The Hardy-Weinberg formulas tell us that and , where is the dominant allele frequency and is the recessive allele frequency. is the frequency of homozygous dominant individuals in a population, is the frequency of heterozygous individuals, and is the frequency of homozygous recessive individuals.

In the question, we are told that 25 people out of 100 are homozygous recessive, meaning that .

Using this, we can find that .

If , and , then .

We are looking for the frequency of heterozygous individuals ().

50% of the population will be heterozygous (carriers) for the trait.

When a population is in Hardy-Weinberg equilibrium, we can predict the genotypic frequencies found in the population using the equation , where  is the frequency of the dominant allele, and  is the frequency of the recessive allele. Since and account for the homozygotes in the population, we can find the frequency of the heterozygotes using the  portion of the equation. We can first solve for using the relationship .

Hardy-Weinberg equilibrium is dependent on five key conditions:

1. The population is very large.

2. Mutations are in equilibrium (no net mutation rate).

3. There is no immigration/emigration that alters the gene pool ratios.

4. There is random mating.

5. There is no selection for the fittest organism taking place in the population.

If any of these conditions are violated in a population, Hardy-Weinberg equilibrium will be disrupted.

This question is testing your knowledge of blood types and of the Hardy-Weinberg equations. Let's first evaluate the blood type portion of the question.

Rh-positive individuals will produce Rh factor. This molecule is an antigen. The given percentage refers to the number of individuals that produce antibodies to the Rh antigen. We can conclude that 75% of the population must be Rh-negative, since they produce antibodies against Rh factor; thus, 75% of the population must be homozygous recessive, since Rh-positive is a dominant allele.

Using this information, we can apply the Hardy-Weinberg equations:

We know that 75% of the population is homozygous recessive, which corresponds to the term in the second Hardy-Weinberg equation.

Use this value to solve for the dominant allele frequency.

The total percentage of the population that is homozygous will be given by the sum of the and terms in the Hardy-Weinberg equation.

This question requires application of the two Hardy-Weinberg equations:

We know that the disease is autosomal recessive, which means that affected individuals must be homozygous. The frequency of the homozygous recessive genotype is given by the term in the second Hardy-Weinberg equation. We can use the given genotype frequency to solve for the recessive allele frequency.

Use the first Hardy-Weinberg equation to determine the dominant allele frequency.

The frequency of the heterozygous genotype (carriers) is given by the term of the second Hardy-Weinberg equation. Use the values of the allele frequencies to calculate this term.

Use equations p + q = 1 and p² + 2pq + q² = 1.

If 64% is homozygous recessive genotype, then  q² = 0.64.

Then solve for p and q.

q = 0.8 and p = 0.2. The frequency of heterozygous individuals is 2pq or 2(0.2)(0.8) = 0.32

The question assumes unusual conditions that must be met for a Hardy-Weinberg genetic equilibrium to exist. The population is large and isolated. The gene is assumed to be stable.  Mating is "random," but only within the population. Here, the affected homozygous recessive people number 2,000—one one hundredth of the population.

In Hardy-Weinberg terms, this means q squared is 0.01 and q (the frequency of the recessive allele) is 0.1.  Since p + q = 1, then p, the frequency of the dominant allele, must be 0.9. The carriers are denoted by 2pq because p + q = 1, and therefore p² + 2pq + q² is also equal to one. Here, 2 (0.9)(0.1) = 0.18, meaning that 18% of the 200,000 population, or 36,000 persons, are heterozygous carriers. The strict conditions for Hardy-Weinberg equilibrium are almost never satisfied in human populations.

We use the formula: 

where p and q are the allelic frequencies.  We know that  is the frequency of homozygous recessive individuals, so q must be 0.4  That means that p must be 0.6, because p and q must add up to 1.  From there, we can just plug in 0.4 a 0.6 to get our answer.

Based on Hardy-Weinberg principles, we can predict the phenotype and genotype frequencies of a given population based on the equation . In this equation, is the recessive allele frequency and is the recessive phenotype frequency, or frequency of homozygous recessive genotypes.

The genotype frequency of , necessary for the development of a recessive disease, is going to be the square of the recessive allele frequency, .

We know that the pink allele is recessive and that one out of every hundred is pink; thus, one out of every hundred rabbits is homozygous recessive. Using Hardy-Weinberg calculations, we should be able to calculate the allele frequency.

If the pink allele frequency is , then is ; will refer to the frequency of homozygous recessive individuals in a population.

Using this set up, we can solve for the recessive allele frequency.