BEJAR, MALAIKA DE RAMOS, HAZEL ANNE PD2F

1. Bejar, Malaika
De ramos, Hazel Anne
PD2F

2.  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

3. for example, the following dihybrid cross:
PpYy x PpYy
normally to solve this we would
2. use FOIL for the gametes, then
3. assemble the Punnet square, then
4. 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):
1 homozyg x homozyg: PP x PP -----------> 100% PP
or pp x pp -–-----------------------------------------> 100% pp
2 heterozyg x homozyg: Pp x PP -----------> ½ Pp, ½ PP
or Pp x pp ---------------------------------------------> ½ Pp, ½ PP
3 heterozyg x heterozyg: Pp x Pp: ¼ PP; ½ Pp; ¼ pp

4. 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

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

6. PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
Yy x Yy will give: ¼ YY
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

7. 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

8. PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
convert all to
Pp x Pp will give:
16ths for
consistency
¼ YY
multiply fractions
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

9. PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
convert all to
Pp x Pp will give:
16ths for
consistency
¼ YY
multiply fractions
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
so – this gives you the same results as the Punnet square – but in some cases might be a faster cleaner way of doing it.