A survey of 10,000 human tissue samples collected randomly in northern California were analyzed for PV92 variation finding: 3,222 (+/+) individuals; 2,010 (–/–) individuals; and 4768 (+/–) individuals.

What is the frequency of the Alu– allele for the PV92 locus in this population?

Question

A survey of 10,000 human tissue samples collected randomly in northern California were analyzed for PV92 variation finding: 3,222 (+/+) individuals; 2,010 (–/–) individuals; and 4768 (+/–) individuals.

What is the frequency of the Alu– allele for the PV92 locus in this population?

BlackTom AI · Accepted Answer

Restating the question and options to begin: A survey of 10,000 human tissue samples analyzed for PV92 variation reports genotype counts: +/+ = 3,222; -/- = 2,010; +/- = 4,768. The task is to determine the frequency of the Alu– allele in this population from these data. Answer choices provided: 0.439, 0.481, 0.402, 0.519, 0.561.

First, tally the total number of alleles surveyed. Each individual carries two alleles for the PV92 locus, so with 10,000 individuals there are 20,000 alleles in total.

Next, count the Alu– alleles contributed by each genotype class:
- -/- individuals have two Alu– alleles each, giving 2 × 2,010 = 4,020 Alu– alleles.
- +/- individuals have one Alu– allele each, giving 1 × 4,768 = 4,768 Alu– alleles.
- +/+ individuals have zero Alu– alleles, contributing 0.

Sum of Alu– alleles = 4,020 + 4,768 = 8,788.

Compute the frequency of the Alu– allele (denoted as p): p = (number of Alu– alleles) / (total alleles) = 8,788 / 20,000 = 0.439.

Now evaluate each option:
- 0.439: This matches the calculated frequency exactly; it is consistent with counting all alleles correctly and dividing by the total number of alleles. The logic here relies on the principle that allele frequency is the proportion of the total alleles that are the allele of interest.
- 0.481: This value would imply a higher count of Alu– alleles than the data support. It might result from miscounting heterozygotes as contributing two Alu– alleles or from using a smaller total allele count. In this dataset, heterozygotes contribute only one Alu– allele each, so this figure is too large.
- 0.402: This would reflect a smaller frequency, potentially arising if one subtracted too many Alu– alleles or divided by an incorrect total (e.g., using 19,999 alleles by mistake). The correct total is 20,000, and the computed count 8,788 leads to 0.439, not 0.402.
- 0.519: This option would correspond to a higher Alu– count or a smaller denominator than correct. It misaligns with the observed genotype counts, given the 4,768 heterozygotes and 2,010 homozygous for the other allele.
- 0.561: This is clearly inconsistent with the data, as it would require an even larger proportion of Alu– alleles than 0.439 while keeping the same genotype counts, which is impossible with the given numbers.

In summary, the frequency calculation hinges on counting alleles from each genotype class and dividing by the total allele pool. The correct algebra yields 8,788 Alu– alleles out of 20,000 total alleles, giving a frequency of 0.439, while the other options reflect miscounts or miscalculations that do not align with the dataset.

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