Mendelian genetics powerpoint massengale

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  • Fig. 5.co
  • 07/24/12
  • 07/24/12
  • Mendelian genetics powerpoint massengale

    1. 1. Mendelelian Genetics 1
    2. 2. Gregor Mendel (1822-1884) Responsible for the Laws governing Inheritance of Traits 2
    3. 3. Gregor Johann MendelAustrian monkStudied theinheritance oftraits in pea plantsDeveloped the lawsof inheritanceMendels work wasnot recognized untilthe turn of the20th century 3
    4. 4. Gregor Johann MendelBetween 1856 and1863, Mendelcultivated andtested some 28,000pea plantsHe found that theplants offspringretained traits ofthe parentsCalled the “Fatherof Genetics" 4
    5. 5. Site ofGregorMendel’sexperimentalgarden in theCzechRepublic 5
    6. 6. Particulate InheritanceMendel stated thatphysical traits areinherited as “particles”Mendel did not knowthat the “particles”were actuallyChromosomes & DNA 6
    7. 7. Genetic Terminology Trait - any characteristic that can be passed from parent to offspring Heredity - passing of traits from parent to offspring Genetics - study of heredity 7
    8. 8. Types of Genetic Crosses Monohybrid cross - cross involving a single trait e.g. flower color Dihybrid cross - cross involving two traits e.g. flower color & plant height 8
    9. 9. Punnett SquareUsed to helpsolve geneticsproblems 9
    10. 10. 10
    11. 11. Designer “Genes” Alleles - two forms of a gene (dominant & recessive) Dominant - stronger of two genes expressed in the hybrid; represented by a capital letter (R) Recessive - gene that shows up less often in a cross; represented by a lowercase letter (r) 11
    12. 12. More Terminology Genotype - gene combination for a trait (e.g. RR, Rr, rr) Phenotype - the physical feature resulting from a genotype (e.g. red, white) 12
    13. 13. Genotype & Phenotype in FlowersGenotype of alleles:R = red flowerr = yellow flowerAll genes occur in pairs, so 2alleles affect a characteristicPossible combinations are:Genotypes RR Rr rrPhenotypes RED RED YELLOW 13
    14. 14. Genotypes Homozygous genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure  Heterozygous genotype - gene combination of one dominant & one recessive allele    (e.g. Rr); also called hybrid 14
    15. 15. Genes and EnvironmentDetermine Characteristics 15
    16. 16. Mendel’s Pea Plant Experiments 16
    17. 17. Why peas, Pisum sativum?Can be grown in asmall areaProduce lots ofoffspringProduce pure plantswhen allowed toself-pollinateseveral generationsCan be artificiallycross-pollinated 17
    18. 18. Reproduction in Flowering PlantsPollen contains sperm Produced by the stamenOvary contains eggs Found inside the flower Pollen carries sperm to the eggs for fertilization Self-fertilization can occur in the same flower Cross-fertilization can occur between flowers 18
    19. 19. Mendel’s Experimental MethodsMendel hand-pollinatedflowers using a paintbrush He could snip the stamens to prevent self-pollination Covered each flower with a cloth bagHe traced traits throughthe several generations 19
    20. 20. How Mendel BeganMendelproducedpurestrains byallowing theplants toself-pollinatefor severalgenerations 20
    21. 21. Eight Pea Plant TraitsSeed shape --- Round (R) or Wrinkled (r)Seed Color ---- Yellow (Y) or  Green (y )Pod Shape --- Smooth (S) or wrinkled (s )Pod Color ---  Green (G) or Yellow (g)Seed Coat Color ---Gray (G) or White (g)Flower position---Axial (A) or Terminal (a)Plant Height --- Tall (T) or Short (t)Flower color --- Purple (P) or white (p ) 21
    22. 22. 22
    23. 23. 23
    24. 24. Mendel’s Experimental Results 24
    25. 25. Did the observed ratio match the theoretical ratio?The theoretical or expected ratio ofplants producing round or wrinkled seedsis 3 round :1 wrinkledMendel’s observed ratio was 2.96:1The discrepancy is due to statisticalerrorThe larger the sample the more nearlythe results approximate to thetheoretical ratio 25
