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These are the user uploaded subtitles that are being translated: 0 00:00:00,000 --> 00:00:01,625 PETER REDDIEN: All right, so let's turn 1 00:00:01,625 --> 00:00:18,420 to phase being unknown and how to deal with this. 2 00:00:18,420 --> 00:00:24,800 3 00:00:24,800 --> 00:00:25,300 OK. 4 00:00:25,300 --> 00:00:28,300 5 00:00:28,300 --> 00:00:31,390 All right, so let's imagine a pedigree here, a simple one. 6 00:00:31,390 --> 00:00:48,100 7 00:00:48,100 --> 00:00:50,460 OK. 8 00:00:50,460 --> 00:00:54,120 And then we have this is our D over plus situation again. 9 00:00:54,120 --> 00:01:06,630 10 00:01:06,630 --> 00:01:08,910 I'll give you the SSR genotypes then. 11 00:01:08,910 --> 00:01:16,650 12 00:01:16,650 --> 00:01:17,150 OK. 13 00:01:17,150 --> 00:01:30,910 14 00:01:30,910 --> 00:01:31,410 OK. 15 00:01:31,410 --> 00:01:35,300 16 00:01:35,300 --> 00:01:39,440 All right, so we've got A over B, A over B, A over 17 00:01:39,440 --> 00:01:44,675 A, B over B, and A over A. Do we have any informative meiosis? 18 00:01:44,675 --> 00:01:53,230 19 00:01:53,230 --> 00:01:53,730 Let's think. 20 00:01:53,730 --> 00:01:56,970 Let's take an individual at a time. 21 00:01:56,970 --> 00:02:01,620 This individual, what did this male offspring 22 00:02:01,620 --> 00:02:05,860 get from this female parent? 23 00:02:05,860 --> 00:02:10,080 It must have gotten A and for the disease gene, 24 00:02:10,080 --> 00:02:13,380 we can tell which went together. 25 00:02:13,380 --> 00:02:19,350 We know that A and D came together from this parent. 26 00:02:19,350 --> 00:02:22,058 That's an informative meiosis. 27 00:02:22,058 --> 00:02:24,600 And you'll see that there are three informative meiosis here. 28 00:02:24,600 --> 00:02:39,480 29 00:02:39,480 --> 00:02:43,420 Do we know the phase of this individual? 30 00:02:43,420 --> 00:02:44,950 No. 31 00:02:44,950 --> 00:02:49,840 We don't have any information about this female's parents. 32 00:02:49,840 --> 00:02:52,590 We have no way of knowing the phase. 33 00:02:52,590 --> 00:02:55,560 OK, so this is a situation where we have data, 34 00:02:55,560 --> 00:02:57,780 but phase is unknown. 35 00:02:57,780 --> 00:02:59,200 What do we do with this data? 36 00:02:59,200 --> 00:03:00,390 Well, we could say, well, we're going 37 00:03:00,390 --> 00:03:01,765 to have a tough time figuring out 38 00:03:01,765 --> 00:03:03,700 recombinant or non-recombinant gametes. 39 00:03:03,700 --> 00:03:05,980 So let's just ignore it. 40 00:03:05,980 --> 00:03:09,770 On the other hand, we want to get every bit of information 41 00:03:09,770 --> 00:03:11,020 we can out of these pedigrees. 42 00:03:11,020 --> 00:03:14,340 It's, kind of, hard to get a lot of data here. 43 00:03:14,340 --> 00:03:18,090 Families tend to be small compared to the type of data 44 00:03:18,090 --> 00:03:20,610 we can get in the lab with model organisms. 45 00:03:20,610 --> 00:03:23,790 And if they kept going a certain direction, 46 00:03:23,790 --> 00:03:25,920 you would intuitively think that's got 47 00:03:25,920 --> 00:03:27,360 to be indicative of something. 48 00:03:27,360 --> 00:03:30,820 It's going to be changing the probability of linkage. 49 00:03:30,820 --> 00:03:33,840 And so we can include these meiosis 50 00:03:33,840 --> 00:03:36,630 in our calculation for the LOD score. 51 00:03:36,630 --> 00:03:41,260 52 00:03:41,260 --> 00:03:52,160 So if the phase is unknown, we're 53 00:03:52,160 --> 00:03:56,562 going to average the probabilities of the data 54 00:03:56,562 --> 00:03:57,770 from the two possible phases. 