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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?. AAbb X aaBB. AaBb Why is this genotype necessary?

natalya-penda
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  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

  5. Observed 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 a b c a b c

  10. The results

  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 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 B A C Before crossing over b b A a c c B a C 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 Corrected Map

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