1 00:00:00,000 --> 00:00:07,320 2 00:00:07,320 --> 00:00:08,530 PROFESSOR: Hello and welcome to the 3 00:00:08,530 --> 00:00:10,660 help session on pedigrees. 4 00:00:10,660 --> 00:00:13,560 Today, we will be working out a problem together. 5 00:00:13,560 --> 00:00:17,050 If you have not yet had a chance to work it on your own, 6 00:00:17,050 --> 00:00:19,600 please do so now, and return to this video 7 00:00:19,600 --> 00:00:20,850 when you are done. 8 00:00:20,850 --> 00:00:29,390 9 00:00:29,390 --> 00:00:32,380 Now that you've had a chance to look at this problem, let's 10 00:00:32,380 --> 00:00:33,780 work it out together. 11 00:00:33,780 --> 00:00:36,780 The first part of this question asks, what is the 12 00:00:36,780 --> 00:00:40,930 mode of inheritance that is observed in this pedigree? 13 00:00:40,930 --> 00:00:43,730 So, we know that there are two main types of modes of 14 00:00:43,730 --> 00:00:44,800 inheritance. 15 00:00:44,800 --> 00:00:47,780 It can either be dominant or recessive. 16 00:00:47,780 --> 00:00:51,030 And from there it can either be autosomal or X-linked. 17 00:00:51,030 --> 00:00:53,770 18 00:00:53,770 --> 00:00:57,720 If a disease follows a dominant inheritance pattern, 19 00:00:57,720 --> 00:01:00,990 generally, it must be present in every generation. 20 00:01:00,990 --> 00:01:04,230 So here we notice that the disease is present in the 21 00:01:04,230 --> 00:01:07,660 first generation, but it's not present in anyone in the 22 00:01:07,660 --> 00:01:09,430 second generation. 23 00:01:09,430 --> 00:01:13,500 However, then it reappears in the third generation. 24 00:01:13,500 --> 00:01:17,170 This suggests that the disease is recessive. 25 00:01:17,170 --> 00:01:27,540 26 00:01:27,540 --> 00:01:30,950 So now, do we think this disease is 27 00:01:30,950 --> 00:01:33,910 autosomal or is it X-linked? 28 00:01:33,910 --> 00:01:37,010 So we look at the pedigree again, and upon closer 29 00:01:37,010 --> 00:01:41,000 inspection, we notice that the affected 30 00:01:41,000 --> 00:01:44,040 individuals are only males. 31 00:01:44,040 --> 00:01:46,620 This is a key characteristic of an 32 00:01:46,620 --> 00:01:49,370 X-linked recessive disease. 33 00:01:49,370 --> 00:01:52,990 However, we're also given more information that tells us that 34 00:01:52,990 --> 00:01:56,460 this individual here does not carry an allele associated 35 00:01:56,460 --> 00:01:59,370 with the affected phenotype. 36 00:01:59,370 --> 00:02:04,100 This rules out the possibility that this disease could be an 37 00:02:04,100 --> 00:02:06,450 autosomal recessive disease. 38 00:02:06,450 --> 00:02:07,890 Thus, the most likely mode of 39 00:02:07,890 --> 00:02:09,795 inheritance is X-linked recessive. 40 00:02:09,795 --> 00:02:17,630 41 00:02:17,630 --> 00:02:21,400 Great, so moving on to the second part of this problem, 42 00:02:21,400 --> 00:02:24,480 we are asked to determine the genotypes of individuals 43 00:02:24,480 --> 00:02:27,280 number one and number three. 44 00:02:27,280 --> 00:02:31,940 So, if we look at this pedigree, we know individual 45 00:02:31,940 --> 00:02:37,400 one is female and not affected by the disease. 