TNBGGA: The No Bullshit Guide to Genetic Analysis

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chapter_18

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chapter_18 [2024/12/24 04:50] – [Migration of individuals between different populations] mikechapter_18 [2026/10/25 01:04] (current) – [The Hardy Weinberg equilibrium] mike
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 </figure> </figure>
  
-Here is an example: $M$ and $N$ are alleles of a gene that specifies different blood antigens. The alleles are codominant, so a simple blood test can distinguish the three possible genotypes of $M/M$, $M/N$, and $N/N$. A survey of the population reveals that 83% of the population has only the $M$ antigen, 1% of the population only has the $N$ antigen, and 16% of the population has both the $M$ and $N$ antigens. In other words:+Here is an example: $M$ and $N$ are alleles of a gene that specifies different blood antigens. The alleles are codominant, so a simple blood test can distinguish between the three possible genotypes of $M/M$, $M/N$, and $N/N$. A survey of the population reveals that 83% of the population has only the $M$ antigen, 1% of the population only has the $N$ antigen, and 16% of the population has both the $M$ and $N$ antigens. In other words:
  
 <figure Fig5> <figure Fig5>
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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_18.1735044639.txt.gz · Last modified: by mike