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.

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 recombinants140

Frequency of recombintion= recombinants/total x100

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

a7 mu c

Explanation:

i think the answer is gold im not entirely sure

zxndjdjxjjxik no f cnnxn k is s dksoozkdkxjxjxjxj