Question

S. no	Gamete types	Number of progenies
1.	ABC	349
2.	Abc	114
3.	abC	124
4.	AbC	5
5.	aBc	4
6.	aBC	116
7.	ABc	128
8.	abc	360

1. What is the name of this test cross?
2. How will you construct gene mapping from the above-given data?
3. Find out the correct order of genes.

Answer 1

1. Three-point test cross.
2. 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 (Abe, abC, AbC, aBc, ABc) are recombinant (R) types.
   
   

   		Recombinant for the alleles
   Gametes	No. of progenies	A and B	A and C	B and C
   ABC	349	-	-	-
   Abc	114	R	R	-
   abC	124	-	R	R
   AbC	5	R	-	R
   aBc	4	R	-	R
   aBC	116	R	R	-
   ABc	128	-	R	R
   abc	360	-	-	-

   
   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 recombinant)"/"(Totaprogenies)" xx100` `(114+5+4+116)/1200xx100 = 19.91%`
   Recombinant frequency for the loci B and C
   The parental form are Be and bC and the recombinant are Be and bC.
   Recombinant frequency of alleles Be and bC = `(4+128+124+5)/1200 xx100 +21.75%`
   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
   
3. The correct gene order is ABC/abc.