Analysis without grandparents
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In family 2, the father carries both A and B, but it is not known whether he inherited them from the same parent or from different parents. Hence, if there is linkage, he may be of constitution AB/ab or Ab/aB. The mother is ab/ab. Of the five children, two are AB/ab, two ab/ab, and one aB/ab. |
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Case I - assume father is AB/ab
then four children would be non-recombinant each with a probability of (1-f) and one would be recombinant with a probability of f.
The combined probability would then be f (1-f)4
Case II - Assume that the father is Ab/aB
then four children are recombinant, each with frequency
of f;
one is non-recombinant = (1-f)
The probability of the observed sequence of 1 non-recombinant and 4 recombinants = (f4)(1-f).
P = [f(1-f)4 +(f4)(1-f)]/2
Text iGenetics by Peter J. Russell
This web site is provided for instruction in Botany and Zoology 342
by Kenneth G. Wilson,
Professor of Botany
Miami University
wilsonkg@muohio.edu