Population Genetics and Hardy-Weinberg
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MCAT Biology › Population Genetics and Hardy-Weinberg
9% of a population of mice suffer from muscle spasms due to an autosomal recessive disease. What percentage of the population are carriers for the disease?
Explanation
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 
Use the first Hardy-Weinberg equation to determine the dominant allele frequency.
The frequency of the heterozygous genotype (carriers) is given by the 
All of the following are important aspects of Hardy-Weinberg equilibrium EXCEPT __________.
temperature
population size
mating patterns
migration
mutations
Explanation
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.
In a population of a particular island, 64% of individuals have a homozygous recessive genotype. Assuming the population correlates with Hardy-Weinberg principles, what percentage of individuals in the next generation will be heterozygous?
Explanation
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
Which of the following is not a necessary condition of Hardy-Weinberg equilibrium?
Migration of individuals
No mutations
Random mating
Large population
No natural selection
Explanation
There are five conditions for Hardy-Weinberg equilibrium.
- No natural selection
- Large population
- Random mating
- No mutations
- No migration
A rare recessive mutation causes rabbits that are normally white to be pink. If one in a hundred rabbits is pink, what is the frequency of the pink allele?
Explanation
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 
Using this set up, we can solve for the recessive allele frequency.
Cryptosporidium is a genus of gastrointestinal parasite that infects the intestinal epithelium of mammals. Cryptosporidium is water-borne, and is an apicomplexan parasite. This phylum also includes Plasmodium, Babesia, and Toxoplasma.
Apicomplexans are unique due to their apicoplast, an apical organelle that helps penetrate mammalian epithelium. In the case of cryptosporidium, there is an interaction between the surface proteins of mammalian epithelial tissue and those of the apical portion of the cryptosporidium infective stage, or oocyst. A scientist is conducting an experiment to test the hypothesis that the oocyst secretes a peptide compound that neutralizes intestinal defense cells. These defense cells are resident in the intestinal epithelium, and defend the tissue by phagocytizing the oocysts.
She sets up the following experiment:
As the neutralizing compound was believed to be secreted by the oocyst, the scientist collected oocysts onto growth media. The oocysts were grown among intestinal epithelial cells, and then the media was collected. The media was then added to another plate where Toxoplasma gondii was growing with intestinal epithelial cells. A second plate of Toxoplasma gondii was grown with the same type of intestinal epithelium, but no oocyst-sourced media was added.
You are conducting a study of an isolated tribe in New Guinea, and you find that there is widespread resistance to cryptosporidium infection. You determine that the gene for resistance is inherited in a recessive fashion. The incidence of resistance in a normal population is 1/900. In New Guinea, it is 1/25. What are the carrier frequencies in the normal population and in New Guinea, respectively? Assume that the populations are in Hardy-Weinberg equilibrium.
58/900, 8/25
2/900, 2/5
58/900, 1/625
58/300, 1/25
58/300, 8/25
Explanation
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:
If forty percent of a Hardy-Weinberg population is phenotypic for an autosomal recessive trait, approximately what percent of the population will be heterozygous?
48%
36%
16%
56%
Explanation
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.
75% of a given population test positive for Rhesus (Rh) factor antibodies. Given that blood type Rh+ is a dominant allele, what percentage of this population is homozygous?
We must know the total population size to solve
Explanation
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 
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 
Cryptosporidium is a genus of gastrointestinal parasite that infects the intestinal epithelium of mammals. Cryptosporidium is water-borne, and is an apicomplexan parasite. This phylum also includes Plasmodium, Babesia, and Toxoplasma.
Apicomplexans are unique due to their apicoplast, an apical organelle that helps penetrate mammalian epithelium. In the case of cryptosporidium, there is an interaction between the surface proteins of mammalian epithelial tissue and those of the apical portion of the cryptosporidium infective stage, or oocyst. A scientist is conducting an experiment to test the hypothesis that the oocyst secretes a peptide compound that neutralizes intestinal defense cells. These defense cells are resident in the intestinal epithelium, and defend the tissue by phagocytizing the oocysts.
She sets up the following experiment:
As the neutralizing compound was believed to be secreted by the oocyst, the scientist collected oocysts onto growth media. The oocysts were grown among intestinal epithelial cells, and then the media was collected. The media was then added to another plate where Toxoplasma gondii was growing with intestinal epithelial cells. A second plate of Toxoplasma gondii was grown with the same type of intestinal epithelium, but no oocyst-sourced media was added.
You are conducting a study of an isolated tribe in New Guinea, and you find that there is widespread resistance to cryptosporidium infection. Upon historical investigation, you find that the population you were studying all derived from a single group of four people that landed on the island 2000 years ago. Which phenomenon is most likely responsible for the observations of cryptosporidum resistance?
Founder effect
Penetrance
Genetic bottleneck
Pleiotropy
Balanced heterozygosity
Explanation
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.
A certain island-nation with a population of 200,000 has laws which severely restrict travel onto or off the island. There is an absolute prohibition against marrying foreigners, but island natives may marry as they wish. A non-lethal recessive genetic condition affects 2,000 of the people. How many carriers of this condition live on the island?
I. 198,000
II. 96,000
III. 36,000
IV. 12,400
V. 9,600
III
I
II
IV
V
Explanation
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.