RankMembers: A function to calculate the likelihood ratio

Description Usage Arguments Value Author(s) Examples

View source: R/RankMembers.R

Description

This function calculates the likelihood ratio for an allele causing a disease asssuming that the allele is extremely rare so that all family members who have the allele got it directly from a common ancestor in the pedigree.

Usage

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RankMembers(ped, affected.vector, gene="BRCA1", legend.location="topleft",
  legend.radius=0.1)

Arguments

ped

A dataframe of a pedigree containing id, momid, dadid, age(Current age or age affected), y.born(year born), female(a logical where 1 is female and 0 is male), and genotype(0 is non-carrier, 1 carrier, and 2 unknown) in any order but with these specific column names.

affected.vector

A vector giving the affection status of the individual. 0 is unaffected, 1 is affected, and 2 is unknown.

gene

A text string (e.g. BRCA1, BRCA2, or MLH1) giving the gene name for analysis.

legend.location

A string indicating the placement of the legend. These can be "topleft", "topright", "bottomleft", "bottomright".

legend.radius

A real number indicating the size of the legend.

Value

likelihood.ratio

The resultant likelihood ratio

separating.meioses

The number of meioses separating all individuals known to have the genotype including those that are are indirectly known to have the genotype.

number.genotypes.found

The number of permissible genotypes found for the pedigree.

Author(s)

John Michael O. Ranola and Brian H. Shirts

Examples

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  ## Not run: 
    #Load all the data included in the CoSeg package.
    data(BRCA1Frequencies.df, package="CoSeg")
    data(BRCA2Frequencies.df, package="CoSeg")
    data(MLH1Frequencies.df, package="CoSeg")
    data(USDemographics.df, package="CoSeg")
    data(ChinaDemographics.df, package="CoSeg")

    #summaries of all the data
    str(BRCA1Frequencies.df)
    str(BRCA2Frequencies.df)
    str(MLH1Frequencies.df)
    str(USDemographics.df)
    str(ChinaDemographics.df)

    #Make a tree with no affection status, g=4 generations above, gdown=2 generations below,
    #seed.age=50, and demographics.df=NULL which defaults to USDemographics.df.
    tree1=MakeTree()

    #Make a tree using Chinese demographics instead.
    tree2=MakeTree(demographics.df=ChinaDemographics.df)

    #Add affection statust to tree2 using BRCA1Frequencies.df which gives the BRCA1
    #penetrance function
    tree1a=AddAffectedToTree(tree.f=tree1,frequencies.df=BRCA1Frequencies.df)

    #make a tree with affection status (same as running MakeTree() and then AddAffectedToTree())
    tree3=MakeAffectedTrees(n=1,g=2,gdown=2,frequencies.df=MLH1Frequencies.df)
    #tree4=MakeAffectedTrees(n=1,g=2,gdown=2,frequencies.df=BRCA2Frequencies.df)


    #Depending on the size of the pedigree generated, probands (defined here as members of the
    #pedigree who are carriers of the genotype with the disease) may not always be present in
    #the pedigree.  To alleviate this problem in this example we manually generate a pedigree.
    #Note that this is from the Mohammadi paper where the Likelihood method originates from.
    ped=data.frame(degree=c(3,2,2,3,3,1,1,2,2,3), momid=c(3,NA,7,3,3,NA,NA,7,NA,8),
      dadid=c(2,NA,6,2,2,NA,NA,6,NA,9), id=1:10, age=c(45,60,50,31,41,68,65,55,62,43),
      female=c(1,0,1,0,1,0,1,1,0,1), y.born=0, dead=0, geno=2, famid=1, bBRCA1.d=0, oBRCA1.d=0,
      bBRCA1.aoo=NA, oBRCA1.aoo=NA, proband=0)
    ped$y.born=2010-ped$age
    ped$geno[c(1,3)]=1
    ped$bBRCA1.d[c(1,3)]=1
    ped$bBRCA1.aoo[1]=45
    ped$bBRCA1.aoo[3]=50
    ped$proband[1]=1

    ped=ped[c(6,7,2,3,8,9,1,4,5,10),]

    #Calculate the likelihood ratio
    CalculateLikelihoodRatio(ped=ped, affected.vector={ped$bBRCA1.d|ped$oBRCA1.d}, gene="BRCA1")

    #Plot the pedigree
    PlotPedigree(ped, affected.vector={ped$bBRCA1.d|ped$oBRCA1.d})

    #Rank and plot the members of the pedigree with unknown genotypes
    RankMembers(ped=ped, affected.vector={ped$bBRCA1.d|ped$oBRCA1.d}, gene="BRCA1")
  
## End(Not run)

CoSeg documentation built on May 29, 2017, 6:10 p.m.