1 00:00:00,000 --> 00:00:07,590 2 00:00:07,590 --> 00:00:09,610 PROFESSOR: Hello, and welcome to a help session on 3 00:00:09,610 --> 00:00:10,910 complementation. 4 00:00:10,910 --> 00:00:12,300 Today, we are going to be working 5 00:00:12,300 --> 00:00:14,120 out a problem together. 6 00:00:14,120 --> 00:00:16,600 If you have not yet had a chance to work out the problem 7 00:00:16,600 --> 00:00:20,370 on your own, please do so now, and then return to this video. 8 00:00:20,370 --> 00:00:31,540 9 00:00:31,540 --> 00:00:33,790 All right, now that you've had a chance to look at the 10 00:00:33,790 --> 00:00:36,850 problem your on, let's work it out together. 11 00:00:36,850 --> 00:00:40,820 So, here we have a series of different fruit flies. 12 00:00:40,820 --> 00:00:44,900 They all have these recessive mini-fly phenotypes. 13 00:00:44,900 --> 00:00:46,460 So, we've noticed that there are lots 14 00:00:46,460 --> 00:00:48,140 of different mutations. 15 00:00:48,140 --> 00:00:49,870 And, now, we would like to put them into 16 00:00:49,870 --> 00:00:51,930 complementation groups. 17 00:00:51,930 --> 00:00:55,990 So, we start off by looking at this nice, large Punnett 18 00:00:55,990 --> 00:00:58,620 square we have, where all the flies have been crossed to 19 00:00:58,620 --> 00:01:00,200 each other. 20 00:01:00,200 --> 00:01:04,690 So, if we look at A and B for instance, when these two flies 21 00:01:04,690 --> 00:01:06,410 are crossed together, they still 22 00:01:06,410 --> 00:01:08,670 have a mini-fly phenotype. 23 00:01:08,670 --> 00:01:13,450 This means that the mutation is in the same gene and that 24 00:01:13,450 --> 00:01:18,300 these two mutants do not complement each other, but 25 00:01:18,300 --> 00:01:20,950 they make up one complementation group. 26 00:01:20,950 --> 00:01:25,630 So if we look farther down this column, A and F together 27 00:01:25,630 --> 00:01:28,980 also have a mini-fly phenotype. 28 00:01:28,980 --> 00:01:34,730 This means A, B, and F, all three of these are in the same 29 00:01:34,730 --> 00:01:35,980 complementation group. 30 00:01:35,980 --> 00:01:44,020 31 00:01:44,020 --> 00:01:47,240 So, now that we've already looked at A and B, let's 32 00:01:47,240 --> 00:01:53,800 continue on to C. So, C has a normal fly in all of these 33 00:01:53,800 --> 00:01:55,710 instances except this one. 34 00:01:55,710 --> 00:02:00,570 When C and E are crossed together, they also have a 35 00:02:00,570 --> 00:02:02,430 mini-fly phenotype. 36 00:02:02,430 --> 00:02:05,920 Thus, C and E also form a complementation group. 37 00:02:05,920 --> 00:02:12,070 38 00:02:12,070 --> 00:02:17,946 So, now we're done with A,B, C, E, and F. So looking at D, 39 00:02:17,946 --> 00:02:21,600 D when crossed to all the other flies, you never get a 40 00:02:21,600 --> 00:02:22,870 mini-fly phenotype. 41 00:02:22,870 --> 00:02:26,760 So, D must be in its own complementation group. 42 00:02:26,760 --> 00:02:31,500 43 00:02:31,500 --> 00:02:35,730 Great, so we've identified all three complementation groups. 44 00:02:35,730 --> 00:02:39,910 So, moving on to the next part of the problem. 45 00:02:39,910 --> 00:02:45,650 We are asked, are mutations B and C in the same gene, or are 46 00:02:45,650 --> 00:02:48,210 they in different genes? 47 00:02:48,210 --> 00:02:55,210 All right, so, if we look over here at B and C together, when 48 00:02:55,210 --> 00:02:58,740 these are crossed, we get a normal fly. 49 00:02:58,740 --> 00:03:01,570 This means that the mutations were able to 50 00:03:01,570 --> 00:03:03,430 complement each other. 51 00:03:03,430 --> 00:03:07,130 Meaning that the two mutations occur in different genes, and, 52 00:03:07,130 --> 00:03:09,500 thus, when you cross the flies, they're able to 53 00:03:09,500 --> 00:03:12,590 compensate for each other. 54 00:03:12,590 --> 00:03:14,395 So, they're in different genes. 55 00:03:14,395 --> 00:03:16,980 56 00:03:16,980 --> 00:03:19,590 All right, now, in the last part of the problem, we're 57 00:03:19,590 --> 00:03:24,260 told that we are introduced to a new fly, or a new mutant 58 00:03:24,260 --> 00:03:29,690 fly, G. When G is crossed to A, we get a 59 00:03:29,690 --> 00:03:31,930 mini-fly phenotype out. 60 00:03:31,930 --> 00:03:38,900 Also, when we cross G to B, we also get a mini-fly phenotype. 61 00:03:38,900 --> 00:03:42,300 So the question is, what does this suggest to us 62 00:03:42,300 --> 00:03:45,480 about mutant G? 63 00:03:45,480 --> 00:03:50,470 Looking over here, we notice that A and B are in the same 64 00:03:50,470 --> 00:03:51,890 complementation group. 65 00:03:51,890 --> 00:03:55,720 When A and B are crossed together, we also get 66 00:03:55,720 --> 00:04:00,690 mini-flies So, this suggests to us that G is yet another 67 00:04:00,690 --> 00:04:03,150 member of this complementation group. 68 00:04:03,150 --> 00:04:19,920 69 00:04:19,920 --> 00:04:23,510 Thus, G also falls into complementation 70 00:04:23,510 --> 00:04:26,280 group number one. 71 00:04:26,280 --> 00:04:30,460 This concludes our help problem on complementation. 72 00:04:30,460 --> 00:04:31,710 Thank you for joining us. 73 00:04:31,710 --> 00:04:33,685