Population Genetics and Hardy-Weinberg - MCAT Biology

Hardy-Weinberg principle mathematically describes how inheritance does not change allele frequency in large populations. This helps explain why dominant and recessive alleles are both found in populations. A change in the predicted genotypes of a population may indicate evolotuion at work.

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.

Hardy-Weinberg equilibrium is acheived when the gene frequencies in a population do not change over time. This means the population is not evolving. The five conditions for this are large population size, no mutations, random mating, no net migration, and equally successful reproduction potential for all genes in the population. Temperature is a directly important aspect of this.

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.

The correct answer is "all of these." The choices listed are all required assumptions for a population to achieve Hardy-Weinberg equilibrium. "No natural selection" is another condition that is not listed here, bringing the total to 5 conditions.

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.

The Hardy-Weinberg equilibrium expression says that p2+2pq+q2 = 1.

We know that the incidence of q2 (getting two recessive alleles, and thus being resistant) is 1/900 in a general population, and 1/25 in New Guinea. The recessive allele frequency, q, will be 1/30 and 1/5, respectively.

The carrier frequency is 2pq, where p = 1-q.

Using this information, we can find the respective carrier frequencies.

General population:

New Guinea:

The founder effect is the abnormal abundance of an allele in a population derived from a small initial population. If, by chance, the initial population had an abnormal abundance of a certain allele, this abnormality will generally persist for future generations.

Hardy-Weinberg equilibrium states that there is random mating, no immigration/emigration, and that there are no mutations.

To meet Hardy-Weinberg criteria, a population must be very large (preferably infinite) and exhibit no mutation, no net migration, no natural selection, and no non-random mating. Of the choices, all break one of these criteria except the large population of wildcats.

If 50% of the population carries the recessive allele, then 50% carry the dominant allele. To determine the genotype breakdown we use the equation p2 + 2pq + q2, where p2 represents homozygous dominant genotype, 2pq represents heterozygous genotype, and q2 represents homozygous recessive genotype. The question tells us the value of allele frequency for the recessive allele, giving us the value of q in this equation. Since p + q = 1, and q is 0.50, p must also be 0.50.

p2 = 0.25

2pq = 0.50

q2 = 0.25

Setting p and q both to 0.50 gives us 25% homozygous dominant, 50% heterozygous, and 25% homozygous recessive; therefore, 75% of the population will display the dominant phenotype (black beak), while 25% will display the recessive phenotype (grey beak). Remember that both homozygous dominant and heterozygous genotypes will display the dominant phenotype.

In Hardy-Weinberg equilibrium, the sum of the dominant allele frequency (p) and the recessive allele frequency (q) is equal to 1.

The question says that 49% of the population consists of mice with the homozygous dominant gene, therefore, the dominant genotype frequency is equal to 0.49.

The question asks us to find the frequency of the recessive allele (q). In order to find the frequency of the recessive allele, we must first find the frequency of the dominant allele (p). According to the Hardy-Weinberg principle, the square root of the homozygous genotype frequency is equal to the allele frequency.

The dominant allele frequency is 0.7. Using this, we can solve for the recessive allele frequency.

By definition, the Hardy-Weinberg principle states that genotype and allele frequencies will remain constant throughout generations. In order for equilibrium to occur, there must be a large, randomly mating population with no selection, genetic drift, migration, or mutation. A small population cannot be in Hardy-Weinberg equilibrium.

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 following are requirements for Hardy-Weinberg.

• Organisms are diploid
• Only sexual reproduction occurs
• Generations are non overlapping
• Mating is random
• Population size is infinitely large
• Allele frequencies are equal in the sexes
• There is no migration, mutation, or selection

Natural selection would affect allele frequency in a population, thus disrupting Hardy-Weinberg equilibrium.

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.

There are five conditions for Hardy-Weinberg equilibrium.

1. No natural selection
2. Large population
3. Random mating
4. No mutations
5. No migration