Chapter 20

If a recessive disease is found in 50 out of 100,000 individuals, what is the frequency of the heterozygote carriers for this disease?

0.043 If q^2 = 0.0005, then q = 0.022 and p = 1 − q = 0.978. The heterozygote frequency is 2pq, or 2 (0.978) (0.022) = 0.043.

Applying Hardy-Weinberg equilibrium, how many genotypes are predicted for a gene that has three alleles?

6 If a gene has three alleles, then six genotypes are predicted by applying the Hardy-Weinberg equilibrium equation. If the alleles are represented by the variables p, q, and r, then the sum of the genotype frequencies is represented by p^2 + 2pq + q^2 + 2pr + r^2 + 2qr = 1. The distribution of alleles in the aforementioned genotypes is determined by the trinomial expansion (p + q + r)^2.

In chickens, having feathers on legs is dominant to having featherless legs. In a population of 200 chickens, 128 have featherless legs. How many chickens are heterozygous for the "feathered leg" allele? Assume the population is in Hardy-Weinberg equilibrium.

64 This problem also requires you to use the square root method to solve for allele frequencies. (Note that this method can only be used when the population is in Hardy-Weinberg equilibrium and the two alleles in question exhibit a dominant and recessive relationship. Also, note that you need to be aware of what the question is asking -- is it asking for a frequency or for a total number of individuals? If it is asking for a total number of individuals that possess a given genotype or allele, be sure your answer is a whole number. So, to solve this problem, you first calculate the frequency of homozygous recessive individuals, f(A2A2) by dividing the total affected by the total population, or 128/200 = 0.64, which is equal to q 2. Then solve for q by taking the square root of 0.64; thus q = 0.8. Using the equation p + q = 1, solve for p, which equals 0.2. To solve for the frequency of heterozygotes, calculate 2pq = 0.32. Note that the problem doesn't end here. Since the question asks for the number of individuals (and not the frequency), you need to multiple the population number by the heterozygote frequency: 200 x 0.32 = 64 chickens.

You want to calculate the frequency of the allele causing sickle cell disease in your patients, and you know the genotypes of every individual. Assuming the population is not in Hardy-Weinberg equilibrium due to population substructure, which method can be used to calculate allele frequency? (A) allele-counting method (B) square root method (C) binomial expansion (D) genotype proportion method (E) chi-square test

D, genotype proportion method A method for estimating allele frequencies in a population by manipulation of genotype frequencies.

In a population of birds in Africa, it was observed that birds with small or large beaks could efficiently crack and eat small or large seeds, respectively. Birds with intermediate beaks had trouble with both types of seeds. What type of selection would be expected to occur in this population if small and large seeds were the only types of food available to these birds?

Disruptive, occurs when both extreme phenotypes are favored over intermediate phenotypes

If the frequency of the M allele in the human MN blood group system is 0.65 in a population at equilibrium, then the frequency of the N allele must be 0.04. T/F?

false The sum of the allele frequencies must equal 1, so the frequency of the N allele must be 0.35

For the ABO blood group in humans, there are three alleles (IA , IB , and i), six possible genotypes (IAIA , IBIB , IAIB , IAi, IBi, and ii), and four possible phenotypes (A, B, AB, and O). Recall that IA is dominant to i, IB is dominant to i, and IA and IB are codominant. In a given population, the allele frequencies are as follows: IA = 0.15 IB = 0.25 i = 0.60

frequency of A phenotype = 0.2025 frequency of B phenotype = 0.3625 frequency of AB phenotype = 0.0750 frequency of O phenotype = 0.3600 The A phenotype is composed of genotypes IAIA and IAi. Thus, the frequency of individuals with phenotype A will be p^2 + 2pr. The B phenotype is composed of genotypes IBIB and IBi. Thus, the frequency of individuals with phenotype B will be q^2 + 2qr. The AB phenotype is determined by one genotype: IAIB . Thus, the frequency of individuals with phenotype AB will be 2pq. The O phenotype is determined by one genotype: ii. Thus, the frequency of individuals with phenotype O will be r^2

Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004. Assuming that this population is under Hardy-Weinberg equilibrium, calculate all allele frequencies and genotype frequencies.

frequency of A1 allele = 0.98 frequency of A2 allele = 0.02 frequency of homozygous dominant individuals = 0.9604 frequency of heterozygous individuals = 0.0392 frequency of homozygous recessive individuals = 0.0004

Galactosemia is an autosomal recessive disorder. If untreated, the disease can cause mental retardation in children. It is characterized by an inability to metabolize galactose (which comprises part of lactose found in milk), due to a deficient enzyme. In one population, the frequency of the normal allele, A1 , is 0.92. The frequency of the galactosemia allele, A2 , is 0.08. Assuming that this population is in Hardy-Weinberg equilibrium, calculate the frequency of homozygous dominant individuals, heterozygous individuals, and homozygous recessive individuals.

frequency of homozygote dominant individuals = 0.8464 frequency of heterozygous individuals = 0.1472 frequency of homozygote recessive individuals = 0.0064

The frequency of carriers for a rare autosomal recessive genetic condition is 0.04 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the allele frequency of the recessive allele?

not enough info is provided Recall that the sum of genotype frequencies is p^2 + 2pq + q^2 = 1.0. Given that the question only gives you the value for 2pq, it is impossible to solve for either p or q, and you would need more information to solve this problem.