Biology 2250 Principles of Genetics

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

2. B2250Readings 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

3. Weekly Online Quizzes Marks Oct. 14 - Oct. 25 Example Quiz 2** for logging in Oct. 21- Oct. 25 Quiz 1 2 Oct. 28 Quiz 2 2 Nov. 4 Quiz 3 2 Nov. 10 Quiz 4 2

4. Weekly Online QuizzesResults Example quiz: Quiz 1: Answers: http://webct.mun.ca:8900/

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

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

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

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

9. Mendel’s Second Law Independent assortment: during gamete formation, the segregation of one gene pair is independent of other gene pairs. Genes independent because they are on different chromosomes

10. Independent Assortment Genotypes AABB AaBb AaBB AABb F1 AaBb X AaBb F2 9 A-B- 3 A-bb 3 aaB- 1 aabb Aabb, AAbb 4 phenotypes aaBb, aaBB

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

12. Independent Assortment Inferred F1 gamete types AB Fig 6-6 ab Ab aB Interchromosomal Recombination

13. (Genes) Meiosis I A Correlation of genes and Chromosomes during meiosis a 4 gamete types b B A A OR a a b B

14. Linkage of Genes - Many more genes than chromosomes - Some genes must be linked on the same chromosome; therefore not independent

15. Complete Linkage X AaBb dihybrid P A B a b F1 A B a b F1 gametes A B a b AB AB ab ab parental

16. Recombinant Gametes ? Crossing over: - exchange between homologous chromosomes

17. Crossing over in meiosis I Meiosis I - homologous chromosomes pair - reciprocal exchange between non-sister chromatids Ch 4 meiosis animation: http://www.whfreeman.com/mga/

18. Crossing over in meiosis I (animation)

19. Gamete Types X F1 A B a b AaBb gametes A B AB Parental a b ab Parental A b Ab Recomb. a B aB Recomb.

20. Two Ways to produce dihybrid 1 Note: Chromatids omitted AABB x aabb  AaBb X A B a b A B a b 2 AAbb x aaBB  AaBb A b a B A b a B X

21. 1. Ways to produce dihybrid Note: Chromatids omitted P X Cis A B a b A B a b A B AaBb a b (dihybrid ) Gametes: AB P ab P Ab R aB R

22. 2. Ways to produce dihybrid P X A b a B A b a B AaBb A b trans (dihybrid ) a B Gametes: P Ab P aB R AB R ab

23. Two ways to produce dihybrid P X X A B a b A b a B A B a b A b a B cis A B AaBb A b trans a b (dihybrid ) a B Gametes: AB P Ab ab P aB Ab R AB aB R ab

24. Independent Assortment Linkage Fig 6-6 Fig 6-11 Interchromosomal Intrachromosomal

25. Example Test Cross AaBb X aabb ab Exp. Obs. AB AaBb 25 10 R Ab Aabb 25 40 P aB aaBb 25 40 P ab aabb 25 10 R 100 100 How to distinguish: Parental high freq. Recombinant low freq.

26. Example (cont.) Gametes: AB R Ab P aB P ab R Therefore dihybrid: A b (trans) a B

27. Linkage Maps Genes close together on same chromosome: - smaller chance of crossovers between them - fewer recombinants Therefore: percentage recombination can be used to generate a linkage map

28. Linkage maps A B large # of recomb. a b C D small number of recombinants c d Alfred Sturtevant (1913)

29. Linkage mapsexample 65 Testcross progeny: P AaBb 2146 R Aabb 43 R aaBb 22 P aabb 2302 Total 4513 1.4 map units = 1.4 % RF 4513 A 1.4 mu B

30. Additivity of map distances separate maps A B A C 7 2 combine maps C A B 2 7 or Locus A C B (pl. loci) 2 5

31. Linkage Deviations from independent assortment Dihybrid gametes 2 parent (noncrossover) common 2 recombinant (crossover) rare % recombinants a function of distance between genes % RF = map distance

32. Linkage maps Drosophila Tomato Linkage group = chromosome

33. Summary Dihybrid Cross (Indep. Assort.): - ratios (9:3:3:1, 1:1:1:1) - linkage (deviation from I.A.) - recombination - linkage maps Mendelian Genetics: Monohybrid cross (segregation): - ratios (3:1, 1:2:1, 1:1) - dominance, recessive - autosomal, sex-linked - probability - pedigrees

34. Gametes Number of Genes Number of Different Gametes monohybrid 1 (Aa) 2 dihybrid 2 (AaBb) 4 trihybrid 3 (AaBbCc) ?

35. Trihybrid Three Point Test Cross AaBbCc X aabbcc ABC ABc abc AbC Abc aBC aBc abC abc 8 gamete types

36. Three Point Test Cross Trihybrid Gametes C ABC B c ABc A C AbC b c Abc a

37. Three Point Test Cross Trihybrid AaBbCc 3 genes: Possibilities: 1. All unlinked 2. Two linked; one unlinked 3. Three linked 1 2 3

38. Three genes Wild (+) mutant 1. Eye colour 2.Wing 3. Wing v cv ct

39. Three Point Test Cross Three recessive mutants of Drosophila: +, v vermilion eyes +, cv crossveinless +, ct cut wing P +/+ cv/cv ct/ct X v/v +/+ +/+

40. Three Point Test Cross P +/+ cv/cv ct/ct x v/v +/+ +/+ Gametes + cv ct v + + F1 trihybrid v/+ cv/+ ct/+

41. Three Point Test Cross F1 v/+ cv/+ ct/+ x v/v cv/cv ct/ct v cv ct 8 gamete types one gamete type

42. 8 gamete types Parental = non crossover (most frequent) F1 v/+ cv/+ ct/+ v + + 580 Parental + cv ct 592 Parental v cv + 45 + + ct 40 v cv ct 89 Recombinant + + + 94 v + ct 3 + cv + 5 1448

43. 8 gamete types Examine two genes at a time Parental F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Recombinant Recombinant Parental

44. 8 gamete types Parental F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Parental Recombinant Recombinant

45. 8 gamete types Parental F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Recombinant Parental Recombinant

46. Calculate Recombination Fraction 1. v - cv R v cv 45 + 89 R + + 40 + 94 268 / 1448 = 18.5 % 2. v - ct R + + 94 + 5 R v ct 89 + 3 191/1448 = 13.2 % 3. ct - cv R ct + 40 + 3 R + cv 45 + 5 93/1448 = 6.4 %

47. Three point test cross Observations: all 3 RF < 50 % 3 genes on same chromosome v-----cv largest distance ct in middle map v-------ct-------cv = cv-------ct-------v 13.2 + 6.4 = 19.6 > 18.5 !! Why ?

48. Three Point Test Cross P +/+ ct/ct cv/cv x v/v +/+ +/+ gametes + ct cv v + + F1 trihybrid v + + + ct cv Correct gene order

49. Three Point Test Cross X X X X Double crossover class rarest: v---cv P v + + v + P + ct cv + cv R v ct + v + R + + cv + cv

50. Three Point test cross 1. Double crossovers not counted in v--cv RF 2. Double crossovers generate P types (with respect to v--cv) 3. Double crossovers not detected as recombinants Consequence: underestimate of v----cv map distance Greater distance of genes  greater error