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They aren't random. You know exactly what they are. Just write out the punnet square. For a triheterozygote, there are 8 possible alleles, and male fly of course only has one kind of gamete.
Here are the 8 possible gametes for the female, the phenotype we'd expect the offspring to have, and the numbers you observed. (put in in excel, so the spacing is right)If all the genes were unlinked, each of those 8 gametes should be equally likely, so each of the 8 offspring types should be equally likely. But we don't see that, hence linkage.
BME bme bmE BMe bMe Bme bME BmE
Barr min, ebony min Barr, ebony ebony Barr, min, ebony wt Barr, min
117 115 111 101 35 31 29 26
Notice how the 4 most common phenotypes have either Barr eyes or min wings. Or, notice that the four most common phenotypes have BM or bm, but that Bm and bM mice are rare. You can see this if you ignore the ebony phenotype, and just count up how many mice fall into each category below:
barr 218
barr, min 57
min 226
wt 64
That means that there's linkage, and that the BM is on one chromosome, and the bm is on the other chromosome in the female parent. If you make those lists as above for the ebony traits as paired with the two phentoypes, you see that there's no linkage, ebony mice are just as likely to be min or Barr as not
barr 143
barr, ebony 132
ebony 150
wt 140
min 137
min, ebony 146
ebony 136
wt 146
So the Barr allele and the min gene are on the same chromosome, and the female has the B allele on the same chromosome as the M allele. The ebony gene is on another chromosome
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