For a gene where p = q = 0.5 you have the following numbers of individuals of different genotypes: A1A1 = 16, A1A2 = 67, A2A2 = 17. Based on your observation of the genotype frequencies, which of the following is correct regarding Hardy-Weinberg equilibrium in this population? Group of answer choices The observed genotype frequencies match what is predicted by Hardy-Weinberg equilibrium. The observed genotype distribution has fewer heterozygotes than predicted by Hardy-Weinberg equilibrium. The observed genotype distribution has more heterozygotes than predicted by Hardy-Weinberg equilibrium.
Added by Sarah C.
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- Total individuals = A1A1 + A1A2 + A2A2 = 16 + 67 + 17 = 100. Show more…
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Imagine that you sampled a population of dung beetles and discovered that the allele frequencies were A1 = 0.24 and A2 = 0.76. You genotyped individuals in your sample and calculated the following genotype frequencies: A1A1 homozygotes = 0.01, A1A2 heterozygotes = 0.44, and A2A2 homozygotes = 0.53. Is your population in Hardy equilibrium? What evidence leads you to arrive at this conclusion? This population is not in Hardy-Weinberg equilibrium because there is an excess of heterozygotes and a deficiency of homozygotes compared to Hardy-Weinberg expectations. The allele frequencies have changed, indicating that the population is not in equilibrium. There is a deficiency of A1A1 homozygotes and an excess of A1A2 heterozygotes compared to Hardy-Weinberg expectations. This population is not in Hardy-Weinberg equilibrium because the observed genotype frequencies do not perfectly match Hardy-Weinberg expectations.
Adi S.
You are helping two of your friends with their homework for another biology class. They are also learning about population genetics and Hardy-Weinberg equilibrium. This is the question they are working on: "There are 200 individuals in the population: 20 are homozygous dominant (A/A), 20 are heterozygous (A/a), and 160 are homozygous recessive (a/a). Is this population in Hardy-Weinberg equilibrium with respect to gene A?" Each of your friends has tried to solve the problem in a different way: Friend 1's Method: # A/A individuals = 20 # A/a individuals = 20 # a/a individuals = 160 Total # of individuals = 200 Observed allele frequencies Observed freq(A) = p = (2x20 + 20) / (2x200) = 60 / 400 = 0.150 Observed freq(a) = q = (2x160 + 20) / (2x200) = 340 / 400 = 0.850 Predicted genotype freq based on observed allele freq If pop' is in H-W eq: Predicted freq(A/A) = p^2 = 0.150^2 = 0.02 Predicted freq(A/a) = 2pq = 2x0.150x0.850 = 0.26 Predicted freq(a/a) = q^2 = 0.850^2 = 0.72 Observed genotype frequencies Based on data provided in the question: Observed freq(A/A) = 20/200 = 0.10 Observed freq(A/a) = 20/200 = 0.10 Observed freq(a/a) = 160/200 = 0.80 Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium. Friend 2's Method: # A/A individuals = 20 # A/a individuals = 20 # a/a individuals = 160 Total # of individuals = 200 If pop' is in H-W eq: Freq(A/A) = p^2 Therefore, assume observed freq(A/A) = p^2 = 20/200 = 0.100 Predicted allele frequencies If pop' is in H-W eq, p^2 = 0.100 Predicted freq(A) = p = sqrt(p^2) = sqrt(0.10) = 0.316 Predicted freq(a) = q = 1 - p = 1 - 0.32 = 0.684 Predicted genotype freq based on predicted allele freq If pop' is in H-W eq: Predicted freq(A/A) = p^2 = 0.316^2 = 0.10 Predicted freq(A/a) = 2pq = 2x0.316x0.684 = 0.43 Predicted freq(a/a) = q^2 = 0.684^2 = 0.47 Observed genotype frequencies Based on data provided in the question: Observed freq(A/A) = 20/200 = 0.10 Observed freq(A/a) = 20/200 = 0.10 Observed freq(a/a) = 160/200 = 0.80 Observed genotype frequencies do not match the genotype frequencies that are predicted under H-W equilibrium; therefore, the population is not in Hardy-Weinberg equilibrium.
Josee P.
When testing these results using a Chi-Square test, you rejected the null hypothesis that this population is in Hardy-Weinberg Equilibrium with p < 0.001. You are interested in understanding what evolutionary mechanism may be most likely responsible for the significant deviations of observed genotype frequencies from Hardy-Weinberg proportions. First, let's describe the pattern of genotypes in relationship to expected genotype frequencies under Hardy-Weinberg. Observed frequency of AA genotype Observed frequency of Aa genotype Observed frequency of aa genotype
Suman K.
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