2. 1 Mendelelian Genetics copyright cmassengale Question: How Are Traits Passed From Parents To Offspring?

3. 2 Genetic Terminology  Genetics - study of heredity  Heredity - passing of traits from parent to offspring  Trait - any characteristic that can be passed from parent to offspring copyright cmassengale

4. Background Blending Theory: 1. A child’s traits are an average mixture of the parents’ traits. 2. The parents’ contributions cannot be passed on to future generations.

5. 4 Gregor Mendel (1822-1884) An Austrian monk… The “father of genetics” The “father of genetics” copyright cmassengale

6. 5 Gregor Johann Mendel Between 1856 and 1863, Mendel cultivated and tested some 28,000 pea plants copyright cmassengale

7. Hypothesis (?) Traits from both parents are passed to each offspring. These traits combine in different ways in each offspring, which can pass the traits on to their offspring These traits combine in different ways in each offspring, which can pass the traits on to their offspring.

8. 7 Mendel’s Pea Plant Experiments copyright cmassengale

9. 8 Why peas (Pisum sativum)?  They are easy to grow  They produce lots of offspring  Can be artificially- pollinated copyright cmassengale

10. 9 Reproduction in Flowering Plants Pollen contains sperm Produced by the stamen Ovary contains eggs Found inside the flower Sperm fertilizes the egg -> seed Self-pollination within one flower Cross-fertilization between flowers copyright cmassengale

11. 10 Mendel’s Experimental Methods Mendel hand-pollinated flowers using a paintbrush He traced traits through the several generations copyright cmassengale

12. 11 1. Mendel produced pure strains via self-pollinaton copyright cmassengale

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15. 14 Mendel’s Experimental Results copyright cmassengale

16. 15 Eight Pea Plant Traits Seed 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 ) copyright cmassengale

17. 16 Generations Parental P 1 Generation F 1 generation (1st filial generation) From breeding individuals from the P 1 generation F 2 generation (2nd filial generation) the second-generation offspring in a breeding experiment. From breeding individuals from the F 1 generation copyright cmassengale

18. 17 Following the Generations Cross 2 Pure Plants Results in all tall plants Cross 2 Tall Plants: 3 Tall & 1 Short copyright cmassengale

19. 18 Did the observed ratio match the theoretical ratio? The theoretical (expected) ratio of plants producing round or wrinkled seeds: 3 round :1 wrinkled (for example) Mendel’s observed ratio was 2.96:1* * Sampling bias… copyright cmassengale

20. 19 Conclusion: “Parents pass on “factors*” to their offspring”  *Alleles - Different forms of a gene  Dominant - stronger of two alleles expressed in the hybrid; capital letter (R)  Recessive - allele that shows up less often in a cross; lowercase letter (r) copyright cmassengale

21. 20 Genotypes Lead To Phenotypes  Genotype - gene combination for a trait (e.g. RR, Rr, rr)  Phenotype - the physical feature resulting from a genotype (e.g. red, white) copyright cmassengale

22. 21 Genotype & Phenotype in Flowers* Genotype of alleles: R = red flower r = yellow flower All genes occur in pairs, so 2 alleles affect a characteristic Possible combinations are: GenotypesRR Rrrr PhenotypesRED RED YELLOW copyright cmassengale

23. 22 Genotypes  Homozygous genotype - “pure” 2 dominant or 2 recessive genes 2 dominant or 2 recessive genes RR or rr RR or rr  Heterozygous genotype - “hybrid” one dominant & one recessive allele Rr copyright cmassengale

24. 23 Mendel’s Laws copyright cmassengale

25. 24 Law of Segregation During the gamete formation, the two alleles separate from each other.* Alleles recombine at fertilization, producing the offspring’s genotype Alleles recombine at fertilization, producing the offspring’s genotype. copyright cmassengale

26. 25 Law of Independent Assortment Law of Independent Assortment Alleles for different traits are distributed to sex cells (& offspring) independently of one another. This law will be illustrated using dihybrid crosses. copyright cmassengale

