1. Biology 2250 Principles of Genetics Announcements Lab 3 Information: B2250 (Innes) webpage Lab 3 Information: B2250 (Innes) webpage download and print before lab. download and print before lab. Virtual fly: log in and practice Virtual fly: log in and practice http://biologylab.awlonline.com/

2. Weekly Online Quizzes Marks Marks Oct. 14 - Oct. 22 Example Quiz 2** for logging in Oct. 21- Oct. 24 Quiz 1 2 Oct. 28 Quiz 2 2 Nov. 4 Quiz 3 2 Nov. 10 Quiz 4 2

3. B2250 Readings and Problems Ch. 4 p. 100 – 112 Prob: 10, 11, 12, 18, 19 Ch. 5 p. 118 – 129 Prob: 1 – 3, 5, 6, 7, 8, 9 Ch. 6 p. 148 – 165 Prob: 1, 2, 3, 10

4. Mendelian Genetics Topics: -Transmission of DNA during cell division -Transmission of DNA during cell division Mitosis and Meiosis Mitosis and Meiosis - Segregation - Segregation - Sex linkage (problem: how to get a white-eyed female) - Sex linkage (problem: how to get a white-eyed female) - Inheritance and probability - Inheritance and probability - Independent Assortment - Independent Assortment - Mendelian genetics in humans - Mendelian genetics in humans - Linkage - Linkage - Gene mapping - Gene mapping - Tetrad Analysis (mapping in fungi) - Extensions to Mendelian Genetics - Gene mutation - Chromosome mutation - Quantitative and population genetics   

5. Sex-linked Inheritance Correlation between inheritance of genes and sex of genes and sex

6. Drosophila melanogaster (T. H. Morgan) X Red eye (wild) White eye (mutant)

7. Red Eye White eye Drosophila

8. Cross A red female X white male F 1 all red F 1 all red F 2 red : white F 2 red : white 3 : 1 3 : 1 white all male white all male red 2 : 1 female : male red 2 : 1 female : male No white- eyed females

9. Cross B white female X red male F 1 females males F 1 females males F 2 females males F 2 females males 1 : 1 : 1 : 1 1 : 1 : 1 : 1 How to obtain a white-eyed female?

10. Cross A w + w + X wY w + w + X wY F 1 ww + w + Y F 1 ww + w + Y ww + X wY ww + X wY w Y w Y w ww wY w ww wY w + ww + w + Y w + ww + w + Y X w+ X w+ Xw YXw YXw YXw Y

11. Mendelian Inheritance Determining mode of inheritance: - single gene or more complicated - single gene or more complicated - recessive or dominant - recessive or dominant - sex linked or autosomal - sex linked or autosomal Approach: cross parents observed progeny observed progeny compare with expected compare with expected

12. Principle of Segregation Implications Answer questions on inheritance: - mode of inheritance (dominant, recessive - mode of inheritance (dominant, recessive sex-linked) sex-linked) - paternity - paternity - hybridization - hybridization

13. Equal segregation of two members of a gene pair Aa ½ A gametes ½ a gametes P(a) = ½ P(A) = ½ Meiosis: diploid nucleus divides diploid nucleus divides produces haploid nuclei produces haploid nuclei Mendel’s First Law

14. Rules of Probability 1.Product rule (AND): probability of 2 independent events occurring probability of 2 independent events occurring simultaneously simultaneously 2. Sum Rule (OR): probability of either one of two mutually probability of either one of two mutually exclusive events exclusive events

15. Probability 1.**Coin toss: P (T) = P(H) = 1/2 P(T, T, T) = P(T) and P(T) and P(T) P(T, T, T) = P(T) and P(T) and P(T) = (½) 3 = ½ x ½ x ½ = (½) 3 = ½ x ½ x ½ 2. One die: P (6) or P (5) = 1/6 + 1/6 **http://shazam.econ.ubc.ca/flip/ http://shazam.econ.ubc.ca/flip/

16. F2F2F2F2 Self F 1 Aa X Aa Self F 1 Aa X Aa 1/2 A 1/2 a 1/2 A 1/2 a 1/2 A 1/4 AA 1/4 Aa P(AA) = ½ * ½ 1/2 A 1/4 AA 1/4 Aa P(AA) = ½ * ½ 1/2 a 1/4 Aa 1/4 aa 1/2 a 1/4 Aa 1/4 aa Prob. (AA or Aa) = 1/4 + 2/4 = ¾ Prob. (aa) = ¼ equalSegregation P(A) = ½ P(a) = ½ eggs sperm

17. Two Characters Monohybrid Cross parents differ for a single character parents differ for a single character (single gene ); seed shape (single gene ); seed shape Dihybrid Cross parents differ for two characteristics parents differ for two characteristics (two genes) (two genes)

