1 00:00:17,820 --> 00:00:20,460 ADAM MARTIN: And so I wanted to start today's lecture 2 00:00:20,460 --> 00:00:22,710 by continuing what we were talking 3 00:00:22,710 --> 00:00:25,380 about in the last lecture. 4 00:00:25,380 --> 00:00:28,540 So I'm just going to hide this real quick. 5 00:00:28,540 --> 00:00:34,410 And so we're talking about the fruit fly and the white gene 6 00:00:34,410 --> 00:00:38,250 and the white mutant, which results in white-eyed flies. 7 00:00:38,250 --> 00:00:46,620 And we talked about how if you take females that have red eyes 8 00:00:46,620 --> 00:00:54,990 and cross them to males, the white-eyed male, 9 00:00:54,990 --> 00:01:01,710 then 100% of the progeny has red eyes in the F1 generation. 10 00:01:01,710 --> 00:01:04,379 And so I asked you guys, would you 11 00:01:04,379 --> 00:01:07,800 get the same results if you did the reciprocal cross? 12 00:01:10,740 --> 00:01:13,680 So what if we took white-eyed females 13 00:01:13,680 --> 00:01:17,670 and mated them to red-eyed males? 14 00:01:17,670 --> 00:01:20,310 So what about this? 15 00:01:23,280 --> 00:01:27,090 Actually, I'm going to move this over to over 16 00:01:27,090 --> 00:01:30,450 here so that maybe it's more visible. 17 00:01:30,450 --> 00:01:34,050 So what if we have white-eyed females 18 00:01:34,050 --> 00:01:36,990 and crossed this to red-eyed males? 19 00:01:41,130 --> 00:01:44,910 So let's unpack this sort of a little bit at a time. 20 00:01:44,910 --> 00:01:50,910 So what's the genotype of these white-eyed females here? 21 00:01:50,910 --> 00:01:51,867 Miles? 22 00:01:51,867 --> 00:01:57,460 AUDIENCE: So if you designate the eye gene as the letter A, 23 00:01:57,460 --> 00:02:00,890 a female would be X lowercase a, X lowercase a. 24 00:02:00,890 --> 00:02:01,950 ADAM MARTIN: Yes. 25 00:02:01,950 --> 00:02:05,700 So Miles is exactly right. 26 00:02:05,700 --> 00:02:08,940 So the dominant phenotype is red eyes, 27 00:02:08,940 --> 00:02:12,090 because the gene encodes for an enzyme that's 28 00:02:12,090 --> 00:02:15,660 important for the production of the red pigment. 29 00:02:15,660 --> 00:02:26,310 And so X lowercase a here would be a recessive mutant that 30 00:02:26,310 --> 00:02:27,060 lacks the pigment. 31 00:02:30,500 --> 00:02:32,730 And because it's a recessive allele-- 32 00:02:32,730 --> 00:02:36,180 because you need only one copy of this gene 33 00:02:36,180 --> 00:02:37,920 to produce the pigment. 34 00:02:37,920 --> 00:02:42,420 So the recessive allele results in the white phenotype. 35 00:02:42,420 --> 00:02:46,080 Therefore, this has to be homozygous recessive. 36 00:02:46,080 --> 00:02:49,710 How about this red-eyed male? 37 00:02:49,710 --> 00:02:51,005 Yeah, Ory? 38 00:02:51,005 --> 00:02:53,502 AUDIENCE: Wouldn't you have a Y and then an X capital A? 39 00:02:53,502 --> 00:02:54,210 ADAM MARTIN: Yes. 40 00:02:54,210 --> 00:02:56,640 So this would be this phenotype, right, 41 00:02:56,640 --> 00:03:01,960 where capital A is the gene that produces-- 42 00:03:01,960 --> 00:03:04,605 is a normal functioning gene that produces the pigment. 43 00:03:07,410 --> 00:03:11,180 So then in your F1 here, are you going 44 00:03:11,180 --> 00:03:15,762 to see something similar to this or something different? 45 00:03:15,762 --> 00:03:17,075 AUDIENCE: Something different. 46 00:03:17,075 --> 00:03:17,700 ADAM MARTIN: Different, great. 47 00:03:17,700 --> 00:03:18,870 Who said different? 48 00:03:18,870 --> 00:03:19,810 Javier? 49 00:03:19,810 --> 00:03:22,175 Do you want to propose what you might see? 50 00:03:22,175 --> 00:03:22,800 AUDIENCE: Yeah. 51 00:03:22,800 --> 00:03:25,110 For the males, they're going to inherit the Y 52 00:03:25,110 --> 00:03:28,740 gene from the father and the [INAUDIBLE].. 53 00:03:28,740 --> 00:03:29,840 ADAM MARTIN: Exactly. 54 00:03:29,840 --> 00:03:33,720 So the males are going to get the Y from the father, 55 00:03:33,720 --> 00:03:36,210 and they're going to get one X from their mother. 56 00:03:36,210 --> 00:03:39,480 So all the males are going to be of this genotype 57 00:03:39,480 --> 00:03:44,910 here, which means they're going to have what color eye? 58 00:03:44,910 --> 00:03:46,860 Javier is exactly right. 59 00:03:46,860 --> 00:03:49,140 That means they're going to have white eyes. 60 00:03:49,140 --> 00:03:53,580 So all the males will have white eyes. 61 00:03:58,380 --> 00:04:02,652 And what about the females? 62 00:04:02,652 --> 00:04:04,090 AUDIENCE: [INAUDIBLE] 63 00:04:04,090 --> 00:04:05,862 ADAM MARTIN: What's that? 64 00:04:05,862 --> 00:04:06,750 AUDIENCE: Red eyes. 65 00:04:06,750 --> 00:04:07,630 ADAM MARTIN: Yeah. 66 00:04:07,630 --> 00:04:10,610 So Ory is saying the males are going to get red eyes, right, 67 00:04:10,610 --> 00:04:11,710 because they're-- 68 00:04:11,710 --> 00:04:14,320 or the females are going to have red eyes, because they're 69 00:04:14,320 --> 00:04:17,740 going to get the X chromosome from their father, which 70 00:04:17,740 --> 00:04:24,010 has the dominant gene that produces the red pigment. 71 00:04:24,010 --> 00:04:27,010 So all the females are going to be heterozygous, 72 00:04:27,010 --> 00:04:29,870 but have a functional copy of this gene. 73 00:04:29,870 --> 00:04:33,640 So all of the females will have red eyes. 74 00:04:38,980 --> 00:04:41,010 OK, does everyone see how-- 75 00:04:41,010 --> 00:04:45,150 now, how would this compare with Mendel's crosses and pea color? 76 00:04:45,150 --> 00:04:47,940 Would there be a difference in Mendel's crosses 77 00:04:47,940 --> 00:04:50,670 if you switch the male versus the female 78 00:04:50,670 --> 00:04:52,890 if these were autosomal traits? 79 00:04:52,890 --> 00:04:55,290 Ory is shaking his head no, and he's right, right? 80 00:04:55,290 --> 00:04:57,400 In that case, it doesn't matter. 81 00:04:57,400 --> 00:04:59,490 You can do the reciprocal crosses, 82 00:04:59,490 --> 00:05:03,930 you get the same result. But because this is sex linked, 83 00:05:03,930 --> 00:05:07,740 which one is the male and which is the female is relevant. 84 00:05:07,740 --> 00:05:10,890 And this actually relates to something 85 00:05:10,890 --> 00:05:14,090 that we just saw on the MIT news. 86 00:05:14,090 --> 00:05:15,870 I just got this email this morning, 87 00:05:15,870 --> 00:05:18,330 but it came out I think yesterday, 88 00:05:18,330 --> 00:05:23,550 which is that biology-related research in the mechanical 89 00:05:23,550 --> 00:05:27,180 engineering department-- specifically, the CAM lab-- 90 00:05:27,180 --> 00:05:32,700 they've been able to design a 3D sort of model 91 00:05:32,700 --> 00:05:37,750 for ALS disease, which is also known as Lou Gehrig's disease. 92 00:05:37,750 --> 00:05:39,570 And so what they've done in the CAM lab 93 00:05:39,570 --> 00:05:44,610 is to take cells from either patients that have ALS 94 00:05:44,610 --> 00:05:47,580 or from normal individuals, and they coax 95 00:05:47,580 --> 00:05:49,710 these cells to become neurons. 96 00:05:49,710 --> 00:05:52,860 Here, you're seeing a neuron in blue and green here. 97 00:05:52,860 --> 00:05:56,220 And you see the neurites extend from this neuron. 98 00:05:56,220 --> 00:05:58,410 And they have a model where this neuron can then 99 00:05:58,410 --> 00:06:00,580 synapse with a muscle. 