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# Gen

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BEJAR, MALAIKA
DE RAMOS, HAZEL ANNE
PD2F

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### Gen

1. 1. Bejar, MalaikaDe ramos, Hazel Anne PD2F
2. 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. 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% pp2 heterozyg x homozyg: Pp x PP -----------> ½ Pp, ½ PP or Pp x pp ---------------------------------------------> ½ Pp, ½ PP3 heterozyg x heterozyg: Pp x Pp: ¼ PP; ½ Pp; ¼ pp
4. 4. PpYy x PpYyso to solve this dihybrid, separate the two traits (since they sort independently):Pp x Pp will give: ¼ PP ½ Pp ¼ pp
5. 5. PpYy x PpYyso 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. 6. PpYy x PpYyso to solve this dihybrid, separate the two traits (since they sort independently): Yy x Yy will give: ¼ YYPp 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. 7. PpYy x PpYyso 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. 8. PpYy x PpYyso to solve this dihybrid, separate the two traits (since they sort independently): convert all toPp 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. 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 1so – this gives you the same results as the Punnet square – but in some cases might be a faster cleaner way of doing it.