Genetics

Topics: Genotype, Gene, Phenotype Pages: 4 (272 words) Published: March 8, 2014
a) Round Yellow (RRYY) and wrinkled green (rryy)

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
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