RNAseq reads covered 5,000 biallelic human autosomal SNPs (all are heterogeneous, assume they are not in LD). Simple R statements plot the distribution of the alternative allelic frequency under sequencing coverage 2x, 5x, 10x, 20x, respectively. Calculate the p-values with the hypothesis that there are no differential allelic expression for the…

binom.R

plot(x=c(0:2)/2, dbinom(c(0:2),2,0.5), pch=20, col="red", type ="b", xlim =c(0,1), ylim =c(0, 0.6), xlab = "AltFreq", ylab="Probability" )
lines (x=c(0:5)/5, dbinom(c(0:5),5,0.5), pch=20, col="brown",type ="b" )
lines (x=c(0:10)/10, dbinom(c(0:10),10,0.5), pch=20, col="green",type ="b")
lines (x=c(0:20)/20, dbinom(c(0:20),20,0.5), pch=20, col="blue",type ="b")
 
legend(0.8, 0.58, legend="2x", col="red",  pch=20,  lty = 1, bty="n" )
legend(0.8, 0.56, legend="5x", col="brown",  pch=20,  lty = 1, bty="n" )
legend(0.8, 0.54, legend="10x", col="green",  pch=20,  lty = 1, bty="n" )
legend(0.8, 0.52, legend="20x", col="blue",  pch=20,  lty = 1, bty="n" )
 
RefAlleleCount:AltAlleleCount = 1:4,  pvalue: 0.375:       binom.test(4,5, 0.5)$p.value     == binom.test(1,5, 0.5)$p.value
RefAlleleCount:AltAlleleCount = 2:8,  pvalue: 0.109375:    binom.test(8,10, 0.5)$p.value    == binom.test(2,10, 0.5)$p.value
RefAlleleCount:AltAlleleCount = 4:16, pvalue: 0.01181793:  binom.test(16,20, 0.5)$p.value   == binom.test(4,20, 0.5)$p.value