You are examining 3 genes A, B, & D. You cross 2 true breeding lines in your par
ID: 63271 • Letter: Y
Question
You are examining 3 genes A, B, & D. You cross 2 true breeding lines in your parental generation to produce a heterozygous F1. You cross an F1 female to a test male to produce an F2 generation. The results of this cross are as follows:
A) What was the genotype of the F1 female? B) Are all 3 genes linked? If any of these genes are linked, what is the map distance between them? C) What were the genotypes of the original “P” generation male & female?
Wildtype 307 b 60 d 305 bd 65 ab 311 a 69 abd 309 ad 74Explanation / Answer
So in F1 generation all offspring were +++ type and crossed to abd male
c.
This can also be represented as
+++ X abd +b+ 60
+++ 307 +bd 65
++d 305 a++ 69
Ab+ 311 a+d 74
Abd 309
Since the result does not follow 1:1:1:1 ratio of test cross, then genes can be linked. But we have to calculate map distance first.
Let say gene a and b are linked then map distance between a and b = 311+60+69+305/1500*100
=745/1500 = 0.4966*100= 49.66CM
Map distance between b and d
=305+311+74+65/1500*100 = 50.33
Since the distance between a and b is almost equal to 50 and so in b and d. It shows that any of the genes are not linked.
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