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A corn plant known to be heterozygous at three loci is testcrossed. The progeny phenotypes and numbers are as follows:+ + + 455a b c 470+ b c 35a + + 33+ + c 37a b + 35+ b + 460a + c 475Total 2,000Give the gene arrangement, linkage relationships, and map distances.

Answer :

abidhussain7972

Answer:

The progeny phenotypes and numbers are as follows:

+ + + 455-------- parental

a b c 470-------- parental

+ b c 35----------- recombinant

a + + 33---------- recombinant

+ + c 37------------ recombinant

a b + 35--------------- recombinant

+ b + 460---------- parental

a + c 475----------- parental

Total 2,000

+ + + 455-------- parental

a b c 470-------- parental

+ b + 460---------- parental

a + c 475----------- parental

It is two point testcross

So the parental ++ (455+460); ac (470+475)------------------++/ac

Gene order is ------------- ++/ac

+ b c 35----------- recombinant

a + + 33---------- recombinant

+ + c 37------------ recombinant

a b + 35--------------- recombinant

sum of recombinants--------------140

Frequency of recombintion= recombinants/total x100

Map distances between a-c = (140/2000) x 100 = 7 mu

a------------------7 mu -----------------c

Explanation:

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