In a population of 200 people, an allele F has a frequency of 84%. What is the frequency of allele f? Using the Hardy-Weinberg equation, estimate the numbers of homozygous dominant, heterozygous, and homozygous recessive genotypes. (Remember that the formula is: p2 + 2pq + q2 = 1, where p represents the dominant allele and q represents the recessive allele.) *Be sure to account for all 200 people in the population.

A given autosomal locus has three possible alleles (A1, A2, and A3). Assuming that the numbers of individuals with the six possible genotypes are as follows, what would be the frequency of the A1 allele in the population?   A1A1 = 30 A1A2  = 27 A2A2 = 12 A1A3 = 17 A2A3   = 10 A3A3 = 4 What would be the frequency of the A1A1 genotype? Assume the population is under Hardy-Weinberg equilibrium.   A. 0.182   B. 0.27   C. 0.107   D. 0.317   E. 0.09   F. 0.03

A given autosomal locus has three possible alleles (A1, A2, and A3). Assuming that the numbers of individuals with the six possible genotypes are as follows, what would be the frequency of the A1 allele in the population?   A1A1 = 30 A1A2  = 27 A2A2 = 12 A1A3 = 17 A2A3   = 10 A3A3 = 4 What is the frequency of A1 allele in the population?   A. 0.375   B. 0.52   C. 0.305   D. 0.175   E. 0.74

In a large, randomly mating human population, the frequencies of the IA, IB, and i alleles are 0.7, 0.2, and 0.1, respectively. Calculate the expected frequencies for each blood type. (show solution)

An autosomal locus has alleles A and a. We are given the frequency of individuals with the autosomal recessive phenotype. Which of the following statements is TRUE?  Choose all that are true. Note: HWE = Hardy-Weinberg equilibrium a) If we assume HWE, we can calculate both allele frequencies b) If we assume HWE, we can calculate the genotype frequencies that we weren't given c) We can calculate both allele frequencies even if we don't assume HWE d) We can calculate q = Freq(a) even if we don't assume HWE e) Even if we don't assume HWE, we can calculate the genotype frequencies that we weren't given

In a large population of squirrels,  0.14 have the recessive genetic condition nlourianfilindr (fictional).  Assume the Hardy-Weinberg distribution.  What is the frequency of the dominant allele in this population? Giver your answer with two digits parts the decimal point.

If a disease determined by autosomal recessive heredity occurs at a frequency of 0.04 in a population, what is the frequency of this disease allele in the population? Assume Hardy-Weinberg equilibrium and use Aand a to symbolize the dominant and recessive alleles, respectively. Show calculation.

Q: In a certain species of plants, violet flower color (V) is dominant over white flower color (v). If p = 0.4 and q = 0.6 in a population of 500 plants, I want you to explain why red is wrong and highlighted is right.

Genotypes of leopard frogs from a population in central Kansas were determined for a locus (M) that encodes the enzyme malate dehydrogenase. The following numbers of genotypes were observed:   Calculate the genotypic and allelic frequencies for the population (round to the nearest thousandth). f(M1M1) = f(M1M2) = f(M2M2) = f(M1M3) = f(M2M3) = f(M3M3) = f(M1) = f(M2) = f(M3) =   What would be the expected number of each genotype (round to the nearest whole number assuming genotype represents an individual) assuming the population was in Hardy Weinberg equilibrium? f(M1M1) = f(M1M2) = f(M2M2) = f(M1M3) = f(M2M3) = f(M3M3) =    Referencing a p-value of 0.05 and a CV of 11.070, state if the population is in HWE. Genotype Observed Expected O–E (O–E)2 (O–E)2/E M1M1           M1M2           M2M2           M1M3           M2M3           M3M3             Chi-squared = Statement:…

Pretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem.  So you hypothesize that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations).  If you carry out chi-square goodness of fit test and calculate a total chi-square value of 0.03 with 1 degree of freedom (see table), what does this mean?