the fork line method

this way you don’t have to write the box 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

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): 1 homozyg x homozyg: PP x PP -----------> 100% PP or pp x pp -–-----------------------------------------> 100% pp 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

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

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

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

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

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

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 so – this gives you the same results as the Punnet square – but in some cases might be a faster cleaner way of doing it.