8) Linkage and recomb

Linkage mapping

• Used to generate chromosome maps in model organisms and humans

SNP

• Single nucleotide polymorphism

• Type of variation most commonly used in mapping experiments’

• So, a SNP is a location in DNA where individuals can differ by one DNA letter.

• They are used as markers fro known location son chromosomes

What does it mean for a mutation to be linked to a SNP

• The closer the mutation and SNP are to each other on a chromosome, the less likely recombination is to separate them.

Parental frequency

• r = {number of recombinants \ total offspring

• (1-r)/2

Recombinant frequency

• r = {number of recombinants \ total offspring

• r/2

How do you evaluate whether the outcome of a cross suggests linkage vs. independent assortment?

• An odds ratio compares the likelihood of observing an outcome under

  two different hypotheses.

• OR = P(outcome A given L) / P(outcome A given no L)

  • A = observed # recombinant and parental offspring

  • L - linkage

lod

• Log of the odds ration

• Used to measure genetic linkage mapping

  • lod > 0

  • lod > 3

lod meanings

• lod > 0

  • the P(observing outcome if linked) > P (Observing outcome in unlinked)

• lod > 3

  • Means its most likely linked

  • Commonly used bu not universal theashold for statistical signification

  • shows that the likelihood of observing these outcomes under linkage hypothesis is 1000x greater than observing if unliked

lod score calculation (ex 1)

• Start with a hypothesis about linkage

  • 10 map units ( r = .1)

  • Probability gamete is recombinant = .10 (or 0.05 for each recombinant class)

  • Probability gamete is parental = .90 (0.45 for each parental class)

• Our alternative hypothesis is that they are not linked:

  • Probability gamete is recombinant = .5 (or 0.25 for each recombinant class)

  • Probability gamete is parental = .5 (or 0.25 for each parental class)

Lod scor caluclation (ex 1)

A lod score can be calculated for any value of r

• Continuing with our example (4 of 13 offspring are recombinant)

  • r = 0.31

  • p(each recomb class) = 0.3/2 = 0.15

  • P(each parental class) 1- 0.3 /2 = 0.35

  • lod90.3) = log10 ( (0.15)^4 (0.35)^9 / (0.25)^13) )

    • = 0.43

  • lod = 0.43 = The value of r with the maximum lod score is the best estimate of r

  • 10^{LOD}=10^{0.43} = 2.69

  • LOD = 0.43 = 2.69× greater likelihood of observing this pedigree with r=0.3 than with r=0.5

So which is better r = 0.3 or r = 0.1

• We observed 4 out of 13 offspring were recombinat

  • 4/13 = 3.08

  • that 32% recobminant offspring

• SO a assumbed recobination freq of 20 is much closer to the observed proportion than an assumed frequency of 10%

FInding an allele that causes malignant melanoma in a family

• For the 15 individuals lavelled “L”

  • Each individuals melanoma phenotype recorded

  • Each individuals genotype at about 10,000 SNPs distributed across all 23 chromomes was determined

  • Maximum lod score was calculated fro each SNPand melanma

How to find the causative variation once a linked SNP is found

• Sequence the genomic region surrounding the causative SNP for variation

• This may include multiple genes

• Use previous knowledge of genes and how variation might affect them to form a hypothesis about which sequence variation is most likely causative

• Then, test hypothesis

Genetic linkae mapping using SNP limitations

• Few offspring per corss = limited number of individuals

• Imperfect pedigree info

  • DNA sequences can only be gather from living generations

  • No controlled crosses

Genetic linkage mapping using SNP pros

• Most effective for Mendelian monogenetic traits, or traits where most of the phenotypic variation is cause by allele of a single gene

• Can be impossible to detect statisticaly significant linkages, esecially for complex multigeneic traits