1. Step one Two gene loci: A & B What will your first cross be in an experiment to test for possible meiotic crossing over? Hint: what condition do you have to have for crossing to be visible if it happens?

2. AAbb X aaBB AaBb Why is this genotype necessary? What is the next cross, and why?

3. AaBb X aabb Why is this the right cross? What are the expected outcomes? In what ratio?

4. Expected Outcomes ABAbaBab AaBbAaBbAabbAabbaaBbaaBbaabbaabb 1/4 250 1/4 250 1/4 250 1/4 250

5. Observed ABAbaBab AaBbAaBbAabbAabbaaBbaaBbaabbaabb 4852015480 What is the null hypothesis here? Do the chi square test on the data. What is the % crossing over? How many map units apart are the loci?

6. To calculate % crossing over % crossing over = (20+15)/(20+15+480+485) = 35/1000 = 0.035 = 3.5% = 3.5 map units

7. Now a triple cross Three gene loci: A, B & C As before, what will your first cross be in an experiment to test for possible meiotic crossing over?

8. AABBCC X aabbcc AaBbCc Why is this genotype necessary? Draw the chromosomal structure of this result. What is the next cross, and why?

9. The Cross A B C a b c X

10. The results ABC400 abc385 Abc10 aBC18 ABc40 abC35 AbC70 aBc85

11. Sums Non-recombinants = (400 + 385) = 785 AB recombinants = (10+18+70+85) = 183 BC recombinants = (40+35+70+85) = 230 AC recombinants = (10+18+40+35) = 103

12. Calculations % Non-recombinants = 785/1043 = 75.3 AB recombinants = 183/1043 = 17.5 BC recombinants = 230/1043 = 22.1 AC recombinants = 103/1043 = 9.9

13. B A C 17.5 mu9.9 mu 22.1 mu Resulting Map

14. Assessment The values yield the correct order of the loci along the chromosome. There is a discrepancy between the two values obtained for the distance between B and C: 22.1 mu and 27.4 mu (17.5+9.1). We assume that the greater of the two values is more accurate. – Mapping is more accurate over shorter distances – The values 17.5 and 9.1 will be more accurate individually than the larger value, 22.1.

15. Explanation The values for long distances along the chromosome are skewed by the frequency of double crossing over. – The greater the distance, the greater the probability of double crossing over – The standard calculations fail to consider the contribution of double crossing over – The value 22.1 underestimates the crossing over frequency between B and C and thus underestimates the BC distance

16. Logic We must correct for the frequency of double crossing over. To do so, we must determine which values represent double crossing over.

17. The Chromosomes Ba C bAc BA C bac Before crossing over After crossing over

18. The Correction Referring to the map, double crossing over occurs between B and C. To correct the BC map distance, we must add in the values for the double cross over species – (10+18) = 28 – However, because each counted individual in the population represents a double crossing over event, we must double these values. 2*(10+18) = 56

19. Corrected Sums Non-recombinants = (400 + 385) = 785 AB recombinants = (10+18+70+85) = 183 BC recombinants = (56+40+35+70+85) = 286 AC recombinants = (10+18+40+35) = 103

20. Corrected Calculations % Non-recombinants = 785/1043 = 75.3 AB recombinants = 183/1043 = 17.5 BC recombinants = 286/1043 = 27.4 AC recombinants = 103/1043 = 9.9

21. B A C 17.5 mu9.9 mu 27.4 mu Corrected Map