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# Genetic Probabilities - PowerPoint PPT Presentation

Genetic Probabilities. Learning Objectives. By the end of this class you should understand: The purpose and nature of dihybrid crosses How to calculate the probability that an unaffected person may be a carrier for a disorder What a rare-allele assumption is for

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## PowerPoint Slideshow about ' Genetic Probabilities' - steven-meyers

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Presentation Transcript

By the end of this class you should understand:

• The purpose and nature of dihybrid crosses

• How to calculate the probability that an unaffected person may be a carrier for a disorder

• What a rare-allele assumption is for

• Identify examples of chromosomal linkage

• A probability is a number that represents the number of outcomes that fit a certain definition

• All probabilities are between 0 and 1

• 0 = never happens, 1 = always happens

• Probabilities may be derived from Punnett Squares

• Number of particular outcomes divided by total number of outcomes

• When two effects do not interact, they are said to be independent

• The assortment of chromosomes during meiosis is independent and follow's Mendel's Law of Independent Assortment

• Two genes on the same chromosome are not independent

• If an individual has a family history of a recessive allele, that individual may be a carrier even if they are healthy

• If we make the rare allele assumption we can assume it has not been introduced by any other pairings

• Probabilities can be influenced by additional knowledge

• If someone's genotype is unknown, you may use each genotype to make a separate Punnett Square

• Assume “Aa” and “AA” for that individual

• Draw separate Punnett Squares for each crossing

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• If an unknown person has no family history of the disorder, you may instead assume they are homozygous dominant

• This is the rare-allele assumption

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• Individual #1 has brown eyes

• Individual 1's father has brown eyes, as does his entire family

• Individual 1's mother has light blue eyes

• Individual #2 has brown eyes

• Individual #2's parents both had brown eyes

• Individual #2's maternal grandfather had blue eyes

• Using the rare allele assumption, what is the probability that #1 x #2 can produce blue eyes?

• A dihybrid cross should have the same probabilities as each individual cross separately

• Independence

• Chromosomal linkage violates the independence pattern

• Closely resembles a single Punnett Square for both alleles

• Why not exact?

• Imagine an X chromosome with both hemophilia and red-green colorblindness

• Use this X chromosome as X' in the following cross:

• XY x X'X

• With crossing over in Meiosis Prophase I, the X woman's X chromosomes trade some genes

• May then become XY x XHXC for hemophilia and colorblindness separately

• Perform a dihybrid cross: AaX'Y x AaX'X

• Assume X' is a recessive defect. What is the probability that a boy will have the disorder? What is the probability that a girl will have the disorder?

• What is the probability that a child will have both?

• The answers were obtainable by using individual Punnett Squares!

• The rules may get more complicated:

• Perform a AaZz x AaZz cross with the following phenotype rules:

• If zz, individual is black

• If has a dominant Z, individual phenotype depends on A:

• If AA, individual is red

• If Aa, individual is brown

• If aa, individual dies at birth

• Will see more polygenic traits in later chapters

• Draw the pedigree for the following information:

• Mother healthy, father afflicted, four children

• 1st child: Boy, healthy, married, two healthy sons

• 2nd child: Girl, healthy, married, one afflicted son, one healthy daughter, one healthy son

• 3rd child: Girl, healthy, married, one afflicted son, two healthy daughters

• 4th child: Boy, healthy, married, one healthy daughter

• What is the pattern of inheritance?

• Everyone choose one of the five patterns and draw your own pedigree chart!

• Be sure it has at least 3 generations and there should be at least five crosses of interest

• Trade with a partner and analyze which pattern(s) it matches!