1. Introduction to Interference By: Nickolay Dovgolevsky Itai Sharon 29/05/03

2. 2 Agenda Biological context Three point analysis Chi-square test Result analysis Exercise execution

3. 3 Biological Context: Meiosis

4. 4 Biological Context: Recombination Recombination (crossing over) occurs between homologous chromosomes

5. 5 Each two loci S and T have recombination rate  ST Biological Context: Recombination

6. 6 Recombination (crossing over) occurs between homologous chromosomes Each two loci S and T have recombination rate  ST If recombination rate between loci A and B is independent of rate between B and C, then  AC =  AB +  BC – 2*  AB *  BC

7. 7 Interference Interference is the phenomenon whereby crossovers do not occur independently along a chromosome This means:  AC   AB +  BC – 2  AB *  BC

8. 8 Measuring Interference Coefficient of Coincidence (c) – c =  AB +  BC –  AC 2  AB *  BC Interference (I) – I = 1 – c c > 1: Negative interference c = 1: No interference 0 < c < 1: Partial interference c = 0: Complete interference

9. 9 Three Point Analysis Assuming three co-linear loci A, B and C, estimate  AB,  BC and  AC by calculating percentage of recombinants for each  (Ott, 1991). No restriction on absence of interference ILINK is capable of analyzing three- point data

10. 10 Three Point Analysis Where:  = 1 -  -  -  Also:  AC =  +  A-B B-C RecombinantsNon- Recombinants Total Recombinants  BC Non- Recombinants  1-  BC Total  AB 1-  AB

11. 11 Results Estimation for Three Point Analysis We can estimate the significance of the results by calculating L(  AB,  BC, c) 2 ln L(  AB,  BC, c=1) ~  2 (1) where L(  AB,  BC, c) – likelihood of  AB,  BC and c L(  AB,  BC, c=1) – likelihood of  AB,  BC and c=1  2 (1) is the chi-square value with df=1

12. 12 The  2 (Chi-Square) Test Given An hypothesis H Expected numbers E according to H Observed numbers O The  2 test gives the probability of seeing numbers greater than, or equal to O instead of E, assuming H

13. 13 The  2 Test: an Example We cross two pure lines of plants, one with yellow petals and one with red. The following numbers are observed on F2: orange:182 yellow: 61 red : 77 Total:320 P F1 F2 yellow X red  orange  yellow, red, orange

14. 14 The  2 Test: an Example Hypothesis: Incomplete dominance 2 alleles, G1 (yellow) and G2 (red) with similar dominance; G1/G2 gives orange. Observed numbersExpected numbers on F2:on F2: orange:182orange:160 yellow: 61yellow: 80 red : 77red : 80 Total:320Total:320

15. 15 The  2 Test: an Example Calculating the  2 value: O E (O – E) 2 / E orange:1821603.0 yellow: 61 80 4.5 red : 77 800.1  2 = 7.6 Degrees of freedom: setting the sizes of two of the three phenotypic classes sets the third. This means: df = 2

16. 16 The  2 Test That’s how it works: Calculate (O – E) 2 /E for each class Calculate the  2 value (  (O – E) 2 /E) Conclude amount of degrees of freedom Find the probability according to the  2 distribution table Accept the hypothesis if p < 0.05

17. 17 The Chi-Square Test

18. 18 Estimating Interference Assume we have 3 loci, with the following  ST calculated:  13 = 0.270  12 = 0.197  23 = 0.314 then 0.270+0.197–0.314 2*0.270*0.197  c = 1.44

19. 19 We know that: Estimating Interference In this example we’ll get:  2 = 878.94 – 878.75 = 0.19  p = 0.66 How significant is this result? ILINK computes –2ln(L(  AB,  BC, c)) We can execute ILINK with “no interference” for –2ln(L(  AB,  BC,c=1))

20. 20 Estimating Interference with sex difference In general – if there are differences between recombination rates among males and females, c T tends to be greater than both It is possible to calculate c m and c f separately It is possible to assume constant sex difference or varying sex difference

21. 21 Remarks Ott (1991) determined that for Kosambi level interference and three equally spaced markers (  = 0.15), 847 meiosis are required to reject the “no human crossover interference” hypothesis Broman et. Al. (2000) found strong evidence for positive interference in the levels implied by Kosambi and Carter-Falconer map functions

22. 22 Exercise

23. 23 Loci Selected 5 – (0.075) – 2 – (0.075) – 8 – (0.225) – 3 – (0.075) – 6 – (0.075) – 4 – (0.075) – 9 – (0.075) – 7 5 – (0.150) – 8 – (0.225) – 3 8 – (0.225) – 3 – (0.150) – 4 8 – (0.225) – 3 – (0.225) – 9 3 – (0.150) – 4 – (0.150) – 7