Biology: Mendelian Inheritance

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You know, the funnest part of genetics is doing some problems. So, here's what were going to do. I'm going to do a problem with you. I'm going to present a genetics problem to you. I'm going to teach you a much simpler technique than I've been using with the sperms and eggs running all over the place. Then I'm going to give you a problem, then we're going to halt this thing. You're going to have to solve it. You're going to have to click on the right answer, or just abandon ship, and then I'm going to show you how to do it the right way.
Step one, the problem. Okay, let me give you one problem. You know what, I also like to use human traits. Plants are nice for Mendel, but let's talk human. A couple has a child with cystic fibroses, CF. Now cystic fibroses is a very sad disease that--we'll talk a lot more about it later on, but the bottom line about cystic fibroses is you have this build up of mucous in your lungs. So, a child with cystic fibroses has to be constantly treated by pounding on the back to break up the clots of mucous in their lungs. So, it's a genetic disease. It's a protein mutation and this couple has a child with cystic fibroses. Neither of the parents have it. Neither parent has CF. They want to have a child and they couldn't use a genetics counselor and they say, "What are the odds of their next child having CF?"
Now this is a little bit different. See, we're working with what we know now. We now have accepted Mendel's law of segregation. We know that there's going to be dominant genes and recessive genes. And you're saying to me, "Okay, but wait a minute. Which gene is dominant?" Which gene is going to mask the other? I'm not going to tell you--well, I've got to tell you, because I told you I'd tell you the answer to this one. You can tell from this problem which gene is dominant. Let's think about it.
A couple has a CF, neither parent has CF. So, what must be true here? Like the wrinkled trait in Mendel's first cross, CF must be hidden. So, therefore, the normal condition must be the dominant. Now, don't go thinking all normal conditions are dominant. I have more problems for you later. That's not always true, but in this case it's true. Let's use n's. N equals normal, n equals CF. Now, what's the second thing we do. We've set up our key, that's step one. Step two--these are good things if you have to do genetics problems, I would recommend you follow this the way I'm doing it.
Step two is we set up our cross. So, we have two parents, both of whom are normal. So, we know that both of these parents have to have an N, don't they? Because normal is dominant. See how I'm doing this. Just detective work, kind of fun. We know that they have a child with CF and so that child, that f[1] child, must be nn and we know that one of these little n's had to come from one parent and the other n had to come from that parent. But that doesn't answer our question. We've only set it up. What are the odds of their next child having CF?
I'm going to teach you a design worked out by a guy named Reginald Punnett in 1905. Something you learned in sixth grade, I'm sure, called the Punnett square. Here's the way a Punnett square works. You know how I've been drawing sperms and eggs and arrows, it gets messy, especially when you get like two or three traits. So, we're going to use a square and here's how we're going to do that square. In that square we're going to make a box like this and on this axis we're going to put one parent. So, I don't care which one you put, there's no convention, male, female. We're going to put one of the parents right here. So, remember what my parents were? Nn crossed with Nn. All right. Ready?
So, we'll put one parent here, Nn and one parent here Nn. Now, all we do is we do a grid. We're going to move this on into here and this one into here. That's going to give me NN. So, it's just like saying the sperm of this, egg of this. We're going to do the same thing here. N, into this part of the grid, n into there, Nn. I don't think you need the arrows anymore, Nn. Let's stay with the right colors, nn. So, what was the question? What are the odds that their next child will have cystic fibroses? What's the phenotype of this one? CF. Phenotype normal, phenotype normal, phenotype normal. That's how you do a simple mono-hybrid cross for a Mendelian cross.
So, ready for your problem? Be careful. These parents have another child. So, the CF child has a brother. He wants to know, "Am I a carrier?" In other words, "What are the odds that I'm a carrier?" There's an interesting word for you. A carrier. We use that a lot in human genetics diseases. What's the right term for that in genetics? To be heterozygous. He wants to know if he is heterozygous or not, if it's hidden. The CF child has a brother, that same CF child. Remember neither of his parents did. The brother of this child wants to know if he's a carrier. What are the odds that he's a carrier?
So, you think you've got it, huh? This was trickier than it sounds. You thought it was easy, but this is tricky. Let me show you the answer to this one. Remember what we have? We have N equals normal, n equals CF. We had the two parents, Nn crossed with Nn. We did our Punnett square. I hope you tried the Punnett square, just for practice. There'll be more of these. We have one parent here and one parent here. We have NN, Nn, Nn, nn. And here's what you did. Let me ask you a question. Did you look at the Punnett square and say, "Well, look, two out of four are carriers, or heterozygous. How dumb. He has a 50% chance of being a carrier."
If you said that, you're wrong and here's why. This child, the brother of the CF, did not have cystic fibroses now did he? So, could he be nn? The kid asking the question, "Am I a carrier or not?" Could he be nn? No. So, we know he's not nn. So now what are his odds? Well, we know he's either NN or Nn or Nn because he's normal. What are the odd that he's a carrier? Two out of three. I got you on that one didn't I?
Mendelian Genetics and Mutation
The Laws of Mendelian Inheritance
Mendelian Inheritance Page [2 of 2]