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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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Mendel’s Interpretations Reinterpretted Mendel’s InterpretationOur Interpretation Traits do not blend but are determined by unchangeable units Genes proteins traits x

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

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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

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

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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

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

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Different traits assort independently Law of Independent Assortment

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

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

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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

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

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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-

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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?

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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?

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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-

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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?

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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:

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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

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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

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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

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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

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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

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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

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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?

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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

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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

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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

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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 -

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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

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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 -

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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 -

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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 -

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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 -

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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?

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

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