100 00:06:00,580 --> 00:06:06,450 And so they're using this 3D sort of tissue model 101 00:06:06,450 --> 00:06:12,690 to model ALS and to look for drugs that might affect 102 00:06:12,690 --> 00:06:17,100 ALS, potentially curing ALS. 103 00:06:17,100 --> 00:06:20,430 And so last night, I started reading about ALS 104 00:06:20,430 --> 00:06:24,540 and was pleased to find that there's actually 105 00:06:24,540 --> 00:06:28,080 a very rare X-linked, dominant form of the disease that 106 00:06:28,080 --> 00:06:31,810 can be passed on from generation to generation. 107 00:06:31,810 --> 00:06:36,780 And the inheritance pattern of this X-linked, dominant version 108 00:06:36,780 --> 00:06:39,930 of ALS would have an inheritance pattern 109 00:06:39,930 --> 00:06:43,580 that's similar to what we observe for the white mutant 110 00:06:43,580 --> 00:06:45,480 in the fruit fly, right? 111 00:06:45,480 --> 00:06:49,000 Whereas, if you have an affected father, 112 00:06:49,000 --> 00:06:52,690 and this is a dominant mutant on the X chromosome, 113 00:06:52,690 --> 00:06:56,010 then all of his daughters will get that X chromosome 114 00:06:56,010 --> 00:07:00,960 and be affected, whereas the sons will all be unaffected. 115 00:07:00,960 --> 00:07:04,170 However, if you have the reciprocal situation, 116 00:07:04,170 --> 00:07:07,080 where you have an affected mother and an unaffected 117 00:07:07,080 --> 00:07:10,770 father, then the sons and daughters 118 00:07:10,770 --> 00:07:13,710 get the disease randomly. 119 00:07:13,710 --> 00:07:18,750 So this is a sort of form of inheritance, which is relevant 120 00:07:18,750 --> 00:07:22,530 if you're considering human disease and some forms of it. 121 00:07:22,530 --> 00:07:26,220 Most versions of ALS, well, are sporadic, 122 00:07:26,220 --> 00:07:29,640 but inherited forms are usually autosomal dominant. 123 00:07:29,640 --> 00:07:33,450 So this is a rare case here. 124 00:07:33,450 --> 00:07:36,090 But I thought it was interesting in that it's relevant to what 125 00:07:36,090 --> 00:07:38,700 we've been talking about. 126 00:07:38,700 --> 00:07:41,060 So now, just to recap-- 127 00:07:41,060 --> 00:07:44,290 here, I'll throw this up so everything's up. 128 00:07:47,340 --> 00:07:49,030 So in the last lecture, we talked 129 00:07:49,030 --> 00:07:51,250 about Mendelian inheritance. 130 00:07:58,380 --> 00:08:02,350 And we talked about when you take two parents that 131 00:08:02,350 --> 00:08:06,610 differ in two traits and you perform a cross, 132 00:08:06,610 --> 00:08:16,155 you get a hybrid individual that is heterozygous for both genes. 133 00:08:19,330 --> 00:08:21,930 And now this is the F1 individual. 134 00:08:21,930 --> 00:08:24,840 Let's say we want to know what types of gametes 135 00:08:24,840 --> 00:08:29,250 this F1 individual produces. 136 00:08:29,250 --> 00:08:34,679 We can perform a type of a cross known as a test cross, where 137 00:08:34,679 --> 00:08:40,440 we cross this individual to another individual that 138 00:08:40,440 --> 00:08:48,240 is homozygous recessive for both these genes, which means 139 00:08:48,240 --> 00:08:51,000 that you know exactly which alleles 140 00:08:51,000 --> 00:08:52,770 are coming from this parent. 141 00:08:52,770 --> 00:08:56,520 And they're both recessive, so you can see whether or not 142 00:08:56,520 --> 00:08:59,010 the gamete produced by this individual 143 00:08:59,010 --> 00:09:01,860 has either the dominant or the recessive allele. 144 00:09:05,130 --> 00:09:05,630 Let me see. 145 00:09:05,630 --> 00:09:06,664 I'll boost this up. 146 00:09:12,890 --> 00:09:15,890 So now we can consider the different types of progeny 147 00:09:15,890 --> 00:09:18,410 that result from this test cross. 148 00:09:18,410 --> 00:09:22,580 And some will have the two dominant alleles 149 00:09:22,580 --> 00:09:25,580 from this parent and will, therefore, 150 00:09:25,580 --> 00:09:30,080 be heterozygous for the A and B gene. 151 00:09:30,080 --> 00:09:35,480 And it would exhibit the dominant A and B phenotype. 152 00:09:35,480 --> 00:09:39,420 I think that is what I'm showing here. 153 00:09:39,420 --> 00:09:42,200 So if the chromosomes, during meiosis I, 154 00:09:42,200 --> 00:09:46,070 align like this, then the two dominant alleles 155 00:09:46,070 --> 00:09:50,180 segregate together, and you get AB gametes. 156 00:09:50,180 --> 00:09:54,800 And you also, reciprocally, get these lowercase a 157 00:09:54,800 --> 00:09:57,060 and b gametes, as well. 158 00:09:57,060 --> 00:10:00,080 So that's the other class here. 159 00:10:00,080 --> 00:10:05,330 So you can get these two classes of progeny. 160 00:10:05,330 --> 00:10:08,270 And the phenotypes of these two classes 161 00:10:08,270 --> 00:10:11,210 will resemble the parents, right? 162 00:10:11,210 --> 00:10:13,970 So these are known as parental gametes. 163 00:10:16,970 --> 00:10:20,030 So these are the parentals. 164 00:10:20,030 --> 00:10:23,030 But you know because Mendel showed 165 00:10:23,030 --> 00:10:26,180 that if you have genes and their alleles 166 00:10:26,180 --> 00:10:29,510 on separate chromosomes, they can assert independently 167 00:10:29,510 --> 00:10:30,780 of each other. 168 00:10:30,780 --> 00:10:36,230 So an alternatively likely scenario 169 00:10:36,230 --> 00:10:38,880 is that the chromosomes align like this, 170 00:10:38,880 --> 00:10:41,420 where now the dominant allele of B 171 00:10:41,420 --> 00:10:44,660 is on the other side of the spindle. 172 00:10:44,660 --> 00:10:47,270 And therefore, these chromosomes are 173 00:10:47,270 --> 00:10:50,960 going to segregate like this during the first meiotic 174 00:10:50,960 --> 00:10:52,340 division. 175 00:10:52,340 --> 00:10:55,960 And that gives rise to gametes that 176 00:10:55,960 --> 00:10:58,885 have a different combination of alleles than the parents. 177 00:11:02,080 --> 00:11:07,610 So you have some that look like this. 178 00:11:07,610 --> 00:11:12,100 So each of these would be different classes of progeny. 179 00:11:12,100 --> 00:11:15,470 And you have one last class that would look like this. 180 00:11:18,220 --> 00:11:23,080 And so neither of these look like the original parents, 181 00:11:23,080 --> 00:11:25,840 and so they're known as non-parental. 182 00:11:30,770 --> 00:11:35,050 And so if these two genes are behaving according 183 00:11:35,050 --> 00:11:37,570 to Mendel's second law, where there's 184 00:11:37,570 --> 00:11:39,070 independent assortment-- 185 00:11:43,600 --> 00:11:46,180 if you have independent assortment, 186 00:11:46,180 --> 00:11:49,390 what's going to be the ratio of parental to non-parental? 187 00:11:55,080 --> 00:11:55,580 Rachel? 188 00:11:55,580 --> 00:11:56,450 AUDIENCE: One to one. 189 00:11:56,450 --> 00:11:58,200 ADAM MARTIN: Yeah, Rachel says one to one, 190 00:11:58,200 --> 00:12:00,530 and I think a number of others also said one to one. 191 00:12:00,530 --> 00:12:05,630 So you have 50% parental, 50% non-parental, right? 192 00:12:05,630 --> 00:12:07,580 Because it's equally likely to get 193 00:12:07,580 --> 00:12:11,030 either of those alignments of the homologous chromosomes 194 00:12:11,030 --> 00:12:15,590 during meiosis I. 195 00:12:15,590 --> 00:12:17,930 So now I'm going to basically break 196 00:12:17,930 --> 00:12:21,230 the rules I just explained to you in the last lecture 197 00:12:21,230 --> 00:12:24,650 and tell you about an exception, which is known as linkage. 