Question 1

If an X- linked recessive trait has a frequency of 5% in males, what percentage of females are carriers? (Assume that the population is in Hardy- Weinberg equilibrium.)
A) 9.5%
B) 4.2%
C) 5%
D) 10%
E) 95%

Accepted Answer

In a population at Hardy-Weinberg equilibrium, the frequency of carriers for an X-linked recessive trait in females can be calculated using the formula 2pq, where p is the frequency of the normal allele and q is the frequency of the recessive allele. Since 5% (or 0.05) of males have the trait, q = 0.05 and p = 1 - q = 0.95. Thus, the frequency of female carriers is 2(0.95)(0.05) = 0.095 or 9.5%.

Question 2

Mexican hairless dogs are heterozygous (Hh), carrying a single copy of the lethal allele. In a given population, the Hh genotypic frequency is 60% and the frequency of each homozygote is 20%. Assuming w = 1 for the Hh and hh dogs, what is the estimated genotypic frequency of the lethal HH genotype after reproduction?
A) 46%
B) 10%
C) 14%
D) 39%
E) 20%

Accepted Answer

C

Question 3

Given the following genetic profile, what is the mean fitness of the population? $$\begin{array} { l l l l } 
\text { Genotype } & A 1 A 1 & A 1 A 2 & A 2 A 2 \
\text { Frequency } & 0.55 & 0.20 & 0.25 \
\text { Number } & 2750 & 1000 & 1250 \
\text { Relative fitness } ( w ) & 1.0 & 0.75 & 0.50
\end{array}$$
A) 1
B) 0.3025
C) 0.825
D) 0.75
E) 0.55

Accepted Answer

The mean fitness of the population can be calculated by multiplying the frequency of each genotype by its relative fitness and then summing these products. Thus, the calculation is: (0.55 * 1.0) + (0.20 * 0.75) + (0.25 * 0.50) = 0.55 + 0.15 + 0.125 = 0.825.

Question 4

You are studying ABO blood groups, and know that 6.25% of the population has genotype I^AI^A and 42.25% of the population has Type O blood. What is the expected frequency of blood type B? A) 14% B) 6.5% C) 51.5% D) 13% E) 1%

Accepted Answer

The expected frequency of blood type B is 14%. This is because the alleles for blood type B are I^B and i. The frequency of the I^B allele can be calculated by subtracting the frequency of the I^A allele (6.25%) and the frequency of the i allele (42.25%) from 100%. This gives us a frequency of 51.5% for the I^B allele. The frequency of the i allele is 42.25%. The expected frequency of blood type B is the product of the frequencies of the I^B and i alleles, which is 14%.

Question 5

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) genotype proportion method
B) binomial expansion
C) chi- square test
D) allele- counting method
E) square root method

Accepted Answer

The answer of You want to calculate the frequency of...

Question 6

During a University of Nevada, Reno blood donation campaign, the frequencies of alleles determining the ABO blood groups were 0.70 for i, 0.2 for I^A and 0.1 for I^B. Assuming random mating, what is the expected frequency of blood type B? A) 0.49 B) 0.18 C) 0.04 D) 0.32 E) 0.15

Accepted Answer

The answer of During a University of Nevada, Reno blood...

Question 7

Which aspect of population genetics can lead to changes in allele frequencies in a new mixed population following migration?
A) founder effect
B) mutation
C) gene flow
D) natural selection
E) bottleneck effect

Accepted Answer

Gene flow refers to the transfer of alleles or genes from one population to another as a result of migration of individuals or gametes, leading to changes in allele frequencies in the new mixed population.

Question 8

The incidence of phenylketonuria (PKU), an autosomal recessive genetic disorder, is approximately 1 in 4,500 in Ireland. Assuming Hardy- Weinberg equilibrium, what is the probability that a phenotypically normal Irish female, with no family history, and a phenotypically normal Irish male, with an affected sister, will have a child with PKU?
A) 0.028%
B) 0.49%
C) 1.96%
D) 0.55%
E) 2.2%

Accepted Answer

To solve this, first, we calculate the frequency of the recessive allele (q) using the incidence of PKU (1 in 4,500 or q^2 = 1/4500). Solving for q gives us the square root of 1/4500, which is approximately 0.015. The frequency of the dominant allele (p) is 1 - q, which is approximately 0.985. A phenotypically normal female with no family history has a 2pq chance of being a carrier, which is 2 * 0.985 * 0.015 ≈ 0.0295 or 2.95%. The male, with an affected sister, has a 2/3 chance of being a carrier because we know he is not affected (thus, he must either be a carrier or not, and given his sister is affected, it's 2 chances out of 3 that he's a carrier if we exclude the possibility of him being affected). The probability that both carriers mate and have an affected child is 1/4. Therefore, the probability that their child will have PKU is 2.95% * 2/3 * 1/4 ≈ 0.49%.

