TBR Bio Book II Section 9 Question 25-27

[Sloppy Toppy]

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Any help on these Hardy-Weinberg questions? Thought I understood how to use the equation, but these are giving a hard time and I don't quite understand where the numbers are coming from in the explanation.

Thanks

 

[aldol16]

Any particular problem? I hate using that equation too and I usually end up just drawing the crosses and/or intuiting through them - which is a simpler way of understanding it if you ask me. Alright, so first things first. Biologists for some reason love to name things. I have no idea why. They take simple concepts and attached names to them so that it just confuses everyone. The Hardy-Weinberg equation is simple. It's just simple statistics. So I'll use a statistical example to help you understand. Okay, I have a deck of cards: 26 red cards, 26 black cards. Assume that the cards are indistinguishable. Now I want you to tell me the probability of drawing two red cards. Well, that's just (26/52)*(26/52) = (1/2)*(1/2) = 1/4. Simple enough. Now, what is the probability of drawing two black cards? It's the same thing because the probability of drawing a red card is equal to the probability of drawing a black card. So 1/4. Now, finally, the only other option left if I want you to draw two cards is getting a black one and a red one. So what's the probability of that? Well, it's just the sum of the probability of drawing a black card then a red card and the probability of drawing a red card and then a black card. In other words, it's (1/2)*(1/2) + (1/2)*(1/2) = 2/4 = 1/2.

Okay, that's all you need for Hardy-Weinberg equilibrium! In that case, you have two alleles in a population (there are various assumptions that go along with it) and you do the same exercise as above with choosing. Let's slap a name on our probabilities this time - let's say the probability of getting the P allele is p and the probability of getting the other allele, Q, is q. These are the only two alleles for that gene. Now, the obvious equation is p + q = 1 - if you only have two alleles, the probability of getting one or the other is unity. Okay. Now the Hardy-Weinberg equation. What are all the possible genotypes? Well, you could have PP, PQ, or QQ. That's it! So then, the probability of getting PP, PQ, or QQ must also equal unity because those are the only options. So then what's the probability of getting PP? Well, it's just p*p, or p^2. Same with Q, but with q^2. What's the probability of getting PQ? Well, you could get P from your dad and Q from your mom, or Q from your dad and P from your mom. So, the probability of that happening must be p*q + q*p = 2pq. Sum these all up and you get the so-called Hardy-Weinberg equation!

 

[Sloppy Toppy]

Any particular problem? I hate using that equation too and I usually end up just drawing the crosses and/or intuiting through them - which is a simpler way of understanding it if you ask me. Alright, so first things first. Biologists for some reason love to name things. I have no idea why. They take simple concepts and attached names to them so that it just confuses everyone. The Hardy-Weinberg equation is simple. It's just simple statistics. So I'll use a statistical example to help you understand. Okay, I have a deck of cards: 26 red cards, 26 black cards. Assume that the cards are indistinguishable. Now I want you to tell me the probability of drawing two red cards. Well, that's just (26/52)*(26/52) = (1/2)*(1/2) = 1/4. Simple enough. Now, what is the probability of drawing two black cards? It's the same thing because the probability of drawing a red card is equal to the probability of drawing a black card. So 1/4. Now, finally, the only other option left if I want you to draw two cards is getting a black one and a red one. So what's the probability of that? Well, it's just the sum of the probability of drawing a black card then a red card and the probability of drawing a red card and then a black card. In other words, it's (1/2)*(1/2) + (1/2)*(1/2) = 2/4 = 1/2.

