FA2011lecture12

FA2011lecture12 - B B E E b b e e B B b b E e E e KEY...

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Unformatted text preview: B B E E b b e e B B b b E e E e KEY CONCEPT: FREQUENCY OF RECOMBINATION BETWEEN TWO GENES IS PROPORTIONAL TO THE DISTANCE BETWEEN THEM purple owers, red owers, X elongated pollen round pollen F1 dihybrid The frequency of recombination (proportion of gametes that have a recombinant chromosome) de nes map distance in centiMorgans (cM) F1 dihybrid ester + gtametes: gametes: NR 44% NR 44% R 6% R 6% (NR = non-recombinant; R = recombinant) F2 progeny: 44% 44% 6% 6% purple, elong red, round purple, round red, elong this result tells you that 12% of gametes were recombinant, so genes are 12 cM (m.u.) apart! Instead of this cross: purple owers, red owers, X elongated pollen round pollen What if cross was like this instead? purple owers, red owers, X round pollen elongated pollen in coupling or in cis F1 dihybrid in repulsion or in trans F1 dihybrid ester + gtametes: gametes: NR 44% NR 44% R 6% R 6% (NR = non-recombinant; R = recombinant) F2 progeny: 44% 44% 6% 6% purple, round red, elong purple, elong red, round Still 12% recombination indicating 12 cM, but which classes are recombinant vs. nonrecombinant are opposite compared to previous cross! Instead of crossing F1 dihybrid with a tester, what if it is selfed or crossed with another F1 instead? F1 dihybrid 1 dihybrid + Fgametes: gametes: NR NR R R 44% 44% 6% 6% 44% 44% 6% 6% F2 progeny: (2 NR) purple, elong (1 NR +1R) purple, elong (2 R) purple, elong . . . (altogether 9 genotypes, 4 phenotypes) Problem: F2 progeny with same phenotype can have different numbers of recombinant chromosomes, so relationship between phenotypic ratios and recombination frequency not straightforward to calculate. Solution: use the following tables! Must know arrangement of alleles in dihybrid (in cis/coupling, as in this example, or in trans/repulsion) and use correct table. Cen$morgans 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 unlinked Expected Propor,ons of F2 Progeny for Two Linked Genes with Complete Dominance Coupling Conforma.on Repulsion Conforma.on A_B_ A_bb aaB_ aabb Cen$morgans A_B_ A_bb aaB_ aabb 0.7450 0.0050 0.0050 0.2450 1 0.5000 0.2500 0.2500 0.0000 0.7401 0.0099 0.0099 0.2401 2 0.5001 0.2499 0.2499 0.0001 0.7352 0.0148 0.0148 0.2352 3 0.5002 0.2498 0.2498 0.0002 BB EE x bb ee 0.7304 0.0196 0.0196 0.2304 4 0.5004 0.2496 0.2496 0.0004 0.7256 0.0244 0.0244 0.2256 5 0.5006 0.2494 0.2494 0.0006 0.7209 0.0291 0.0291 0.2209 6 0.5009 0.2491 0.2491 0.0009 0.7162 0.0338 0.0338 0.2162 7 0.5012 0.2488 0.2488 0.0012 0.7116 0.0384 0.0384 0.2116 8 0.5016 0.2484 0.2484 0.0016 0.7070 0.0430 0.0430 0.2070 9 0.5020 0.2480 0.2480 0.0020 0.7025 0.0475 0.0475 0.2025 10 0.5025 0.2475 0.2475 0.0025 0.6980 0.0520 0.0520 0.1980 11 0.5030 0.2470 0.2470 0.0030 0.6936 0.0564 0.0564 0.1936 12 0.5036 0.2464 0.2464 0.0036 matches actual results, conclude distance = 12 cM 0.6892 0.0608 0.0608 0.1892 13 0.5042 0.2458 0.2458 0.0042 0.6849 0.0651 0.0651 0.1849 14 0.5049 0.2451 0.2451 0.0049 0.6806 0.0694 0.0694 0.1806 15 0.5056 0.2444 0.2444 0.0056 0.6764 0.0736 0.0736 0.1764 16 0.5064 0.2436 0.2436 0.0064 0.6722 0.0778 0.0778 0.1722 17 0.5072 0.2428 0.2428 0.0072 0.6681 0.0819 0.0819 0.1681 18 0.5081 0.2419 0.2419 0.0081 0.6640 0.0860 0.0860 0.1640 19 0.5090 0.2410 0.2410 0.0090 0.6600 0.0900 