    26. 26. Generation “Gap”Parental P1 Generation = the parental generation in a breeding experiment.F1 generation = the first-generation offspring in a breeding experiment. (1st filial generation) From breeding individuals from the P1 generationF2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation) From breeding individuals from the F1 generation 26
    27. 27. Following the GenerationsCross 2 Results Cross 2 Hybrids Pure in all get Plants Hybrids 3 Tall & 1 ShortTT x tt Tt TT, Tt, tt 27
    28. 28. Monohybrid Crosses 28
    29. 29. P1 Monohybrid CrossTrait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Wrinkled seeds RR x rr Genotype: Rr r r Phenotype: Round Phenotype R Rr Rr Genotypic Ratio: All alike R Rr Rr Phenotypic Ratio: All alike 29
    30. 30. P1 Monohybrid Cross Review Homozygous dominant x Homozygous recessive Offspring all Heterozygous (hybrids) Offspring called F1 generation Genotypic & Phenotypic ratio is ALL ALIKE 30
    31. 31. F1 Monohybrid CrossTrait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Round seeds Rr x Rr Genotype: RR, Rr, rr R r Phenotype: Round & Phenotype RR Rr wrinkled R G.Ratio: 1:2:1 r Rr rr P.Ratio: 3:1 31
    32. 32. F1 Monohybrid Cross Review Heterozygous x heterozygous Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr Offspring called F2 generation Genotypic ratio is 1:2:1 Phenotypic Ratio is 3:1 32
    33. 33. What Do the Peas Look Like? 33
    34. 34. …And Now the Test CrossMendel then crossed a pure & a hybrid from his F2 generationThis is known as an F2 or test crossThere are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid 34
    35. 35. F2 Monohybrid Cross (1st)Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Round seeds RR x Rr Genotype: RR, Rr R r Phenotype: Round Phenotype R RR Rr Genotypic Ratio: 1:1 R RR Rr Phenotypic Ratio: All alike 35
    36. 36. F2 Monohybrid Cross (2nd)Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Wrinkled seeds x Round seeds rr x Rr R r Genotype: Rr, rr Phenotype: Round & Phenotype r Rr rr Wrinkled G. Ratio: 1:1 r Rr rr P.Ratio: 1:1 36
    37. 37. F2 Monohybrid Cross Review Homozygous x heterozygous(hybrid) Offspring: 50% Homozygous RR or rr 50% Heterozygous Rr Phenotypic Ratio is 1:1 Called Test Cross because the offspring have SAME genotype as parents 37
    38. 38. Practice Your CrossesWork the P1, F1, and bothF2 Crosses for each of the other Seven Pea Plant Traits 38
    39. 39. Mendel’s Laws 39
    40. 40. Results of Monohybrid CrossesInheritable factors or genes are responsible for all heritable characteristicsPhenotype is based on GenotypeEach trait is based on two genes, one from the mother and the other from the fatherTrue-breeding individuals are homozygous ( both alleles) are the same 40
    41. 41. Law of DominanceIn a cross of parents that arepure for contrasting traits, onlyone form of the trait will appear inthe next generation.All the offspring will beheterozygous and express only thedominant trait.RR x rr yields all Rr (round seeds) 41
    42. 42. Law of Dominance 42
    43. 43. Law of SegregationDuring the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other.Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring. 43
    44. 44. Applying the Law of Segregation 44
    45. 45. Law of Independent AssortmentAlleles for different traits are distributed to sex cells (& offspring) independently of one another.This law can be illustrated using dihybrid crosses. 45
    46. 46. Dihybrid CrossA breeding experiment that tracks the inheritance of two traits.Mendel’s “Law of Independent Assortment”a. Each pair of alleles segregates independently during gamete formationb. Formula: 2n (n = # of heterozygotes) 46
