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Download Lecture Outline 9/15 Chi-square Test for Independence Chi
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Chi-square Test for Independence Lecture Outline 9/15 • Linkage of genes on chromosomes • Test Cross AaBb x aabb • Results: – Estimating recombination fraction • Chi-square tests for independence – Predicting offspring phenotypes when genes are linked – Three-point mapping • • • • A-BA-bb aaBaabb Total = 200 I have posted a new homework assignment on the web page Are these loci independent? If not, How far apart are they? Chi-square Test for Independence Chi-square Test for Independence • 2 way table of gamete frequencies for each locus • Test if genes assort independently into GAMETES • Results of test cross: • • • • A-BA-bb aaBaabb 70 20 25 85 What did the parental gametes have to be? 70 20 25 85 A a total B 70 25 95 b 20 85 105 total 90 110 200 Total = 200 Expect? What is the Prob( A and b), if the loci are independent? Number of gametes that had big A and little b Chi-square Test for Independence Chi-square Test for Independence • 2 way table of gamete frequencies for each locus Expected value = Row total * Column total / Grand Total B A a total 70 25 95 b 20 total 90 47.25 85 105 110 200 Expected NUMBER = 0.236 * 200 = 47.25 What is the Prob( A and b), if the loci are independent? P(A) = 90/200 P(b) = 105/200 Prob (A and b) =90/200 * 105/200 = 0.236 B b total A a total 70 25 95 E=95*90/200 E=42.75 E=95*110/200 E=52.25 20 85 E=47.25 E=57.75 90 110 easy formula 105 200 1 Chi-square Test for Independence Chi-square Test for Independence Degrees of freedom= (Rows-1) * (Cols-1) X2=Sum((O-E)2/E) A B b total a total 70 25 95 E=42.75 E=52.25 20 85 E=47.25 E=57.75 90 110 B 105 b 200 total A a total 70 25 95 E=42.75 E=52.25 20 85 E=47.25 E=57.75 90 110 X2= • • • • • • Are these loci independent? A-B- 70 A-bb 20 aaB- 25 aabb 85 Total = 200 – No (s ignificant lack of fit: 60.14 >> 3.84) • How far apart are they? 17.37 +14.21+15.71+12.85 = 105 200 60.14 Compare this to the table of X2 for 1 df Predicting phenotypes in crosses with linked genes Testcross AaBb x aabb • Results: <-- This formula is the same as before • In maize, the genes g and r are 20 cM apart on Chromosome 10. What are the expected phenotypic classes in the cross: – what fraction of the gametes are recombinant? gr/GR X gr/gr • What would you expect if the map distance was zero? • What would you expect if they were unlinked? So, • What if r=0.20? 45/200 = 0.225 = 22.5 cM Predicting phenotypes in crosses with linked genes Three-point crosses Cross: Possible Offspring Genotypes Gamete type GgRr Ggrr ggRr Ggrr Parental Recombinant Recombinant Parental Map distance = 20 cM Gray body,Red-eye,normal wing x yellow body, white eye, miniature wing (+++ x ywm) Total recombinants = 20% Shorthand notation for phenotypes Results: +++ ++m +w+ +wm y++ y+m yw+ ywm 758 401 1 16 12 0 317 700 --What is the order of these genes on the X chromosome? --What are the map distances? 2 Three-point crosses 1. Find reciprocal pairs Results: +++ ++m +w+ +wm y++ y+m yw+ ywm 758 401 1 16 12 0 317 700 Three-point crosses 3. Compare parentals and double recombinants to find the middle gene Results: 2. Which are the parentals? single recombinants? double recombinants? +++ ++m +w+ +wm y++ y+m yw+ ywm 758 401 1 16 12 0 317 700 + + + + y w + + m y w m See fig 4-13 Three-point crosses 4. Calculate the recombination fraction between neighboring pairs of loci Results: +++ ++m +w+ +wm y++ y+m yw+ ywm 758 401 1 16 12 0 317 700 • Find reciprocal pairs • Use double recombinants to find gene order – Which are the double recombinants? y ? w m Recombination fraction between y and w =single crossovers + double crossovers =(16+12+1+0)/2205 =1.3% Now you try it: dan da+ d+n d++ +++ ++n +a+ +an Three point crosses 299 5 131 65 308 8 124 60 Total = 1000 1. Which are the parental classes? 2. Which are the single recombinants? 3. Which are the double recombinants? 4. What is the gene order? • Then estimate recombination between the two pairs of loci to get map distances – Don’t forget to add in the double crossovers! Interference • You can predict the frequecy of double crossovers: – it should be the product of the two single crossovers • R(ab and bc) = R(ab) * R(bc) Why is it a product? • Often the observed number is less than expected due to interference (rewrite in correct gene order if necessary) 5. What are the recombination distances for each neighboring pair of genes? Coincidence coefficient = observed double crossovers / expected double crossovers • Interference can differ among organisms – In Drosophila, there is complete interference (no double crossovers) to about 10 cM; – in yeast, only partial interference at 3 cM – There is essentially no interference beyond 20 or 30 cm in most organisms 3