0.0900 0.1600 20 0.5100 0.2400 0.2400 0.0100 0.6560 0.0940 0.0940 0.1560 21 0.5110 0.2390 0.2390 0.0110 0.6521 0.0979 0.0979 0.1521 22 0.5121 0.2379 0.2379 0.0121 0.6482 0.1018 0.1018 0.1482 23 0.5132 0.2368 0.2368 0.0132 0.6444 0.1056 0.1056 0.1444 24 0.5144 0.2356 0.2356 0.0144 0.6406 0.1094 0.1094 0.1406 25 0.5156 0.2344 0.2344 0.0156 0.5625 0.1875 0.1875 0.0625 unlinked 0.5625 0.1875 0.1875 0.0625 (9/16) (3/16) (3/16) (1/16) Expected Propor,ons of F2 Progeny for Two Linked Genes with Complete Dominance Coupling Conforma.on Repulsion Conforma.on Cen$morgans A_B_ A_bb aaB_ aabb Cen$morgans A_B_ A_bb aaB_ 1 0.7450 0.0050 0.0050 0.2450 1 0.5000 0.2500 0.2500 2 0.7401 0.0099 0.0099 0.2401 2 0.5001 0.2499 0.2499 3 0.7352 0.0148 0.0148 0.2352 3 0.5002 0.2498 0.2498 4 0.7304 0.0196 0.0196 0.2304 4 0.5004 0.2496 0.2496 5 0.7256 0.0244 0.0244 0.2256 5 0.5006 0.2494 0.2494 6 0.7209 0.0291 0.0291 0.2209 6 0.5009 0.2491 0.2491 7 0.7162 BB ee x bb EE 0.0338 0.0338 0.2162 7 0.5012 0.2488 0.2488 8 0.7116 0.0384 0.0384 0.2116 8 0.5016 0.2484 0.2484 9 0.7070 0.0430 0.0430 0.2070 9 0.5020 0.2480 0.2480 10 0.7025 0.0475 0.0475 0.2025 10 0.5025 0.2475 0.2475 11 0.6980 0.0520 0.0520 0.1980 11 0.5030 0.2470 0.2470 12 0.6936 0.0564 0.0564 0.1936 12 0.5036 0.2464 0.2464 13 0.6892 0.0608 0.0608 0.1892 13 0.5042 0.2458 0.2458 14 0.6849 0.0651 0.0651 0.1849 14 0.5049 0.2451 0.2451 15 0.6806 0.0694 0.0694 0.1806 15 0.5056 0.2444 0.2444 16 0.6764 0.0736 0.0736 0.1764 16 0.5064 0.2436 0.2436 17 17 0.5072 0.2428 0.2428 Notice that 0.6722 0.0778 proportions the progeny 0.0778 0.1722 18 18 0.5081 0.2419 0.2419 observed in0.6681 purple elongated X red the 0.0819 0.0819 0.1681 19 0.6640 0.0860 0.0860 0.1640 19 0.5090 0.2410 0.2410 round cross0.6600 0.0900 here because this are not found 0.0900 0.1600 20 20 0.5100 0.2400 0.2400 is the wrong table! 0.0940 0.0940 0.1560 21 0.6560 21 0.5110 0.2390 0.2390 22 0.6521 0.0979 0.0979 0.1521 22 0.5121 0.2379 0.2379 23 0.6482 0.1018 0.1018 0.1482 23 0.5132 0.2368 0.2368 24 0.6444 0.1056 0.1056 0.1444 24 0.5144 0.2356 0.2356 25 0.6406 0.1094 0.1094 0.1406 25 0.5156 0.2344 0.2344 unlinked 0.5625 0.1875 0.1875 0.0625 unlinked 0.5625 0.1875 0.1875 (9/16) (3/16) (3/16) aabb 0.0000 0.0001 0.0002 0.0004 0.0006 0.0009 0.0012 0.0016 0.0020 0.0025 0.0030 0.0036 0.0042 0.0049 0.0056 0.0064 0.0072 0.0081 0.0090 0.0100 0.0110 0.0121 0.0132 0.0144 0.0156 0.0625 (1/16) Incredibly easy five step program for conducting… Chi Squared Test For Mendelian Ratios Linkage 1.  State “null hypothesis”: Mendelian ratio is observed (expected if phenotype is controlled by alternate alleles of a single gene, or two genes, exhibiting simple dominant/recessive relationship) Genes are unlinked (assort independently) 2. Calculate expected number of progeny in each class predicted by Mendel’s law of independent assortment, e.g. 3:1 or 9:3:3:1 ratio (if dihybrid selfed) or 1:1:1:1 (if dihybrid testcrossed) (O - E)2 3. Calculate Chi squared = 4. Translate chi squared value into a p value = probability that observed results would be obtained if the hypothesis is true (need degrees of freedom = # of progeny classes minus 1) 5. Draw a conclusion about your hypothesis (accept if p>0.05 and conclude that Mendelian ratios are observed genes are unlinked; reject if p<0.05 and conclude that data do not fit Mendelian ratios genes are linked). E ...
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