The cross between them
RY
RY
RY
RY
ry
RrYy
RrYy
RrYy
RrYy ry RrYy
RrYy
RrYy
RrYy
ry
RrYy
RrYy
RrYy
RrYy ry RrYy
RrYy
RrYy
RrYy
So all the progeny are Round, Yellow with the genotype RrYy (F1 plants)
Now for the F2 generation,
RrYy is mated with itself (RrYy)
RY
Ry
rY ry RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYY
RrYy rrYY rrYy ry RrYy
Rryy
rrYy rryy Therefore, the genotypic ratio is: RRYY: RRYy: RRyy: RrYY: RrYy: rrYY:Rryy:rrYy:rryy = 1:2:1:2:4:1:2:2:1
And…
Phenotypic ratio = Round Yellow: Round Green: Wrinkled Yellow: Wrinkled Green = 9:3:3:1 b) Now the actual number of plants seen upon crossing RrYy with rryy is:
Round yellow - 445
Round green - 48
Wrinkled yellow - 52
Wrinkled green – 455
RY
Ry
rY ry ry
RrYy
Rryy rrYy rryy ry RrYy
Rryy
rrYy rryy ry
RrYy
Rryy rrYy rryy ry RrYy
Rryy
rrYy rryy Where as the ratio should ideally be 1:1:1:1. So as follows: - Parental types - Round Yellow (RrYy) and Wrinkled green (rryy) - Recombinant types - Round Green (Rryy) and Wrinkled Yellow (rrYy) c)Recombinant frequency = Number of recombinants / Total = 0.1 = 10% recombination d) Map distance = recombination frequency = 10 cM
e) R = 0.1 Then, the gamete frequencies are RY – ½ -1/2( r)
Ry – ½(r) rY - ½(r) ry – ½ - ½ (r) Therefore; RRYY - 1/4 - 1/2(r) + 1/4(r^2) = 0.2025
RRYy - 1/2(r) - 1/2(r^2) = 0.045
RRyy - 1/4(r^2) = 0.0025
RrYY - 1/2(r) - 1/2(r^2) = 0.045
RrYy - 1/2 - r + r^2 = 0.41
Rryy - 1/2(r) - 1/2(r^2) = 0.045 rrYY - 1/4(r^2) = 0.0025 rrYy - 1/2(r) - 1/2(r^2) = 0.045 rryy - 1/4 - 1/2(r) + 1/4(r^2) = 0.2025 Thus the overall modified phenotypic ratio will be: 202.5 : 45 : 2.5 : 410 : 45: 2.5: 45 : 202.5 = 81: 18: 1: 164: 18: 1 :18: 81