(i) Three point test cross.
(ii) Construction of gene map :
To construct the gene map, the recombinant frequency (RF) of the alleles has to be calculated.
From the given data it is clear that ABC and abc are parental (P) types and the others (Abc, abC , AbC, aBc, ABc) are recombinant (R) type.
Lets analyse the loci of two alleles at a time starting with A and B. Since the genes AB and ab are parental type, the recombinants will be Ab and aB.
Therefore
Recombinant frequency of alleles Ab and aB = `("No. of recombinants")/("Total progenies")xx100`
`=(114+5+4+116)/(1200)xx100=19.91%` Recombinant frequency for the loci A and C
The parental form are AC and ac and the recombinants are Ac and aC
Recombinant frequency of alleles Ac and aC `=(114+128+124+116)/(1200)xx100=40.16%` Recombinant frequency for the loci B and C
The parental form are Bc and bC and the recombinant are Bc and bC.
Recombinant frequency of alleles Bc and bC = `(4+128+124+5)/1200xx100=21.75%`
Since the Since the recombinant frequency of the alleles A and C shown highest frequency , they must be the farthest apart and alleles B must lie in between A and C. So the gene map can be constructed as follows
(iii) The correct gene order is ABC/abc.
