1. Today: some things Mendel did not tell us… plus Mapping and Epigenetics –Exam #3 W 7/30 in class (bonus #2 due)–

2. Single genes controlling a single trait are unusual. Inheritance of most genes/traits is much more complex… Dom.Rec. Dom.

3. PhenotypeGenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple.

4. Fig 4.4

5. Fig 4.7 Sickle-cell anemia is caused by a point mutation

6. Sickle and normal red blood cells Fig 4.7

7. Mom = HSDad = HS H or S HH HSSS HS possible offspring 75% Normal 25% Sickle-cell Mom Dad S=sickle-cell H=normal Sickle-Cell Anemia: A dominant or recessive allele? Fig 4.7

8. Coincidence of malaria and sickle-cell anemia Fig 24.14

9. Mom = HSDad = HS H or S HH HSSS HS possible offspring Oxygen transport: 75% Normal 25% Sickle-cell Malaria resistance: 75% resistant 25% susceptible Mom Dad Sickle-Cell Anemia: A dominant or recessive allele? S=sickle-cell H=normal Fig 4.7

10. The relationship between genes and traits is often complex Complexities include: Complex relationships between alleles

11. Sex determination is normally inherited by whole chromosomes or by number of chromosomes. Fig 3.18

12. X/Y chromosomes in humans

13. The X chromosome has many genes; the Y chromosome only has genes for maleness.

14. Human sex chromosomes (includes Mic2 gene) Fig 4.14

15. Sex-linked traits are genes located on the X chromosome

16. Color Blind Test

17. Sex-linked traits: Genes on the X chromosome No one affected, female carriers A= normal; a= colorblind colorblind normal similar to Fig 4.13

18. Sex-linked traits: Genes on the X chromosome 50% of males affected, 0 % females affected A= normal; a= colorblind normal similar to Fig 4.13

19. Sex-linked traits: Genes on the X chromosome 50% males affected, 50% females affected A= normal; a= colorblind colorblind normal similar to Fig 4.13

20. Sex-linked traits: Genes on the X chromosome No one affected, female carriers 50% of males affected, 0 % female affected 50% males affected, 50% females affected A= normal ; a= colorblind similar to Fig 4.13

21. males and females may have different numbers of chromosomes Fig 3.18

22. Tbl 7.1 dosage compensation

23. At an early stage of embryonic development The epithelial cells derived from this embryonic cell will produce a patch of white fur While those from this will produce a patch of black fur Fig 7.4

24. Promotes compaction Prevents compaction Mammalian X-inactivation involves the interaction of 2 overlapping genes.

25. The Barr body is replicated and both copies remain compacted Barr body compaction is heritable within an individual

26. A few genes on the inactivated X chromosome are expressed in the somatic cells of adult female mammals –Pseudoautosomal genes (Dosage compensation in this case is unnecessary because these genes are located both on the X and Y) –Up to a 25% of X genes in humans may escape full inactivation The mechanism is not understood

27. Epigenetics: http://www.pbs.org/wgbh/nova/sciencenow/3411/02.html http://www.pbs.org/wgbh/nova/sciencenow/3411/02.html Lamarck was right? Sort of… Image from: http://www.sparknotes.com/biology/evolution/lamarck/section2.rhtml

28. Genomic Imprinting Genomic imprinting is a phenomenon in which expression of a gene depends on whether it is inherited from the male or the female parent Imprinted genes follow a non-Mendelian pattern of inheritance –Depending on how the genes are “marked”, the offspring expresses either the maternally- inherited or the paternally-inherited allele ** Not both

29. Genomic Imprinting: Methylation of genes during gamete production.

30. A hypothetical example of imprinting A=curly hair a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females a B* a B* A* b A* b

31. A hypothetical example of imprinting A=curly hair a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females A*a bB* A*a bB* a B* a B* A* b A* b

32. A hypothetical example of imprinting A=curly hair a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females A*a bB* A*a bB* A*a bB Aa bB* a B* a B* A* b A* b

33. A hypothetical example of imprinting A=curly hair a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females A*a bB* A*a bB* A*a bB Aa bB* A*b, A*B, ab, aB Ab, AB*, ab, aB* a B* a B* A* b A* b similar to Fig 7.10

34. Thus genomic imprinting is permanent in the somatic cells of an animal –However, the marking of alleles can be altered from generation to generation

35. Genomic imprinting must involve a marking process At the molecular level, the imprinting is known to involve differentially methylated regions –They are methylated either in the oocyte or sperm Not both Imprinting and DNA Methylation

36. For most genes, methylation results in inhibition of gene expression –However, this is not always the case

37. Haploid female gametes transmit an unmethylated gene Haploid male gametes transmit a methylated gene Fig 7.11 Changes in methylation during gamete development alter the imprint

38. To date, imprinting has been identified in dozens of mammalian genes Tbl 7.2

40. Imprinting plays a role in the inheritance of some human diseases: Prader-Willi syndrome (PWS) and Angelman syndrome (AS) –PWS is characterized by: reduced motor function, obesity, mental deficiencies –AS is characterized by: hyperactivity, unusual seizures, repetitive muscle movements, mental deficiencies Usually, PWS and AS involve a small deletion in chromosome 15 –If it is inherited from the mother, it leads to AS –If it is inherited from the father, it leads to PWS

41. AS results from the lack of expression of UBE3A (encodes a protein called EA-6P that transfers small ubiquitin molecules to certain proteins to target their degradation) –The gene is paternally imprinted (silenced) PWS results (most likely) from the lack of expression of SNRNP (encodes a small nuclear ribonucleoprotein that controls gene splicing necessary for the synthesis of critical proteins in the brain) –The gene is maternally imprinted (silenced)

42. Fig 7.12 The deletion is the same in males and females, but the expression is different depending on who you received the normal version from.