Question 9

Jeffrey Pines from Mt. Rose were genotyped at a resistance gene for a type of fungi that affects pines. Genotypes of 1000 of these Jeffrey Pines were: 300 R1R1, 500 R1R2, and 200 R2R2. Use a chi- square (χ2) test to determine if this population is in Hardy- Weinberg equilibrium? The critical chi- square value at P=.05 for 1 degree of freedom is 3.84.
A) 0.10; fail to reject
B) 10; fail to reject
C) 0.10; reject
D) 10; reject
E) 37.5; reject

Accepted Answer

To determine if the population is in Hardy-Weinberg equilibrium, we first calculate the allele frequencies. Let p be the frequency of R1 and q be the frequency of R2. From the genotypes, we have:- 300 R1R1: 600 R1 alleles- 500 R1R2: 500 R1 alleles and 500 R2 alleles- 200 R2R2: 400 R2 allelesTotal R1 alleles = 600 + 500 = 1100Total R2 alleles = 500 + 400 = 900Total alleles = 1100 + 900 = 2000p = 1100 / 2000 = 0.55q = 900 / 2000 = 0.45Expected frequencies under Hardy-Weinberg equilibrium are:- R1R1: p^2 * 1000 = (0.55)^2 * 1000 = 302.5- R1R2: 2pq * 1000 = 2 * 0.55 * 0.45 * 1000 = 495- R2R2: q^2 * 1000 = (0.45)^2 * 1000 = 202.5Chi-square (χ2) calculation:χ2 = [(observed - expected)^2 / expected] for each genotype= [(300 - 302.5)^2 / 302.5] + [(500 - 495)^2 / 495] + [(200 - 202.5)^2 / 202.5]= 0.021 + 0.051 + 0.031= 0.103Since the calculated χ2 value (0.10) is less than the critical value of 3.84, we fail to reject the null hypothesis, indicating that the population is in Hardy-Weinberg equilibrium.

Question 10

Data for an autosomal recessive trait shows that heterozygous individuals have the highest relative fitness. Those who are unaffected and homozygous for the dominant A allele have a relative fitness of 84% compared to heterozygous individuals, but only about 30% of those who are affected survived to reproduce. What is the estimated equilibrium frequency for allele A in this population?
A) 0.700
B) 0.160
C) 0.186
D) 0.814
E) 0.357

Accepted Answer

The equilibrium frequency of allele A (p) in a population where heterozygotes have the highest fitness can be estimated using the formula p = sqrt(w_AA / w_AA + s), where w_AA is the relative fitness of the homozygous dominant genotype and s is the selection coefficient against the homozygous recessive genotype. Given that homozygous dominant individuals have a relative fitness of 84% (0.84) compared to heterozygotes, and only 30% of homozygous recessive individuals survive to reproduce, the selection coefficient against them is 1 - 0.30 = 0.70. Plugging these values into the formula gives an equilibrium frequency for allele A of approximately 0.186.

Question 11

The incidence of Tay Sachs, an autosomal recessive genetic disorder, is approximately 1 in 3,500 in a certain population of Ashkenazi Jews. Assuming Hardy- Weinberg equilibrium, what is the frequency of carriers for the Tay Sachs allele in this population?
A) 3.3%
B) 98.5%
C) 1.5%
D) 2.9%
E) 1.7%

Accepted Answer

The frequency of carriers for the Tay Sachs allele in a population in Hardy-Weinberg equilibrium can be calculated using the formula:q^2 = frequency of recessive homozygotesp^2 + 2pq + q^2 = 1where p is the frequency of the dominant allele and q is the frequency of the recessive allele.In this case, the incidence of Tay Sachs is 1 in 3,500, which means that q^2 = 1/3,500 = 0.000286.Therefore, q = 0.017 and p = 1 - q = 0.983.The frequency of carriers is 2pq = 2 * 0.017 * 0.983 = 0.0333, or 3.3%.

