1 / 9

the fork line method

the fork line 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

penningtong
Download Presentation

the fork line method

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related