43. The relationship between genes and traits is often complex Complexities include: Multiple genes controlling one trait

44. Two genes control coat color in mice Fig 4.21

45. The interaction of these two proteins explains their affect on a single trait (in fruit flies).

46. Variation in Peas Fig 3.2

47. Fig 2.8 Inheritance of 2 independent genes

48. Y y rR Gene for seed color Gene for seed shape Approximate position of seed color and shape genes in peas Chrom. 1/7Chrom. 7/7

49. There must be a better way… Fig 2.9

50. Inheritance can be predicted by probability Section 2.2, pg 30-32

51. Sum rule The probability that one of two or more mutually exclusive events will occur is the sum of their respective probabilities Consider the following example in mice Gene affecting the ears –De = Normal allele –de = Droopy ears Gene affecting the tail –Ct = Normal allele –ct = Crinkly tail

52. If two heterozygous (Dede Ctct) mice are crossed Then the predicted ratio of offspring is –9 with normal ears and normal tails –3 with normal ears and crinkly tails –3 with droopy ears and normal tails –1 with droopy ears and crinkly tail These four phenotypes are mutually exclusive –A mouse with droopy ears and a normal tail cannot have normal ears and a crinkly tail Question –What is the probability that an offspring of the above cross will have normal ears and a normal tail or have droopy ears and a crinkly tail?

53. Applying the sum rule –Step 1: Calculate the individual probabilities 9(9 + 3 + 3 + 1)= 9/16P (normal ears and a normal tail) = 1(9 + 3 + 3 + 1)= 1/16P (droopy ears and crinkly tail) = –Step 2: Add the individual probabilities 9/16 + 1/16 = 10/16 10/16 can be converted to 0.625 –Therefore 62.5% of the offspring are predicted to have normal ears and a normal tail or droopy ears and a crinkly tail

54. Product rule The probability that two or more independent events will occur is equal to the product of their respective probabilities Note –Independent events are those in which the occurrence of one does not affect the probability of another

55. Consider the disease congenital analgesia –Recessive trait in humans –Affected individuals can distinguish between sensations However, extreme sensations are not perceived as painful –Two alleles P = Normal allele p = Congenital analgesia Question –Two heterozygous individuals plan to start a family –What is the probability that the couple’s first three children will all have congenital analgesia?

56. Applying the product rule –Step 1: Calculate the individual probabilities This can be obtained via a Punnett square 1/4P (congenital analgesia) = –Step 2: Multiply the individual probabilities 1/4 X 1/4 X 1/4 = 1/64 1/64 can be converted to 0.016 –Therefore 1.6% of the time, the first three offspring of a heterozygous couple, will all have congenital analgesia

57. Crossing- over Meiosis I Meiosis II 4 Haploid cells, each unique (Ind. Assort.) Different genes are not always independent

58. The haploid cells contain the same combination of alleles as the original chromosomes The arrangement of linked alleles has not been altered Fig 5.1

59. These haploid cells contain a combination of alleles NOT found in the original chromosomes These are termed parental or non- recombinant cells This new combination of alleles is a result of genetic recombination These are termed recombinant cells Fig 5.1

60. Linked alleles tend to be inherited together

61. Crossing over produces new allelic combinations

62. Recombinants are produced by crossing over

63. For linked genes, recombinant frequencies are less than 50 percent

64. Homologous pair of chromosomes

65. Does this pedigree show recombination or linkage?

67. Longer regions have more crossovers and thus higher recombinant frequencies

68. Some crosses do not give the expected results

69. =25% 8%9%41%42%

70. These two genes are on the same chromosome

72. By comparing recombination frequencies, a linkage map can be constructed = 17 m.u.

73. Linkage map of Drosophila chromosome 2: This type of map, with mapping units more than 50, can only be put together by making comparisons of linked genes.

74. Today: some things Mendel did not tell us… plus Mapping and Epigenetics –Exam #3 W 7/30 in class (bonus #2 due)– Lecture ended here, but I am leaving in the following material so you can get a preview of the mapping problem we will work on to start class on M 7/28.

75. A much greater proportion of the two types found in the parental generation Fig 5.2

76. The probability of crossing over can be used to determine the spatial relationship of different genes

77. Double recombinants arise from two crossovers Recombinant

78. Double recombinants can show gene order

79. similar to Fig 5.3, also see Fig 5.9, and pg 115-117 What is the relationship between these 3 genes? What order and how far apart?

80. similar to Fig 5.3 What is the relationship between these 3 genes? What order and how far apart?

81. Double crossover

82. Which order produces the double crossover?

84. We have the order. What is the distance?

85. Recombinants between st and ss: (50+52+5+3)/755 =14.6%

86. Recombinants between ss and e: (43+41+5+3)/755 =12.2%

87. stsse 14.6 m.u. 26.8 m.u. 12.2 m.u. Put it all together…

88. Drosophila linkage map

89. Linkage map of Drosophila chromosome 2

90. Yeast chromosome 3 physical distance linkage map Recombination is not completely random.

91. Alignment of physical and recombination maps

92. PhenotypeGenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple.

93. For life to exist, the information (genes) must be passed on. {Mitosis: producing more cells} {Meiosis: producing gametes}