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--- chapter_18 [2024/09/10 08:04] – [An example of calculating allele frequency in humans: albinism] mike
+++ chapter_18 [2024/12/24 04:51] (current) – [An example of calculating allele frequency in humans: albinism] mike
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 <typo fs:x-large> Chapter 18. The Hardy-Weinberg equilibrium</typo>
-Until now, we have been carrying out genetic analysis of individuals, This approach works well when you are studying a model organism, such as Drosophila or mice. It doesn't work well when you are interested in the genetics of organisms that cannot be studied in this way. For the next several chapters, we will consider genetics from the point of view of groups of individuals, or populations. We will treat this subject entirely from the perspective of human population studies where population genetics is used to get the type of information that would ordinarily be obtained by breeding experiments in model organisms.
+In earlier chapters, we have been carrying out genetic analysis using controlled crosses where the genotypes are known and we can breed organisms however we want. This approach works well when you are studying a model organism, such as Drosophila or mice. It doesn't work well when you are interested in the genetics of organisms that cannot be studied in this way. For the next several chapters, we will consider genetics from the point of view of populations. We will treat this subject entirely from the perspective of human population studies where population genetics is used to get the type of information that would ordinarily be obtained by breeding experiments in model organisms.
 In the next several chapters, we will use a substantial amount of math. To avoid confusion with fractions, we will write genotypes as $A/a$ instead of $\frac{A}{a}$.
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 </table>
-Based on the observed genotype frequencies of population 1, we can calculate $p$ and $q$, and we can further derive $p^2=0.09$, $2pq=0.42$, and $q^2=0.49$. This clearly does not match the observed genotype frequencies of $f(M/M)=0.2$, $f(M/N)=0.2$, and $f(N/N)=0.6$. $p^2$ is greater than the observed genotype frequency $f(M/M)$ — suggesting something is happening to cause the $M$ allele to be underrepresented after a generation of breeding. Population 1 is not in a Hardy-Weinberg equilibrium.
+Based on the observed genotype frequencies of population 1, we can calculate $p$ and $q$, and we can further derive $p^2=0.09$, $2pq=0.42$, and $q^2=0.49$. This clearly does not match the observed genotype frequencies of $f(M/M)=0.2$, $f(M/N)=0.2$, and $f(N/N)=0.6$. $p^2$ is greater than the observed genotype frequency $f(M/M)$ — suggesting something is happening to cause the $M$ allele to be underrepresented after a generation of breeding. Population 1 is not in Hardy-Weinberg equilibrium.
-We can do the same analysis for population 2. Even though the observed genotype frequencies are different, the allele frequencies $p$ and $q$ wind up being the same as population 1. And when we now use the allele frequencies to derive $p^2$, $2pq$, and $q^2$, we find they match the observed genotype frequencies. Therefore, population 2 is in a Hardy-Weinberg equilibrium.
+We can do the same analysis for population 2. Even though the observed genotype frequencies are different, the allele frequencies $p$ and $q$ wind up being the same as population 1. And when we now use the allele frequencies to derive $p^2$, $2pq$, and $q^2$, we find they match the observed genotype frequencies. Therefore, population 2 is in Hardy-Weinberg equilibrium.
-Here is a graph showing the relationship between allele and genotype frequencies for genes that are in a Hardy-Weinberg equilibrium:
+Here is a graph showing the relationship between allele and genotype frequencies for genes that are in Hardy-Weinberg equilibrium:
 <figure Fig10>
 {{ :hw_equilibrium.png?400 |}}
 <caption>
-Relationship between allele frequency $p$ (and $q$) and genotype frequency for genes in a Hardy-Weinberg equilibrium.
+Relationship between allele frequency $p$ (and $q$) and genotype frequency for genes in Hardy-Weinberg equilibrium.
 </caption>
 </figure>
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 ===== Hardy-Weinberg vs. the real (human) world =====
-Our discussion above of the Hardy-Weinberg equilibrium assumes that there is random mating in human populations. That is to say, in order for a population to be in a Hardy-Weinberg equilibrium, there must be random mating within the population. How good is the random mating assumption in actual human populations? These are some of the conditions that affect random mating assumption and therefore may affect whether a population can be in a Hardy-Weinberg equilibrium:
+Our discussion above of the Hardy-Weinberg equilibrium assumes that there is random mating in human populations. That is to say, in order for a population to be in a Hardy-Weinberg equilibrium, there must be random mating within the population. How good is the random mating assumption in actual human populations? These are some of the conditions that affect random mating assumption and therefore may affect whether a population can be in Hardy-Weinberg equilibrium:
 ==== Genotypic effects on choice of mating partner ====
-Examination of allele frequencies and genotype frequencies for most genes in the human populations reveals that they closely fit a Hardy-Weinberg equilibrium. The implication is that in general, humans choose their mates at random with respect to individual genes and alleles. This may seem odd given that personal experience says that choosing a mate is anything but random. However the usual criteria for choosing mates such as character, appearance, and social position are largely not determined genetically and, to the extent that they are genetically determined, these are all very complex traits that are influenced by a large number of different genes. The net result is that our decision of with whom we have children does not in general systematically favor some alleles over others.