198 00:12:28,280 --> 00:12:29,570 Gesundheit. 199 00:12:29,570 --> 00:12:34,970 And in the abstract sense, linkage 200 00:12:34,970 --> 00:12:37,580 is simply when you have two traits that 201 00:12:37,580 --> 00:12:40,670 tend to be inherited together. 202 00:12:40,670 --> 00:12:43,040 So just considering probability. 203 00:12:43,040 --> 00:12:46,730 So you have traits inherited together. 204 00:12:46,730 --> 00:12:49,390 They're exhibiting what is known as linkage. 205 00:12:52,310 --> 00:12:54,860 But that's an abstract way to think about it. 206 00:12:54,860 --> 00:12:57,980 It's just based on probability, right? 207 00:12:57,980 --> 00:13:01,640 So a physical model for what linkage is, 208 00:13:01,640 --> 00:13:03,680 is that you have chromosomes. 209 00:13:03,680 --> 00:13:05,780 The genes are on the chromosomes. 210 00:13:05,780 --> 00:13:09,020 And for two genes to be linked, those genes 211 00:13:09,020 --> 00:13:12,920 are physically near each other on the chromosome. 212 00:13:12,920 --> 00:13:16,700 So the physical model is that two genes 213 00:13:16,700 --> 00:13:19,130 are near each other on the chromosome. 214 00:13:30,810 --> 00:13:33,750 OK, so let's consider again these generic genes, 215 00:13:33,750 --> 00:13:38,480 A and B. If A and B resulting from this cross-- 216 00:13:38,480 --> 00:13:42,260 if these two happen to be on the same chromosome, 217 00:13:42,260 --> 00:13:45,570 now they're physically coupled to each other. 218 00:13:45,570 --> 00:13:49,460 Then they're going to tend to be inherited together. 219 00:13:49,460 --> 00:13:52,400 No matter how these align, they're 220 00:13:52,400 --> 00:13:54,080 always going to go together during 221 00:13:54,080 --> 00:13:56,070 the first meiotic division. 222 00:13:56,070 --> 00:14:00,770 And that's just going to only give you the parental gametes. 223 00:14:00,770 --> 00:14:04,880 So if there's linkage, you're going to have-- 224 00:14:04,880 --> 00:14:07,325 let's consider the case where you have complete linkage. 225 00:14:10,520 --> 00:14:14,570 If you have complete linkage, 100% of the gametes 226 00:14:14,570 --> 00:14:16,790 are going to be parentals, and you're 227 00:14:16,790 --> 00:14:19,190 going to have 0% non-parental. 228 00:14:23,120 --> 00:14:26,520 That's if the genes are really, really, 229 00:14:26,520 --> 00:14:28,610 really close to each other, and maybe you 230 00:14:28,610 --> 00:14:30,200 don't count so many progeny. 231 00:14:30,200 --> 00:14:33,950 You won't see any mixing between the two. 232 00:14:33,950 --> 00:14:37,790 But there is a phenomenon that can separate these genes, 233 00:14:37,790 --> 00:14:39,440 and it's known as crossing over. 234 00:14:42,760 --> 00:14:46,920 And another term to describe it is recombination. 235 00:14:46,920 --> 00:14:48,980 So the alleles are getting recombined 236 00:14:48,980 --> 00:14:50,180 between the chromosomes. 237 00:14:53,570 --> 00:14:54,590 Recombination. 238 00:14:57,620 --> 00:15:01,970 And what crossing over or recombination is, is it's 239 00:15:01,970 --> 00:15:08,700 a mixing of the chromosomes, if you will. 240 00:15:08,700 --> 00:15:11,510 Or it's an exchange of DNA. 241 00:15:11,510 --> 00:15:14,570 So there's a physical exchange of DNA from one 242 00:15:14,570 --> 00:15:18,230 of the homologous chromosomes to the other, OK? 243 00:15:18,230 --> 00:15:26,660 So you can think of this as an exchange of DNA 244 00:15:26,660 --> 00:15:30,945 between the homologous chromosomes, OK? 245 00:15:30,945 --> 00:15:31,820 And that's important. 246 00:15:31,820 --> 00:15:35,120 It's not an exchange between the sister chromatids, 247 00:15:35,120 --> 00:15:37,252 but between the homologous chromosomes 248 00:15:37,252 --> 00:15:38,585 that have the different alleles. 249 00:15:43,940 --> 00:15:47,520 And what's shown here is a micrograph showing you 250 00:15:47,520 --> 00:15:50,970 a picture of the process of crossing over. 251 00:15:50,970 --> 00:15:54,660 You can see the centromeres are the dark structures there. 252 00:15:54,660 --> 00:15:57,780 And you can see how the homologous chromosomes 253 00:15:57,780 --> 00:15:59,490 intertwine. 254 00:15:59,490 --> 00:16:02,700 And there are regions where it looks like there's a cross. 255 00:16:02,700 --> 00:16:06,570 Those are the homologous chromosomes crossing over 256 00:16:06,570 --> 00:16:11,700 and exchanging DNA such that one part of that chromosome 257 00:16:11,700 --> 00:16:15,270 gets attached to the other centromere. 258 00:16:15,270 --> 00:16:20,580 So I'll just show you in sort of my silly cartoon form 259 00:16:20,580 --> 00:16:23,170 how this is, just to make it clear. 260 00:16:23,170 --> 00:16:25,800 So let's say, again, you have these A and B genes, 261 00:16:25,800 --> 00:16:28,530 and they're physically linked on the chromosome. 262 00:16:28,530 --> 00:16:31,740 During crossing over, you can get 263 00:16:31,740 --> 00:16:36,030 an exchange of these alleles, such as a bit of one chromosome 264 00:16:36,030 --> 00:16:40,290 goes to the other homologous chromosome and vice versa, OK? 265 00:16:40,290 --> 00:16:42,750 So now you have the dominant A allele 266 00:16:42,750 --> 00:16:45,910 with the recessive b allele and vice versa. 267 00:16:45,910 --> 00:16:50,190 So now during meiosis I-- 268 00:16:50,190 --> 00:16:52,230 after meiosis II, this will give rise 269 00:16:52,230 --> 00:16:57,420 to two types of gametes, one of which is non-parental. 270 00:16:57,420 --> 00:17:00,150 And the same for the one down here. 271 00:17:00,150 --> 00:17:02,370 You get two types of gametes. 272 00:17:02,370 --> 00:17:06,264 One is non-parental-- the lowercase a, uppercase B 273 00:17:06,264 --> 00:17:06,764 gamete. 274 00:17:11,930 --> 00:17:14,930 So this happens if there's incomplete linkage. 275 00:17:14,930 --> 00:17:17,720 That means there can be a recombination event that 276 00:17:17,720 --> 00:17:19,550 separates the two genes. 277 00:17:22,680 --> 00:17:27,079 And I'm going to give you an example of a case where 278 00:17:27,079 --> 00:17:30,470 data was collected with what fraction of each class there 279 00:17:30,470 --> 00:17:31,280 is. 280 00:17:31,280 --> 00:17:33,590 So now we're considering an example 281 00:17:33,590 --> 00:17:35,610 where you have a linkage. 282 00:17:35,610 --> 00:17:40,980 So A and B are on the same chromosome. 283 00:17:40,980 --> 00:17:43,910 And so we'll consider a case where, in this class, 284 00:17:43,910 --> 00:17:47,600 there are 165 members. 285 00:17:47,600 --> 00:17:49,800 For this one, there is 191. 286 00:17:49,800 --> 00:17:51,896 So I'm kind of-- 287 00:17:51,896 --> 00:17:54,110 line down like this. 288 00:17:54,110 --> 00:17:56,720 And then for the first recombinant class, 289 00:17:56,720 --> 00:17:58,710 23 individuals. 290 00:17:58,710 --> 00:18:02,510 And for the last, there are 21 individuals. 291 00:18:02,510 --> 00:18:05,810 So you can see there are many more of the parental class 292 00:18:05,810 --> 00:18:07,790 than the recombinant class, but we 293 00:18:07,790 --> 00:18:15,320 can calculate a frequency, or recombination frequency, 294 00:18:15,320 --> 00:18:16,415 between these two genes. 