For a number of human hereditary conditions, genetic tes...

Question

For a number of human hereditary conditions, genetic testing is available to identify heterozygous carriers. Some heterozygous carrier testing programs are community-based, often as part of an organized effort targeting specific populations in which a disease and carriers of a disease are relatively frequent. For example, carrier genetic testing programs for Tay–Sachs disease target Ashkenazi Jewish populations and sickle cell disease carrier testing programs target African American populations. The testing is usually free or available at minimal cost, the wait time for results is short, and the results are confidential and unavailable to third parties such as insurance companies. Neither the Tay–Sachs nor the sickle cell allele produces serious consequences for heterozygous carriers.

Do you personally think you would participate in the kind of carrier genetic testing described if you were a member of a population targeted for such testing?

Accepted Answer

Welcome back, everyone. Let's look at our next question. Sachs disease is an autism or recessive disorder characterized by nerve cell destruction in the brain and spinal cord. A couple's child has been diagnosed with the disorder. What is the probability that their next child will not be affected by the disorder? So in this case, we can reason backwards to the parents genotype when you have an affected child with an autism or recessive disorder, that child has to have inherited a copy of the affected gene from each parent. Since neither parent is affected with the disorder, that must mean that both parents Are Hetero six carriers of the Tsx gene. So our question asked us to consider what is the proper believability that their next child will not be affected by the disorder? So we can look at our Punnett Square here. We know that both parents are heterosexual. We'll call the gene T with big T meaning the normal gene and little T being the recessive Tay Sachs gene. So both parents must be big T little T. So we put across the top and on the left of our Punnett square, big T little T and fill in the genotype of the offspring. And as we expect from a mono hybrid cross, we have one out of four homocide is dominant, two out of four hatteras Vegas and one out of four homicide is recessive. So in this case, one out of four future Children will be affected by the disorder. But our question is saying, what is the possibility the child, next child will not be affected? Well, that would be 3/4 not affected that referring to the phenotype of being affected by the disease. Um Two, there's a, you know, two out of four chance that the child will be ahead of us carrier like the parents. But the question is asking the probability of being affected by the disease. So our answer choice, oops, we'll be over here. Choice. C 3/4 is the probability that their next child will not be affected by the disorder. See you in the next video.

Key Concepts

Heterozygous Carriers

Genetic Testing

Population Screening Programs

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