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.

The second child has reduced penetrance. Reduced penetrance is found when a given genotype (the presence of the mutation) fails to "penetrate" the phenotype. In other words, even though the second child has the dysfunctional allele, he does not have a clinical disease.

Penetrance refers to the level of dominance. These alleles can still be described as dominant or recessive, but penetrance describes the degree of dominance along a spectrum.

In this example, some individuals with a certain genotype do not express the expected phenotype (symptoms of the disease) at all. The only term that properly describes this effect is reduced penetrance. Penetrance refers to the percent of individuals with a specific genotype who express the associated phenotype. In most common examples given in genetics courses, autosomal dominant diseases are 100% penetrant, meaning that all individuals with one disease allele will show symptoms to some extent. Here, however, the disease appears to show 80% penetrance.

One term often confused with penetrance is expressivity. This refers to the extent that the phenotype is expressed, and is only applicable when penetrance is 100%. If all of the individuals showed symptoms of the disease, but some showed slightly different defects than others, the disease would have variable expressivity. The other answer choices refer to unrelated genetics concepts.

The phenotype for a Hardy-Weinberg population can be defined as p + q =1, where p and q represent dominant and recessive frequencies. The genotypes for the population can be defined as p²+2pq+q²=1.

In this formula, p² and q² represent the homozygous dominant and homozygous recessive genotype frequencies respectively, and 2pq represents the heterozygous genotype frequency in the population.

Since 40% of the population is phenotypic recessive, we know that 60% of the population is phenotypic dominant.

q = 0.4

p + q =1; p = 0.6

p² and q²  are equal to 0.16 and 0.36 respectively. Solving for 2pq gives us 0.48, so approximately 48% of the population will be heterozygous.