27. 26 Law of Dominance Homozygous dominant x homozygous recessive --> All offspring are heterozygous. Dominant traits are expressed in heterozygotes. RR x rr yields all Rr (round seeds) copyright cmassengale

28. 27 Genes and Environment Determine Characteristics copyright cmassengale

29. 28 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 copyright cmassengale

30. 29 Monohybrid Crosses copyright cmassengale

31. 30 Trait: Seed Shape Alleles: R – Roundr – Wrinkled Cross: Round seeds x Wrinkled seeds RR x rr P 1 Monohybrid Cross R R rr Rr Genotype:Rr Genotype: Rr PhenotypeRound Phenotype: Round Genotypic Ratio:All alike Genotypic Ratio: All alike Phenotypic Ratio: All alike copyright cmassengale

32. 31 P 1 Monohybrid Cross Review  Homozygous dominant x Homozygous recessive  Offspring all Heterozygous (hybrids)  Offspring called F 1 generation  Genotypic & Phenotypic ratio is ALL ALIKE copyright cmassengale

33. 32 Trait: Seed Shape Alleles: R – Roundr – Wrinkled Cross: Round seeds x Round seeds Rr x Rr F 1 Monohybrid Cross R r rR RR rrRr Genotype:RR, Rr, rr Genotype: RR, Rr, rr PhenotypeRound & wrinkled Phenotype: Round & wrinkled G.Ratio:1:2:1 G.Ratio: 1:2:1 P.Ratio: 3:1 copyright cmassengale

34. 33 F 1 Monohybrid Cross Review  Heterozygous x heterozygous  Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr  Offspring called F 2 generation  Genotypic ratio is 1:2:1  Phenotypic Ratio is 3:1 copyright cmassengale

35. 34 …And Now the Test Cross Mendel then crossed a pure & a hybrid from his F 2 generation This is known as an F 2 or test cross There are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid copyright cmassengale

36. 35 Trait: Seed Shape Alleles: R – Roundr – Wrinkled Cross: Round seeds x Round seeds RR x Rr F 2 Monohybrid Cross (1 st ) R R rR RR RrRR Rr Genotype:RR, Rr Genotype: RR, Rr PhenotypeRound Phenotype: Round Genotypic Ratio:1:1 Genotypic Ratio: 1:1 Phenotypic Ratio: All alike copyright cmassengale

37. 36 Trait: Seed Shape Alleles: R – Roundr – Wrinkled Cross: Wrinkled seeds x Round seeds rr x Rr F 2 Monohybrid Cross (2nd) r r rR Rr rrRr rr Genotype:Rr, rr Genotype: Rr, rr PhenotypeRound & Wrinkled Phenotype: Round & Wrinkled G. Ratio:1:1 G. Ratio: 1:1 P.Ratio: 1:1 copyright cmassengale

38. 37 F 2 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 copyright cmassengale

39. 38 Practice Your Crosses Work the P 1, F 1, and both F 2 Crosses for each of the other Seven Pea Plant Traits copyright cmassengale

40. 39 Results of Monohybrid Crosses Inheritable factors or genes are responsible for all heritable characteristics Phenotype is based on Genotype Each trait is based on two genes, one from the mother and the other from the father True-breeding individuals are homozygous ( both alleles) are the same copyright cmassengale

41. 40 Applying the Law of Segregation copyright cmassengale

42. 41 Dihybrid Cross A breeding experiment that tracks the inheritance of two traits. Mendel’s “Law of Independent Assortment” a. Each pair of alleles segregates independently during gamete formation b. Formula: 2 n (n = # of heterozygotes) copyright cmassengale

43. 42 Question: How many gametes will be produced for the following allele arrangements? Remember: 2 n (n = # of heterozygotes) 1.RrYy 2.AaBbCCDd 3.MmNnOoPPQQRrssTtQq copyright cmassengale

44. 43 Answer: 1. RrYy: 2 n = 2 2 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2 n = 2 3 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2 n = 2 6 = 64 gametes copyright cmassengale

45. 44 Dihybrid Cross Traits: Seed shape & Seed color Alleles: Alleles: R round r wrinkled Y yellow y green RrYy x RrYy RY Ry rY ry All possible gamete combinations copyright cmassengale

