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GENETICS

GENETICS. Genetics. The study of heredity . Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breeding garden peas. P generation – true breeding parents F 1 generation – hybrid offspring of P generation F 2 generation – offspring of hybrids.

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GENETICS

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  1. GENETICS

  2. Genetics • The study of heredity. • Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breedinggarden peas.

  3. P generation – true breeding parents • F1 generation – hybrid offspring of P generation • F2 generation – offspring of hybrids

  4. 1.Alternative versions of genes called alleles • (genes – factors passed from parent to offspring) • Account for variations • Ex: white flower allele or purple flower allele

  5. 2. Inherit two alleles for each trait 3. Principle of Dominance If two alleles differ, one determines appearance, dominant allele; the other has no noticeable affect, recessive allele 4. Segregation – the two alleles separate during meiosis • Gametes get only1 allele

  6. eye color locus B = brown eyes eye color locus b = blue eyes Paternal Maternal Homologous Chromosomes This person would have brown eyes (Bb)

  7. B sperm B B Bb haploid (n) b b diploid (2n) b meiosis II meiosis I Meiosis - eye color

  8. Genetic Vocabulary • Homozygous - pair of identical alleles • Heterozyous - two different alleles for a gene • Phenotype –organism’s trait (physical appearance) • Genotype – genetic makeup

  9. Alleles homozygous dominant genotype PP phenotype - purple flowers homozygous recessive genotype pp phenotype – white flowers Heterozygous genotype Pp phenotype – purple flowers

  10. Punnett square and Probability • A Punnett square is used to show the possiblecombinations of gametesor probability of genotypes • Monohybrid Cross • tracks the inheritance of a single trait.

  11. P P p p Breed the P generation • purple (PP) vs. white (pp) flowers

  12. P P produces the F1 generation Pp Pp p Pp Pp p All Pp purple (heterozygous) purple (PP) vs. white (pp) flowers

  13. P p P p Breed the F1 generation • purple (Pp) vs. Purple (Pp) flower plants

  14. P p produces the F2 generation Pp PP P 1/4 (25%) = PP 1/2 (50%) = Pp 1/4 (25%) = pp Pp pp p 1:2:1 genotype 3:1 phenotype Purple (Pp) vs. purple (Pp) flower plants

  15. B b male gametes B Bb x Bb b female gametes Monohybrid Cross • Example: Cross between two heterozygotesfor brown eyes (Bb) BB = brown eyes Bb = brown eyes bb = blue eyes

  16. B b 1/4 = BB - brown eyed 1/2 = Bb - brown eyed 1/4 = bb - blue eyed BB Bb B Bb x Bb b Bb bb 1:2:1 genotype 3:1 phenotype Monohybrid Cross

  17. Dihybrid Cross • A breeding experiment that tracks the inheritance of two traits. • Mendel’s “principle of independent assortment” a. each pair of alleles segregates independently during gamete formation (metaphase I) b. formula: 2n (n = # of heterozygotes)

  18. Independent Assortment • Question: How many gametes will be produced for the following allele arrangements? • Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq

  19. Answer: 1. RrYy: 2n = 22= 4 gametes RY Ry rY ry 2. AaBbCCDd: 2n = 23= 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26= 64 gametes

  20. RY Ry rY ry x RY Ry rY ry possible gametes produced Dihybrid Cross • Example: cross between round and yellow heterozygous pea seeds. R = round r = wrinkled Y = yellow y = green RrYy x RrYy

  21. RY Ry rY ry RY Ry rY ry Dihybrid Cross

  22. RY Ry rY ry Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio Dihybrid Cross

  23. bC b___ bc Test Cross • A mating between an individual ofunknown genotypeand a homozygous recessive individual. • Example:bbC__ x bbcc BB = brown eyes Bb = brown eyes bb = blue eyes CC = curly hair Cc = curly hair cc = straight hair

  24. bC b___ C bC b___ c bc bbCc bbCc or bc bbCc bbcc Test Cross • Possible results:

  25. R R W W Incomplete Dominance • F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. • Example:snapdragons (flower) • red (RR) x white (WW) RR = red flower WW= white flower

  26. R R produces the F1 generation RW RW W W RW RW All RW = pink (heterozygous pink) Incomplete Dominance

  27. Codominance • Two alleles are expressed (multiple alleles) in heterozygous individuals. • Example: blood 1. type A = IAIA or IAi 2. type B = IBIB or IBi 3. type AB = IAIB 4. type O = ii

  28. IB IB IAIB IAIB IA 1/2 = IAIB 1/2 = IBi i IBi IBi Codominance • Example: homozygous male B (IBIB) x heterozygous female A (IAi)

  29. IA IB IAi IBi i 1/2 = IAi 1/2 = IBi i IAi IBi Codominance • Example:male O (ii) x female AB (IAIB)

  30. Codominance • Question: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents. • boy - type O (ii) X girl - type AB (IAIB)

  31. IA i IAIB IB i ii Codominance • Answer: Parents: genotypes = IAi and IBi phenotypes = A and B

  32. Sex-linked Traits • Traits (genes) located on the sex chromosomes • Example:fruit flies (red-eyed male) X (white-eyed female)

  33. fruit fly eye color XX chromosome - female Xy chromosome - male Sex-linked Traits Sex Chromosomes

  34. XR y Xr Xr Sex-linked Traits • Example: fruit flies (red-eyed male) X (white-eyed female) • Remember: the Y chromosome in males does not carry traits. RR = red eyed Rr = red eyed rr = white eyed Xy = male XX = female

  35. XR y XR Xr Xr y Xr Xr XR Xr Xr y Sex-linked Traits 1/2 red eyed and female 1/2 white eyed and male

  36. Population Genetics • The study of genetic changes in populations. • The science of microevolutionary changes in populations. • Hardy-Weinberg equilibrium: the principle that shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic makeup of a population. • Hardy-Wienberg equation: 1 = p2 + 2pq + q2

  37. Question: • How do we get this equation? Answer: “Square” 1 = p + q  12 = (p + q)2  1 = p2 + 2pq + q2

  38. Hardy-Wienberg equation • Five conditions are required for Hardy-Wienberg equilibrium. 1. large population 2. isolated population 3. no net mutations 4. random mating 5. no natural selection

  39. Important • Need to remember the following: p2 = homozygous dominant 2pq = heterozygous q2 = homozygous recessive

  40. Question: • Iguanas with webbed feet (recessive trait) make up 4% of the population. What in the population is heterozygous and homozygousdominant.

  41. q2 = .04 2. then use 1 = p + q 3. for heterozygous use 2pq 4. For homozygous dominant use p2 Answer: 1. q2 = 4% or .04 q = .2 1 = p + .2 1 - .2 = p .8 = p 2(.8)(.2) = .32 or 32% .82 = .64 or 64%

  42. Hardy-Wienberg equation 1 = p2 + 2pq + q2 • 64% = p2 = homozygous dominant • 32% = 2pq = heterozygous • 04% = q2 = homozygous recessive • 100%

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