1 / 37

Topics for Today

Topics for Today. Review of Dihybrid Cross - Relationship with molecular/chromosomal behavior Deductions from Pedigrees Calculation of Genetic Probabilities. Mendel’s Interpretations Reinterpretted. Mendel’s Interpretation Our Interpretation.

ardice
Download Presentation

Topics for Today

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. Topics for Today • Review of Dihybrid Cross- Relationship with molecular/chromosomal behavior • Deductions from Pedigrees • Calculation of Genetic Probabilities

  2. Mendel’s InterpretationsReinterpretted Mendel’s Interpretation Our Interpretation Traits do not blend but are determined by unchangeable units Genes proteins traits x

  3. Mendel’s InterpretationsReinterpretted Mendel’s Interpretation Our Interpretation Each trait is determined by two units Two homologous chromosomes

  4. Mendel’s InterpretationsReinterpretted Mendel’s Interpretation Our Interpretation A a The two units may or may not be identical Genes come in different forms, alleles, which make different protein

  5. Mendel’s InterpretationsReinterpretted Mendel’s Interpretation Our Interpretation One character form is recessive to or dominant over another P > p p > P

  6. Mendel’s InterpretationsReinterpretted Mendel’s Interpretation Our Interpretation The two character forms carried by a heterozygote are passed to progeny with equal likelihood Law of Segregation

  7. Mendel’s InterpretationsReinterpretted Mendel’s Interpretation Our Interpretation Different traits assort independently Law of Independent Assortment

  8. Different traits assort independentlyLaw of Independent Assortment

  9. Different traits assort independentlyLaw of Independent Assortment

  10. Deductions from Pedigrees • Pedigree with ephemeral trait (Fig. 2) • Pedigrees with other kinds of traits (next week) • Genetic counseling (later today)

  11. Deductions from PedigreesHow is the trait inherited? A- A- A-? A-? Try dominant

  12. Deductions from PedigreesHow is the trait inherited? aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa A- aa A- aa Any problem?

  13. Deductions from PedigreesHow is the trait inherited? aa aa Try recessive

  14. Deductions from PedigreesHow is the trait inherited? A- A- A- A- A- A- A- A- A- A- A- A- A- A- A- A- A- aa A- aa A- Can we get more?

  15. Deductions from PedigreesHow is the trait inherited? Which one gave a? Which one gave a? A- A- A- A- A- A- A- A- A- A- A- A- A- A- A- Aa Aa A- A- aa A- aa A- Can we get more?

  16. Deductions from PedigreesHow is the trait inherited? Which one gave a? Which one gave a? A- A- What about outsiders? What about outsiders? A- Aa A- Aa A- A- A- A- A- Aa Aa A- A- aa A- aa A- Can we get more?

  17. Deductions from PedigreesHow is the trait inherited? A- A- AA Aa A- Aa AA A- A- A- A- Aa AaAA A- aa A- aa A-

  18. Genetic Counseling Make the problem concrete Will our children be normal ? ? ? What’s the probability that a child of III.5 x III.6 will have CS? What’s the probability that a child of III.5 x III.6 will be aa?

  19. Genetic Counseling A- aa Parse the problem (start simple) AA A- A- A- AA A- A- A- A- A- A- A- Child will be aa if: AND III5a AND III6a III5 is Aa AND III6 is Aa Probability:

  20. Genetic Counseling A- aa Solve each segment AA A- A- A- AA A- A- A- A- A- A- A- Child will be aa if: AND III5a AND III6a III5 is Aa AND III6 is Aa Probability: II2 is Aa AND II2a 1/2 1/2 1 1/2

  21. Genetic Counseling A- aa Solve each segment AA A- A- A- AA A- A- A- A- A- A- A- Child will be aa if: AND III5a AND III6a III5 is Aa AND III6 is Aa Probability: 1/2 1/2 1/2 1/2

  22. Genetic Counseling A- aa Put partstogether AA A- A- A- AA A- A- A- A- A- A- A- Child will be aa if: AND III5a AND III6a III5 is Aa AND III6 is Aa Probability: 1/2 1/2 1/2 1/2 Multiply? Add? … union… mutually exclusive… more possibilities … intersection … independent… fewer possibilities

  23. Union of possibilities Probability that progeny of Aa x Aa has A phenotype Gets A from female OR gets A from male Rule of additionunionmutually exclusive Gets aA OR AA OR Aa P(A-) = 1/4 + 1/4 + 1/4 = 3/4

  24. Intersection of possibilities Probability that progeny of Aa x Aa has a phenotype Gets a from female AND gets a from male Rule of multiplicationintersectionindependent Gets a from female AND gets a from male P(aa) = 1/2 x 1/2 = 1/4

  25. Genetic Counseling A- aa Put partstogether AA A- A- A- AA A- A- A- A- A- A- A- Child will be aa if: AND III5a AND III6a III5 is Aa AND III6 is Aa x x x Probability: 1/2 1/2 1/2 1/2 Multiply? Add? … intersection … independent… fewer possibilities … union… mutually exclusive… more possibilities 1/16

  26. Example illustrating Rule of Complementation • Suppose there are two genes (A, B) that are required for dark hair • A defect in any one of them will produce light hair • What is the probability that a person will have light hair? Make problem concrete: Light hair if A-OR B- Parse problem: P(A-OR B-) = P(A-) + P(B-) Is Rule of Addition valid here? Is possession of A- and possession of B-mutually exclusive?

  27. How to Calculate P(A- OR B-)Probability of light hair P(A-) P(B-) P(A-) + P(B-)? Not mutually exclusive.P(A-B-) added twice

  28. How to Calculate P(A- OR B-)Probability of light hair P(A-) + P(B-)? Not mutually exclusive.P(A-B-) added twice P(A-) x P(B-)? Gives intersection, not union

  29. How to Calculate P(A- OR B-)Probability of light hair P(A-) P(B-) P(A-): probability of possessing defective allele of gene A

  30. How to Calculate P(A- OR B-)Probability of light hair not A- P(not A-) = 1 - P(A-) probability of not possessing defective allele of gene A

  31. How to Calculate P(A- OR B-)Probability of light hair P(A-) P(B-) P(B-): probability of possessing defective allele of gene B

  32. How to Calculate P(A- OR B-)Probability of light hair not B- P(not B-) = 1 - P(B-) probability of not possessing defective allele of gene B

  33. How to Calculate P(A- OR B-)Probability of light hair not A- P(not A-) = 1 - P(A-) probability of not possessing defective allele of gene A

  34. How to Calculate P(A- OR B-)Probability of light hair not A-ANDnot B- P(not A- and not B-) = [1 - P(A-)] x [1 - P(B-)] probability of not possessing either defective allele

  35. How to Calculate P(A- OR B-)Probability of light hair A-OR B- not A-ANDnot B- P(A- or B-) = 1 - [1 - P(A-)] x [1 - P(B-)] probability of possessing either defective allele

  36. How likely to get hemophilia? • There are five known alleles for the clotting factor protein Factor VIII (H1-, H2-, H3-, H4-, H5-). • Three of them (H3-, H4-, H5-) cause hemophilia. H3- hemophilia H4- hemophilia H5- hemophilia H4+ ??? • What is the probability that a person will have one of the three defective alleles and thus get hemophilia? P(H3-, H4-, OR H5-) = P(H3-) + P(H4-) + P(H5-) Rule of addition Union? … but is possession of the alleles mutually exclusive?

  37. Union of possibilities P(A-) = 0.4 + 0.4 + 0.1 = 0.9

More Related