Okay, that's all you need for Hardy-Weinberg equilibrium! In that case, you have two alleles in a population (there are various assumptions that go along with it) and you do the same exercise as above with choosing. Let's slap a name on our probabilities this time - let's say the probability of getting the P allele is p and the probability of getting the other allele, Q, is q. These are the only two alleles for that gene. Now, the obvious equation is p + q = 1 - if you only have two alleles, the probability of getting one or the other is unity. Okay. Now the Hardy-Weinberg equation. What are all the possible genotypes? Well, you could have PP, PQ, or QQ. That's it! So then, the probability of getting PP, PQ, or QQ must also equal unity because those are the only options. So then what's the probability of getting PP? Well, it's just p*p, or p^2. Same with Q, but with q^2. What's the probability of getting PQ? Well, you could get P from your dad and Q from your mom, or Q from your dad and P from your mom. So, the probability of that happening must be p*q + q*p = 2pq. Sum these all up and you get the so-called Hardy-Weinberg equation!

Thanks for that. I followed it perfectly. I get the Hardy-Weinberg equation and what it's telling me, but these particular questions are hard to decipher for me. Would you like me to take a pic of the passage and questions if you don't have TBR book or PDF?

 

[aldol16]

Thanks for that. I followed it perfectly. I get the Hardy-Weinberg equation and what it's telling me, but these particular questions are hard to decipher for me. Would you like me to take a pic of the passage and questions if you don't have TBR book or PDF?

Yeah, I don't have the TBR book. But I can give you just general guidance - either use the equation as is, or use a cross to figure out the problem. Those two methods will cover all such questions you will encounter.

 

[Sloppy Toppy]

Here's the question if you are willing to do it. Thanks for your help. Questions 25-27. Similar premise.

 

[aldol16]

Okay, for 25, you first have to determine the allele frequencies. The allele is X-linked, so you want to look at the males. You want to look at the males because if a male gets the mutant, he will be color-blind (assuming full penetrance). In other words, the probability of a male being color-blind is the same probability as that of encountering the mutant allele in the gene pool, which, in this case, is 0.08. Now, the question is asking you what the probability is for a female to be color-blind. For a female to be color-blind, she must have both mutant copies because it's recessive. So what's the probability of that? Well, it's 0.08*0.08 = 0.64%.

I'll post the others later - I have to attend to something else now.

 

[Sloppy Toppy]

Okay, for 25, you first have to determine the allele frequencies. The allele is X-linked, so you want to look at the males. You want to look at the males because if a male gets the mutant, he will be color-blind (assuming full penetrance). In other words, the probability of a male being color-blind is the same probability as that of encountering the mutant allele in the gene pool, which, in this case, is 0.08. Now, the question is asking you what the probability is for a female to be color-blind. For a female to be color-blind, she must have both mutant copies because it's recessive. So what's the probability of that? Well, it's 0.08*0.08 = 0.64%.

I'll post the others later - I have to attend to something else now.

Thanks for that. The other questions are basically the same thing so you don't have to look into it. I understand your explanation, and that's how the book explained it. The one small problem I have that I need explained is for a female to be colorblind she needs to be homozygous recessive. Meaning one mutant X chromosome from dad, and one from mom. When you do 0.08 X 0.08 isn't that 0.08 in regards to the father. So aren't you saying both are coming from dad? Hopefully I'm verbalizing this clearly.

Get back to me whenever you are free. Really appreciate it.

 

[aldol16]

Thanks for that. The other questions are basically the same thing so you don't have to look into it. I understand your explanation, and that's how the book explained it. The one small problem I have that I need explained is for a female to be colorblind she needs to be homozygous recessive. Meaning one mutant X chromosome from dad, and one from mom. When you do 0.08 X 0.08 isn't that 0.08 in regards to the father. So aren't you saying both are coming from dad? Hopefully I'm verbalizing this clearly.

It doesn't mean that in this case because, remember, the probability that the dad has it is equal to the allele frequency in the entire gene pool. Why? Because if dad gets a single allele of the mutant, he will be color-blind. So the probability of him being color-blind is equal to the probability of him getting the allele. Put another way, the probability of him getting the allele from the gene pool is equal to the probability of him being color-blind. So if there's a 0.08 chance of him being color-blind, then that also must mean there was a 0.08 chance of him getting it from the overall allele pool, which means that the allele frequency must be 0.08.