46 00:02:37,400 --> 00:02:44,860 Female individuals contain two X chromosomes, and since she 47 00:02:44,860 --> 00:02:50,345 is not affected she must contain at least one large R. 48 00:02:50,345 --> 00:02:53,170 So we'll call the allele associated with the 49 00:02:53,170 --> 00:02:54,420 disease small r. 50 00:02:54,420 --> 00:02:59,850 51 00:02:59,850 --> 00:03:03,950 We do not yet know what her second X chromosome could be. 52 00:03:03,950 --> 00:03:09,050 It could be either another large R or it 53 00:03:09,050 --> 00:03:10,100 could be a small r. 54 00:03:10,100 --> 00:03:13,140 And she also would still not be affected by the disease. 55 00:03:13,140 --> 00:03:21,230 56 00:03:21,230 --> 00:03:23,930 This male right here is affected by the disease. 57 00:03:23,930 --> 00:03:26,410 So he must have an X with a small r. 58 00:03:26,410 --> 00:03:30,480 59 00:03:30,480 --> 00:03:33,320 Remember that males have one X chromosome and one Y 60 00:03:33,320 --> 00:03:35,650 chromosome. 61 00:03:35,650 --> 00:03:39,250 So now, making our way down to individual number three. 62 00:03:39,250 --> 00:03:43,120 Individual number three does not have the disease and is 63 00:03:43,120 --> 00:03:45,520 female, so again, must have two X chromosomes. 64 00:03:45,520 --> 00:03:51,270 65 00:03:51,270 --> 00:03:54,060 One of the X chromosomes must come from her father and the 66 00:03:54,060 --> 00:03:55,940 other from her mother. 67 00:03:55,940 --> 00:03:59,070 The only X chromosome that her father can pass along to her 68 00:03:59,070 --> 00:04:02,940 is X small r. 69 00:04:02,940 --> 00:04:05,930 In order for her to not be affected with the disease, her 70 00:04:05,930 --> 00:04:13,480 other X chromosome must have a large R. 71 00:04:13,480 --> 00:04:16,860 She was able to get this genotype with her mother 72 00:04:16,860 --> 00:04:20,420 having either of these two genotypes. 73 00:04:20,420 --> 00:04:22,780 To make sure that both of these genotypes are possible 74 00:04:22,780 --> 00:04:27,320 for this mother, we need to examine her other children. 75 00:04:27,320 --> 00:04:29,300 Her son here is unaffected. 76 00:04:29,300 --> 00:04:39,540 So his genotype must have been X large R, Y. It's possible 77 00:04:39,540 --> 00:04:42,640 for him to get this genotype when the mother has either of 78 00:04:42,640 --> 00:04:45,560 these two genotypes. 79 00:04:45,560 --> 00:04:49,635 Similarly, this daughter could be unaffected just as this 80 00:04:49,635 --> 00:04:51,000 daughter was unaffected. 81 00:04:51,000 --> 00:04:53,970 So the mother can have either of these two genotypes. 82 00:04:53,970 --> 00:04:56,770 And we're unable to rule out one of them. 83 00:04:56,770 --> 00:04:59,620 So let me just write up the answer over here. 84 00:04:59,620 --> 00:05:19,020 85 00:05:19,020 --> 00:05:21,030 All right, so far, so good. 86 00:05:21,030 --> 00:05:24,060 Next, we're asked to calculate the probability that 87 00:05:24,060 --> 00:05:27,720 individual A is affected. 88 00:05:27,720 --> 00:05:32,960 Individual A is over here at the bottom of the pedigree. 89 00:05:32,960 --> 00:05:37,520 In order for individual A to be affected, she must have the 90 00:05:37,520 --> 00:05:38,770 following genotype. 91 00:05:38,770 --> 00:05:45,380 92 00:05:45,380 --> 00:05:48,460 The only way for her to get two little r's is for her to 93 00:05:48,460 --> 00:05:51,100 get one from her father and one from her mother. 94 00:05:51,100 --> 00:05:53,970 We know this is possible because her father is affected 95 00:05:53,970 --> 00:05:59,970 with the following genotype of Y, X small r. 96 00:05:59,970 --> 00:06:01,790 Her mother is not affected. 