The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (that no evolution is occurring) the following five conditions must be met: 1. No mutation must occur so that new alleles do not enter the population, 2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population), 3. Random mating must occur (i.e. individuals must pair by chance), 4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change, and 5. No selection can occur so that certain alleles are not selected for, or against.

Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some of all of these types of forces all act on living population at various times, and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized: p squared + 2pq + q squared and p + q = 1.

P= frequency of the dominant allele in the population

Q= frequency of the recessive allele in the population.

p squared = percentage of homozygous dominant individuals

q squared = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Problem #1- You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

A. The frequency of the "aa" genotype.

B. The frequency of the "a" allele.

C. The frequency of the "A" allele.

D. The frequencies of the genotypes "AA" and "Aa".

E. The frequencies of the two possible phenotypes if "A is completely dominant over "a."

aa=36% or 0.36 <=This the frequency of the aa genotype

let aa=p^2

0.36=p^2

p=sqrt*[0.36]

p=0.6 <=This is the frequency of the a allele

p+q=1

So, 1-p=q

1-0.6=0.4=q

q=0.4 <=This is the frequency of the A allele

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

(0.6)^2+2(0.6)(0.4)+(0.4)^2=1

0.36+0.48+0.16=1

2pq=0.48 <= This is the frequency of the Aa genotype

q^2=0.16 <= This is the frequency of the AA genotype

Aa+AA= 2pq +q^2=0.48+0.16=0.64

1-0.64=0.36

Frequencies of the two possible phenotypes:

0.64 for dominant A phenotype and 036 for a phenotype.

To calculate the frequencies of genotypes and alleles in the population, we can use the given information and the Hardy-Weinberg formulas. Let's solve each part of the problem step-by-step:

A. The frequency of the "aa" genotype:
Given that the percentage of the homozygous recessive genotype (aa) is 36%, we can convert it to a frequency by dividing it by 100:
Frequency of aa = 36% / 100% = 0.36

B. The frequency of the "a" allele:
The "a" allele is present in both the homozygous recessive (aa) and heterozygous (Aa) genotypes.
Since the frequency of aa is 0.36, we know that q squared (q^2) = 0.36.
Therefore, the frequency of the "a" allele (q) is the square root of q squared:
q = √(0.36) = 0.6

C. The frequency of the "A" allele:
Since the frequency of the "a" allele is 0.6 (q), and p + q = 1, we can calculate the frequency of the "A" allele (p) by subtracting q from 1:
p = 1 - 0.6 = 0.4

D. The frequencies of the genotypes "AA" and "Aa":
The frequency of aa genotype (q squared) has already been calculated as 0.36, so we can subtract it from 1 to find the combined frequency of "AA" and "Aa" genotypes:
Frequency of AA + Aa = 1 - 0.36 = 0.64

To determine the frequencies of "AA" and "Aa" individually, we can use the formula 2pq for the heterozygous genotype (Aa):
2pq = frequency of Aa
2 * 0.4 * 0.6 = 0.48

Therefore:
Frequency of AA = Frequency of AA + Aa - Frequency of Aa
= 0.64 - 0.48 = 0.16

Frequency of Aa = 0.48

E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a":
Since "A" is completely dominant over "a," individuals with either AA or Aa genotypes will have the same phenotype. The dominant phenotype, which is the expression of "A," will be present in individuals with either AA or Aa genotypes.
Therefore, the frequency of the dominant phenotype = Frequency of AA + Frequency of Aa = 0.16 + 0.48 = 0.64.

The frequency of the recessive phenotype, which is the expression of "a," can be calculated by using the frequency of the aa genotype:
Frequency of recessive phenotype = Frequency of aa = 0.36

So, the frequencies of the two possible phenotypes are:
Frequency of dominant phenotype = 0.64
Frequency of recessive phenotype = 0.36

To solve this problem using the Hardy-Weinberg formulas, we need to know the percentage of the homozygous recessive genotype (aa), which is given as 36%.

A. The frequency of the "aa" genotype can be calculated by converting the percentage to a decimal and taking the square root. Since the percentage is given as 36%, we convert it to a decimal by dividing by 100: 36% / 100 = 0.36. Taking the square root of 0.36 gives us the frequency of the "aa" genotype: √0.36 = 0.6.

B. The frequency of the "a" allele can be calculated by subtracting the frequency of the "aa" genotype from 1. Since the frequency of the "aa" genotype is 0.6, we can calculate the frequency of the "a" allele as 1 - 0.6 = 0.4.

C. The frequency of the "A" allele can be calculated by subtracting the frequency of the "a" allele from 1. Since the frequency of the "a" allele is 0.4, we can calculate the frequency of the "A" allele as 1 - 0.4 = 0.6.

D. The frequencies of the genotypes "AA" and "Aa" can be calculated using the formula p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. We already know that the frequency of the "a" allele (q) is 0.4. From this information, we can calculate the frequency of the "A" allele (p) as 1 - 0.4 = 0.6.

Using the values of p and q, we can calculate the frequencies of the genotypes:

- Frequency of "AA" genotype: p^2 = (0.6)^2 = 0.36 (or 36%)
- Frequency of "Aa" genotype: 2pq = 2 * 0.6 * 0.4 = 0.48 (or 48%)

E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a" can be determined using the genotypic frequencies.

Since "A" is completely dominant over "a," individuals with either the "AA" or "Aa" genotype will exhibit the dominant phenotype. Therefore, the frequency of the dominant phenotype is the sum of the frequencies of the "AA" and "Aa" genotypes: 0.36 + 0.48 = 0.84 (or 84%).

The frequency of the recessive phenotype (aa genotype) can be calculated as the complement to the dominant phenotype frequency. Therefore, it is 1 - 0.84 = 0.16 (or 16%).

So, the frequencies of the two possible phenotypes are 84% dominant and 16% recessive.