46. 45 Dihybrid Cross RYRyrYry RYRy rY ry copyright cmassengale

47. 46 Dihybrid Cross RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 phenotypic ratio RYRyrYryRY Ry rY ry copyright cmassengale

48. 47 Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 copyright cmassengale

49. 48 Test Cross A 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 CC = curly hair Cc = curly hair cc = straight hair bCb___bc copyright cmassengale

50. 49 Test Cross Possible results: bCb___bcbbCc CbCb___bc bbccor c copyright cmassengale

51. 50 Summary of Mendel’s laws LAW PARENT CROSS OFFSPRING DOMINANCE TT x tt tall x short 100% Tt tall SEGREGATION Tt x Tt tall x tall 75% tall 25% short INDEPENDENT ASSORTMENT RrGg x RrGg round & green x round & green 9/16 round seeds & green pods 3/16 round seeds & yellow pods 3/16 wrinkled seeds & green pods 1/16 wrinkled seeds & yellow pods copyright cmassengale

52. 51 Incomplete Dominance and Codominance copyright cmassengale

53. 52 Incomplete Dominance F1 hybrids in betweenphenotypes F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example:snapdragons (flower) Example: snapdragons (flower) red (RR) x white (rr) RR = red flower rr = white flower R R rr copyright cmassengale

54. 53 Incomplete Dominance RrRrRrRr R Rr All Rr = pink (heterozygous pink) produces the F 1 generation r copyright cmassengale

55. 54 Incomplete Dominance copyright cmassengale

56. 55 Codominance Two alleles are expressed (multiple alleles) in heterozygous individuals. Example: blood type 1.type A= I A I A or I A i 2.type B= I B I B or I B i 3.type AB= I A I B 4.type O= ii copyright cmassengale

57. 56 Codominance Problem Example:homozygous male Type B (I B I B ) x heterozygous female Type A (I A i) IAIBIAIB IBiIBi IAIBIAIB IBiIBi 1/2 = I A I B 1/2 = I B i IBIB IAIA i IBIB copyright cmassengale

58. 57 Another Codominance Problem Example:Example: male Type O (ii) x female type AB (I A I B ) IAiIAiIBiIBi IAiIAiIBiIBi 1/2 = I A i 1/2 = I B i i IAIA IBIB i copyright cmassengale

59. 58 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 (I A I B ) copyright cmassengale

60. 59 Codominance Answer: IAIBIAIB ii Parents: genotypes genotypes = I A i and I B i phenotypes phenotypes = A and B IBIB IAIA i i copyright cmassengale

61. 60 Sex-linked Traits Traits (genes) located on the sex chromosomes Sex chromosomes are X and Y XX genotype for females XY genotype for males Many sex-linked traits carried on X chromosome copyright cmassengale

62. 61 Sex-linked Traits Sex Chromosomes XX chromosome - femaleXy chromosome - male fruit fly eye color Example: Eye color in fruit flies copyright cmassengale

63. 62 Sex-linked Trait Problem Example: Eye color in fruit flies (red-eyed male) x (white-eyed female) X R Y x X r X r Remember: the Y chromosome in males does not carry traits. RR = red eyed Rr = red eyed rr = white eyed XY = male XX = female XRXR XrXr XrXr Y copyright cmassengale

64. 63 Sex-linked Trait Solution: X R X r X r Y X R X r X r Y 50% red eyed female 50% white eyed male XRXR XrXr XrXr Y copyright cmassengale

65. 64 Female Carriers copyright cmassengale

66. 65 Genetic Practice Problems copyright cmassengale

67. 66 Breed the P 1 generation tall (TT) x dwarf (tt) pea plants T T tt copyright cmassengale

68. 67 Solution: T T tt Tt All Tt = tall (heterozygous tall) produces the F 1 generation tall (TT) vs. dwarf (tt) pea plants copyright cmassengale

69. 68 Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plants T t Tt copyright cmassengale

70. 69 Solution: TT Tt tt T t Tt produces the F 2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype 3:1 phenotype tall (Tt) x tall (Tt) pea plants copyright cmassengale

71. 70copyright cmassengale