search for: homozygot

...)) [1] 1.542047 >DirichNormHWE(nvec=c(2,10,88), c(1,1))/DirichNormSat(nvec=c(2,10,88), c(1,1,1)) [1] 1.542047 Could you explain why the HWETriangBF2 is returining completely different values of Bayes Factor?? -- View this message in context: http://r.789695.n4.nabble.com/HWEBayes-swapping-the-homozygotes-genotype-frequencies-tp3886313p3886313.html Sent from the R help mailing list archive at Nabble.com.

...classes and methods for creating, representing, and manipulating genotypes (unordered allele pairs) and haplotypes (ordered allele pairs). Genotypes and haplotypes can be annotated with chromosome, locus, gene, and marker information. Utility functions compute genotype and allele frequencies, flag homozygotes or heterozygotes, flag allele carriers of certain alleles, count the number of a specific allele carried by an individual, extract one or both alleles, estimate and generate confidence intervals for measures of single-marker disequlibrium, and test for departure from Hardy-Weinberg equilibrium....

...classes and methods for creating, representing, and manipulating genotypes (unordered allele pairs) and haplotypes (ordered allele pairs). Genotypes and haplotypes can be annotated with chromosome, locus, gene, and marker information. Utility functions compute genotype and allele frequencies, flag homozygotes or heterozygotes, flag allele carriers of certain alleles, count the number of a specific allele carried by an individual, extract one or both alleles, estimate and generate confidence intervals for measures of single-marker disequlibrium, and test for departure from Hardy-Weinberg equilibrium....

...currently totally confused by the many R packages available to do such analysis. Here is my case: I've got a list of genes, and a number of case-control population pairs, and for each population and gene, the various genotypes that have been found. I've got both aggregate data (ex. gene1: homozygote wildtype: 201, heterozygote mutation carrier: 34, homozygote mutation carrier: 5) and per-gene data (i.e. for gene1 a list of e.g. "V/V", "V/I", "II" etc). The question asked is whether there is a difference in the mutation pattern between the case and the control gr...

...- matrix(0,16,2) result.df<-data.frame(result) result.df[,1] <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4) result.df[,2] <- c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4) names(result.df)[1]<-"L(A)a(i)" names(result.df)[2]<-"L(B)a(j)" result.df ### The loop written to find Double Homozygotes for (a in seq(1,(ncol(X)-3), by=2)){ for (b in seq(3,(ncol(X)-1), by=2)){ for (i in a){ j <- a+1 for (k in b){ l <- b+1 match.rows <- ((X [,i] == X [, j] ) & ( X [,k] == X [, l])) double_homo_i <- X [match.rows, i] double_homo_k <- X [match.rows, k]...

...- matrix(0,16,2) result.df<-data.frame(result) result.df[,1] <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4) result.df[,2] <- c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4) names(result.df)[1]<-"L(A)a(i)" names(result.df)[2]<-"L(B)a(j)" result.df ### The loop written to find Double Homozygotes for (a in seq(1,(ncol(X)-3), by=2)){ for (b in seq(3,(ncol(X)-1), by=2)){ for (i in a){ j <- a+1 for (k in b){ l <- b+1 match.rows <- ((X [,i] == X [, j] ) & ( X [,k] == X [, l])) double_homo_i <- X [match.rows, i] double_homo_k <- X [match.rows, k]...

Hi, I have a genotype data for both case and controls and would like to calculate the HW p-value. However, since the number of one genotype is 0, I got wired result. Would someone help me to figure it out? Or confirm it's right? Thanks a lot. ============ > library( "genetics" ) NOTE: THIS PACKAGE IS NOW OBSOLETE. The R-Genetics project has developed an set of enhanced