zuikaku Posted March 7, 2021 Share Posted March 7, 2021 (edited) Hello, Gene for cataract is located on X chromosome. This disease is gonosomally recessive. Let's say two healthy parents have twins (both males - sons). I am supposed to decide whether those twins are monozygotic or dizygotic if: A) both sons have cataract; B) one son is healthy, the other one has cataract; C) both sons are healthy Also could you decide in this case about twins being monozygotic/dizygotic if they were females? Thank you for your help Edited March 7, 2021 by zuikaku grammar Link to comment Share on other sites More sharing options...
VenusPrincess Posted March 7, 2021 Share Posted March 7, 2021 (edited) 59 minutes ago, zuikaku said: Hello, Gene for cataract is located on X chromosome. This disease is gonosomally recessive. Let's say two healthy parents have twins (both males - sons). I am supposed to decide whether those twins are monozygotic or dizygotic if: A) both sons have cataract; B) one son is healthy, the other one has cataract; C) both sons are healthy Also could you decide in this case about twins being monozygotic/dizygotic if they were females? Thank you for your help If any of the sons has cataracts and the parents are healthy then the parents must be heterozygotes. Now if the twins are dizygotic the probability that both sons will have cataracts is 1/16, the probability that only one son will have cataracts is 3/8, and the probability that both sons will be healthy is 9/16. If however the twins are monozygotic then the probability that both sons will have cataracts is 1/4, the probability that only one son will have cataracts is 0, and the probability that both sons will be healthy is 3/4. Then P(monozygotic | both sons have cataracts) = P(both sons have cataracts | monozygotic) / P(both sons have cataracts) and by Bayes' theorem: P(monozygotic | both sons have cataracts) = 1/4 / (1/4 * P(monozygotic) + 1/16 * P(dizygotic)) and by pattern matching: P(dizygotic | both sons have cataracts) = 1/16 / (1/4 * P(monozygotic) + 1/16 * P(dizygotic)) so that it's 4 times more likely that they are monozygotic. I was able to do these deductions and calculations almost instantly despite not being an expert in genetics, so you should be able to figure it out if you have decent mathematical aptitude. Edit: Actually since it's located on the X chromosome and the mother is healthy the sons will always be healthy unless there are mutations... so the question makes no sense. Edited March 7, 2021 by VenusPrincess Link to comment Share on other sites More sharing options...
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