When examined each F2 yellow family separately, Mendel found
- 2/3 of the F2 yellows give 1/2 yellow, 1/2 green
- 1/3 of F2 yellows gave all yellow progeny
Mendel' s explanation
- Genetic information exists as discrete units occurring in pairs
- YY is the genotype of the pure Yellow line
- yy the genotype of pure Green line
- Y dominant to y (y is recessive to Y)
- YY, Yy (denoted by Y-) = Yellow
- yy = green
- Here, the discrete units segregate out in the F2
Additional test of Mendel
Can we account for the observed 2:1 ratio
for F2 Yellows X pure green?
- Mendel's answer: 2/3 of F2 Yellows are Yy, 1/3 are YY.
- Why? Use conditional probabilities
- Pr(Yellow F2 is Yy) = Pr(Yy in F2) / Pr(F2 is Yellow) = (1/2)/(3/4) =
2/3
- Pr(Yellow F2 is YY) = Pr(YY in F2) / Pr(F2 is Yellow) = (1/4)/(3/4) =
1/3
- Hence in the F2 Yellow X green cross,
- 2/3 of F2 are Yy x yy crosses, giving 1/2 yellow, 1/2 green
- 1/3 are YY x yy crosses, giving all yellow
- Hence, total yellow = (1/2)(2/3) + 1/3 = 2/3, total green =
(1/2)(2/3) = 1/3
Genetic Notation
- Character : Seed color
- Yellow phenotype is dominant to green phenotype
- Genotypes
- YY homozygous dominant
- Yy heterozygous
- yy homozygous recessive
- Alleles
- Y is dominant
- y is recessive
Dihybrid crosses
- Crossing pure-breeding lines differing in two characters
- Example: for Seed shape
- RR, Rr = round
- rr = wrinkled
- Cross yellow, wrinkled pure line with round green
Mendel observed Independent Assortment
- How are these ratios computed?
- Pr(R-) = Pr(RR) + Pr(Rr) = 1-Pr(rr) = 3/4
- Same for Pr(Y-) = 3/4
- Pr(R-,Y-) = Pr(R-)*Pr(Y-)
- = 3/4 * 3/4 = 9/16
- Assumes R and Y are independent, i.e., Independent assortment
- Pr(R-,yy) = 3/4 * 1/4 = 3/16
- Pr(rr, Y-) = 1/4 * 3/4 = 3/16
- Pr(rr, yy) = 1/4 * 1/4 = 1/16
Mendel cheated (or was very lucky)
The seven characters he used to show independent assortment were all
on different chromosomes. Hence they are not linked.