Biology 2250 Principles of Genetics

1. Biology 2250Principles of Genetics Announcements Lab 4 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 – Oct. 31 Quiz 2 2 Nov. 4 Quiz 3 2 Nov. 10 Quiz 4 2

4. 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 

5. 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

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

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

8. 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

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

10. Crossing over in meiosis I (animation)

11. 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.

12. 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

13. 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.

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

15. 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

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

17. 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

18. 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

19. 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

20. Linkage maps Drosophila Tomato Linkage group = chromosome

21. Practice Questions: 1. Gene A and gene B are linked. A test cross produces 10 AaBb progeny out of a total of 100. The estimated map distance between gene A and B is: a. 10 b. 20 c. 30 d. 40 e. 50 2. For the pedigree, indicate the most probably mode of inheritance for the rare trait. 3. For the pedigree, what is the probability that the indicated female will produce an affected child?

22. Practice Questions: 1. Mode of inheritance: every generation; father to daughter: sex-linked dominant 2. Probability that the indicated female will produce an affected child? Aa x aY  ¼ Aa ¼ aa ¼ AY ¼ aY Prob. = 1/2

23. Quiz 2 Questions Quiz 2 Answers: http://webct.mun.ca:8900/

24. Using Linkage to Hunt for Human Disease Genes Basic approach: Collect pedigree information on disease Collect blood samples from individuals Correlate genetic marker information with disease Use recombinants to map gene and marker

25. Huntington’s Disease

26. Huntington’s Disease Not sex-linked Autosomal Dominant

27. Huntington’s DiseaseLinked marker

29. Huntington’s Disease As a result of the gene discovery, a direct genetic test for Huntington's disease has replaced the indirect linkage marker test. While the Huntington's disease gene discovery alters the technical aspects of predictive testing for Huntington's disease, there is still no cure for Huntington's disease and no available treatment to delay its onset or to slow, stop or reverse the disease's relentless progression.

30. Hunting for Human Disease Genes Newfoundland Population Small founding population – high freq. of allele Isolated – little gene flow Inbreeding – increased chance of “aa” Good pedigree records

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

32. Rare autosomal recessive BBS1 Chr - 11

33. B10 Aa Aa Chromosome 11 aa aa Aa AA Aa Aa Aa AA Aa AA aa aa AA 3 (2.5) Aa 5 (5) aa 2 (2.5)

34. Genetic Diseases dominant

35. Gene Discovery Genetic marker and linkage analysis Narrow location of gene (chromosome and region) Genome sequencing  identify gene Genetic counseling, gene therapy??

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

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

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

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

40. Three genes Wild (+) mutant 1. Eye colour 2.Wing 3. Wing v vermillion cv crossveinless ct cut wing

41. 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 +/+ +/+

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

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

44. 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

45. 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

46. 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

47. 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

48. 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 %

49. 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

50. Three point test cross 13.2 6.4 18.5 Observations: map v ct cv 13.2 + 6.4 = 19.6 > 18.5 !! Why ?