1. Predicting Genotypes and Phenotypes

2. Punnett Squares
-used to predict genotypes and phenotypes of offspring based on the parents genotype
-parental generation= parents genotypes
-F1 generation= 1st generation offspring
-F2 generation= 2nd generation offspring

3. Punnett Squares
Monohybrid crosses- 1 heterozygous trait is crossed (Bb X Bb)
Dihybrid crosses- 2 heterozygous traits are crossed (BbSs X BbSs)

4. Punnett Squares (parent generation)
Eye Color mom BB
B B b
dad bb
Offspring 1
Offspring 2
Offspring 3
Offspring 4 Bb Bb Bb Bb
Genotypic Ration=? Phenotypic Ratio?

5. Punnett Squares (F1) Eye Color mom Bb B b B dad Bb b BB Bb Bb bb
Offspring 1
Offspring 2
Offspring 3
Offspring 4 BB Bb Bb bb
Genotypic Ration=? Phenotypic Ratio?

6. Punnett Squares Eye Color and Height mom BbSs BS Bs bS bs BS Dad
Genotypic Ration=? Phenotypic Ratio?

7. Laws of Probability Used to predict genotype or phenotype
-Multiplication rule-to determine the probability of different events happening in combination, multiply the probability of each event.
-Addition rule-to determine the probability of the same events happening add the probability of each event.
See examples

8. Forkline Method
this is just another way to be able to predict genotype and phenotype ratios in dihybrid problems
this way you don’t have to write the box
but it does require you to know the basic ratios that arise from monohybrids
based on the idea that: in a dihybrid, the two traits sort INDEPENDENTLY of one another
i.e. what happens with one trait is completely unrelated to what happens with the other trait

9. for example, the following dihybrid cross:
PpYy x PpYy
normally to solve this we would
use FOIL for the gametes, then
assemble the Punnet square, then
count up the genotypic and phenotypic ratios.
However, we can make use of two simple concepts:
the traits (flower color and seed color) sort out independently of each other
there are essentially only three different ratios that can result in a monohybrid cross (it doesn’t matter what the traits are; I’ve used P here, but it could be anything):
homozyg x homozyg: PP x PP > 100% PP
or pp x pp -– > 100% pp 1
heterozyg x homozyg: Pp x PP > ½ Pp ½ PP
or Pp x pp > ½ Pp ½ PP 2 3
heterozyg x heterozyg: Pp x Pp: ¼ PP ½ Pp ¼ pp

10. PpYy x PpYy
so to solve this dihybrid, separate the two traits
(since they sort independently):
Pp x Pp will give:
¼ PP
½ Pp
¼ pp

11. PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
similarly,
Yy x Yy will give:
Pp x Pp will give:
¼ YY
½ Yy
¼ yy
¼ PP
½ Pp
¼ pp

12. PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
¼ YY
Yy x Yy will give:
Pp x Pp will give:
½ Yy
¼ yy
multiply fractions
¼ YY
1/16 PPYY
¼ PP
½ Yy
1/8 PPYY
¼ yy
1/16 PPYY
½ Pp
¼ pp

13. PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
Pp x Pp will give:
multiply fractions
¼ YY
1/16 PPYY
¼ PP
½ Yy
1/8 PPYy
¼ yy
1/16 PPyy
¼ YY
1/8 PpYY
½ Pp
½ Yy
1/4 PpYy
¼ yy
1/8 Ppyy
¼ YY
1/16 ppYY
¼ pp
½ Yy
1/8 ppYy
¼ yy
1/16 ppyy

14. convert all to 16ths for consistency
PpYy x PpYy
so to solve this dihybrid, separate the two traits
(since they sort independently):
convert all to 16ths for consistency
Pp x Pp will give:
multiply fractions
¼ YY
1/16 PPYY 1
¼ PP
½ Yy
1/8 PPYy 2
¼ yy
1/16 PPyy 1
¼ YY
1/8 PpYY 2
½ Pp
½ Yy
1/4 PpYy 4
¼ yy
1/8 Ppyy 2
¼ YY
1/16 ppYY 1
¼ pp
½ Yy
1/8 ppYy 2
¼ yy
1/16 ppyy 1