295 00:18:23,750 --> 00:18:26,690 And in this case, the recombination frequency 296 00:18:26,690 --> 00:18:34,730 is 44/400, which is equal to 11%, OK? 297 00:18:34,730 --> 00:18:38,570 So 11% of the progeny from this cross 298 00:18:38,570 --> 00:18:42,928 had some sort of crossing over between the A and B alleles. 299 00:18:42,928 --> 00:18:43,970 It would've been up here. 300 00:18:50,350 --> 00:18:55,900 Now, this frequency is interesting, 301 00:18:55,900 --> 00:19:02,230 because it is proportional to the distance that 302 00:19:02,230 --> 00:19:05,500 separates these two genes. 303 00:19:05,500 --> 00:19:07,450 So this recombination frequency is 304 00:19:07,450 --> 00:19:12,760 proportional to the linear distance along the chromosome 305 00:19:12,760 --> 00:19:13,600 between the genes. 306 00:19:26,800 --> 00:19:29,740 Now, it also depends on the recombination frequency 307 00:19:29,740 --> 00:19:33,710 in a given organism or in a given part of the chromosome. 308 00:19:33,710 --> 00:19:37,570 So when you're comparing recombination frequencies 309 00:19:37,570 --> 00:19:39,460 between different organisms, there's 310 00:19:39,460 --> 00:19:42,522 actually differences in the real different-- 311 00:19:42,522 --> 00:19:43,480 they're not equivalent. 312 00:19:43,480 --> 00:19:46,120 You can't compare them, basically. 313 00:19:46,120 --> 00:19:48,310 And also, there are regions of the chromosome 314 00:19:48,310 --> 00:19:51,740 where recombination happens less frequently than others. 315 00:19:51,740 --> 00:19:56,020 And so, again, you can't compare distances along those. 316 00:19:56,020 --> 00:20:00,460 But overall, you can use this as a distance in order 317 00:20:00,460 --> 00:20:08,350 to map genes along the linear axis of a chromosome. 318 00:20:08,350 --> 00:20:13,490 And maps are useful, because you can see where stuff is, right? 319 00:20:13,490 --> 00:20:17,770 So in this example here, I'll highlight a couple of places. 320 00:20:17,770 --> 00:20:19,700 Here's Rivendell. 321 00:20:19,700 --> 00:20:21,880 Here's Lonely Mountain. 322 00:20:21,880 --> 00:20:24,070 Here's Beorn's house. 323 00:20:24,070 --> 00:20:27,970 So let's say we are able to determine 324 00:20:27,970 --> 00:20:31,750 the distance between Rivendell and Lonely Mountain, 325 00:20:31,750 --> 00:20:35,260 and the distance between Lonely Mountain and Beorn's house, 326 00:20:35,260 --> 00:20:38,280 and the distance between Rivendell and Beorn's house. 327 00:20:38,280 --> 00:20:42,520 You'd be able to get a relative picture of where 328 00:20:42,520 --> 00:20:47,410 all of these places are in relation to each other. 329 00:20:47,410 --> 00:20:49,840 So this is a two-dimensional map I'm showing here. 330 00:20:49,840 --> 00:20:51,850 It's not one dimensional, but chromosomes 331 00:20:51,850 --> 00:20:56,590 are one dimensional, so it's a bit more accurate, OK? 332 00:20:56,590 --> 00:21:01,000 So this idea that recombination frequency can 333 00:21:01,000 --> 00:21:04,660 be used to measure distances between genes 334 00:21:04,660 --> 00:21:08,690 and that this could be used to generate a map 335 00:21:08,690 --> 00:21:12,040 is an idea that an undergraduate had 336 00:21:12,040 --> 00:21:16,030 while working in Thomas Hunt Morgan's lab back in 1911. 337 00:21:16,030 --> 00:21:19,360 And what I find fascinating about the story 338 00:21:19,360 --> 00:21:22,120 is this guy basically blew off his homework 339 00:21:22,120 --> 00:21:25,880 to produce the first genetic map in any organism. 340 00:21:25,880 --> 00:21:29,680 So the person who did it was Alfred Sturtevant, 341 00:21:29,680 --> 00:21:32,020 and he was an undergraduate at Columbia 342 00:21:32,020 --> 00:21:34,360 working for Thomas Hunt Morgan. 343 00:21:34,360 --> 00:21:37,120 And I'll just paraphrase this quote here. 344 00:21:37,120 --> 00:21:40,750 In 1911, he was talking with his advisor, Morgan, 345 00:21:40,750 --> 00:21:42,850 and he realized that the variations 346 00:21:42,850 --> 00:21:45,760 in the strength of linkage attributed by Morgan 347 00:21:45,760 --> 00:21:49,090 to differences in the separation of genes-- 348 00:21:49,090 --> 00:21:52,060 so Morgan had already made this connection, 349 00:21:52,060 --> 00:21:55,270 that the recombination frequency reflects 350 00:21:55,270 --> 00:21:58,180 the distance between the genes. 351 00:21:58,180 --> 00:22:00,700 But then Sturtevant realized that this 352 00:22:00,700 --> 00:22:03,910 offered the possibility of determining sequences 353 00:22:03,910 --> 00:22:07,440 in the linear dimension of the chromosome between the genes, 354 00:22:07,440 --> 00:22:08,740 OK? 355 00:22:08,740 --> 00:22:11,080 So then-- this is my favorite part-- 356 00:22:11,080 --> 00:22:13,090 "I went home and spent most of the night, 357 00:22:13,090 --> 00:22:15,490 to the neglect of my undergraduate homework, 358 00:22:15,490 --> 00:22:19,120 in producing the first chromosome map." 359 00:22:19,120 --> 00:22:20,440 And this is it. 360 00:22:20,440 --> 00:22:24,350 So the first chromosome map was of the Drosophila X chromosome, 361 00:22:24,350 --> 00:22:25,900 which we've been talking about. 362 00:22:25,900 --> 00:22:27,550 There's the white gene, which we've 363 00:22:27,550 --> 00:22:30,070 been talking about in the context of eye color. 364 00:22:30,070 --> 00:22:31,870 There's a yellow body gene here. 365 00:22:31,870 --> 00:22:34,750 There's vermilion, miniature, rudimentary, right? 366 00:22:34,750 --> 00:22:39,200 These are all visible phenotypes that you can see in the fly. 367 00:22:39,200 --> 00:22:41,200 And you can measure recombination 368 00:22:41,200 --> 00:22:43,765 between various alleles of these different genes. 369 00:22:48,990 --> 00:22:51,650 All right, so now I want to go through with you 370 00:22:51,650 --> 00:22:57,150 an example of how you can make one of these genetic maps. 371 00:22:57,150 --> 00:22:59,690 And it's essentially the same conceptually 372 00:22:59,690 --> 00:23:02,030 as to what Sturtevant did. 373 00:23:02,030 --> 00:23:04,700 And it involves what is known as a three-point cross. 374 00:23:07,220 --> 00:23:08,465 So a three-point cross. 375 00:23:14,500 --> 00:23:17,030 So there are going to be three genes, all of which 376 00:23:17,030 --> 00:23:19,340 are going to be hybrid, and I'll start 377 00:23:19,340 --> 00:23:21,560 with the parental generation that 378 00:23:21,560 --> 00:23:25,280 is little a, capital B, capital D. 379 00:23:25,280 --> 00:23:28,580 And we'll cross this fly or organism 380 00:23:28,580 --> 00:23:34,770 to an organism that is capital A, lower case b, lowercase d. 381 00:23:34,770 --> 00:23:35,501 Yes, Carmen? 382 00:23:35,501 --> 00:23:37,543 AUDIENCE: So when you write the gametes up there, 383 00:23:37,543 --> 00:23:40,598 does that imply that they were analogous parents? 384 00:23:40,598 --> 00:23:42,140 ADAM MARTIN: So what I'm writing here 385 00:23:42,140 --> 00:23:44,910 is the phenotype, basically. 386 00:23:44,910 --> 00:23:50,270 And so these are homozygous for each of these, yes. 387 00:23:50,270 --> 00:23:51,950 I could also write this as-- 388 00:23:57,350 --> 00:24:00,410 but I'm not going to draw the chromosomes, because it 389 00:24:00,410 --> 00:24:04,730 kind of gets more confusing. 