In: Biology
groucho 518
rough 471
groucho, rough 6
wild-type 5
1000
A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).
S0, P1- groucho (gro gro ro+ ro+) X rough (ro ro gro+ gro+)
the F1 progenies will be= gro+ gro ro+ ro
A test cross is when F1 progenies is crossed with the individual have a homozygous recessive allele for all genes.
Therefore,
gro+ gro ro+ ro X gro gro ro ro (test cross)
The gametes of the F1 hybrids are - gro+ ro, gro ro+ (parental types) and gro + ro +, gro ro (recombinant types)
gametes of the homozygous recessive plants are - gro ro
With this cross, the number of offsprings are following-
groucho (gro gro ro+ ro) 518 (parental type 1)
rough (gro+ gro ro ro) 471 (parental type 2)
groucho, rough (gro gro ro ro) 6 (recombinant type 1)
wild-type (gro+ gro ro+ ro) 5 (recombinant type 2)
The frequency of the recombinant types is equal to the map distance between two genes
So, (6+5)/1000 = 0.011 = 1.1 %
a. so the distance between two genes is 1.1 cM
b. See the following figures
C. If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would be
Wild : groucho : rough : groucho, rough = 9:3:3:1
Wild = 562.5
groucho = 187.5
rough = 187.5
grouch, rough = 62.5
(* the number of offspring can not be fractioned. It's only shown for the calculation purpose)