# Estimating Population Abundance and Allele Frequencies

## 1. Estimating Sloth Population Abundance

**You are wanting to know how many sloths are in a population in Costa Rica using the Lincoln Index (capture-mark-recapture) Method. The first time you go out, you capture 15 sloths and mark each with a band. The second time you go out, you capture 10 but 7 of them are recaptures. What is the estimated abundance of this population?**

### Final answer:

The estimated sloth population using the Lincoln Index Method is 21 sloths.

### Explanation:

The estimated abundance of the sloth population in Costa Rica using the Lincoln Index (capture-mark-recapture) Method can be calculated by multiplying the number of sloths captured the first time by the number captured the second time, and then dividing the result by the number of recaptured sloths. In this case, it would be (15 sloths × 10 sloths) ÷ 7 recaptured sloths, which equals 150 ÷ 7, resulting in an estimated population of approximately 21 sloths.

## 2. Allele and Genotype Frequencies in a Frog Population

There are 400 frogs in a population. 250 are brown (dominant, BB/Bb) and the rest are green (recessive, bb).

**What are the allele and genotype frequencies?**

### Final answer:

The allele frequencies for the frogs are 0.625 for 'B' and 0.375 for 'b'. Genotype frequencies cannot be determined without additional information.

### Explanation:

For the frog population, to determine the allele and genotype frequencies, we'll start by recognizing that there are 400 frogs total, with 250 being brown (dominant phenotype). Since the brown frogs can have genotypes BB or Bb, we can't directly determine their exact genotype frequencies without additional information, but we can determine the frequency of the recessive allele 'b' from the green frogs. Since all 150 green frogs are 'bb', we know that they contribute 300 b alleles to the gene pool. Since there are 400 frogs total and each frog has two alleles for this gene, there are 800 alleles in the population. Thus, the frequency of 'b' is 300/800, or 0.375. The frequency of the dominant allele 'B' would be 1 - 0.375 = 0.625. These are the allele frequencies, but without more information, we can't accurately calculate the genotype frequencies for Bb and BB.