GENETIC MAPPING III

1. GENETICMAPPING III

2. The problem of double crossovers in genetic mapping experiments

3. Consider a cross to map 2 genes, a and bThey are some distance apart, but mappableThe heterozygote is in tetrad stage:a+ b+/ab

4. A single crossover generates recombinant chromosomes Which give recombinant gametes and eventually recombinant progeny

5. A 2-strand, double crossover restores the original arrangement of the marker genes So all progeny are scored as parental, with no recombinants

6. It looks exactly as if there has been no crossing over There have been two crossover events which will be uncounted

7. And since recombination frequency is a measure of map distance, this means that the distance between the genes will be underestimated

8. How can we avoid these errors?Two general ways:

9. 1. Map closely linked genesDouble crossovers rarely occur within map distances < 10 cM

10. 2. Do three-point testcrosses, rather than two-pointThese involve 3 genes within a relatively short section of chromosome

11. The rationale for using these is illustrated in the next slide.

12. As before, a 2-strand double crossover gives gametes that are nonrecombinant for genes a and b

13. BUT, notice that the resulting gametes are recombinant with respect to c

14. The gene in the middle reveals the occurrence of a double crossover 3-point crossovers are routinely used for mapping, because they allow us to correct for double crossovers, and determine the gene order

15. Suppose we want to map 3 genes in a plantFruit color: p = purple; p+ = yellowFruit shape: r = round; r+ = elongatedJuiciness: j = juicy; j+ = dry

16. What is the order, and map distances, of these 3 genes?

17. We set up our testcross with a triply heterozygous parent, in coupling phase (in this case) and count the offspring

18. We know that is the genes were unlinked, we would expect eight phenotypic classes of progeny

19. For this kind of trihybrid cross, we expect the same classes, but not in the same proportions

20. Because of linkage, some phenotypic classes may have 0 individuals; if so, that’s important to note

21. Here are the eight phenotypic classes of progeny

22. These are parentals. Note that they are in approximately equal numbers

23. These recombinants both involve the p gene Notice that they are in about equal numbers, and are rarer than the parentals

24. These recombinants involve the r gene They are rarer still

25. These recombinants are the rarest. Gene j is involved

26. We expect double crossovers to be rarer than single crossovers So it follows that recombinants due to double crossovers will be the rarest class

27. We can use this fact to help us order the genes. How? Recall our earlier example:

28. Notice how the double crossover restored the outside genes to the parental arrangement, but the middle gene has its orientation changed

29. So the gene which is in a recombinant arrangement in the rarest, double crossover class of progeny, must be the middle gene.

30. We can see that p and r are in their parental configuration, but j is in a new arrangement So, j must be the gene in the middle The order must be p , j , r

31. Now that we know the correct gene order, we can interpret the data to generate map distances:

33. For the p - j distance, we need to add together all the recombinant progeny resulting from crossovers in Region I

34. This includes both single crossovers and the double crossovers (since they also involve this region of the chromosome)

35. So, the percentage of recombinants = [(52+46) + (4+2)]/500 x 100% = 104/500 x 100% = 20.8% So, p and j are 20.8 cM apart

36. We do the same sort of calculations to find the distance between j and r

37. We again add together the single crossovers (this time from Region II) and the double crossovers

38. [(22+22) + (4+2)]/500 x 100% = 50/500 x 100% = 10.0% So, j and r are 10.0 cM apart

39. Our linkage map now looks like this. To get the distance between p and r, we simply add the inner distances = 30.8 cM