User:Hussein Alasadi/Notebook/stephens/2013/10/03: Difference between revisions
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<math>X_j</math> = frequency of "1" allele at SNP j in the pool (i.e. the true frequency of the 1 allele in the pool) | <math>X_j</math> = frequency of "1" allele at SNP j in the pool (i.e. the true frequency of the 1 allele in the pool) | ||
*'''Data:''' <math> (n_j^0, n_j^1) </math> = number of "0", "1" alleles at SNP j (<math> n_j = n_j^0 + n_j^1 </math>) | *'''Data:''' | ||
<math> (n_j^0, n_j^1) </math> = number of "0", "1" alleles at SNP j (<math> n_j = n_j^0 + n_j^1 </math>) | |||
Revision as of 15:56, 13 October 2013
 analyzing pooled sequenced data with selection
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Notes from MeetingConsider a single lineage for now. [math]\displaystyle{ X_j }[/math] = frequency of "1" allele at SNP j in the pool (i.e. the true frequency of the 1 allele in the pool)
[math]\displaystyle{ (n_j^0, n_j^1) }[/math] = number of "0", "1" alleles at SNP j ([math]\displaystyle{ n_j = n_j^0 + n_j^1 }[/math])
[math]\displaystyle{ n_j^1 }[/math] ~ [math]\displaystyle{ Bin(n_j, X_j) \approx N(n_jX_j, n_jX_j(1-X_j)) }[/math] Normal approximation to binomial [math]\displaystyle{ \frac{n_j^1}{n_j} \approx N(X_j, \frac{X_j(1-X_j)}{n_j}) }[/math] The variance of this distribution results from error due to binomial sampling. To simplify, we just plug in [math]\displaystyle{ \hat{X_j} = \frac{n_j^1}{n_j} }[/math] for [math]\displaystyle{ X_j }[/math] [math]\displaystyle{ \implies \frac{n_j^1}{n_j} | X_j \approx N(X_j, \frac{\hat{X_j}(1-\hat{X_j})}{n_j}) }[/math]
[math]\displaystyle{ f_{i,k,j} = }[/math] frequency of reference allele in group i, replicate and SNP j. [math]\displaystyle{ \bar{f_{i,k}} = }[/math] vector of frequencies Without loss of generality, we assume that the putative selected site is site [math]\displaystyle{ j = 1 }[/math]
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