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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. Genetics. Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits.

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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. Genetics • Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits. 3. Dominant alleles (TT - tall pea plants) a. homozygous dominant 4. Recessive alleles (tt - dwarf pea plants) a. homozygous recessive 5. Heterozygous (Tt - tall pea plants)

  4. Phenotype • Outward appearance • Physical characteristics • Examples: 1. tall pea plant 2. dwarf pea plant

  5. Genotype • Arrangement of genes that produces the phenotype • Example: 1. tall pea plant TT = tall (homozygous dominant) 2. dwarf pea plant tt = dwarf (homozygous recessive) 3. tall pea plant Tt = tall (heterozygous)

  6. Punnett square • A Punnett square is used to show the possible combinations of gametes.

  7. T T t t Breed the P generation • tall (TT) vs. dwarf (tt) pea plants

  8. T T produces the F1 generation Tt Tt t Tt Tt t All Tt = tall (heterozygous tall) tall (TT) vs. dwarf (tt) pea plants

  9. T t T t Breed the F1 generation • tall (Tt) vs. tall (Tt) pea plants

  10. T t produces the F2 generation Tt TT T 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt Tt tt t 1:2:1 genotype 3:1 phenotype tall (Tt) vs. tall (Tt) pea plants

  11. Monohybrid Cross • A breeding experiment that tracks the inheritance of a single trait. • Mendel’s “principle of segregation” a. pairs of genes separate during gamete formation (meiosis). b. the fusion of gametes at fertilization pairs genes once again.

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

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

  14. 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

  15. 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

  16. 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)

  17. 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

  18. 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

  19. 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

  20. RY Ry rY ry RY Ry rY ry Dihybrid Cross

  21. 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

  22. 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

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

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

  25. R R produces the F1 generation Rr Rr r r Rr Rr All Rr = pink (heterozygous pink) Incomplete Dominance

  26. 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

  27. 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)

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

  29. 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)

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

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

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

  33. 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

  34. 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

  35. 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

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

  37. 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

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

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

  40. 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%

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

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