390 00:24:04,730 --> 00:24:07,600 I'll draw the chromosomes here in F1, 391 00:24:07,600 --> 00:24:12,080 because we have now, basically, a tri-hybrid with one 392 00:24:12,080 --> 00:24:14,570 chromosome that looks like this, right? 393 00:24:14,570 --> 00:24:17,930 They got that chromosome from this individual here. 394 00:24:17,930 --> 00:24:19,820 And another chromosome will look like this. 395 00:24:22,950 --> 00:24:23,810 See? 396 00:24:23,810 --> 00:24:30,110 So this F1 fly is heterozygous for these three genes, 397 00:24:30,110 --> 00:24:34,430 and it has these two parental chromosomes. 398 00:24:37,130 --> 00:24:40,580 So now we can look at the gametes that result from this 399 00:24:40,580 --> 00:24:47,240 fly by doing a test cross, just like we did before. 400 00:24:47,240 --> 00:24:49,940 And so we want to cross this to a fly that's 401 00:24:49,940 --> 00:24:52,610 homozygous recessive for each of these genes. 402 00:24:56,400 --> 00:24:59,120 And now we can look at the progeny. 403 00:24:59,120 --> 00:25:01,430 And just by looking at the phenotype, 404 00:25:01,430 --> 00:25:03,650 we're going to know the genotype, because we 405 00:25:03,650 --> 00:25:07,070 know all of the flies from this cross 406 00:25:07,070 --> 00:25:11,150 have a chromosome from this individual that 407 00:25:11,150 --> 00:25:15,860 has recessive alleles for each gene. 408 00:25:15,860 --> 00:25:20,270 So we can consider now this first one here. 409 00:25:20,270 --> 00:25:24,170 That's one potential class of progeny. 410 00:25:24,170 --> 00:25:27,230 Another class would be this one. 411 00:25:27,230 --> 00:25:31,160 And these two, you can see, resemble the parents, right? 412 00:25:31,160 --> 00:25:36,027 So these are the parental classes of the progeny. 413 00:25:36,027 --> 00:25:36,860 So this is parental. 414 00:25:41,900 --> 00:25:45,760 All right, now you can consider all other combinations 415 00:25:45,760 --> 00:25:47,790 of alleles. 416 00:25:47,790 --> 00:25:50,650 And so I'll quickly write them down. 417 00:25:50,650 --> 00:25:52,360 You could have something-- 418 00:25:52,360 --> 00:25:56,380 progeny that look like this and this. 419 00:25:56,380 --> 00:25:59,380 These are just kind of reciprocal from each other. 420 00:25:59,380 --> 00:26:05,130 You could have progeny that look like this and this. 421 00:26:05,130 --> 00:26:14,350 And the last class would be this and this. 422 00:26:14,350 --> 00:26:17,290 So all of these progeny that I drew down here 423 00:26:17,290 --> 00:26:18,820 are recombinant, because they don't 424 00:26:18,820 --> 00:26:22,840 resemble the parents, right? 425 00:26:22,840 --> 00:26:25,210 Because there are three genes, now there's 426 00:26:25,210 --> 00:26:29,860 many more ways to get recombinant progeny, as opposed 427 00:26:29,860 --> 00:26:32,710 to having just two genes, right? 428 00:26:32,710 --> 00:26:34,630 So you can have many different combinations 429 00:26:34,630 --> 00:26:36,550 of these different alleles. 430 00:26:39,580 --> 00:26:41,260 And so now I'm going to give you data 431 00:26:41,260 --> 00:26:44,780 from a cross with three such genes. 432 00:26:44,780 --> 00:26:48,100 So you might get 580 individuals that 433 00:26:48,100 --> 00:26:59,830 look like this, 592 like this, 45 and 40, 89, 94, 3, and 5. 434 00:26:59,830 --> 00:27:04,720 So this is data that is, I believe, from fly genes. 435 00:27:04,720 --> 00:27:06,970 I've just ignored the fly nomenclature, 436 00:27:06,970 --> 00:27:11,710 because it's confusing, and just given them lettered names, OK? 437 00:27:11,710 --> 00:27:16,900 But this reflects data from some cross somewhere. 438 00:27:16,900 --> 00:27:20,110 So now we want to know-- 439 00:27:20,110 --> 00:27:22,960 let's go back to our map. 440 00:27:22,960 --> 00:27:25,090 We want to make a map, OK? 441 00:27:25,090 --> 00:27:26,720 And so to make our map, we're going 442 00:27:26,720 --> 00:27:30,490 to want to consider all pairwise distances 443 00:27:30,490 --> 00:27:33,350 between different genes. 444 00:27:33,350 --> 00:27:36,310 So we'll start with the A and B gene. 445 00:27:36,310 --> 00:27:38,120 I'll write over here. 446 00:27:38,120 --> 00:27:42,910 So let's consider the A/B distance. 447 00:27:42,910 --> 00:27:44,830 And remember, to get a distance, we're 448 00:27:44,830 --> 00:27:47,110 looking at the number of the frequency which 449 00:27:47,110 --> 00:27:51,250 there is recombination between these two genes, OK? 450 00:27:51,250 --> 00:27:55,720 So we now have to look through all of these recombinant class 451 00:27:55,720 --> 00:27:58,180 of progeny and figure out the ones that 452 00:27:58,180 --> 00:28:01,840 have had a recombination between A and B, right? 453 00:28:01,840 --> 00:28:05,980 So on the parent chromosome, you see little lowercase a 454 00:28:05,980 --> 00:28:11,290 started out with capital B and vice versa. 455 00:28:11,290 --> 00:28:13,690 So any case where we don't have lowercase 456 00:28:13,690 --> 00:28:19,030 a paired with capital B, there's been some type of exchange. 457 00:28:19,030 --> 00:28:21,850 So here, lowercase a's with lowercase b. 458 00:28:21,850 --> 00:28:23,440 So that's a recombinant. 459 00:28:23,440 --> 00:28:26,590 Here, uppercase A's with uppercase B. 460 00:28:26,590 --> 00:28:28,180 That's a recombinant, too. 461 00:28:28,180 --> 00:28:30,130 So we have to add all these up. 462 00:28:30,130 --> 00:28:33,700 So 45 plus 40. 463 00:28:33,700 --> 00:28:34,390 How about here? 464 00:28:34,390 --> 00:28:37,440 Recombination here or no? 465 00:28:37,440 --> 00:28:37,940 Yes. 466 00:28:37,940 --> 00:28:38,780 I'm hearing yes. 467 00:28:38,780 --> 00:28:39,860 That's correct. 468 00:28:39,860 --> 00:28:43,640 Here, recombination, yes or no? 469 00:28:43,640 --> 00:28:44,900 Carmen's shaking her head no. 470 00:28:44,900 --> 00:28:46,490 She's exactly right. 471 00:28:46,490 --> 00:28:49,505 So we just have to-- these are all the recombinants between A 472 00:28:49,505 --> 00:28:54,770 and B. So it's 45 plus 40 plus 89 473 00:28:54,770 --> 00:29:04,430 plus 94, which equals 268 over a total progeny of 1,448. 474 00:29:04,430 --> 00:29:09,050 And that gives you a map distance of 18.5%. 475 00:29:09,050 --> 00:29:12,320 Because this method was developed in Morgan's lab, 476 00:29:12,320 --> 00:29:15,570 this measurement is also known as a centimorgan. 477 00:29:15,570 --> 00:29:18,350 It was named in honor of Morgan. 478 00:29:18,350 --> 00:29:21,605 So that's what I refer to when I have lowercase c capital 479 00:29:21,605 --> 00:29:23,960 M. That's a centimorgan. 480 00:29:23,960 --> 00:29:28,340 So you can also use centimorgan here. 481 00:29:28,340 --> 00:29:30,440 All right, so that's A and B, but now we 482 00:29:30,440 --> 00:29:33,960 have to consider other distances. 483 00:29:33,960 --> 00:29:35,615 So how about the A/D distance? 484 00:29:39,200 --> 00:29:42,590 And again, we have to go through and figure out 485 00:29:42,590 --> 00:29:48,260 where the alleles for A and D have been recombined, OK? 486 00:29:48,260 --> 00:29:54,080 So little a is with capital D, and upper case A 487 00:29:54,080 --> 00:29:56,210 is with lowercase d. 488 00:29:56,210 --> 00:29:59,950 So we have to find all the cases where that's not the case. 489 00:29:59,950 --> 00:30:02,750 Here, this is lowercase a with capital D 490 00:30:02,750 --> 00:30:05,510 and capital A with lowercase d. 491 00:30:05,510 --> 00:30:08,625 This is parental from the respect of just the A and D 492 00:30:08,625 --> 00:30:13,610 genes, but all the rest of these guys are recombinants, OK? 493 00:30:13,610 --> 00:30:18,920 So this is 89 plus 94 plus 3 plus 5, which 494 00:30:18,920 --> 00:30:26,870 comes out to be 191 over 1,448. 