A 'wrongful birth' case was recently brought before a court in wh...

Question

A 'wrongful birth' case was recently brought before a court in which a child with Smith–Lemli–Opitz syndrome was born to apparently healthy parents. This syndrome is characterized by a cluster of birth defects including cleft palate, and an array of problems with the reproductive and urinary organs. Originally considered by their physician as having a nongenetic basis, the parents decided to have another child, who was also born with Smith–Lemli–Opitz syndrome. In the role of a genetic counselor, instruct the court about what occurred, including the probability of the parents having two affected offspring, knowing that the disorder is inherited as a recessive trait.

Calculate the probability that both the male and female are carriers for TSD.

Accepted Answer

Hi everyone, welcome back, let's look at our next problem. It says cystic fibrosis is a hereditary disease. It is due to a gene defect that interferes with how salt and water enter and exit sells. This can lead to a build up of thick sticky mucus in the body's tubes and passages, notably the lungs and digestive system along with recurring infections. What is the probability of having a child that is affected by cystic fibrosis? If both parents are carriers big c little C. For cystic fibrosis, note the normal allele is big C. And the aloe for cystic fibrosis is little C. So this is straightforward mono hybrid cross where you have two heterosexual parents. And we may just recall off the top of our heads that the genotype ratio of across like this is 1-2- of one being homeless. I guess dominant To being hetero cigarettes and one being homesick it's recessive. Or another way we know the ratio is 3 to 1 showing the dominant versus the recessive trait. In this case cystic fibrosis is the recessive trait the individual that has, it must be homogeneous recessive. So we know that the percent chance of having a child with cystic fibrosis when both parents are carriers is choice D 1/4 or 25%. Um If you want to be thorough and just look at the punnett square here to see where these ratios come from. Put up that straightforward square. We have our two hetero zegas parents. So we see here's our one homicide is dominant. Here are two heterocyclic offspring and then r One homogeneous recessive offspring. So here is our one in four chance of having a child with cystic fibrosis. And that's again choice D. As our answer. See you in the next video.

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