97 00:06:01,790 --> 00:06:05,840 So for her to get this small r, her mother must be a 98 00:06:05,840 --> 00:06:10,100 carrier for the disease, and have the genotype X 99 00:06:10,100 --> 00:06:14,860 large R, X small r. 100 00:06:14,860 --> 00:06:19,940 So in order to determine the probability that A is affected 101 00:06:19,940 --> 00:06:23,850 with both X small r's, we must also know the probability that 102 00:06:23,850 --> 00:06:26,570 her mother was a carrier for the disease. 103 00:06:26,570 --> 00:06:30,190 Let's call her mother individual C. 104 00:06:30,190 --> 00:06:33,820 So, we know the genotypes of her parents. 105 00:06:33,820 --> 00:06:35,410 Her mother was a carrier. 106 00:06:35,410 --> 00:06:38,190 And her father did not have the disease, so his genotype 107 00:06:38,190 --> 00:06:45,420 was Y with X large R. So to calculate the probability that 108 00:06:45,420 --> 00:06:48,680 individual C is a carrier for the disease, let's go back 109 00:06:48,680 --> 00:06:50,731 over here and draw out a Punnett square. 110 00:06:50,731 --> 00:07:00,240 111 00:07:00,240 --> 00:07:02,410 On one side, we're going to write the genotype of her 112 00:07:02,410 --> 00:07:09,080 father, which was Y because he's a male. 113 00:07:09,080 --> 00:07:13,210 And then, he was not affected by the disease, so it was X 114 00:07:13,210 --> 00:07:19,410 with a large R. Her mother was a carrier. 115 00:07:19,410 --> 00:07:24,330 So she had one copy that was X large R, and one copy that was 116 00:07:24,330 --> 00:07:26,770 x little r. 117 00:07:26,770 --> 00:07:30,630 So next, we want to know the probability 118 00:07:30,630 --> 00:07:33,270 that she was a carrier. 119 00:07:33,270 --> 00:07:35,990 The only ways you can get females is by looking at this 120 00:07:35,990 --> 00:07:37,240 top row here. 121 00:07:37,240 --> 00:07:44,190 122 00:07:44,190 --> 00:07:47,170 So, for now, we're going to ignore the bottom row. 123 00:07:47,170 --> 00:07:51,090 Of the two possible ways to generate a female, only one of 124 00:07:51,090 --> 00:07:52,960 them is a carrier. 125 00:07:52,960 --> 00:07:57,260 So there's a 50% chance that a female will be a carrier. 126 00:07:57,260 --> 00:08:05,680 127 00:08:05,680 --> 00:08:09,660 So this is a 50% chance that individual C is a carrier, 128 00:08:09,660 --> 00:08:14,330 which we will denote with this half filled in circle. 129 00:08:14,330 --> 00:08:17,530 All right, so now that we know the probability that C was a 130 00:08:17,530 --> 00:08:20,610 carrier, we need to know the probability that her progeny, 131 00:08:20,610 --> 00:08:23,290 A, was affected with the disease. 132 00:08:23,290 --> 00:08:28,570 Again, looking back over here, her progeny needs to inherit 133 00:08:28,570 --> 00:08:31,030 the X small r from the mother and the X 134 00:08:31,030 --> 00:08:33,150 small r from the father. 135 00:08:33,150 --> 00:08:36,130 So, once again, we can draw a Punnett square to better 136 00:08:36,130 --> 00:08:37,380 understand this. 137 00:08:37,380 --> 00:08:46,840 138 00:08:46,840 --> 00:08:49,700 Individual A's father was indeed 139 00:08:49,700 --> 00:08:51,340 affected by the disease. 140 00:08:51,340 --> 00:08:56,960 So his genotype was X small r and Y. Her mother, we just 141 00:08:56,960 --> 00:09:01,730 calculated the probability of her being a carrier, so she is 142 00:09:01,730 --> 00:09:04,690 X large R, X small r. 143 00:09:04,690 --> 00:09:07,500 144 00:09:07,500 --> 00:09:10,550 Now, we're looking at individual A, who is a female, 145 00:09:10,550 --> 00:09:12,500 so again, we can ignore the bottom row because 146 00:09:12,500 --> 00:09:14,760 this will be males. 