495 00:30:26,870 --> 00:30:30,670 And this is 13.2 centimorgans. 496 00:30:30,670 --> 00:30:39,570 So that's the distance between A and D. Distance between A-B, 497 00:30:39,570 --> 00:30:42,060 distance between A-D. 498 00:30:42,060 --> 00:30:48,490 So the last combination, then, is just B and D. 499 00:30:48,490 --> 00:30:53,360 So if we consider the B/D distance, 500 00:30:53,360 --> 00:30:57,080 again, we have to look for all cases in which lowercase 501 00:30:57,080 --> 00:31:00,140 b and d become separated and uppercase 502 00:31:00,140 --> 00:31:03,380 B and D become separated. 503 00:31:03,380 --> 00:31:05,600 Here, they're separated. 504 00:31:05,600 --> 00:31:08,150 Here, they're separated. 505 00:31:08,150 --> 00:31:10,700 Wait, no, not here, sorry. 506 00:31:10,700 --> 00:31:12,260 Here, they're not separated. 507 00:31:12,260 --> 00:31:13,430 Here, they are separated. 508 00:31:15,950 --> 00:31:18,720 Everyone see how I'm doing this? 509 00:31:18,720 --> 00:31:22,618 Are there any questions about it? 510 00:31:22,618 --> 00:31:24,660 You can just shout it out if you have a question? 511 00:31:27,520 --> 00:31:30,430 So this distance is 6.4 centimorgans. 512 00:31:34,170 --> 00:31:38,730 Everyone see how I'm considering every pairwise combination 513 00:31:38,730 --> 00:31:40,800 of genes and then just ignoring the other one 514 00:31:40,800 --> 00:31:42,870 and looking for where there's been 515 00:31:42,870 --> 00:31:46,740 a recombination in the progeny? 516 00:31:46,740 --> 00:31:49,260 So now that we have our distances, 517 00:31:49,260 --> 00:31:51,810 we can make our map, right? 518 00:31:51,810 --> 00:31:57,600 So the two genes that are farthest apart are A and B, OK? 519 00:31:57,600 --> 00:32:01,320 So that's kind of like here, Rivendell and Lonely Mountain. 520 00:32:01,320 --> 00:32:04,560 Those are the two genes that are at the extremities. 521 00:32:04,560 --> 00:32:07,860 So I'll draw this out. 522 00:32:07,860 --> 00:32:10,950 It doesn't matter which way you put it. 523 00:32:10,950 --> 00:32:14,340 We're just mapping these genes relative to each other. 524 00:32:14,340 --> 00:32:18,460 But B and A are the farthest apart from each other. 525 00:32:18,460 --> 00:32:24,960 Now if we consider the distance between B and D, 526 00:32:24,960 --> 00:32:28,380 that's 6.4 centimorgans. 527 00:32:28,380 --> 00:32:31,020 So it appears that D is closer to B 528 00:32:31,020 --> 00:32:37,050 than it is to A, because it's 13 centimorgans away from A, OK? 529 00:32:37,050 --> 00:32:40,510 So D is kind of like Beorn's house here. 530 00:32:40,510 --> 00:32:43,140 It's closer to Rivendell than it is from Lonely Mountain. 531 00:32:47,560 --> 00:32:49,720 So we'll put that in there. 532 00:32:49,720 --> 00:32:52,680 This distance is 6.4 centimorgans. 533 00:32:52,680 --> 00:32:59,170 And then the distance here is 13.2 centimorgans. 534 00:32:59,170 --> 00:33:03,340 As far as I know, no one of the field of genetic mapping 535 00:33:03,340 --> 00:33:05,950 has been assaulted by large spiders, 536 00:33:05,950 --> 00:33:08,800 but the field is still young. 537 00:33:08,800 --> 00:33:13,780 So one thing that should be maybe bothering you right now 538 00:33:13,780 --> 00:33:17,680 is if you add up the distance between B and D 539 00:33:17,680 --> 00:33:22,780 and D and A, you don't, in fact, get 18.4 centimorgans. 540 00:33:22,780 --> 00:33:28,042 Instead, you get 19.6 centimorgans. 541 00:33:28,042 --> 00:33:32,800 This is 19.6 centimorgans. 542 00:33:32,800 --> 00:33:37,870 So it seems like, somehow, we are underestimating 543 00:33:37,870 --> 00:33:39,520 this distance here, OK? 544 00:33:39,520 --> 00:33:41,800 So we seem to be underestimating this. 545 00:33:47,950 --> 00:33:52,380 So why is it that we are underestimating this distance? 546 00:33:52,380 --> 00:33:55,930 Well, to consider that, you have to sort of look 547 00:33:55,930 --> 00:34:00,930 at how all of these classes were generated. 548 00:34:00,930 --> 00:34:02,680 So now I'm going to go through each class, 549 00:34:02,680 --> 00:34:05,140 and we'll look at how it was generated. 550 00:34:05,140 --> 00:34:07,690 I'm also going to-- 551 00:34:07,690 --> 00:34:10,030 well, I'll just draw new chromosomes. 552 00:34:10,030 --> 00:34:12,460 So we have to draw this order now. 553 00:34:12,460 --> 00:34:14,830 We have B, D, a. 554 00:34:14,830 --> 00:34:22,020 So the first chromosome is B, D, a. 555 00:34:22,020 --> 00:34:25,270 That's right, B, D, a. 556 00:34:25,270 --> 00:34:33,730 The other chromosome is b, d, A. So now we 557 00:34:33,730 --> 00:34:36,040 look and see how this recombinant class was 558 00:34:36,040 --> 00:34:38,260 generated. 559 00:34:38,260 --> 00:34:40,510 So this is lowercase a. 560 00:34:40,510 --> 00:34:48,810 We'll start with b-- lowercase b, lowercase a, but capital D. 561 00:34:48,810 --> 00:34:52,139 So it's capital D, lowercase a. 562 00:34:52,139 --> 00:34:55,690 So that recombinant results from this chromosome here, where 563 00:34:55,690 --> 00:34:58,080 there's crossing over, and little b gets 564 00:34:58,080 --> 00:35:01,315 hooked up to big D and little a. 565 00:35:01,315 --> 00:35:02,970 See how I did that? 566 00:35:02,970 --> 00:35:05,250 And then if we consider this class, 567 00:35:05,250 --> 00:35:13,230 this is uppercase B, lowercase d, capital A. So these two 568 00:35:13,230 --> 00:35:17,760 classes of progeny result from a single crossover 569 00:35:17,760 --> 00:35:19,710 between B and D, OK? 570 00:35:19,710 --> 00:35:24,760 So this is a single crossover between B and D genes. 571 00:35:24,760 --> 00:35:29,080 And now we can go through and look at how this is generated. 572 00:35:29,080 --> 00:35:32,940 So to get all recessive alleles on the same chromosome, 573 00:35:32,940 --> 00:35:35,970 there would be a crossover here. 574 00:35:35,970 --> 00:35:42,400 And so this is a single crossover between D and A. 575 00:35:42,400 --> 00:35:49,320 So a single crossover between D and A. Now, 576 00:35:49,320 --> 00:35:51,450 these last couple classes of progeny 577 00:35:51,450 --> 00:35:55,950 are interesting in that they're the least frequent class. 578 00:35:55,950 --> 00:35:59,430 And so when we consider how they're generated, 579 00:35:59,430 --> 00:36:02,400 we'll start with uppercase B. 580 00:36:02,400 --> 00:36:08,720 Let's see if I can get rid of this. 581 00:36:08,720 --> 00:36:16,700 So uppercase B, lowercase d, lowercase a. 582 00:36:16,700 --> 00:36:20,570 And so what this last class is, is actually a double crossover. 583 00:36:26,530 --> 00:36:28,570 So this is a double crossover. 584 00:36:28,570 --> 00:36:31,240 And it's least frequent because there's 585 00:36:31,240 --> 00:36:35,260 a lower probability of getting two crossovers in this region. 