1. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings PowerPoint Lectures for Biology, Seventh Edition Neil Campbell and Jane Reece Lectures by Chris Romero Chapter 14 Mendel and the Gene Idea

2. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Gregor Mendel – experiments with garden peas Figure 14.1

3. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mendel’s Experimental Approach Why peas? = available in many varieties = could strictly control mating

4. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mendel’s Experimental Approach Stamens (Male) Carpel (Female)

5. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mendel’s Experimental Approach

6. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mendel’s Experimental Approach

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8. Genetics Vocabulary Alternative versions of genes = Alleles

9. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Organism inherits 2 alleles: 1 from mom, 1 from dad – A genetic locus is represented twice Genetics Vocabulary

10. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Genetics Vocabulary If the two alleles at a locus differ… Dominant allele = determines appearance Recessive allele = no noticeable effect on appearance

11. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Genetic Vocabulary: Homozygous vs. Heterozygous Homozygous for a particular gene – Identical pair of alleles for that gene Ex: PP (2 purple flower alleles) True-breeding - Homozygous dominant (PP) - Homozygous recessive (pp)

12. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Homozygous vs. Heterozygous Homozygous for a particular gene – Identical pair of alleles for that gene Ex: PP (2 purple flower alleles) True-breeding Heterozygous for a particular gene – Has a pair of alleles that are different for that gene Ex: Pp (1 purple allele, 1 white allele)

13. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Homozygous or Heterozygous?

14. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Genetics Vocabulary An organism’s genotype (EX: Pp, PP, pp) – genetic makeup An organism’s phenotype (Ex: Purple or white) – physical appearance

15. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Phenotype versus genotype

16. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Phenotype versus genotype

17. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mendel used : Characters that varied in an “either-or” manner Varieties that were “true-breeding”

18. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Typical Mendelian Experiment Parental Generation Hybridization F1 Generation F1 self-pollinate F2 generation

19. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings All Purple Hybrids 3:1 Purple : White

20. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Does Mendel’s segregation model account for the 3:1 ratio observed in the F 2 generation? – We can answer this question using a Punnett square

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24. Other pea plant characters

25. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Construct a Punnett Square for the following crosses: Seed color: Y = Yellow, y = green – YY X Yy – Expected ratio observed in offspring? Seed shape: R = Round, r = wrinkled – Rr X rr – Expected ratio observed in offspring?

26. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings The Testcross In pea plants with purple flowers – Genotype is not obvious (Pp or PP)? = Perform testcross

27. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings The testcross

28. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings The testcross

29. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Monohybrid Cross Mendel Followed a single trait (ex: flower color) The P = true-breeding (PP or pp) The F 1 offspring = monohybrids (heterozygous for one character) (Pp)

30. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Dihybrid Cross Mendel followed 2 characters at the same time P generation = Cross two, true-breeding parents differing in two characters – YYRR X yyrr

31. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Character 1 Y =YELLOW y =green Character 2 R=ROUND r = wrinkled

32. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mendel followed 2 characters at the same time P generation = Cross two, true-breeding parents differing in two characters – YYRR X yyrr F1 generation = Produces dihybrids (heterozygous for both characters) – YyRr

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34. 2. Independent Assortment of Chromosomes

35. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 2. Independent Assortment of Chromosomes Homologous orient randomly at metaphase I of meiosis

36. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings How are two characters transmitted from parents to offspring? 1. As a package? (Ex: yellow and round YR) =Dependent Assortment 2. Independently? =Independent Assortment

37. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings A dihybrid cross Only YR and yr as inherited from P generation? YR Yr yR yr ? Make a punnett square for each case

38. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

39. Concept 14.2: The rules of probability govern Mendelian inheritance Multiplication Rule Addition Rule

40. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings The Multiplication and Addition Rules Applied to Monohybrid Crosses The multiplication rule – Probability that two or more independent events will occur together – Ex: coin toss – Heads ½ X Heads ½ = ¼

41. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Ex: Probability in a monohybrid cross

42. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Rule of Addition Probability that any one of two or more exclusive events will occur Ex: Heterozygotes: ¼Rr + ¼rR = ½

43. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings A multi-character cross = two or more independent monohybrid crosses occurring simultaneously Calculate the chances for various genotypes: 1. Consider each character separately 2. Go back to question being asked 3. Multiply individual probabilities together 4. Use Rule of addition (if necessary)

44. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3 characters = trihybrid cross – Purple flowers (Pp), Yellow (Yy), Round (Rr) – Purple flowers (Pp), green (yy), wrinkled (rr) PpYyRr X Ppyyrr Question: What percentage of the offspring from this cross would be predicted to have purple flowers and green and wrinkled seeds?

45. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 1. Consider each character separately (make a punnett square for each character) PpYyRr X Ppyyrr: – Pp X Pp = – Yy X yy = – Rr X rr =

46. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 1. Consider each character separately (make a punnett square for each character) – PpYyRr X Ppyyrr – Pp X Pp = ¼ PP, ½ pP, ¼ pp – Yy X yy = ½ Yy, ½ yy – Rr X rr = ½ Rr, ½ rr

47. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 2. Go back to the original Question PpYyRr X Ppyyrr – Pp X Pp = ¼ PP, ½ pP, ¼ pp – Yy X yy = ½ Yy, ½ yy – Rr X rr = ½ Rr, ½ rr Question: What percentage of the offspring from this cross would be predicted to have purple flowers and green and wrinkled seeds?  Start by listing all genotypes that fulfill this condition:

48. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 2. Go back to the original Question PpYyRr X Ppyyrr – Pp X Pp = ¼ PP, ½ Pp, ¼ pp – Yy X yy = ½ Yy, ½ yy – Rr X rr = ½ Rr, ½ rr Question: What percentage of the offspring from this cross would be predicted to have purple flowers and green and wrinkled seeds?  Start by listing all genotypes that fulfill this condition: Ppyyrr, PPyyrr

49. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3. Calculate probability for each genotype – Pp X Pp = ¼ PP, ½ pP, ¼ pp – Yy X yy = ½ Yy, ½ yy – Rr X rr = ½ Rr, ½ rr Ppyyrr ½ X ½ X ½ = 2/16 Ppyyrr

50. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3. Calculate probability for each genotype – Pp X Pp = ¼ PP, ½ pP, ¼ pp – Yy X yy = ½ Yy, ½ yy – Rr X rr = ½ Rr, ½ rr Ppyyrr ½ X ½ X ½ = 2/16 PPyyrr ¼ X ½ X ½ =1/16

51. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 4. Rule of addition 2/16 Ppyyrr +1/16 Ppyyrr 3/16 = chance that the offspring from this cross would have purple flowers and green and wrinkled seeds

52. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings A multi-character cross #2 = two or more independent monohybrid crosses occurring simultaneously Calculate the chances for various genotypes: 1. Consider each character separately 2. Go back to question being asked 3. Multiply individual probabilities together 4. Use Rule of addition

53. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3 characters = trihybrid cross #2 – white flowers (pp), Yellow (Yy), wrinkled (rr) – Purple flowers (Pp), green (yy), Round (Rr) ppYyrr X PpyyRr Question: What percentage of the offspring from this cross would be predicted to have white flowers and green and wrinkled seeds?

54. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 1. Consider each character separately (make a punnett square for each character) ppYyRr X Ppyyrr: – pp X Pp = ½ Pp, ½ pp – Yy X yy = ½ Yy, ½ yy – rr X Rr = ½ Rr, ½ rr

55. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 2. Go back to the original Question ppYyRr X Ppyyrr: – pp X Pp = ½ Pp, ½ pp – Yy X yy = ½ Yy, ½ yy – rr X Rr = ½ Rr, ½ rr Question: What percentage of the offspring from this cross would be predicted to have white flowers and green and wrinkled seeds?  Start by listing all genotypes that fulfill this condition:

56. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 2. Go back to the original Question ppYyRr X Ppyyrr: – pp X Pp = ½ Pp, ½ pp – Yy X yy = ½ Yy, ½ yy – rr X Rr = ½ Rr, ½ rr Question: What percentage of the offspring from this cross would be predicted to have white flowers and green and wrinkled seeds?  Start by listing all genotypes that fulfill this condition: ppyyrr

57. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3. Calculate probability for each genotype ppYyRr X Ppyyrr: – pp X Pp = ½ Pp, ½ pp – Yy X yy = ½ Yy, ½ yy – rr X Rr = ½ Rr, ½ rr ppyyrr ½ pp X ½ yy X ½ rr= 1/8

58. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Concept 14.3 Inheritance patterns are often more complex than predicted by simple Mendelian genetics The relationship between genotype (Ex: Pp) and phenotype (Ex: purple) is rarely simple

59. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings The Spectrum of Dominance Complete dominance – Phenotypes of the heterozygote and dominant homozygote are identical

60. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Codominance – Two dominant alleles affect the phenotype in separate, distinguishable ways Ex: human blood group MN MM = RBC with M molecules NN = RBC with N molecules MN = ?

61. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Incomplete dominance F 1 hybrid phenotype is between the phenotypes of the two parental varieties Figure 14.10

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64. Dominance and Phenotype Dominant and recessive alleles – Do not “interact” – Different alleles = synthesis of different proteins that produce a phenotype

65. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Dominance and Phenotype Dominant and recessive alleles – Do not “interact” – Different alleles = synthesis of different proteins that produce a phenotype Ex: flower color – White (W) vs. Red (R) – W= protein that produces white pigment – R = protein that produces red pigment

66. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Multiple Alleles Most genes exist in populations – In more than two allelic forms 1 2 3 12

67. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings The ABO blood group in humans – Is determined by multiple alleles: 3 different alleles for enzyme I – IA = attaches the A carbohydrate – IB = attaches the B carbohydrate – i = attaches neither A nor B

68. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Table 14.2

69. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Complex inheritance patterns Codominance Incomplete dominance Multiple alleles – Mendel’s fundamental laws still apply!

70. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Concept 14.4: Human traits follow Mendelian patterns of inheritance Humans = not convenient subjects for genetic research How can we study Human Genetics?

71. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Concept 14.4: Human traits follow Mendelian patterns of inheritance Humans = not convenient subjects for genetic research How can we study Human Genetics? = Pedigree analysis

72. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Male = Female =

73. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Follow Attached earlobe = ff

74. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Carriers? Disease condition = aa No disease symptoms = Aa or AA

75. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Mating of Close Relatives Mating between relatives – Can increase the probability of the appearance of a genetic disease CcCC Cc cc

76. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Albinism- recessive phenotype Only aa

77. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Albinism- recessive phenotype

78. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Human achondroplasia phenotype The phenotype is determined by a dominant allele = AA or Aa

79. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Human achondroplasia: Dominant allele disease

80. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings PHENYLKETONURIAPHENYLKETONURIA - [PKU] pp Autosomal recessive disorder Gene for phenylalanine hydroxylase (PAH), found on chromosome 12 mutated PAH converts the amino acid phenylalanine to tyrosine No PAH = concentration of phenylalanine in the body can build up to toxic levels

81. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings PKU: Recessive disease (pp) Pppp Pp Pp/PP

82. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Review! Mendel’s Pea experiments: Experimental method Typical Mendelian experiment: – P, F1, F2 Monohybrid cross vs. Dihybrid cross Law of Segregation and Law of Independent assortment

83. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Review! Solving Multi-hybrid crosses with probability More complex inheritance patterns: – Co-dominance, Incomplete dominance, Multiple alleles Pedigree Analysis