18. Dihybrid Two Characters: 1. Seed colour yellow green Y y Y y 2. Seed shape Round wrinkled R r R r 4 phenotypes

19. Dihybrid RRyy X rrYY Ry rY Ry rY RrYy DIHYBRID RrYy DIHYBRID P F1F1F1F1 Gametes

20. F 1 Dihybrid ----->F 2 F 1 RrYy RrYy X RrYy RrYy X RrYy F 2 9 315 round, yellow 3 108 round, green 3 108 round, green 3 101 wrinkled, yellow 3 101 wrinkled, yellow 1 32 wrinkled, green 1 32 wrinkled, green Total 556 Total 556

21. Individual Characters 1. Seed shape round : wrinkled 423 : 133 423 : 133 3 : 1 (¾ : ¼) 3 : 1 (¾ : ¼) 2. Seed colour yellow : green 416 : 140 416 : 140 3 : 1 3 : 1

22. Conclusion * 3 : 1 monohybrid ratio for each character * 3 : 1 monohybrid ratio for each character * 9 : 3 : 3 : 1 phenotypic ratio a random * 9 : 3 : 3 : 1 phenotypic ratio a random combination of 2 independent 3:1 ratios combination of 2 independent 3:1 ratios

23. Two Independent Genes F 2 seed shape F 2 seed shape 3/4 1/4 3/4 1/4 colour round wrinkled yellow 3/4 9/16 3/16 yellow 3/4 9/16 3/16 green 1/4 3/16 1/16 green 1/4 3/16 1/16 F 2 Phenotypes

24. Applying Probability to Genetics Dihybrid: RrYy Hypothesis: mechanism for putting R or r into a gamete is independent of the mechanism for putting Y or y into a gamete mechanism for putting R or r into a gamete is independent of the mechanism for putting Y or y into a gamete

25. Gametes from Dihybrid Gametes from Dihybrid Dihybrid: RrYy (F 1 ) Principle of segregation during gamete formation: Yy -------> P(Y) = P(y) = 1/2 Yy -------> P(Y) = P(y) = 1/2 Rr ------->P(R) = P(r) = 1/2 Rr ------->P(R) = P(r) = 1/2

26. Gametes from dihybrid RrYy : probability probability Y and R 1/2 * 1/2 = ¼ YR Y and R 1/2 * 1/2 = ¼ YR Y and r 1/2 * 1/2 = ¼ Yr Y and r 1/2 * 1/2 = ¼ Yr y and R 1/2 * 1/2 = ¼ yR y and R 1/2 * 1/2 = ¼ yR y and r 1/2 * 1/2 = ¼ yr y and r 1/2 * 1/2 = ¼ yr 4 gamete types

27. F 1 gametes produce F 2 YyRr X YyRr YyRr X YyRr ¼ YR Yr yR yr ¼ YR Yr yR yr ¼ YR 1/16 RRYY ¼ YR 1/16 RRYY Yr Yr yR yR yr yr F1F1F1F1 F2F2F2F2 gametes gametes

28. Fig. 6-7 F 2 F 2 4 Gametes 4 Gametes 9 Genotypes 9 Genotypes 4 Phenotypes 4 PhenotypesSpermEggs F2F2F2F2

29. Mendel’s Second Law Independent assortment: during gamete formation, the segregation of one gene pair is independent of other gene pairs. during gamete formation, the segregation of one gene pair is independent of other gene pairs.

30. A a B b A a b B Meiosis I OR(Genes) Correlation of genes and Chromosomes during Meiosis I A a

31. Producing the F 2 YyRr X YyRr YyRr X YyRr 1. F 1 Gametes  produce F 2 1. F 1 Gametes  produce F 2 2. Genotypes 2. Genotypes 3. Phenotypes 3. Phenotypes F1F1F1F1 F2F2F2F2

32. Independent Assortment Two gene systems: 1. Gametes from dihybrid 4 x 4 = 16 YyRr: YyRr: ¼ YR Yr yR yr ¼ YR Yr yR yr ¼ YR 1/16YYRR ¼ YR 1/16YYRR Yr Yr yR yR yr yr Male gametes Femalegametes F2F2F2F2

33. Independent Assortment 2. F 2 Genotypes 3 x 3 = 9 ¼ RR ½ Rr ¼ rr ¼ RR ½ Rr ¼ rr ¼ YY 1/16 YYRR ¼ YY 1/16 YYRR ½ Yy ½ Yy ¼ yy ¼ yy F2F2F2F2 YyRr X YyRr

34. Independent Assortment 3. F 2 Phenotypes 2 x 2 = 4 ¾ R- ¼ rr ¾ R- ¼ rr ¾ Y- 9/16 R-Y- ¾ Y- 9/16 R-Y- ¼ yy ¼ yy YyRr X YyRr

35. F1F1 9 Genotypes4 phenotypes YY RR YY Rr Yy RR Yy Rr YY rr Yy rr yy RR yy Rr yy rr YyRr x YyRr Y-R- Y-rr yyR- yyrr

36. Independent Assortment F 1 AaBb X AaBb F 1 AaBb X AaBb F 2 9 A-B- F 2 9 A-B- 3 A-bb 3 A-bb 3 aaB- 3 aaB- 1 aabb 1 aabb 4 phenotypes

