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

Question

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

Genetics: Eye Color Each of two parents has the genotype brown/blue, which consists of the pair of alleles that determine eye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one brown allele, that color will dominate and the eyes will be brown. (The actual determination of eye color is more complicated than that.)

b. What is the probability that a child of these parents will have the blue/blue genotype?

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. Genetics, hair texture. Each of the two parents has the genotype curly slash straight, consisting of the alleles that influence hair texture. Each parent passes on one allele to the offspring. Assume that the presence of at least one curly allele results in curly hair texture. For example, curly is dominant over straight. What is the probability that the offspring of these parents will have the straight slash straight genotype? Awesome. So it appears for this particular prom, we're ultimately trying to determine what the probability is that the offspring, considering these two parents that are described to us by the prom itself, will have the straight slash straight genotype. So now that we know we're ultimately trying to figure out this specific probability, let's read off our multiple choice answers to see what our final answer might be. A is 0.75, B is 0.50, C is 0.25, and D is 0.05. Awesome. So first off, in order to solve this particular problem, we need to determine the parent genotypes. So, as we should note and recall, both parents have the genotype curly slash straight. So let's denote curly as C and let us denote straight as S. So both parents are CS. So now we need to recall and use the dominance rule. So we need to recall a note, as we're told by the problem itself. The curly allele C is dominant over the straight allele S. So let's make a note that C is dominant, which I'll call capital D for dominant. And So once again, curly, the curly allele is gonna be the dominant trait versus the straight allele, which is not dominant. So this will mean that CC or CS will result in curly hair, because C is dominant and an SS combination will result in straight hair, because there is Only going to be the straight, so the child's only gonna have straight hair, the offspring is only gonna have straight hair if there is no C allele present. So it has to be SS. It can't have a C within the combination, or else they will have curly hair because curly hair is a dominant trait. So, with that said, we can now solve for this particular problem using a punnit square, which, as we should recall, is a diagram that is used to predict genotype distribution. Of offspring from the parental alleles from the parent alleles. So now, what we need to do is we need to determine the alleles each parent can pass on. So both parents are CS, so they can either pass along either C or S alleles to their offspring. So, when we go to construct our punnit square, which you can draw a square shape for your pun and square, so if we draw a little square here for a pun and square, and we divide it into three separate columns here. And split it in half. So we have a C for parent 2, which I'll call it P2 for parent 2. So we have a C allele for parent two, we also have an S allele. For a parent too. We also have AC For parent one, which I'll call it P1, awesome. And we also have a S for parent two, which I'll call P2, so S. P2, so P2 means parent 2, P1 means parent one. Fantastic. So, moving right along here, and it's important to note I said P2 up top. So it's we have a C for parent one, so you have a C allele for parent 1, and we have an S for parent 1 as well. So now we could do our combinations, so we can have, so we're when a parent one has a C, and a parent two has a C allelia of a C slash C combination. And for a parent one who has an S allele and a parent two that has a C allele, you'll have a CS combination. And then for C for parent one and an S for parent 2, we will have a CS combination, and then finally, if a parent one has an S allele and parent 2 has an S allele, they'll have an S slash S combination. So now we can identify all the possible genotypes. So we can have C C, we can also have CS, we can also have CS, and we can also have S S. So now we need to count how many out of the 4 combinations are straight slash straight combinations. So as you can see, we have 4 different combinations, but only 1, which I'll put a red box, and only 1 results in the straight genotype or the straight trait where the where the offspring will have straight hair. So that means that only 1 out of the 4 times will be this S slash Strait. So that will mean that the probability of having an S slash. S slash S, so basically the probability, so P of S-S is gonna be equal to 1 divided by 4 because there's only one opportunity out of the four possible outcomes. So 1 divided by 4 gives us 0.25. And that is our final answer. Hooray, we did it. So the probability that the offspring will have the SS genotype and will therefore have straight hair as a result will be 0.25, and this corresponds with multiple choice answer letter C, which once again is 0.25. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Genotype and Alleles

Dominance in Genetics

Punnett Square

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