3-point test crosses in genetics x-linked?

we are doing a cross in which an F1 female and F1 male are crossed to produce.


p + bt -9 - double cross over


+ w + - 6 - double cross over


p w Bt- 90 - single cross over


+ + + - 95 - single cross over


+ + bt - 424 - parental types


p w + - 376 - parental types





these are only male offspring. The question asked you to construct a linkage map (all traits are recessive and X-linked)


you are missing the other 2 single crossovers ( p + + and + w bt)





p is the center on the map...





map units from white eyes (located at 1.5 MU) to p is 20 MU





The question is if you don't have the other S.C.O how do you determine the distance from p to bt???

3-point test crosses in genetics x-linked?
Assuming no interference, then p(double) should be equal to p(single 1) x p(single2), so this would mean that p(single2) =p(double)/p(single 1). However, we don't have a direct way of measuring these since it requires knowing the total number of offspring. This will be 1000 (observed) + missing singles(N). As a rough estimate it will be .015/.185=.081. More precisely, the value is N/(100+N)= (15/(100+N))/(185/(1000+N))





However: if the distance w--p is given as 18.5 (20-1.5) and the calculated value 185/1000 gives the same, then it would seem that there are *no* other singles, implying a map distance of zero. But, it that is true, then you would not expect any double cross overs either.





Bottom line: there is something odd about the cross, or about the presented data. it can't all be right.
Reply:I believe you must ask your extra y chromosome.