DIP Episode 375 - Hardy-Weinberg Made Easy (for Step 1-3)

Okay, welcome. My name is Divine. This is episode 375 of the Divine Intervention podcast. In this podcast is going to be extremely short but it's going to be like extremely high yield to a much like a single concept that's tested on the US Mily exams. This is actually something that shows up a lot on step one. It's almost like every step one exam has like one or 12 these questions. And then this is something that also shows up a lot on step two. Many times it's just a question or two when many people get it wrong. And this relates to genetic probabilities and the hardy Weinberg equation. So we're going to address this. I think the best way to address this is a practice problem. But as I discussed the practice problem, I'll discuss a few more high yield things. Now if you're going to be taking the US Mily step two CK or step three exams or complex level two or three exams within the month of March, I mean within the month of March or like the first three weeks of April, I'd like to invite you to the MBME Test Ticking Strategies course I have. That's going to be taking place on the 21st of March from 2 to 4 30 PM Pacific Standard Time. And then I also have a 24 hour review course that's going to be taking place from the 22nd to the 25th of March from 70 AM to 1 PM Pacific Standard Time. Again, these are courses that have been taken by tons of people and they've done extremely well on the exams.

And then if you're taking the US Mily exams in the summer, step two CK, step three or complex level two or three, I have a 75 hour step two CK school that's going to be taking place within the first two weeks of May. That I'm keeping a limit of 40 people. If you want to sign up, go ahead and send me for any of these courses really. Go ahead and send me an email through the website and I'll give you some more information. The 75 hour school has a cup because I really want to deeply invest in everyone that is attending. Okay, so let's just go ahead and get started with this. So the Hollywood in British something many, many people struggle with. And the thing is, it's actually simple to understand if you know exactly. Okay, so let's drop right into the question. So hemophilia C is a bleeding disorder characterized by deep tissue, muscle and joint bleeds. This disease is seen in approximately one out of 25 Ashkenaz situation individuals. Right? Answer the following questions. So the first question I'm going to ask here is what is the carrier frequency of hemophilia C in this population? So the thing is when you see questions like this, to be honest with you, again, like many things in life, it's always good to be systematic. If you're systematic, you'll never get this question wrong. If you're not systematic, you'll forward into the NV Me's hands. What do I mean by that?

The thing is you need to know what the Hadewymeberg equation deals with and what you do in each situation on the exam. So many of us know that there are two equations for the Heide Weinberg business, right? There is a P plus Q equals one and then there's a P squared plus two PQ plus Q squared equals one. So now one common area of confusion is people are like people wonder divine. When do I use P? When do I use P squared? When do I use Q? When do I use Q squared? How do I know which one I've been given in the question? Let me tell you the trick. The trick to all of this is remembering that the letters without squares refer to the frequency of the allele in the population. So if they give you the frequency of the allele and they will mention it specifically, you know that you're dealing with your P or with your Q, right? The P refers to the frequency of the dominant allele. The Q refers to the frequency of the recessive allele. But if you're looking at that P squared plus two PQ plus Q squared equation, the P squared refers to the frequency of the people that are homozygous dominant. So like the P squared refers to actual phenotypes. It refers to the population. P squared refers to people that are always more dominant, right? Like you know dominant homozygous, Q squared refers to the people that are homozygous recessive, right? And then two PQ refers to people that are heterozygous, right? Once you again can define these, you'll be in good shape.

So you just always need to ask yourself, what am I given in the question? And then you'll be set. So like for example, in this question I say that all this disease is seen in approximately one in twenty-five Ashkenazi Jewish individuals, right? So that means all the diseases seen in one in twenty-five Ashkenazi Jewish individuals. So basically, what am I saying? They're saying, oh, this is the population. Noticing the question, I did not see anything about the number of alleles, the frequency of alleles. I'm talking about the actual population, about the actual phenotype, right? So that means I'm talking about people that have disease, right? Remember hemophilia, C is a little more recessive. That means I'm basically giving you Q squared in this question. I'm giving you Q squared in this question. Since again, I'm referring to the population. Again, remember the P squared homozygous dominant Q squared homozygous recessive two PQ are the heterozygous. So if we know that Q squared is one in twenty-five, then if we're looking for Q, right? Because we're trying to find the carrier frequency. Carriers, remember we say this two PQ. So if we have Q squared, then we're looking for Q. Well, if Q squared is one out of twenty-five, then Q is one out of five, right? Remember, one over five, multiply by one over five is one over twenty-five. So Q has to be one over five, which is zero point two, right? Well, again, let's just say Q is one over five.