Question 12

All Old Order Amish families from Lancaster County with Ellis- van Creveld syndrome can trace their genealogies to Mr. and Mrs. Samuel King, who immigrated to Lancaster County in 1744. This is an example of___________ .
A) Hardy- Weinberg equilibrium
B) founder effect
C) mutation
D) gene flow
E) bottleneck effect

Accepted Answer

The founder effect occurs when a new population is established by a very small number of individuals from a larger population, leading to a reduced genetic variation and a non-random sample of the genes present in the original population. In this case, the Old Order Amish families with Ellis-van Creveld syndrome can trace their ancestry back to a single couple, indicating that the genetic condition proliferated in the population due to the limited genetic pool of the initial founders.

Question 13

Phenylketonuria (PKU) is extremely common in Ireland, affecting approximately 1 in 4500 live births. Assuming that the population is in Hardy- Weinberg equilibrium, what is the frequency of PKU carriers?
A) 2.9%
B) 1.49%
C) 2.2%
D) 98.5%
E) 0.022%

Accepted Answer

The frequency of PKU (q^2) is 1/4500. To find the carrier frequency (2pq), first find q by taking the square root of 1/4500, which gives q ≈ 0.0149. Assuming p + q = 1, p ≈ 1 - 0.0149 = 0.9851. The carrier frequency is 2pq = 2 * 0.9851 * 0.0149 ≈ 0.029, or 2.9%.

Question 14

In an insect species, survival in agricultural fields regularly sprayed with insecticide is determined by the genotype for a detoxification enzyme encoded by a gene with two alleles, A and D. Which of the following statements is consistent with the data? $$\begin{array} { | l | l | } 
\hline \text { Genotype } & \text { Relative Fitness } \
\hline \mathrm { AA } & 1.00 \
\hline \mathrm { AD } & 0.70 \
\hline \mathrm { OD } & 0.05 \
\hline
\end{array}$$
A) Directional selection will increase the frequency of the A allele at a pace determined by the intensity of natural selection.
B) The D allele frequency will change faster as it gets less frequent.
C) Eventually, natural selection should lead to a balanced polymorphism.
D) This will likely lead to the frequency of the A allele being eliminated (f=0.0) and the D allele being fixed (f=1.0).
E) Both the A and D alleles should reach stable equilibrium frequencies that are maintained in a steady state.

Accepted Answer

Given the relative fitness values, individuals with the AA genotype have the highest fitness (1.00), indicating that the A allele confers a survival advantage in the presence of insecticide. Therefore, directional selection will favor the A allele, increasing its frequency in the population over time. The pace of this increase will depend on the strength of the selection pressure, i.e., the intensity of natural selection.

Question 15

Researchers collect data on an island population of frogs and find a mutation rate of the recessive allele, d, is 1 × 10^{- 6}. The relative fitness of the individuals homozygous from this allele is 0.7. What is the allele frequency of this recessive allele at equilibrium? A) 3.0 × 10^{- 7} B) 1.4 × 10^{- 6} C) 7.0 × 10^{- 7} D) 1.2 × 10^{- 3} E) 1.8 × 10^{- 3}

Accepted Answer

The answer of Researchers collect data on an island population...

Question 16

In a theoretical population where f (A₁) = 0.95 and f (A₂) = 0.05, µ = 1 × 10^{- 6}, and v = 1 × 10^{- 7}, what is Oq? A) 5.26 × 10^{- 3} B) 9.55 × 10^{- 7} C) 9.45 × 10^{- 7} D) 5.00 × 10^{- 9} E) 9.50 × 10^{- 7}

Accepted Answer

Oq = 2µv/(µ + v) = 2(1 × 10^{- 6})(1 × 10^{- 7})/(1 × 10^{- 6} + 1 × 10^{- 7}) = 9.45 × 10^{- 7}

Question 17

A researcher is studying a newly discovered gene that causes increased body weight in domesticated chickens. In a mainland population, the frequency of the A₁ allele is 0.2, for this gene with two alleles. If 100 of these mainland chicken are transported on a ship to an isolated island with a population of 200 A₁A₁ chickens, 400 A₁A₂ chickens, and 400 A₂A₂chickens, what would the frequency of the A₁ allele in the admixed population? A) 0.764 B) 0.618 C) 0.745 D) 0.236 E) 0.382

Accepted Answer

The frequency of the A1 allele in the mainland population is 0.2. In the island population, the frequency of A1 is (2*200 + 400)/(2*(200+400+400)) = 0.3. The admixed population has 100 mainland chickens and 1000 island chickens, so the frequency of A1 is (0.2*100 + 0.3*1000)/(100+1000) = 0.382.

