait of interest, and they are correlated with our trait of interest
  • They are our QTLs
18
Q

Flaws with QTL mapping:

A
  • The markers are unlikely to be the causal variants, but they will be linked to the causal variants
19
Q

Thoday’s method for mapping QTLs within a chromosome:

4

A
  • Use mapping stock with two closely linked phenotypic markers (a and b) on a chromosome containing the QTLs.
  • Cross this to a stock that is WT at the two phenotypic markers (A and B) and differs from marker stock in the trait of interest (eg. has a higher number of bristles).
  • The QTLS can be outside the phenotypic markers to between them
  • We want to know which it is
20
Q

If the QTL is outside the markers:

2

A
  • Progeny with the parental marker types will exhibit intermediate trait means with high variance (ignoring double cross overs)
  • Recombinant type will be either high or low with little variance
21
Q

We can quantify where the QTL is in the interval:

A
  • How many individuals are in the Ab low class? (5)
  • How many individuals are in the aB high class? (3)
  • There are 8 individuals that result from a recombination in region 1.
  • Do the same for region two.. 12 from recombination in region two.
  • The distance between our gene of interest and marker A is closer than that of our gene and marker B.
22
Q

We can quantify the effect size is as well..

A
  • From a graph comparing trait score and marker class calculate the mean of each class and average it.

Evolutionary Genetics and Genomics (34 decks)

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