A newspaper story on AIDS reports that an allele (R) confers complete resistance to HIV when homozygous and that 1% of the human population is resistant (RR). The story also states that 20% of the population carries one copy of the resistance allele (i.e., is heterozygous). Is this a guess, an approximation, or an exact frequency, assuming that the population is at Hardy Weinberg equilibrium?

Oh, Hardy Weinberg equilibrium, you sly dog! Let's break it down, shall we?

If we assume that the population is at Hardy Weinberg equilibrium (which means no genetic forces like mutation, selection, or migration are at play), we can determine the frequency of the resistance allele (R).

Let's call the frequency of the resistance allele 'p'. So, according to the given information, 20% of the population carries one copy of the allele. Therefore, if we divide 20% by 2 (since heterozygotes carry one copy), we get 10% or 0.1.

Now, to find the frequency of the non-resistant allele (r), we subtract the frequency of the resistance allele (p) from 1. So, that would be 1 - 0.1 = 0.9.

Since we know that the resistant allele (R) is completely resistant to HIV when homozygous (RR), its frequency would be the square of the frequency of the resistance allele (p). Therefore, the frequency of homozygous resistance (RR) would be 0.1^2 = 0.01, or 1%.

So, to answer your question, based on the assumptions of Hardy Weinberg equilibrium, the frequency of the human population that is resistant (RR) is an exact 1%.

To determine whether the frequency of the resistant allele (R) in the population is a guess, an approximation, or an exact frequency, we can use the Hardy-Weinberg equilibrium equation. The Hardy-Weinberg equilibrium is a principle in population genetics that states that the frequencies of alleles in a population will remain constant from generation to generation in the absence of other forces, such as mutation, migration, genetic drift, or natural selection.

The equation for Hardy-Weinberg equilibrium is expressed as follows:

p^2 + 2pq + q^2 = 1

Where:
- p^2 represents the frequency of the homozygous dominant genotype (RR)
- 2pq represents the frequency of the heterozygous genotype (Rr)
- q^2 represents the frequency of the homozygous recessive genotype (rr)
- p represents the frequency of the dominant allele (R)
- q represents the frequency of the recessive allele (r)

In this case, the newspaper story states that the frequency of the RR genotype (homozygous resistant) is 1% of the population, which can be expressed as 0.01. Therefore, p^2 = 0.01.

The story also states that 20% of the population carries one copy of the resistance allele (heterozygous). This can be expressed as 0.20. Therefore, 2pq = 0.20.

By rearranging and substituting values into the equation, we can solve for q:

q^2 = 1 - p^2 - 2pq
q^2 = 1 - 0.01 - 0.20
q^2 = 0.79
q = √0.79 ≈ 0.89

Now that we have the value of q (frequency of the recessive allele), we can calculate the frequency of the dominant allele (R) using p = 1 - q:
p = 1 - 0.89
p ≈ 0.11

Therefore, the resistant allele (R) occurs in the population with a frequency of approximately 0.11 or 11%.

In conclusion, the frequency of the resistant allele (R) provided in the newspaper story is not a guess or an approximation; it is an exact frequency based on the assumption that the population is at Hardy-Weinberg equilibrium.