Now, let's see if we can use the Hardy-Weinberg law to illustrate the case of a human recessive disease. As mentioned already, we do expect selection against disease traits, but we can still use the Hardy-Weinberg law to make some predictions. But first, in humans, unlike the examples with flowers, we often do not know the frequency of a recessive gene in the population. The disease trait is phenotypically expressed for a recessive trait only when an individual is homozygous recessive. Thus, the information we can gather is the number of affected individuals in the population, that is, the proportion of ho-mozygous recessive individuals in the population under study. How can we use this information?

Let's again apply the Punnett square (figure 10.4), using as an example the human recessive disease PKU that we learned about in chapter 3. We can fill in the genes and genotypes (figure 10.4.A). Now, we know from medical records that roughly 1 in 10,000 U.S. Caucasians suffer from PKU. This is 0.0001 in decimals, or in scientific notation, 10-4. This number represents the proportion of homozygous recessive individuals so we'll associate that number with the genotype aa (figure 10.4.B). The proportion of the a gene in the population is thus the square root of the proportion of aa, since the number of a multiplied by itself gives aa. The square root of 0.0001, or 10-4, is 10-2, or 0.01, or 1 percent. The frequency of the normal A gene is thus 1 -0.01 = 0.99, or 100 percent - 1 percent = 99 percent. We now have the full information necessary to fill out this Punnett square for estimating the frequency of the PKU gene in the population (figure 10.4.D). Therefore, from only knowing approximately how many individuals suffer from PKU, we can estimate the frequency of the PKU gene in the U.S. Caucasian population. This frequency is 1 percent.

An even more useful piece of information we can estimate is the proportion of carrier individuals in the population. As you see from the Punnett square (figure 10.4.D), this proportion is twice 10-2, or 0.02, or, to put it another way, 2 out of 100 individuals (2 percent) are carriers. This may seem to be a surprisingly large number for a disease with a frequency of only 1 out of 10,000 people, but it is correct. This is one very useful aspect of the Hardy-Weinberg law of population genetics: it allows us to calculate the carrier frequency, something that cannot easily be determined in any other way.

A |
A |
a |

A |
AA |
Aa |

a |
Aa |
aa |

c |
1 A |
0.01 a |

1 A |
AA |
Aa |

0.01 a |
Aa |
0.0001 aa |

A |
a | |

A |
AA |
Aa |

a |
Aa |
0.0001 aa |

D |
I A |
0.01 a |

1 A |
AA |
0.01 Aa |

0.01 a |
0.01 Aa |
0.0001 aa |

Figure 10.4 A Punnett Square Used to Calculate the Frequency of Recessive Disease Genes in a Population. A. Punnett square with just the genes showing. B. As A, with the frequency of aa individuals shown. C. As B, with the calculated frequency of the a gene calculated. D. Complete Punnett square. To simplify, we write "1 A" rather than ".99 A."

The PKU example also illustrates why it is valid to use this law even when the trait is clearly selected against. (PKU individuals, until newborn PKU testing became widespread, were mentally retarded and unlikely to reproduce). Let us calculate what percentage of the PKU gene in the population is carried by affected homozygous individuals, as compared to the percentage carried by phenotypically normal heterozygous individuals. We know at the outset that there is a proportion of 1 in 10,000, or 0.0001, homozygous recessive PKU individuals, and we calculated the percentage of carrier individuals as 0.02. So the proportion of PKU genes carried by homozygous individuals who have 2 copies of the PKU gene is 2 X 0.0001 = 0.0002, and that carried by heterozygous individuals is 0.02. The total is 0.0202. Thus the proportion of the gene carried by homozygous recessive individuals is 0.0002 divided by 0.0202, or ~1 percent. This means that PKU individuals carry less than 1 percent of the total PKU genes in the population, while greater than 99 percent of the PKU gene in the population is carried by phenotypically normal heterozygous individuals. So, even if none of the PKU individuals passed their PKU genes to the next generation, 99 percent of the PKU genes will still be passed onto the next generation by heterozygous carrier parents.

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