586 00:36:35,260 --> 00:36:38,560 But now you see that even though it doesn't 587 00:36:38,560 --> 00:36:42,310 look like there was recombination between A and D, 588 00:36:42,310 --> 00:36:43,720 in fact, there was. 589 00:36:43,720 --> 00:36:46,060 There were two crossovers, and it just 590 00:36:46,060 --> 00:36:48,940 looks like there was no recombination, if you didn't 591 00:36:48,940 --> 00:36:55,630 see the behavior of gene D. 592 00:36:55,630 --> 00:36:57,910 So if we take into account that there are actually 593 00:36:57,910 --> 00:37:03,550 double crossovers between B and A, 594 00:37:03,550 --> 00:37:05,980 then if we add that into our calculation 595 00:37:05,980 --> 00:37:11,770 here, where you add in 3 plus 5 multiplied by 2 596 00:37:11,770 --> 00:37:16,060 because these are both double crossover events, then 597 00:37:16,060 --> 00:37:19,690 you get the 19.6 centimorgans that you 598 00:37:19,690 --> 00:37:23,590 would expect by adding up the other recombinations, OK? 599 00:37:27,420 --> 00:37:28,320 How is that? 600 00:37:28,320 --> 00:37:30,660 Is that clear to everybody? 601 00:37:30,660 --> 00:37:35,970 You're going to have to make maps like this on the problem 602 00:37:35,970 --> 00:37:38,020 set and possibly the test. 603 00:37:38,020 --> 00:37:40,976 So make sure you can given-- yeah, Ory? 604 00:37:40,976 --> 00:37:43,184 AUDIENCE: I realized that you immediately [INAUDIBLE] 605 00:37:43,184 --> 00:37:45,017 overestimated the difference between A and B 606 00:37:45,017 --> 00:37:48,590 and not overestimated B to D or D to A? 607 00:37:48,590 --> 00:37:50,360 ADAM MARTIN: It's because when you 608 00:37:50,360 --> 00:37:53,690 have two genes that are very far apart, 609 00:37:53,690 --> 00:37:56,000 you can have multiple crossovers. 610 00:37:56,000 --> 00:38:00,050 And when you have sort of crossovers 611 00:38:00,050 --> 00:38:02,580 that are in pairs of two, then it's 612 00:38:02,580 --> 00:38:05,447 going to go from one strand back to the other, 613 00:38:05,447 --> 00:38:08,030 and so you're not going to see a recombination between the two 614 00:38:08,030 --> 00:38:10,280 alleles. 615 00:38:10,280 --> 00:38:14,720 So it's an underestimate, because if you have multiples 616 00:38:14,720 --> 00:38:16,610 of two in terms of crossovers, you're 617 00:38:16,610 --> 00:38:20,367 going to miss the recombination events. 618 00:38:20,367 --> 00:38:21,200 You see what I mean? 619 00:38:25,480 --> 00:38:28,330 You understand that you can miss the double crossover events? 620 00:38:28,330 --> 00:38:28,960 AUDIENCE: Yeah, I get that. 621 00:38:28,960 --> 00:38:30,002 ADAM MARTIN: Yeah, right? 622 00:38:30,002 --> 00:38:31,660 So then you're going to underestimate 623 00:38:31,660 --> 00:38:34,330 the number of crossovers that actually happened 624 00:38:34,330 --> 00:38:35,500 in that genetic region. 625 00:38:38,860 --> 00:38:45,520 All right, now I want to end with an experiment that, again, 626 00:38:45,520 --> 00:38:51,610 makes the point that genes are these entities that 627 00:38:51,610 --> 00:38:54,550 are on chromosomes. 628 00:38:54,550 --> 00:38:56,710 So just like you can have linkage 629 00:38:56,710 --> 00:38:59,380 between two genes on the same chromosome, 630 00:38:59,380 --> 00:39:02,290 you can also have linkage between genes 631 00:39:02,290 --> 00:39:05,350 and physical structures on chromosomes, 632 00:39:05,350 --> 00:39:06,400 like the centromere. 633 00:39:09,640 --> 00:39:12,370 So you could have genes like A and B 634 00:39:12,370 --> 00:39:14,980 here that are present on the chromosomes 635 00:39:14,980 --> 00:39:16,930 and present very near the centromere 636 00:39:16,930 --> 00:39:20,470 of those chromosomes, OK? 637 00:39:20,470 --> 00:39:24,100 So they could be right on top of the centromere, OK? 638 00:39:24,100 --> 00:39:27,220 And to show you how this manifests itself, 639 00:39:27,220 --> 00:39:31,390 I have to tell you about another organism, which 640 00:39:31,390 --> 00:39:35,620 is a unicellular organism called yeast. 641 00:39:35,620 --> 00:39:38,050 And yeast is special and that it can 642 00:39:38,050 --> 00:39:41,830 exist in both a haploid and a diploid form. 643 00:39:41,830 --> 00:39:44,050 So it has a lifecycle that involves 644 00:39:44,050 --> 00:39:47,485 it going both as a haploid and as a diploid. 645 00:39:50,650 --> 00:39:54,490 And so you can take yeast-- 646 00:39:54,490 --> 00:39:58,870 and we'll take two haploid yeast cells. 647 00:39:58,870 --> 00:40:03,220 And much like gametes, these can fuse to form a zygote. 648 00:40:03,220 --> 00:40:05,260 So, in this case, I'm taking-- 649 00:40:05,260 --> 00:40:09,370 again, we'll consider two generic genes, A and B. 650 00:40:09,370 --> 00:40:14,160 And we'll make a diploid yeast cell that is heterozygous, 651 00:40:14,160 --> 00:40:17,440 or hybrid, for A and B. 652 00:40:17,440 --> 00:40:20,350 And what's great and special about yeast 653 00:40:20,350 --> 00:40:25,030 and why I'm telling you about this is because as opposed 654 00:40:25,030 --> 00:40:30,040 to flies and us and other organisms, 655 00:40:30,040 --> 00:40:33,040 the product of a single meiosis is 656 00:40:33,040 --> 00:40:40,040 packaged in this single package, if you will. 657 00:40:40,040 --> 00:40:44,500 So the yeast can undergo meiosis, 658 00:40:44,500 --> 00:40:47,500 and the product of a single meiosis 659 00:40:47,500 --> 00:40:51,640 is present in this case, where each of these 660 00:40:51,640 --> 00:40:59,680 would represent a haploid cell that then can divide and make 661 00:40:59,680 --> 00:41:00,790 many cells. 662 00:41:00,790 --> 00:41:04,480 But this is the product of a single meiosis in a package, 663 00:41:04,480 --> 00:41:05,170 OK? 664 00:41:05,170 --> 00:41:09,280 So you can actually see the direct result 665 00:41:09,280 --> 00:41:12,350 of a meiotic division, a single meiotic division. 666 00:41:12,350 --> 00:41:17,425 So this is the product of a single meiotic division. 667 00:41:22,280 --> 00:41:25,810 And that's special because when we make gametes, 668 00:41:25,810 --> 00:41:27,370 we have individual cells. 669 00:41:27,370 --> 00:41:30,010 All the products of meiosis are split up, 670 00:41:30,010 --> 00:41:33,190 and then just one randomly finds an egg and fertilizes it. 671 00:41:33,190 --> 00:41:35,800 So you don't know which of the gametes 672 00:41:35,800 --> 00:41:37,570 are from the product of a single meiosis. 673 00:41:41,350 --> 00:41:45,190 And so being able to see the product of a single meiosis 674 00:41:45,190 --> 00:41:47,980 allows us to see things like genes being 675 00:41:47,980 --> 00:41:50,800 linked to physical structures on the chromosome 676 00:41:50,800 --> 00:41:52,680 like the centromere. 677 00:41:52,680 --> 00:41:55,540 So if we consider this case, these two genes 678 00:41:55,540 --> 00:41:58,030 are both linked to the centromere. 679 00:41:58,030 --> 00:41:59,980 And during metaphor phase of meiosis I, 680 00:41:59,980 --> 00:42:03,270 they could align like this, in which case 681 00:42:03,270 --> 00:42:07,750 you would get spores that are parental 682 00:42:07,750 --> 00:42:10,300 for both dominant alleles or parental 683 00:42:10,300 --> 00:42:12,880 for both recessive alleles. 684 00:42:12,880 --> 00:42:15,760 So each of these cells is known as a spore. 685 00:42:15,760 --> 00:42:18,950 So I'll label spore numbers here. 686 00:42:18,950 --> 00:42:20,960 So this is spore number. 