1. © 2006 Jones & Bartlett Publishers Chapter 2 classical Mendelian genetics

2. http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm Gregor Mendel monk gardener careful observer experimenter

3. http://biology.clc.uc.edu/Fankhauser/Travel/Berlin/for_web/Mendel_in_Brno.html

4. Mendel at that time (1850’s-1860’s) it was thought that the traits passed on by the parents were blended in the offspring red + white = ? pink purple + white =purple or white

5. Mendel parents contributed to offspringfactors factors remained unchanged set out to trace their movements looked at phenotypeappearance looked at ratio’s

6. Mendel used peas access to different varieties usually self-pollinated could manipulate pollinization

7. female male

8. Mendel started with true-breeding plants round wrinkled parent 1 offspring x x

9. Mendel started with true-breeding plants seed shaperound vs. wrinkled seed coloryellow vs. green

10. Mendel seed shaperound vs. wrinkled round x wrinkled wrinkled x round round P pollen P flower F1F1

11. © 2006 Jones & Bartlett Publishers Fig. 2.2. Reciprocal crosses of true-breeding pea plants

12. Mendel seed shaperound vs. wrinkled round x wrinkled wrinkled x round round P pollen P flower F1F1 hybridsdominant vs. recessive

13. © 2006 Jones & Bartlett Publishers Fig. 2.3. Traits studied in peas by Mendel dominant recessive

14. F 1 cross round hybrid round hybrid X ?

15. © 2006 Jones & Bartlett Publishers Fig. 2.7.

16. F 1 cross round hybrid round hybrid X each hybrid has two parents therefore each hybrid has two “factors”

17. F 1 cross round hybrid round hybrid X To solve: define terms (use consistently) determine parent genotype determine gamete genotype Punnett square

18. To solve: R= round ?= wrinkled capital letter=dominant lower case letter-recessive r define terms R and r are alleles: alternative forms of a gene

19. To solve: original cross: roundvs.wrinkled (pure-breeding) RR determine parent genotypes parents: rr (homozygous)

20. To solve: original cross: roundvs.wrinkled (pure-breeding) RR rr determine gamete genotypes R or R r or r parents: gametes:

21. To solve: Punnett square RR Rr r r

22. all F1 would be Rr (heterozygous)

23. F 1 cross round hybrid round hybrid X Rr parents: gametes: R or r back to the

24. Punnett square Rr RRRr rr R r What would the ratio of round to wrinkled be? #1

25. © 2006 Jones & Bartlett Publishers Table 2.1. Results of Mendel’s monohybrid experiments

26. Mendel’s first law… …The Law of Segregation The two “factors” in the adult separate from each other during the production of the gametes. homologous pairs of chromosomes separate during meiosis I. #2

27. A problem: Suppose you have a plant with purple flowers, but unknown ancestry. What is its’ genotype? Is it homozygous dominant or heterozygous? PP Pp 2 P’s or not…..

28. A solution: cross it with homozygous recessive P?xpp do it!

29. A solution: if PP x pp if Pp x pp all offspring would be Pp (heterozygous) (purple flowers) some offspring would be Pp some offspring would be pp (purple) (white)

30. The solution: Geneticists call this a “test cross” (pg. 44) pause for ? #3

31. Mendel also looked at two traits at once round vswrinkled yellowvs green RR, rr YY, yy pure-breeding round and yellow pure-breeding green and wrinkled

32. pure-breeding round and yellow pure-breeding green and wrinkled RRYYrryy parental genotype gametes genotype ? RYry

33. RY ry RrYy all heterozygous; phenotype =?

34. all F 1 were round, yellow hybrids RrYy (dihybrids) RrYy parental genotype gametes genotype ? Mendel then did a dihybrid cross (F 1 cross)

35. metaphase I cell R y r Y R r y Y split class

36. ry RY ry RY RRYY RrYy rryy rY Ry rY Ry RRyy RrYy rrYY 3 round, yellow 1 wrinkled, green 1 round, green 2 round, yellow 1 wrinkled, yellow

37. round, yellow round, green wrinkled, yellow wrinkled, green 315 108 101 32 556 9 3 3 1 Mendel’s results: #4 (dihybrid cross)

38. © 2006 Jones & Bartlett Publishers Fig. 2.11. Independent segregation of the Ww and Gg allele pairs

39. Fig. 2.12. Dihybrid cross

40. Mendel’s second law… …The Law of Independent Assortment (in modern language) How a pair of homologous chromosomes align at Metaphase I is independent of how all the other pairs of chromosomes align at metaphase I #5

41. metaphase I cell y r R Y R r y Y RY ry Ry rY

42. 2.4probability a fraction (ratio; like 1/4) between0 and 1 will never happen will always happen

43. Yy YYYy yy Y y Yy x Yy yellow

44. yellow 2001 green 6022 yellow 2001 (6022 + 2001) = 1 4.01 Yy YYYy yy Y y What is the probability of getting plants with green seed ?

