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Genetics: Complex Inheritance, Sex Linkage, X-Inactivation

Genetics: Complex Inheritance, Sex Linkage, X-Inactivation. AP Biology Unit 3. Incomplete Dominance. Heterozygous phenotype is a blend of the 2 homozygous phenotypes Ex. Red flower crossed with white flower  heterozygous flower is pink. Epistasis.

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Genetics: Complex Inheritance, Sex Linkage, X-Inactivation

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  1. Genetics:Complex Inheritance, Sex Linkage, X-Inactivation AP Biology Unit 3

  2. Incomplete Dominance • Heterozygous phenotype is a blend of the 2 homozygous phenotypes • Ex. Red flower crossed with white flower  heterozygous flower is pink

  3. Epistasis • When one gene product affects the expression of another gene. • B and b are fur color alleles • bb = brown fur • Fur color will only be expressed if the C gene is also present

  4. Polygenic Inheritance • Two or more genes work together to create a single phenotype • Eye color, skin color are good examples • Opposite is pleiotropy (where one gene affects several different phenotypes) ex. PKU

  5. Nature vs. Nurture • Virtually all human diseases have some genetic component

  6. Pedigrees • Used to trace the genotypes for a particular trait in a family • Can help determine the probability that future offspring will have a trait.

  7. Sample Pedigrees Widow’s Peak = Dominant Trait W = widow’s peak • What is the grandfather’s genotype? • Ww

  8. Sample Pedigrees • Attached earlobes = Recessive Trait F = unattached f = attached • What is the genotype of these grandparents? • Both Ff

  9. Sex Linkage • When a trait is carried on the X or Y chromosomes, it is called a sex-linked trait • Don’t confuse this with linked genes = when 2 genes are on the same chromosome XY XX X Y X XX XY

  10. Sex-linked genes and Punnett Squares • You have to include the X and Y chromosomes in the Punnett Square • Superscripts on the X and Y denote which allele is present • XA, Xa

  11. Red-Green Colorblindness • Gene that controls this (opsin gene) is on the X chromosome • Colorblindness is caused by a recessive allele (mutation in the opsin gene) • Who is more likely to be color blind– men or women? • Men: only 1 X chromosome – if they have the recessive allele they don’t have another X to make up for it.

  12. Sex-limited Trait • Trait whose expression depends on the sex of the individual • Not found on the X or Y chromosome- NOT the same thing as sex-linked trait. • Ex. Milk production in females, pattern baldness in males (triggered by hormones)

  13. X-inactivation in female mammals • In females, one of the two X chromosomes in each cell becomes inactive during embryonic development • Why would one X chromosome inactivate itself in females? • Cells of females and males would have same effective dose of genes on the X chromosome • Inactive X chromosomes are called Barr bodies

  14. Example: Calico (Tortoiseshell) cats • animation

  15. Example in humans • Anhidrotic dysplasia • X linked mutation prevents the development of sweat glands • A woman who is heterozygous will have patches of normal skin and patches of skin without sweat glands • Difficulty controlling body temperature

  16. Practice Problem #5 • A normal (not colorblind male) marries a woman who is a carrier for the colorblindness allele • What are the chances their son will be color blind? • What are the chances their daughter will be colorblind?

  17. Answer #5 • The man is XBY, the woman is XBXb. • 50% chance that son will be colorblind • 0% chance that daughter will be colorblind, but she could be a carrier.

  18. A B Practice Problem #6 • Is this trait on the X or Y chromosome? • Is it dominant or recessive? • If individual A marries an unaffected male, what are the % chances her children will have the trait? • If individual B marries an unaffected woman, what are the chances their children will have the trait?

  19. Answer #6 • Trait must be on the X chromosome (X*). only men would have it if on Y chromosome. • Dominant trait– determine the genotypes of males  look at their mother’s to help determine recessive or dominant. • Individual A must be X*X (since her mother was unaffected). 50% chance children will have trait (girls or boys). • Individual B must be X*Y. All daughters will have trait, none of sons will.

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