In Exercises 21-28, find the probability and answer the questi...

Question

In Exercises 21-28, find the probability and answer the questions.

X-Linked Genetic Disease Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective X chromosome that occurs without a paired X chromosome that is not defective. In the following, represent a defective X chromosome with lowercase x, so a child with the xY or Yx pair of chromosomes will have the disease and a child with XX or XY or YX or xX or Xx will not have the disease. Each parent contributes one of the chromosomes to the child.

c. If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Males have XY chromosomes and females have XX chromosomes. X-linked recessive traits such as red green color blindness appear when a defective X chromosome is not paired with a healthy X. Represent the defective X chromosome with lowercase x, so a child with lowercase x Y or X will express the trait. A child with X X X where Y is capital Y X X X or X X will not express the trait. Each parent passes one of their chromosomes to the child. What is the probability that a son will exhibit the trait if the mother has one faulty X chromosome and the father has normal XY chromosomes? Awesome. So it appears for this particular problem, we're ultimately trying to figure out what the probability is that a son will exhibit a trait, this specific trait for color blindness. If the mother has one faulty capital X chromosome and the father has a normal XY chromosome. So what is the probability. That the son, considering their parents, will have this color blindness trait. So with that said, let's read off our multiple choice sensors to see what our final answer might be. A is 0, B is 0.25, C is 0.50, D is 0.75. Awesome. So, first off, our first step we need to take is we need to identify the parent's chromosomes. So we need to know that the mother has a capital X X chromosomes because one is normal and one is a defective capital X chromosome. And this is once again for the mother, which I'll denote as M, and for our father, they, they have a capital X capital Y. Chromosomes, which means that it is normal chromosomes, and this is for the dad. So once again, M for mom, D for dad, fantastic. So now our next step is we need to determine how a son gets the chromosomes. So we know that the son, which I'll call S, the son, will receive a Y from the father. Whereas the sun will also receive either an X. Or lowercase x. So a capital X, which I'll add little lines to the top so we know that's a capital X. So the sun will either receive a capital X chromosome or a lowercase X chromosome which once again denotes our defective. But the detect the defective trait for the chromosome from the mother. Awesome. So once again, the son can receive a capital Y chromosome from the father and the son can receive either a capital X chromosome or a lowercase X chromosome from the mother. And our third step is we need to list the possible combinations that the son can receive chromosome-wise. So, if the mother gives a capital X chromosome, that will mean that the son will have a XY. Chromosomes, so that will mean that it will have no defective traits, so it won't have color blindness, it'll be a result in a no. However, if the mother gives a lowercase X, that will mean that the son will have a lowercase X Y, which will mean that yes, it will have this defective trait, which means that they will be born with color blindness. So, that will mean as a result that there are 2 equally likely outcomes for sons, and these two equally likely outcomes will be X. Capital Y. Or lowercase X and Y. So now our fourth step we need to calculate the probability that the sun, this specific sun in question, has the trait. So the probability P, that the sun has a trait, so sun has trait, I'll call it SHT for short, so P of S. So the probability of the sun having the trait will be equal to 1 divided by 2, where 2 is the total amount of outcome. So it's 1 divided by 2, which will mean that the probability that the sun has the trait will be equal to 0.5 and 0.5 is our final answer. Hooray, we did it. We solved for this problem. So looking at our multiple choice answers, the correct answer has to be multiple choice answer letter C, which is 0.50. And that's it, 0.50. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

��app

Key Concepts

X-Linked Inheritance

Probability in Genetics

Punnett Square

Watch next