Question 18

The incidence of Tay- Sachs, an autosomal recessive genetic disorder, is approximately 1 in 3,500 in a certain population of Ashkenazi Jews. Assuming that this population is in Hardy- Weinberg equilibrium, what is the probability that a phenotypically normal female, with no family history, and a phenotypically normal male, who had an affected sister, will have a child with Tay- Sachs?
A) 1.96%
B) 0.028%
C) 0.55%
D) 2.2%
E) 0.49%

Accepted Answer

The answer of The incidence of Tay- Sachs, an autosomal...

Question 19

In a group of 500 people, the frequency of genotype NN is 40%. Assuming both autosomal inheritance and that the population is in Hardy- Weinberg equilibrium, how many individuals would you expect to have the MN genotype?
A) 68
B) 300
C) 200
D) 258
E) 232

Accepted Answer

The answer of In a group of 500 people, the...

Question 20

In the same population of patients you examined for sickle cell allele frequencies, you now want to test whether the disease symptoms are related to blood type. Since blood type is a codominant trait and the alleles are easily identifiable by the phenotype (blood type) of the individuals, which method is most helpful for identifying blood type allele frequency?
A) square root method
B) chi- square test
C) genotype proportion method
D) allele- counting method
E) binomial expansion

Accepted Answer

The allele-counting method is most helpful for identifying blood type allele frequency because it involves directly counting the alleles for each blood type in the population and then calculating their frequencies. This method is straightforward and effective for traits like blood type, where phenotypes directly correspond to genotypes due to codominance.

Quiz 20: Genetic Analysis of Quantitative Traits

If an X- linked recessive trait has a frequency of 5% in males, what percentage of females are carriers? (Assume that the population is in Hardy- Weinberg equilibrium.)

You are studying ABO blood groups, and know that 6.25% of the population has genotype IAIA and 42.25% of the population has Type O blood. What is the expected frequency of blood type B?

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?

During a University of Nevada, Reno blood donation campaign, the frequencies of alleles determining the ABO blood groups were 0.70 for i, 0.2 for IA and 0.1 for IB. Assuming random mating, what is the expected frequency of blood type B?

Which aspect of population genetics can lead to changes in allele frequencies in a new mixed population following migration?

The incidence of Tay Sachs, an autosomal recessive genetic disorder, is approximately 1 in 3,500 in a certain population of Ashkenazi Jews. Assuming Hardy- Weinberg equilibrium, what is the frequency of carriers for the Tay Sachs allele in this population?

All Old Order Amish families from Lancaster County with Ellis- van Creveld syndrome can trace their genealogies to Mr. and Mrs. Samuel King, who immigrated to Lancaster County in 1744. This is an example of___________ .

Phenylketonuria (PKU) is extremely common in Ireland, affecting approximately 1 in 4500 live births. Assuming that the population is in Hardy- Weinberg equilibrium, what is the frequency of PKU carriers?

Researchers collect data on an island population of frogs and find a mutation rate of the recessive allele, d, is 1 × 10- 6. The relative fitness of the individuals homozygous from this allele is 0.7. What is the allele frequency of this recessive allele at equilibrium?

In a theoretical population where f (A1) = 0.95 and f (A2) = 0.05, µ = 1 × 10- 6, and v = 1 × 10- 7, what is Oq?

In a group of 500 people, the frequency of genotype NN is 40%. Assuming both autosomal inheritance and that the population is in Hardy- Weinberg equilibrium, how many individuals would you expect to have the MN genotype?

You are studying ABO blood groups, and know that 6.25% of the population has genotype I^AI^A and 42.25% of the population has Type O blood. What is the expected frequency of blood type B?

During a University of Nevada, Reno blood donation campaign, the frequencies of alleles determining the ABO blood groups were 0.70 for i, 0.2 for I^A and 0.1 for I^B. Assuming random mating, what is the expected frequency of blood type B?

Researchers collect data on an island population of frogs and find a mutation rate of the recessive allele, d, is 1 × 10^{- 6}. The relative fitness of the individuals homozygous from this allele is 0.7. What is the allele frequency of this recessive allele at equilibrium?

In a theoretical population where f (A₁) = 0.95 and f (A₂) = 0.05, µ = 1 × 10^{- 6}, and v = 1 × 10^{- 7}, what is Oq?