    47. 47. Question: How many gametes will be producedfor the following allele arrangements? Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq 47
    48. 48. Answer:1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64gametes 48
    49. 49. Dihybrid CrossTraits: Seed shape & Seed colorAlleles: R round r wrinkled Y yellow y green RrYy x RrYy RY Ry rY ry RY Ry rY ry All possible gamete combinations 49
    50. 50. Dihybrid Cross RY Ry rY ryRYRyrYry 50
    51. 51. Dihybrid Cross RY Ry rY ryRY RRYY Round/Yellow: 9 RRYy RrYY RrYyRy RRYy Round/green: 3 RRyy RrYy Rryy wrinkled/Yellow: 3rY RrYY RrYy rrYY rrYy wrinkled/green: 1ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio 51
    52. 52. Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 52
    53. 53. Test CrossA mating between an individual of unknown genotype and a homozygous recessive individual.Example: bbC__ x bbcc BB = brown eyes Bb = brown eyes bb = blue eyes bC b___ bc CC = curly hair Cc = curly hair cc = straight hair 53
    54. 54. Test CrossPossible results: bC b___ C bC b___ cbc bbCc bbCc or bc bbCc bbcc 54
    55. 55. Summary of Mendel’s laws PARENT LAW OFFSPRING CROSSDOMINANCE TT x tt 100% Tt tall x short tall Tt x Tt 75% tallSEGREGATION tall x tall 25% short 9/16 round seeds & green pods RrGg x RrGgINDEPENDENT 3/16 round seeds & yellow round & green podsASSORTMENT x 3/16 wrinkled seeds & green pods round & green 1/16 wrinkled seeds & yellow pods 55
    56. 56. Incomplete Dominance and Codominance 56
    57. 57. Incomplete DominanceF1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties.Example: snapdragons (flower)red (RR) x white (rr) r r RR = red flower R rr = white flower R 57
    58. 58. Incomplete Dominance r r produces theR Rr Rr F1 generationR Rr Rr All Rr = pink (heterozygous pink) 58
    59. 59. Incomplete Dominance 59
    60. 60. CodominanceTwo alleles are expressed (multiple alleles) in heterozygous individuals.Example: blood type 1. type A = IAIA or IAi 2. type B = IBIB or IBi 3. type AB = IAIB 4. type O = ii 60
    61. 61. Codominance ProblemExample: homozygous male Type B (IBIB) x heterozygous female Type A (IAi) IA i IB IAIB IBi 1/2 = IAIB 1/2 = IBi IB IAIB IBi 61
    62. 62. Another Codominance Problem• Example: male Type O (ii) x female type AB (IAIB) IA IB i IAi IBi 1/2 = IAi 1/2 = IBi i IAi IBi 62
    63. 63. CodominanceQuestion: 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) 63
    64. 64. CodominanceAnswer: IA iIB IAIB Parents: genotypes = IAi and IBi phenotypes = A and Bi ii 64
    65. 65. Sex-linked TraitsTraits (genes) located on the sex chromosomesSex chromosomes are X and YXX genotype for femalesXY genotype for malesMany sex-linked traits carried on X chromosome 65
    66. 66. Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes fruit fly eye colorXX chromosome - female Xy chromosome - male 66
    67. 67. Sex-linked Trait ProblemExample: Eye color in fruit flies (red-eyed male) x (white-eyed female) XRY x XrXrRemember: the Y chromosome in males does not carry traits. Xr XrRR = red eyedRr = red eyed XRrr = white eyedXY = maleXX = female Y 67
    68. 68. Sex-linked Trait Solution: Xr Xr 50% red eyedXR X X X X R r R r female 50% white eyedY Xr Y Xr Y male 68
    69. 69. Female Carriers 69
    70. 70. Genetic Practice Problems 70
    71. 71. Breed the P1 generationtall (TT) x dwarf (tt) pea plants t t T T 71
    72. 72. Solution:tall (TT) vs. dwarf (tt) pea plants t t Tt Tt produces theT F1 generationT Tt Tt All Tt = tall (heterozygous tall) 72
    73. 73. Breed the F1 generationtall (Tt) vs. tall (Tt) pea plants T t T t 73
    74. 74. Solution: tall (Tt) x tall (Tt) pea plants T t produces the TT Tt F2 generationT 1/4 (25%) = TT Tt tt 1/2 (50%) = Ttt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype 74
    75. 75. 75

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