# we are looking at the Hardy weinberg equilibrium theory and we were given fifty problems to work over the break and even though i have answered most of them there is one that has be taken a long time because I am not sure if it can be solved. here is the question i hope you can help me.

I have three populations of birds.

Population 1 has ten birds, of which one is brown (a recessive trait) and nine are red.

Population 2 has 100 birds ten of which are brown.

Population 3 has 30 birds and 3 are brown.

Which one has the highest frequency of the allele for brown feathers.

Please tell me how to approach this problem.

## They all have the same frequency.

http://en.wikipedia.org/wiki/Hardy-Weinberg_principle#Graphical_representation

Small populations have bottleneck tendencies.

## thank you very much

## To determine which population has the highest frequency of the allele for brown feathers, we can use the Hardy-Weinberg equation. The equation allows us to calculate the expected frequencies of the alleles in a population under certain conditions, assuming it is in Hardy-Weinberg equilibrium.

The Hardy-Weinberg equation is as follows:

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

Where:

- p^2 represents the frequency of individuals homozygous for the dominant allele (in this case, red feathers).

- 2pq represents the frequency of heterozygous individuals (those carrying one copy of the dominant allele and one copy of the recessive allele).

- q^2 represents the frequency of individuals homozygous for the recessive allele (brown feathers).

- The sum of p^2, 2pq, and q^2 is equal to 1, representing the total population.

Here's how we can approach the problem step-by-step:

1. Calculate the allele frequencies for population 1:

- Since the recessive trait is brown feathers, the recessive allele (q) is responsible for brown feathers.

- In population 1, there are 10 birds, of which 1 is brown. Therefore, the frequency of the recessive allele (q) can be calculated as 1/10 = 0.1.

- Since there are no brown homozygous individuals (q^2), the frequency of red homozygotes (p^2) can be calculated as (10 - 1)/10 = 0.9.

- Because the total population size is 10, the frequency of heterozygotes (2pq) can be calculated as 1 - (0.1 + 0.9) = 0.

2. Calculate the allele frequencies for population 2:

- In population 2, there are 100 birds, of which 10 are brown. Thus, the frequency of the recessive allele (q) can be calculated as 10/100 = 0.1, the same as population 1.

- As there are 10 brown individuals, the frequency of brown homozygotes (q^2) can be calculated as 10/100 = 0.1.

- Similarly, the frequency of red homozygotes (p^2) can be calculated as (100 - 10)/100 = 0.9.

- The frequency of heterozygotes (2pq) can be calculated as 1 - (0.1 + 0.1) = 0.8.

3. Calculate the allele frequencies for population 3:

- In population 3, there are 30 birds, of which 3 are brown. Thus, the frequency of the recessive allele (q) can be calculated as 3/30 = 0.1, still the same as the previous populations.

- With 3 brown individuals, the frequency of brown homozygotes (q^2) can be calculated as 3/30 = 0.1.

- The frequency of red homozygotes (p^2) can be calculated as (30 - 3)/30 = 0.9.

- The frequency of heterozygotes (2pq) can be calculated as 1 - (0.1 + 0.9) = 0.

4. Compare the allele frequencies for the populations:

As we calculated, all three populations have the same allele frequency for the brown feather allele (q) of 0.1.

Therefore, population 1, population 2, and population 3 all have the same highest frequency of the allele for brown feathers.

In conclusion, all three populations have the same frequency of the allele for brown feathers, suggesting that there is no significant difference in the frequency of the brown feather allele among the populations.

## To solve this problem, you need to calculate the frequency of the allele for brown feathers in each population. The frequency of an allele is the number of times the allele is present in a population divided by the total number of alleles in that population.

First, calculate the total number of alleles for each population:

Population 1: Since there are 10 birds, each with two alleles (one from each parent), there are a total of 20 alleles in this population.

Population 2: Since there are 100 birds, each with two alleles, there are a total of 200 alleles in this population.

Population 3: Since there are 30 birds, each with two alleles, there are a total of 60 alleles in this population.

Next, calculate the number of brown alleles in each population:

Population 1: There is 1 brown bird, which means there is 1 brown allele.

Population 2: There are 10 brown birds, which means there are 20 brown alleles.

Population 3: There are 3 brown birds, which means there are 6 brown alleles.

Now, calculate the frequency of brown alleles by dividing the number of brown alleles by the total number of alleles in each population:

Population 1: frequency of brown alleles = 1 brown allele / 20 total alleles = 0.05

Population 2: frequency of brown alleles = 20 brown alleles / 200 total alleles = 0.1

Population 3: frequency of brown alleles = 6 brown alleles / 60 total alleles = 0.1

Comparing the frequencies, both Population 2 and Population 3 have the highest frequency of the allele for brown feathers, which is 0.1. Therefore, both Population 2 and Population 3 have the highest frequency of the allele for brown feathers.

I hope this helps you solve the problem! Let me know if you have any further questions.