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These are the user uploaded subtitles that are being translated: 0 00:00:00,000 --> 00:00:00,660 1 00:00:00,660 --> 00:00:04,530 PETER REDDIEN: So let's turn now to informative meiosis 2 00:00:04,530 --> 00:00:08,910 and phase in order to be able to calculate our LOD scores. 3 00:00:08,910 --> 00:00:27,940 4 00:00:27,940 --> 00:00:30,822 OK, I'll start with informative meiosis. 5 00:00:30,822 --> 00:00:32,530 I have it defined on the screen, and I'll 6 00:00:32,530 --> 00:00:35,060 come to that in a moment. 7 00:00:35,060 --> 00:00:37,130 So let's think about what we're trying to do. 8 00:00:37,130 --> 00:00:39,860 We're trying to do some mapping, essentially, here. 9 00:00:39,860 --> 00:00:44,720 And, when we've gone through this before, 10 00:00:44,720 --> 00:00:47,150 one way we've looked at it is, for example, 11 00:00:47,150 --> 00:00:49,900 in a test cross where we have some scenario like this where 12 00:00:49,900 --> 00:00:52,180 an individual is heterozygous for two alleles 13 00:00:52,180 --> 00:00:53,620 of the A gene and the B gene. 14 00:00:53,620 --> 00:00:58,075 15 00:00:58,075 --> 00:00:59,980 We crossed with some tester strain. 16 00:00:59,980 --> 00:01:01,060 So this is a test cross. 17 00:01:01,060 --> 00:01:05,370 18 00:01:05,370 --> 00:01:06,970 This test cross was nice and handy. 19 00:01:06,970 --> 00:01:09,630 We set this up in a way where we can determine 20 00:01:09,630 --> 00:01:13,260 the genotype of every gamete produced 21 00:01:13,260 --> 00:01:16,990 by this individual and their offspring. 22 00:01:16,990 --> 00:01:25,240 So the gametes that we could infer, if you recall, 23 00:01:25,240 --> 00:01:30,670 would have been big A big B, little a little B, big A little 24 00:01:30,670 --> 00:01:36,497 b, little a big B. And we can see in all of the offspring 25 00:01:36,497 --> 00:01:38,080 what the identity of the gametes where 26 00:01:38,080 --> 00:01:40,840 that came from this individual just by the phenotype 27 00:01:40,840 --> 00:01:42,600 of the offspring. 28 00:01:42,600 --> 00:01:45,640 And we figured out that these were non-recombinant 29 00:01:45,640 --> 00:01:49,590 and these were recombinant. 30 00:01:49,590 --> 00:01:51,210 And then we used that information 31 00:01:51,210 --> 00:01:53,640 to get a map distance by the frequency-- 32 00:01:53,640 --> 00:01:55,560 the fraction of the gametes that were 33 00:01:55,560 --> 00:01:58,520 recombinant over the total. 34 00:01:58,520 --> 00:02:01,130 OK, so that's what we've done in the past. 35 00:02:01,130 --> 00:02:03,380 Now, we want to try to do something similar here 36 00:02:03,380 --> 00:02:06,770 with data in the pedigree, but the challenge 37 00:02:06,770 --> 00:02:10,639 is, as I've been talking about, we 38 00:02:10,639 --> 00:02:12,740 can't design the ideal cross. 39 00:02:12,740 --> 00:02:14,370 We have to live with what we've got. 40 00:02:14,370 --> 00:02:17,390 So we just have to look at the meiosis we have 41 00:02:17,390 --> 00:02:20,600 and figure out if we can infer this kind of information 42 00:02:20,600 --> 00:02:23,180 from the data we have. 43 00:02:23,180 --> 00:02:24,950 So that's the extra challenge that we 44 00:02:24,950 --> 00:02:27,380 have to go through here. 45 00:02:27,380 --> 00:02:28,880 So, just to put it in context, let's 46 00:02:28,880 --> 00:02:39,800 imagine a different kind of cross like this. 47 00:02:39,800 --> 00:02:42,520 48 00:02:42,520 --> 00:02:45,020 Could we get any information out of this cross about the map 49 00:02:45,020 --> 00:02:47,000 distance between A and B? 50 00:02:47,000 --> 00:02:47,810 What do you think? 51 00:02:47,810 --> 00:02:53,790 52 00:02:53,790 --> 00:02:55,810 I've got a thumbs down in the back. 53 00:02:55,810 --> 00:02:57,930 Why not? 54 00:02:57,930 --> 00:03:00,550 So what kind of gametes could we get here? 55 00:03:00,550 --> 00:03:05,190 So the gametes from this individual 56 00:03:05,190 --> 00:03:10,260 would be big A big B, little a big B. 57 00:03:10,260 --> 00:03:12,480 Can we tell whether recombination happened 58 00:03:12,480 --> 00:03:14,700 between the A gene and the B gene? 59 00:03:14,700 --> 00:03:15,990 We can't tell. 60 00:03:15,990 --> 00:03:18,300 So are we going to get any information here 61 00:03:18,300 --> 00:03:20,220 that allows us to tell recombinant fraction? 62 00:03:20,220 --> 00:03:22,860 No. 63 00:03:22,860 --> 00:03:25,170 And the problem is, we may have many meiosis like this 64 00:03:25,170 --> 00:03:27,810 in a pedigree where you might want to get information, 65 00:03:27,810 --> 00:03:29,640 but you can't. 66 00:03:29,640 --> 00:03:33,660 I've described informative meiosis on the screen here. 67 00:03:33,660 --> 00:03:37,140 They're meiosis where we have two alleles of some marker 68 00:03:37,140 --> 00:03:38,940 and two alleles for some trait gene, 69 00:03:38,940 --> 00:03:42,190 and we can determine which we're transmitted together. 70 00:03:42,190 --> 00:03:44,190 And we're going to need that kind of information 71 00:03:44,190 --> 00:03:47,700 in order to identify recombinant and non-recombinant gametes. 72 00:03:47,700 --> 00:03:51,680 So, to do that, we must have an informative meiosis. 5201

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