+Examination of allele frequencies and genotype frequencies for most genes in the human populations reveals that they closely fit Hardy-Weinberg equilibrium. The implication is that in general, humans choose their mates at random with respect to individual genes and alleles. This may seem odd given that personal experience says that choosing a mate is anything but random. However the usual criteria for choosing mates such as character, appearance, and social position are largely not determined genetically and, to the extent that they are genetically determined, these are all very complex traits that are influenced by a large number of different genes. The net result is that our decision of with whom we have children does not in general systematically favor some alleles over others.
-One of the exceptional conditions that produce a population that is not in a Hardy-Weinberg equilibrium is known as assortative mating, which means preferential mating between similar individuals. For example, individuals with inherited deafness have a relatively high probability of having children together. But even this type of assortative mating will only affect the genotype frequencies related to deafness.
+One of the exceptional conditions that produce a population that is not in Hardy-Weinberg equilibrium is known as assortative mating, which means preferential mating between similar individuals. For example, individuals with inherited deafness have a relatively high probability of having children together. But even this type of assortative mating will only affect the genotype frequencies related to deafness.
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 ==== Migration of individuals between different populations ====
-When individuals from populations with different allele frequencies mix, the combined population will be in a Hardy-Weinberg equilibrium after one generation of random mating. The combined population will be out of equilibrium to the extent that mating is assortatative.
+When individuals from populations with different allele frequencies mix, the combined population will be in Hardy-Weinberg equilibrium after one generation of random mating. The combined population will be out of equilibrium to the extent that mating is assortatative.
 ===== An example of calculating allele frequency in humans: albinism =====
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 Therefore, $f(A/A)=p^2 \approx 1$, $f(A/a)=2pq \approx 2q$, and $f(a/a)=q^2$. Since most genetic diseases are rare, these approximations are valid for many of the population genetics calculations that are of medical importance. Note that by convention, the $q$ symbol is used to represent rare alleles.
-Let's look at a real-life example. Albinism occurs in approximately 1 in 20,000 individuals in humans. Let's say that this condition is due to a recessive allele $a$ of a single gene that is in a Hardy-Weinberg equilibrium. Based on this information, we can derive the allele frequency for $a$:
+Let's look at a real-life example. Albinism occurs in approximately 1 in 20,000 individuals in humans. Let's say that this condition is due to a recessive allele $a$ of a single gene that is in Hardy-Weinberg equilibrium. Based on this information, we can derive the allele frequency for $a$:
 $$f(a/a) = \frac{1}{20000} = 5\times10^{-5}=q^2\\
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 $$f(A/a)=2pq\approx 2q = 1.4\times10^{-2}$$
-In other words, approximately 1 in 140 humans are carriers for albinism. We can next calculate the fraction of alleles for albinism that are in individuals that are actually albinos (i.e., their genotype is $a/a$. If we let $\text{N}$=population size, then the number of alleles in homozygotes will be $2\times\text{N}\cdot q^2$. The number of alleles in heterozygotes (carriers) will be $1\times\text{N} \cdot 2pq \approx \text{N}(2q)$. Therefore, the fraction of $a$ alleles in homozygotes is:
+In other words, approximately 1 in 140 humans are carriers for albinism. We can next calculate the fraction of alleles for albinism that are in individuals that are actually albinos (i.e., their genotype is $a/a$). If we let $\text{N}$ = population size, then the number of alleles in homozygotes will be $2\times\text{N}\cdot q^2$. The number of alleles in heterozygotes (carriers) will be $1\times\text{N} \cdot 2pq \approx \text{N}(2q)$. Therefore, the fraction of $a$ alleles in homozygotes is:
 $$ \frac{2\times\text{N}\cdot q^2}{2\times\text{N}\cdot q^2+\text{N}(2q)}\\