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Streamlining Dihybrid Cross Calculations: Fork Line Method

A simple way to predict genotype & phenotype ratios in dihybrid problems by separating traits independently. Learn the method with three key ratios. Master the technique with an example of PpYy x PpYy cross calculation step by step. Avoid the Punnet square and use the innovative approach for quicker results.

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Streamlining Dihybrid Cross Calculations: Fork Line Method

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  1. the fork line method

  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 • 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

  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): similarly, Yy x Yy will give: ¼ YY Pp x Pp 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): ¼ YY Yy x Yy will give: Pp x Pp will give: ½ Yy ¼ yy multiply fractions 1/16 PPYY ¼ YY ½ Yy 1/8 PPYY ¼ PP 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 1/16 PPYY ¼ YY ½ Yy 1/8 PPYy ¼ PP ¼ yy 1/16 PPyy 1/8 PpYY ¼ YY ½ Pp ½ Yy 1/4 PpYy ¼ yy 1/8 Ppyy 1/16 ppYY ¼ YY 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 16ths for consistency Pp x Pp will give: multiply fractions 1 1/16 PPYY ¼ YY 2 ½ Yy 1/8 PPYy ¼ PP ¼ yy 1/16 PPyy 1 1/8 PpYY 2 ¼ YY ½ Pp ½ Yy 1/4 PpYy 4 ¼ yy 1/8 Ppyy 2 1/16 ppYY 1 ¼ YY 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 16ths for consistency Pp x Pp will give: multiply fractions 1 1/16 PPYY ¼ YY 2 ½ Yy 1/8 PPYy ¼ PP ¼ yy 1/16 PPyy 1 1/8 PpYY 2 ¼ YY ½ Pp ½ Yy 1/4 PpYy 4 ¼ yy 1/8 Ppyy 2 1/16 ppYY 1 ¼ YY 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.

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