Similar Question
"These are the results of a three point test cross in Drosophila for the three loci ABC: aCB = 167, ACb = 22, AcB = 145, ACB = 432, aCb = 122, acB = 15, Acb = 155, acb = 400. What is the map distance to 4 significant figures between the A and B loci? Hint: Establish gene order first."
The best way to solve these problems is to develop a systematic approach. First, determine which of the the genotypes are the parental gentoypes. The genotypes found most frequently are the parental genotypes that is acb=400 and ACB = 432
Next we need to determine the order of the genes. Once we have determined the parental genotypes, we use that information along with the information obtained from the double-crossover.
The double-crossover gametes are always in the lowest frequency. ACb=22 and acB=15. Now double crossover places the middle allele from one sister chromatid to the other. Which in this case has happened with B and hence B is the middle one. Hence clearly the order is ABC/abc.
Now for map distances we would need the total of aCB = 167, ACb = 22, AcB = 145, ACB = 432, aCb = 122, acB = 15, Acb = 155, acb = 400.
Which comes out to be 1458. For distance between A and B we need to have the ones where Single-crossover between genes A and B occurs and the double cross overs i.e. 155,167 ,15 and 22.
To calculate the distance between A and B sum them up divide by 1458 and multiply by 100. A--B = 24.62cM
And distance between B and C can be obtained by
((122+145+15+22)/1458)*100= 20.85cM.
So clearly you can have a map A-----------B-----------C
24.62cm 20.85cM
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