687 00:42:20,960 --> 00:42:27,340 And in this case, you get two spores 688 00:42:27,340 --> 00:42:30,040 that are dominant for both alleles and two spores that 689 00:42:30,040 --> 00:42:33,040 are recessive for both alleles. 690 00:42:33,040 --> 00:42:36,020 Because there are two types, it's known as a ditype. 691 00:42:36,020 --> 00:42:42,790 And this is a parental ditype, because you have two types 692 00:42:42,790 --> 00:42:43,930 of spores . 693 00:42:43,930 --> 00:42:53,580 And they are both parental 694 00:42:53,580 --> 00:42:57,030 An alternative scenario is that these chromosomes 695 00:42:57,030 --> 00:42:58,950 would align differently, right? 696 00:42:58,950 --> 00:43:02,190 So you get parental spores there. 697 00:43:02,190 --> 00:43:06,030 However, alternatively, you could have this configuration, 698 00:43:06,030 --> 00:43:08,550 where this is now flipped. 699 00:43:08,550 --> 00:43:13,380 And during meiosis I here, these chromosomes move together. 700 00:43:13,380 --> 00:43:16,510 And they, again, produce two types of spores, 701 00:43:16,510 --> 00:43:17,910 so it's a ditype. 702 00:43:17,910 --> 00:43:20,940 But in this case, all the spores are non-parental. 703 00:43:23,760 --> 00:43:25,830 So another scenario is you get this. 704 00:43:30,540 --> 00:43:34,350 And because there are two types and they're non-parental, 705 00:43:34,350 --> 00:43:36,300 this is known as non-parental ditype. 706 00:43:45,940 --> 00:43:47,200 That's a non-parental ditype. 707 00:43:49,750 --> 00:43:55,330 And if these genes are linked to the centromere completely, 708 00:43:55,330 --> 00:44:02,590 then you can only get these two classes of packages, OK? 709 00:44:02,590 --> 00:44:04,630 So if these genes are unlinked-- 710 00:44:04,630 --> 00:44:10,900 so the two genes are unlinked, but both 711 00:44:10,900 --> 00:44:22,410 linked to the centromere, then you get parental ditype-- 712 00:44:22,410 --> 00:44:28,420 50% parental ditype, type 50% non-parental ditype. 713 00:44:28,420 --> 00:44:32,890 So I'm abbreviating parental ditype PD 714 00:44:32,890 --> 00:44:34,720 and non-parental ditype NPD. 715 00:44:37,960 --> 00:44:43,180 So what has to happen to get another type of spore? 716 00:44:43,180 --> 00:44:45,190 And another type of spore would be-- 717 00:44:45,190 --> 00:44:48,160 you could have spores that are all different genotypes 718 00:44:48,160 --> 00:44:54,670 from each other and that you have A cap dominant A/B; 719 00:44:54,670 --> 00:44:59,500 dominant A, recessive b; recessive a, dominant B; 720 00:44:59,500 --> 00:45:01,510 and lowercase a and b. 721 00:45:01,510 --> 00:45:05,605 And this is known as tetratype, because there are four types. 722 00:45:08,590 --> 00:45:10,570 So how do you get this tetratype? 723 00:45:10,570 --> 00:45:11,580 Anyone have an idea? 724 00:45:15,800 --> 00:45:16,800 Yeah, Jeremy? 725 00:45:16,800 --> 00:45:18,900 AUDIENCE: You're crossing over. 726 00:45:18,900 --> 00:45:26,760 So one of A and B would switch in one of the [INAUDIBLE].. 727 00:45:26,760 --> 00:45:29,130 ADAM MARTIN: Where would the crossing over happen? 728 00:45:29,130 --> 00:45:34,050 AUDIENCE: Between the two [INAUDIBLE] one [INAUDIBLE].. 729 00:45:41,220 --> 00:45:46,486 ADAM MARTIN: Between the allele and what? 730 00:45:46,486 --> 00:45:47,360 AUDIENCE: Sorry? 731 00:45:47,360 --> 00:45:50,235 ADAM MARTIN: The crossing over would occur between the gene-- 732 00:45:50,235 --> 00:45:51,210 AUDIENCE: Oh, and the centromere. 733 00:45:51,210 --> 00:45:52,918 ADAM MARTIN: And the centromere, exactly. 734 00:45:52,918 --> 00:45:54,090 Jeremy is exactly right. 735 00:45:54,090 --> 00:45:57,450 So Jeremy said that in order to get a tetratype, 736 00:45:57,450 --> 00:45:59,850 you have to have a recombination event, 737 00:45:59,850 --> 00:46:02,340 but this time, not between two genes, 738 00:46:02,340 --> 00:46:05,130 but between a gene and the centromere. 739 00:46:05,130 --> 00:46:07,620 So at least one of the genes has to be 740 00:46:07,620 --> 00:46:09,475 unlinked to the centromere. 741 00:46:12,580 --> 00:46:16,020 And in that case, now you get a meiotic event 742 00:46:16,020 --> 00:46:18,870 that gives rise to four spores. 743 00:46:18,870 --> 00:46:21,990 And there are four different ways to get this. 744 00:46:21,990 --> 00:46:31,560 So if you have two genes unlinked and at least one 745 00:46:31,560 --> 00:46:40,650 is unlinked to the centromere, then you 746 00:46:40,650 --> 00:46:43,350 get a pattern where you have a 1 to 1 747 00:46:43,350 --> 00:46:46,840 to 4 ratio between all these different events. 748 00:46:46,840 --> 00:46:51,630 So you have a 1 to 1 to 4 ratio between parental ditype, 749 00:46:51,630 --> 00:46:53,850 non-parental ditype, and tetratype. 750 00:46:56,980 --> 00:46:58,440 And we can see this in yeast. 751 00:46:58,440 --> 00:47:01,710 If you have two genes that are linked to the centromere, 752 00:47:01,710 --> 00:47:05,330 you only get parental ditypes and non-parental ditypes, 753 00:47:05,330 --> 00:47:08,910 where virtually everything else gives rise to tetratypes, 754 00:47:08,910 --> 00:47:10,445 except if they're linked. 755 00:47:10,445 --> 00:47:11,820 What happens if the two genes are 756 00:47:11,820 --> 00:47:17,730 linked to each other, irregardless of the centromere? 757 00:47:17,730 --> 00:47:20,910 If you have two genes that are linked, what's going to be-- 758 00:47:20,910 --> 00:47:23,859 what are your progeny going to look like? 759 00:47:23,859 --> 00:47:25,340 AUDIENCE: Parentals. 760 00:47:25,340 --> 00:47:26,630 ADAM MARTIN: You're only going to get the parentals, 761 00:47:26,630 --> 00:47:28,380 or you're going to get a lot of parentals. 762 00:47:28,380 --> 00:47:30,420 Javier is exactly right, right? 763 00:47:30,420 --> 00:47:33,800 If the two genes are linked, the parental ditypes 764 00:47:33,800 --> 00:47:36,320 are going to be much greater than any of the other classes. 765 00:47:41,330 --> 00:47:44,180 Now, this might seem esoteric, but I like the idea 766 00:47:44,180 --> 00:47:48,380 that you can have linkage between a gene and something 767 00:47:48,380 --> 00:47:50,810 that's just the place on the chromosome that's 768 00:47:50,810 --> 00:47:52,730 getting physically pulled. 769 00:47:52,730 --> 00:47:54,680 It all makes it much more physical, 770 00:47:54,680 --> 00:47:59,630 which I think is nice to think about. 771 00:47:59,630 --> 00:48:01,530 All right, we're almost done. 772 00:48:01,530 --> 00:48:03,560 I just have-- yes, Natalie? 773 00:48:03,560 --> 00:48:06,000 AUDIENCE: Can you go over what the PD [INAUDIBLE]?? 774 00:48:08,930 --> 00:48:09,740 ADAM MARTIN: Yes. 775 00:48:09,740 --> 00:48:11,960 So PD is parental ditype. 776 00:48:11,960 --> 00:48:13,990 So this is the parental ditype. 777 00:48:13,990 --> 00:48:18,020 It's a class of product here where 778 00:48:18,020 --> 00:48:22,130 you get four spores that are each of these genotypes, OK? 779 00:48:22,130 --> 00:48:25,520 So each of these 1, 2, 3, and 4 would 780 00:48:25,520 --> 00:48:31,160 represent one of these cells from a single meiotic event. 781 00:48:31,160 --> 00:48:33,802 Does that make sense, Natalie? 782 00:48:33,802 --> 00:48:35,260 Does everyone see what I did there? 783 00:48:39,350 --> 00:48:44,660 So these 1 2 and 3 are the spores of the meiotic event 784 00:48:44,660 --> 00:48:46,390 right here.