Match each statement (a–e) with the best answer from the follo...

Question

Match each statement (a–e) with the best answer from the following list: consultand, 50%, prior probability, 66.7%, obligate carrier, 100%.

The Mendelian risk that a person is a heterozygous carrier of a recessive condition

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. It says There is a 100% chance that a female will be a carrier for color blindness. When and we have choice. A, both her parents are colorblind choice. B both her parents have normal vision choice C. Her father is color blind and her mother is not a carrier and has normal vision. And Joyce Dee, her father has normal vision and her mother is a carrier and has normal vision. So we've got a lot to think through here. But let's start with just looking at the different genotypes and what they mean, we know that color blindness is an X linked recessive trait. So a female could be a carrier, which is what we're asked about here where her genotype would be X A normal X and an X superscript C with a gene for color blindness. That would mean she's a carrier. A colorblind female would have to have two copies of this gene. So a colorblind female would be X superscript C X super C and that would mean she's color blind. And when we think about males, a meal has Only one x chromosome. So the male who has one copy, even though this is the recessive trait. So X superscript C, but then Y as his other sex chromosome is color blind. A male cannot be a carrier since he only has one X chromosome. So our question is asking us about a scenario where A female has a 100% chance of being a carrier. So she's not color blind but she carries that all. So that means her genotype must be that X X X X superscript C. So one normal X one X with the color bind gene. So where did these come from? Well, let's look through our answer choices and think about the different possibilities. Choice A was both her parents are colorblind. Well, in that case, her colorblind father would have to be X superscript C Y and her mother would have to have two copies X superscript C X superscript C. But in that case, the female offspring we're asking about would not be a carrier. She would actually Have a 100% chance of being color blind herself because all of the X chromosomes that belong to her parents all have that gene for color blindness. So she would inherit her father's colorblind gene. And since her mother's both chromosomes have a gene for color blindness, she will inherit an affected gene from her mother as well. So choice A is not a correct answer because that would result in a female who has all the female offspring will be color blind choice. B says both her parents have normal vision. Well, a normal vision female could be a carrier or could not have the gene at all. But a male with normal vision has an unaffected. And then A Y. So we're asking about again, a female with a 100% chance that a female will be a carrier. Well, this wouldn't be the case here because all we know about the mother is that she has normal vision if the mother does not carry the gene for color blindness, so mother could be normal X normal X. If that's the case, the daughter will not be a carrier. And again, we're focusing on that chan that specification that there has to be a 100% chance the daughter will be a carrier. So choice B is not correct in choice C, it says her father is colorblind and her mother is not a carrier and has a normal vision. Well, this is actually what we're looking for here because in, I'm gonna bring this down here. It doesn't fit next to the choice. So at the bottom, I'm going to write father color blind. So he must have X superscript C Y and the mother is not a carrier and has normal vision. So we know in this case for sure that the mother's genotype is X both unaffected. So since we're told the mother is not a carrier and has normal vision. We know for certain, unlike in Choice B where it just tells us the mother has normal vision, what the mother's genotype must be. In this case, the father will always Pass on the colorblind gene. So 100 of the time an X superscript C will come from the father And the mother, both her x chromosomes are normal. So 100% of the time we have a normal from the mother. So 100% of the time the daughter will be a carrier. So choice C is our correct answer. Just to be thorough. Let's look at choice D and choice die. It says her father has normal vision and her mother is a carrier and has normal vision. Well, in this case, the father is unaffected X Y, the mother is a carrier. So she has one unaffected X and one X superscript C. But in that case, the mother is going to donate, the father will always give a normal X, the mother will give a normal X 50% of the time and the color by gene, 50% of the time. So in that case, in choice D, the daughter has just a 50% chance of being a carrier Since 50% of the time she'll inherit a normal gene from both parents. So that's why choice D is incorrect. Since we're looking for a 100% chance of the daughter being a carrier. So again, There is a 100% chance that a female will be a carrier for color blindness. When choice c her father is colorblind and her mother is not a carrier and has normal vision. See you in the next video.

Match each statement (a–e) with the best answer from the following list: consultand, 50%, prior probability, 66.7%, obligate carrier, 100%.
The Mendelian risk that a person is a heterozygous carrier of a recessive condition

Key Concepts

Mendelian Inheritance

Carrier Status

Prior Probability

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