147 00:09:14,760 --> 00:09:16,655 Here are the progeny which are female. 148 00:09:16,655 --> 00:09:22,560 149 00:09:22,560 --> 00:09:27,150 Of the two possibilities, only the one on the right, X small 150 00:09:27,150 --> 00:09:30,900 r, X small r is affected by the disease. 151 00:09:30,900 --> 00:09:35,130 So there is a 50% chance that A will be affected by the 152 00:09:35,130 --> 00:09:38,220 disease given that her mother is a carrier of the disease. 153 00:09:38,220 --> 00:09:43,950 154 00:09:43,950 --> 00:09:48,330 Now to finish up this part of the problem, we need to 155 00:09:48,330 --> 00:09:50,760 multiply these two probabilities together because 156 00:09:50,760 --> 00:09:53,280 we're looking at the probability that C is a 157 00:09:53,280 --> 00:10:00,930 carrier, and the probability that A is 158 00:10:00,930 --> 00:10:02,180 affected with the disease. 159 00:10:02,180 --> 00:10:07,400 160 00:10:07,400 --> 00:10:11,170 This gives us our answer of one fourth. 161 00:10:11,170 --> 00:10:14,310 All right, so, moving on to the last part of the problem. 162 00:10:14,310 --> 00:10:17,590 We need to calculate the probability that B is affected 163 00:10:17,590 --> 00:10:18,910 with the disease. 164 00:10:18,910 --> 00:10:21,840 So this isn't that hard now that we've drawn out all of 165 00:10:21,840 --> 00:10:23,820 these Punnett squares already. 166 00:10:23,820 --> 00:10:29,700 So, let's take a look back at our pedigree over here. 167 00:10:29,700 --> 00:10:33,520 If B is going to be affected with the disease, he must have 168 00:10:33,520 --> 00:10:38,760 the genotype Y, X with a small r. 169 00:10:38,760 --> 00:10:43,450 He gets the Y from his father and he must get this X with a 170 00:10:43,450 --> 00:10:45,770 small r from his mother. 171 00:10:45,770 --> 00:10:49,600 So again, we need the mother, C, to be a carrier. 172 00:10:49,600 --> 00:10:54,000 So let's go back to our Punnett squares over here. 173 00:10:54,000 --> 00:10:56,510 We know that there is a 50% chance that 174 00:10:56,510 --> 00:10:59,310 individual C is a carrier. 175 00:10:59,310 --> 00:11:03,220 Now we need to look at individual B being affected by 176 00:11:03,220 --> 00:11:03,810 the disease. 177 00:11:03,810 --> 00:11:06,090 So we can go ahead and fill in the bottom part of this 178 00:11:06,090 --> 00:11:07,340 Punnett square. 179 00:11:07,340 --> 00:11:14,010 180 00:11:14,010 --> 00:11:18,910 So of these two males, that are possible, only one of them 181 00:11:18,910 --> 00:11:20,620 is affected with the disease. 182 00:11:20,620 --> 00:11:24,510 So again, there is a 50% probability that that child 183 00:11:24,510 --> 00:11:25,760 will be affected with the disease. 184 00:11:25,760 --> 00:11:30,560 185 00:11:30,560 --> 00:11:32,340 Again, we are going to multiply these 186 00:11:32,340 --> 00:11:33,620 probabilities together. 187 00:11:33,620 --> 00:11:38,270 So the probability that the mother is a carrier, and that 188 00:11:38,270 --> 00:11:46,470 the son is affected with the disease, gives us one quarter. 189 00:11:46,470 --> 00:11:49,270 That concludes our problem on pedigrees. 190 00:11:49,270 --> 00:11:50,520 Thank you for joining us. 191 00:11:50,520 --> 00:11:52,447