Well, if we know Q is one over five, then we can find P, right? So this one over five refers to the frequency of the recessive allele in the Ashkenazi Jewish population. So if, for example, wanted to find the frequency of the dominant allele in the Ashkenazi Jewish population, we'll be trying to look for P, right? So we know that P plus Q equals one, and if Q is one over five, then that means P is going to be one minus one over five, which is going to be four over five, right? So now we know what P is, we know what Q is. Q is one over five, P is four over five. So if we're trying to find the carrier frequency of hemophilia C in the population, we just do two PQ. So we do two, multiply by four over five, multiply by one over five, right? And that's basically eight over twenty-five, okay? That's basically eight over twenty-five if I'm doing my math well, right? Two times four, you multiply the numerators together, that's eight, and then five times five at the bottom, that's going to be twenty-five. So it's going to be eight over twenty-five, okay? It's going to be eight over twenty-five. It's going to be eight over twenty-five, where you can say, point two times point eight, right? Which is zero point one, six, multiply by two, that's zero point three, two, okay? So zero point three, two is two PQ. So that's the carrier frequency, right? Of hemophilia C in the population, right? It's going to be eight in every twenty-five individuals or zero point three, two.

Now the second question I want to ask here is, a medical student in a genetic sortation is attending to a family who have one son with hemophilia C and another son who is unaffected. The present for genetic canceling after a urine pregnancy test was found to be positive ten weeks ago. What is the probability of this fetus being affected by hemophilia C? Okay, well let's look at this. So this family, we have a son that has hemophilia C. We know that hemophilia C is a deficiency of factory 11, but we also know that hemophilia C is an autosomal recessive disorder. So the fact that we have a son that has the disease, that tells us automatically that both parents are carriers, right? For you to have a child that has an autosomal recessive disease, then the most I've got in one of the affected and those from mom and one from dad, right? So we know that the parents are carriers. Well since the parents are carriers, right? If you do the ponet square of the cross, right? Let's say like the mom is big a little a and that is big a little a. If you work out the ponet square you will get like one big a, one big a. So one big a big a, two big a's, little a's and then one little a little a, right? So the chance of, because it's an autosomal recessive disease, you need both recessive alleles. So the chance there, right? Since there's one little a little a out of four different possibilities, it's going to be a one in four chance.

So the probability of the a fetus being affected by hemophilia C is going to be one in four. Now the final question I have here is the family asks the medical student what he thinks about their unaffected sons probability of having a child with hemophilia C. If he marries within their local close knit community. So let's say marry some other Ashkenazi Jewish person. What is the most appropriate medical student response in this situation? Okay. So the thing is if you want to estimate these probabilities, right? You need to know certain bits of information, again, this medical student, again, he's attending to this family, this family, again, they have one son that has hemophilia C. They have another son that does not have hemophilia C. Okay. Now, this other son, if he decides to marry another person, right? Marry a wife, right from within Ashkenazi Jewish population, are trying to estimate, oh, what is the probability that you have a child that has hemophilia C? Well, the thing is when you do these problems, the best way to accomplish is to look at both, look at each parent separately and then combine everything, right? We know that, again, hemophilia C is an autosomal recessive disease, right? So for you to have a child that has hemophilia C, both parents have to be carriers. So before we can start figuring out the probabilities, we need to figure out first, what is the probability of that being a carrier? So let's do that first.

So the probability of that being a carrier, remember that is from a family where both of these parents are carriers, right? So if we look at it, we know that that is unaffected phenotypically, right? We look at that, we're like, oh, wow, that does not seem to look like a person that has, you know, that phenotypically does not have hemophilia C, he doesn't have the bleeding, the joint bleeds, the hematrosis, or the muscle belly bleeds. It doesn't have any of those states. But remember, the fact that you have phenotypically normal, right? It could mean one of two genotypic possibilities. You could either be big A, big A, so you don't have any of the recessive alleles at all, right? So you don't have any phenotype of hemophilia C, or you could be big A little A, you could be a carrier, because hemophilia C is autosomal recessive. By being a carrier, you're not necessarily showcasing any symptoms, right? Because you have one dominant a little that's helping you out, right? So basically, the chance of the child be, because if you remember that point, it's where we said, oh, the parents being carriers, big A little A, multiply by big A little A, we have a big A big A, that's one big A big A, two big A little A's and one little A little A, right? We know that the child is unaffected, so we can scrub out little A little A, there's no child that there's no chance that this one affected, so genotype is little A little A, if not it will be affected, right?

So the options we have, we have one big A big A and two little A, two big A little A's, right? So obviously, big A little A is the carrier, right? So if we know that, oh, there are three possibilities, and two of those are big A little A the carriers, that means the probability of that being a carrier is two out of three, okay? The probability of that being a carrier is two out of three. Now, what is the probability of mom being a carrier, right? The thing is mom, we don't know anything about her family, since we don't know anything about her family, we need to use the population data for the Ashkenazi Jewish population, right? So we can say, oh, you know what? Well, what's the probability of mom being a carrier? A probability of mom being a carrier is the answer to the very first question I posed here, the carrier frequency of hemophilia seen in the population is going to be eight out of 25, right? It's going to be eight out of 25, right? So we know that probability, right, of being a carrier is two thirds, mom's probability of being a carrier is eight out of 25, but we know that if you have two carriers and they are passing and you know, they are both carriers, the probability of their child having an autosomal recessive disease is one quarter, right? Already proved that earlier. So all you just need to do is multiply the probabilities of that and mom, you get an answer and they multiply that answer by one quarter, right?

