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

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Presentation on theme: "Topics for Today Review of Dihybrid Cross - Relationship with molecular/chromosomal behavior Deductions from Pedigrees Calculation of Genetic Probabilities."— 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 Interpretations Reinterpretted Mendel’s InterpretationOur Interpretation Traits do not blend but are determined by unchangeable units Genes proteins traits x

3 Mendel’s Interpretations Reinterpretted Mendel’s InterpretationOur Interpretation Each trait is determined by two units Two homologous chromosomes

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

5 Mendel’s Interpretations Reinterpretted Mendel’s InterpretationOur Interpretation One character form is recessive to or dominant over another P > p p > P

6 Mendel’s Interpretations Reinterpretted Mendel’s InterpretationOur Interpretation The two character forms carried by a heterozygote are passed to progeny with equal likelihood Law of Segregation

7 Mendel’s Interpretations Reinterpretted Mendel’s InterpretationOur Interpretation Different traits assort independently Law of Independent Assortment

8 Different traits assort independently Law of Independent Assortment

9

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 Pedigrees How is the trait inherited? Try dominant A-? A-

12 Deductions from Pedigrees How is the trait inherited? Any problem? aa aa aa aa aa aa aa aa aa aa aa aa aa aa A- aa A- aa aa

13 Deductions from Pedigrees How is the trait inherited? Try recessive aa

14 Deductions from Pedigrees How is the trait inherited? Can we get more? A- A- A- A- A- A- A- A- A- A- A- A- A- A- aa A- aa A- A-

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

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

17 Deductions from Pedigrees How is the trait inherited? A- AA Aa A- Aa AA A- A- A- A- Aa Aa AA A- aa A- aa A-

18 Genetic Counseling Will our children be normal ? ? ? Make the problem concrete 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- A- A- A- A- A- A- AA A- A- A- AA A- aa Child will be aa if: III 5 is Aa AND III 6 is Aa AND III 5 a AND III 6 a Parse the problem (start simple) Probability:

20 Genetic Counseling A- A- A- A- A- A- A- AA A- A- A- AA A- aa Child will be aa if: III 5 is Aa AND III 6 is Aa AND III 5 a AND III 6 a Solve each segment Probability:1/2 II 2 is Aa AND II 2 a 1/2 1

21 Genetic Counseling A- A- A- A- A- A- A- AA A- A- A- AA A- aa Child will be aa if: III 5 is Aa AND III 6 is Aa AND III 5 a AND III 6 a Solve each segment Probability:1/2

22 Genetic Counseling A- A- A- A- A- A- A- AA A- A- A- AA A- aa Child will be aa if: III 5 is Aa AND III 6 is Aa AND III 5 a AND III 6 a Put parts together Probability:1/2 Add?… union… mutually exclusive… more possibilitiesMultiply?… intersection … independent… fewer possibilities

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

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

25 Genetic Counseling A- A- A- A- A- A- A- AA A- A- A- AA A- aa Child will be aa if: III 5 is Aa AND III 6 is Aa AND III 5 a AND III 6 a Put parts together Probability:1/2 Add?… union… mutually exclusive… more possibilitiesMultiply?… intersection … independent… fewer possibilities x x x 1/16

26 Example illustrating Rule of Complementation Suppose there are two genes (A, B) that are required for dark hair P(A - OR B - ) = P(A - ) + P(B - ) Is possession of A - and possession of B - mutually exclusive? 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: Is Rule of Addition valid here?

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 P(not A - ) = 1 - P(A - ) probability of not possessing defective allele of gene A not 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 P(not B - ) = 1 - P(B - ) probability of not possessing defective allele of gene B not B -

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

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

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

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

37 Union of possibilities P(A-) = = 0.9


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