55 00:03:57,770 --> 00:04:16,670 56 00:04:16,670 --> 00:04:23,900 OK, well, we could say it could have been with A with D, 57 00:04:23,900 --> 00:04:25,760 or it could have been A with plus. 58 00:04:25,760 --> 00:04:27,500 Since we don't know, we'll just say 59 00:04:27,500 --> 00:04:32,943 there's a 50% chance of either scenario. 60 00:04:32,943 --> 00:04:34,610 And then we'll average the probabilities 61 00:04:34,610 --> 00:04:37,040 of getting these data. 62 00:04:37,040 --> 00:04:40,535 OK, so our LOD score then-- 63 00:04:40,535 --> 00:04:51,330 64 00:04:51,330 --> 00:04:55,050 OK, just like before, it's going to be now 65 00:04:55,050 --> 00:05:08,260 the probability of data given linkage in phase one 66 00:05:08,260 --> 00:05:11,320 plus the probability of getting the data with linkage 67 00:05:11,320 --> 00:05:11,935 in phase two. 68 00:05:11,935 --> 00:05:33,050 69 00:05:33,050 --> 00:05:43,295 OK, and then we're going to take the average of this 70 00:05:43,295 --> 00:05:45,170 and then divide by the probability of getting 71 00:05:45,170 --> 00:05:46,420 the data if they are unlinked. 72 00:05:46,420 --> 00:05:57,142 73 00:05:57,142 --> 00:05:57,642 OK. 74 00:05:57,642 --> 00:06:00,600 75 00:06:00,600 --> 00:06:04,920 So let's go through that with this new example 76 00:06:04,920 --> 00:06:09,130 where we have three informative meiosis here. 77 00:06:09,130 --> 00:06:10,560 OK, so our LOD score then-- 78 00:06:10,560 --> 00:06:29,060 79 00:06:29,060 --> 00:06:31,550 OK, so one scenario I'm going to give you 80 00:06:31,550 --> 00:06:33,770 a theta for this example, just going 81 00:06:33,770 --> 00:06:41,830 to give you a theta of 0.1. 82 00:06:41,830 --> 00:06:51,850 So our data here would be 0.45 cubed plus 0.05 cubed divided 83 00:06:51,850 --> 00:06:58,980 by 0.25 cubed. 84 00:06:58,980 --> 00:07:01,500 OK. 85 00:07:01,500 --> 00:07:09,300 All right, so some phase where they were A with D 86 00:07:09,300 --> 00:07:12,000 would have three non-recombinant gametes, 87 00:07:12,000 --> 00:07:13,680 as you'll see if you walk it through, 88 00:07:13,680 --> 00:07:15,960 and then some phase where it was A with plus 89 00:07:15,960 --> 00:07:17,730 would have three recombinant gametes. 90 00:07:17,730 --> 00:07:20,960 91 00:07:20,960 --> 00:07:24,840 And with theta 0.1, then we'd have this probability 92 00:07:24,840 --> 00:07:27,600 of a given non-recombinant gamete and this probability 93 00:07:27,600 --> 00:07:29,760 of a recombinant gamete, and we'll 94 00:07:29,760 --> 00:07:37,680 see that we get a LOD score here of 0.465. 95 00:07:37,680 --> 00:07:41,140 OK, so you'll see that it is positive. 96 00:07:41,140 --> 00:07:46,080 So these data do contribute it-- do contribute to a LOD score. 97 00:07:46,080 --> 00:07:50,640 They contribute, in this case, a positive value to a LOD score. 98 00:07:50,640 --> 00:07:53,880 What you'll see is that if you have one individual, one 99 00:07:53,880 --> 00:07:57,300 meiosis where the phase is not known like in the example 100 00:07:57,300 --> 00:08:01,380 we had up here if you do this, your LOD score 101 00:08:01,380 --> 00:08:04,240 will be 0 for one meiosis. 102 00:08:04,240 --> 00:08:05,970 So if you just have one individual coming 103 00:08:05,970 --> 00:08:08,850 from one meiosis where the phase was unknown in the parents, 104 00:08:08,850 --> 00:08:11,050 you don't have-- you can ignore it. 105 00:08:11,050 --> 00:08:13,500 But if you have multiple individuals where 106 00:08:13,500 --> 00:08:16,300 the parents had phase unknown like in this scenario, 107 00:08:16,300 --> 00:08:19,040 then they do contribute to the LOD score. 7271

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