So we're going to do two thirds times eight over 25, right? So two times three at the top, I mean two times eight at the top is 16, and then three times 25 at the bottom is 75. So it's going to be 16 over 75 and then we multiply that number by a quarter, multiply that number by a quarter, right? So again 16 divided by 75, multiply by a quarter, right? And basically we'll end up with four divided by 75, right? So there is a four out of 75 chance that this one affected son will have a child that has hemophilia C. So I'm really, really hoping that this makes sense, but again, don't forget in your Hattie-Winberg equation, it's P plus Q equals one and P squared plus 2 PQ plus Q squared equals one. The P refers to the frequency of the dominant allele. The Q refers to the frequency of the recessive allele. P squared refers to the frequency of people that are homozygous dominant in the population. Look at the linksialio in the population. Q squared refers to the frequency of both that are homozygous recessive in the population and then two PQ refers to the probability of people that are the frequency of people that are carriers that are heterozygous for that disease in the population. Now the final thing I want to say here is when you say when you're talking about Hattie-Winberg, they are just certain assumptions you need to make. These things are just things unfortunately you need to come into memory.

So remember that in Hattie-Winberg calculations, you assume that there's no mutation, you assume that there's no selection going on, you assume that there's no migration of the population. Those are the big, big, big things. You want to keep at the back of your mind as assumptions when you're dealing with Hattie-Winberg. The assumptions they tend to test it more in step one than step two CK step three. They tend to assume that. Now one thing I would say that is maybe a little different. I talked about an Orozomer recessive disease. To be honest with you, everything that I said for Orozomer recessive diseases applies to like ex-lingu-processive diseases. But there's a little twist there. Because remember in an ex-lingu-processive disease, being a guy, you have to be XY. So you literally have only one X chromosome. You have one X-lingu-sense, but you have only one X chromosome. The thing is, there's no point doing the Q-sweat conversion in guys. Like literally the frequency. Remember I said that O and Hattie-Winberg math, the population frequency is given by the population frequency of the recessive allele. I mean the population frequency of people that are homozygous recessive. Sorry, the population frequency of Orozomer recessive is Q-sweat. The thing is, emails from an ex-lingu-processive disease since they have only one X chromosome. The population frequency of the homozygous recessive trait, you know, is Q. So you don't need to worry about Q-sweating emails.

But you do that Q-sweat way referring to females. You do that Q-sweat way referring to females. So that Q-sweat dichotomy applies in females because females have two X chromosomes. But in males for ex-lingu-processive diseases, that Q-sweat dichotomy does not apply. Just whatever value you get for Q is it. That's it. That Q represents the population frequency for being homozygous recessive. Again, if you really think about it, that should make sense because for a guy that is X-Y, if you have the recessive allele, you have the recessive allele. It's not like you have two options because you have only one X chromosome. So that's the only one minor tweak you need to make with the Hadi-Wineberg calculations when you're dealing with an X-Linu-processive disease in GANS. But if you're dealing with females, the calculations, everything are perfectly the same. So thank you for listening to me. I do at the end of every podcast. I offer one I want you to rainfall all the USML exams, step one to step three, preclinical medical exams, 30-shelf exams. And then I have these podcasts on all the major podcasts apps, at least the most 3,750. If you want everything from episode one to 3,75, if you go to my website, divineinterventionpodcasts.com. And whenever I, if you have a Word Press account, I subscribe to my website. Whenever I make a new podcast, you're going to get an email notification. And then I also have a new website called the divine intervention life lessons.com.

If you go on there, you're going to find my life lessons podcasts. I post about two every week. You know, that uses Bible-based teaching to address common problems faced by humanity. Most of them are about 10 to 15 minutes long. And then if you want, again, if you want to sign up for any of my review courses, should be an email through the website. And I'll be happy to give you some more information. So thank you for listening to me. Have a wonderful rest of your day. God bless you and thank you. Good night for now.

P represents the frequency of the dominant allele in the population.

$2pq$ represents the frequency of heterozygous individuals (carriers) in the population.

No mutation, no selection, and no migration (gene flow).

For males (XY), you do not worry about $q^2$; for females (XX), you must use the $q^2$ method.

They could be homozygous dominant ($AA$) or heterozygous carrier ($Aa$).

The probability is $2/3$. (The affected genotype $aa$ is excluded from the possibilities).

Allele frequency: $p + q = 1$; Genotype frequency: $p^2 + 2pq + q^2 = 1$.

The frequency of the recessive allele.

Both parents must be carriers (heterozygous).

Males are hemizygous ($X^aY$), meaning they only have one X chromosome and thus cannot be homozygous recessive.

$2/3$.

Mutation, selection, or migration (gene flow).

Allele frequency: $p + q = 1$; Genotype frequency: $p^2 + 2pq + q^2 = 1$.

The frequency of the recessive allele.

Both parents must be carriers (heterozygous).

Males are hemizygous ($X^aY$), meaning they only have one X chromosome and thus cannot be homozygous recessive.

$2/3$.

Mutation, selection, or migration (gene flow).