37. Independent Assortment Test Cross AaBb X aabb AaBb X aabb gametes ab gametes ab 1/4 AB AaBb 1/4 AB AaBb 1/4 Ab Aabb 1/4 Ab Aabb 1/4 aB aaBb 1/4 aB aaBb 1/4 ab aabb 1/4 ab aabb 4 phenotypes 4 genotypes

38. Fig 6-6 Independent Assortment Interchromosomal Recombination AB AB ab ab Ab Ab aB aB Inferred F 1 gamete types

39. Independent Assortment Any number of independent genes: Genes Phenotypes Genotypes 1 2 3 1 2 3 2 4 9 2 4 9 3 8 27 3 8 27 n 2 n 3 n n 2 n 3 n

40. Mendelian Genetics in Humans Determining mode of inheritance Problems: 1. long generation time 1. long generation time 2. can not control mating 2. can not control matingAlternative: * information from matings that have already occurred “Pedigree” * information from matings that have already occurred “Pedigree”

41. Human Pedigrees Pedigree analysis: trace inheritance of disease or condition trace inheritance of disease or condition provide clues for mode of inheritance provide clues for mode of inheritance (dominant vs. recessive) (dominant vs. recessive) (autosomal vs. sex linked) (autosomal vs. sex linked) however, some pedigrees ambiguous however, some pedigrees ambiguous

42. Human Pedigrees 1. Ambiguous: 2. Unambiguous: 1. Ambiguous: 2. Unambiguous: Affectedfemale Normalmale Normalfemale

43. Clues (non sex-linked) Recessive: 1. individual expressing trait has two 1. individual expressing trait has two normal parents normal parents 2. two affected parents can not have an 2. two affected parents can not have an unaffected child. unaffected child.

44. Rare Recessive A- A- (AA or Aa) Cousins(inbreeding) Rare = AA

45. Clues Dominant: 1. every affected person has at least one 1. every affected person has at least one affected parent affected parent 2. each generation will have affected 2. each generation will have affected individuals individuals

46. Dominant All genotypes known Not AA

47. Examples Recessive: - phenylketonuria (PKU) - phenylketonuria (PKU) - hemophilia (sex linked) - hemophilia (sex linked) - cystic fibrosis - cystic fibrosis - albinism - albinismDominant: - huntingtons chorea - huntingtons chorea - brachydactyly (short fingers) - brachydactyly (short fingers) - polydactyly (extra fingers) - polydactyly (extra fingers) - achondroblasia (dwarf) - achondroblasia (dwarf)

48. http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?db=OMIM

50. Brachydactyly Bb short fingers bb normal Bbbb

51. http://www.biology.arizona.edu/mendelian_genetics/mende lian_genetics.html http://www.biology.arizona.edu/mendelian_genetics/mende lian_genetics.html Online Tutorial:

52. Solving Genetics Problems 1.Don’t panic! 2.Carefully read the problem 3.What information is given? Know the terms used. 4.What aspect of genetics does the problem address?

53. X-linked Dominant 1. affected male ---> all daughters affected no sons no sons aa x AY ----> Aa, aY aa x AY ----> Aa, aY 2. affected female ----> ½ sons, ½ daughters affected affected Aa x aY ----> AY, aY, aa, Aa Aa x aY ----> AY, aY, aa, Aa * * * * Sex Linked Inheritance

54. X-Linked Dominant 1. 2. All daughters affected, no sons 1/2 daughters affected, 1/2 sons affected

55. X-linked Inheritance X-linked recessive: 1. more males than females show recessive phenotype recessive phenotype 2. affected female ------> both mother and father have recessive allele and father have recessive allele A a x a Y --------> a a A a x a Y --------> a a

56. X-linked Inheritance X-linked recessive: 3. affected male ----> mother carries allele A a x AY -----> a Y A a x AY -----> a Y 4. affected male -----> no affected offspring AA x a Y ----> AY, Aa AA x a Y ----> AY, Aa carrier

57. X-Linked Recessive Mothercarrier

58. Sex Linked Inheritance (examples) X linked genes Humans: - colour blindness Humans: - colour blindness - hemophilia - hemophilia More common in males (hemizygous aY) More common in males (hemizygous aY) X linked recessives expressed X linked recessives expressed

59. X-linked recessive hemophilia Queen Victoria (carrier) QE II Hemophilic male Carrier female

60. X – linked disease genes

61. Mendelian Genetics Topics: -Transmission of DNA during cell division -Transmission of DNA during cell division Mitosis and Meiosis Mitosis and Meiosis - Segregation (Monohybrid) - Segregation (Monohybrid) - Sex linkage - Sex linkage - Inheritance and probability - Inheritance and probability - Independent Assortment (Dihybrid) - Independent Assortment (Dihybrid) - Mendelian genetics in humans (Pedigree) - Mendelian genetics in humans (Pedigree)