45. Yy x Yy yellow What is the probability of getting plants with yellow peas? YY Yy Yy YYYy yy Y y 1/4 2/4 prob [YY or Yy] = prob [YY] + prob [Yy] = 1/4 + 2/4 = 3/4 prob [YY] + prob [Yy] #6

46. RrYy x RrYy What is the probability of getting plants with round, yellow peas? Yy YYYy yy Y y prob [round and yellow] = prob [round] x prob [yellow] = 3/4 x 3/4 = 9/16 Rr RRRr rr R r 9/16 prob [round] x prob [yellow] #7 YY or Yy and RR or Rr 3/4

47. ?

48. 2.6dominance ? RR Rr R gene codes for starch branching enzyme 1 (SBEI) same phenotype ?

49. © 2006 Jones & Bartlett Publishers Fig. 2.20. Three attributes of phenotype affected by Mendel's alleles W and w, which determine round versus wrinkled seeds. dominance is not necessarily all or none

50. RR and Rr same phenotype in peas, but really different at the molecular level Are there some cases where homozygous dominant (RR) looks different than heterozygous (Rr) ?

51. © 2006 Jones & Bartlett Publishers Fig. 2.21. Incomplete dominance in snapdragons.

52. Incomplete dominance When the heterozygote is intermediate between the homozygous phenotypes Seen with traits that are quantitative (can be measured on a continuous scale) as opposed to a discrete trait (appears to be all or none)

53. Blood typing(ABO) Codominance both traits are expressed Human blood types: A B AB O

54. Blood typing(ABO) IAIA IBIB i three alleles (multiple alleles) make “A” carbohydrate make “B” carbohydrate make neither carbohydrate

55. Fig. 2.22. The ABO antigens on the surface of human red blood cells are carbohydrates.

56. Blood typing(ABO) Human blood types: A B AB O phenotype: genotype: I A I A, I A i I B I B, I B i IAIBIAIB ii codominant

57. Blood typing(ABO) Our immune system makes proteins called antibodies to attack foreign molecules called antigens.

58. Blood typing(ABO) For someone with type A blood (someone with the I A allele): A carbohydrate“self” B carbohydrate“non-self” or foreign antigen

59. Blood typing(ABO) For someone with type B blood (someone with the I B allele): A carbohydrate“non-self” or foreign antigen B carbohydrate“self”

60. Blood typing(ABO) For someone with type O blood (someone with ii alleles): A carbohydrate“non-self” or foreign antigen B carbohydrate“non-self” or foreign antigen

61. Blood typing(ABO) For someone with type AB blood (someone with I A and I B alleles): A carbohydrate“self” B carbohydrate“self”

62. Blood typing(ABO) What kind of antibodies would they make? Type A B AB O - Bantibodies - Aantibodies - neither A nor B antibodies - A and B antibodies

63. © 2006 Jones & Bartlett Publishers Table 2.3. Genetic control of the ABO blood groups ABuniversal recipient Ouniversal donor

64. © 2006 Jones & Bartlett Publishers Fig. 2.23. Antibody against type- A antigen will agglutinate red blood cells carrying the type-A antigen.

65. Other reasons why Mendel rules aren’t always observed incomplete dominance multiple alleles variable expressivity same mutation-different results penetrance complete 100% incomplete< 100% polygenic traits ? (phenotype = expected) lung cancer example

66. 2.7Epistasis e.g., When the expected 9:3:3:1 ratio of a dihybrid cross is altered non-allelic genes interacting to affect the same trait Merriam Webster:suppression of the effect of a gene by a nonallelic gene

67. 2.7Epistasis Cpurple flowers cwhite flowers plants with C- and P- genotypes have purple flowers (wt) plants with cc or pp have ? flowers Given: Ppurple flowers pwhite flowers Two different genes C and P for flower color white flowers

68. 2.7Epistasisaside: How could that happen? (hypothetically) ABCDpurple pigment mutation #1 mutation #3 E1E2E3E4

69. 2.7Epistasis CC ppxcc PP phenotype gamete genotype F 1 genotype F 1 phenotype white flowers CpcP CcPp all purple

70. © 2006 Jones & Bartlett Publishers Fig. 2.24. A cross showing epistasis in the determination of flower color in peas. F 1 cross what gametes ?

71. CP Cp cP cp CP Cp cP cp 9:7 ratio purple to white epistasis When the expected 9:3:3:1 ratio of a dihybrid cross is altered CCPP CCPp CcPP CcPp CCPc CCpp CcPp Ccpp CcPP CcPp ccPP ccPp CcPp Ccpp ccPp ccpp

72. © 2006 Jones & Bartlett Publishers Fig. 2.25. Modified F 2 dihybrid ratios.

73. 2.8Complementation mutations c and p show complementation CCppxccPPwhite CcPp purple

74. © 2006 Jones & Bartlett Publishers Fig. 2.26. Complementation reveals whether two recessive mutations are alleles of different genes.

75. 2.8Complementation means that mutations affect different genes means that mutations affect the same gene Lack of complementation

76. © 2006 Jones & Bartlett Publishers Fig. 2.27. Results of complementation tests among six mutant strains of peas.

77. Fig. 2.28. A method for interpreting results of complementation tests.