1. Linkage and Mapping Bonus #2 due now

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

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

4. Two genes control coat color in mice Fig 4.21

5. Variation in Peas Fig 3.2

6. Fig 2.8 Inheritance of 2 independent genes

7. 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

8. There must be a better way… Fig 2.9

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

10. 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

11. 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?

12. 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

13. 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

14. 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?

15. 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

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

17. 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

18. 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

19. Linked alleles are more likely to be inherited together than non-linked alleles

20. Recombinants are produced by crossing over

21. Crossing over produces new allelic combinations

22. Only 2 chromosomes cross-over, and so the maximum number of recombinants that can be produced is 50%.

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

24. Homologous pair of chromosomes

25. Does this show recombination? D/d M1/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2 D/d M2/M2 d/d M2/M2

26. Does this show recombination? D/d M1/M2 d/d M1/M2 D/d M1/M2 d/d M2/M2 D/d M2/M2 d/d M2/M2

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

28. Some crosses do not give the expected results

29. =25% 8%9%41%42%

30. These two genes are on the same chromosome

32. By comparing recombination frequencies, a linkage map can be constructed

33. = 17 m.u.

34. 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.

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

36. 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?

37. Homologous pair of chromosomes Linkage can be used to determine distance

38. Double recombinants arise from two crossovers Recombinant

39. Double recombinants can show gene order

40. 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?

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

42. Double crossover

43. Which order produces the double crossover?

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

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

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

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

49. Drosophila linkage map

50. Linkage map of Drosophila chromosome 2

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

52. Alignment of physical and recombination maps

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