TNBGGA: The No Bullshit Guide to Genetic Analysis

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chapter_18

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chapter_18 [2024/12/24 04:42] – mikechapter_18 [2024/12/24 04:51] (current) – [An example of calculating allele frequency in humans: albinism] mike
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 ==== Migration of individuals between different populations ==== ==== 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 ===== ===== 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.   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 $a$:
  
 $$f(a/a) = \frac{1}{20000} = 5\times10^{-5}=q^2\\ $$f(a/a) = \frac{1}{20000} = 5\times10^{-5}=q^2\\
chapter_18.1735044176.txt.gz · Last modified: 2024/12/24 04:42 by mike