If you survey the genotype frequencies at locus E, which has alleles E (frequency =p) and e (frequency=q), and find that p2 + 2pq + q2 is much greater than 1, you can confidently conclude that Group of answer choices you made a math error the allele frequencies must also sum to greater than 1 the population is not at HWE with respect to locus E the population is not at HWE with respect to any loci

If you survey the genotype frequencies at locus E, which has alleles E (frequency =p) and e (frequency=q), and find that p2 + 2pq + q2 is much greater than 1, you can confidently conclude that Group of answer choices you made a math error the allele frequencies must also sum to greater than 1 the population is not at HWE with respect to locus E the population is not at HWE with respect to any loci

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 2QP: How Can We Measure Allele Frequencies in Populations? The MN blood group is a single-gene,...

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How Can We Measure Allele Frequencies in Populations? Drawing on your newly acquired understanding of the HardyWeinberg equilibrium law, point out why the following statement is erroneous: Because most of the people in Sweden have blond hair and blue eyes, the genes for blond hair and blue eyes must be dominant in that population.
Using the HardyWeinberg Law in Human Genetics In a given population, the frequencies of the four phenotypic classes of the ABO blood groups are found to be A = 0.33, B = 0.33, AB = 0.18, and i = 0.16. What is the frequency of the i allele?
In a population of 123 individuals, a locus has two alleles: E and e. If 30 individuals have the ee genotype, and the locus is at Hardy-Weinberg equilibrium, what is the frequency of the EE genotype? Round your answer to the second decimal place.
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?
Consider the B locus which has two alleles in a population: B and b. Researchers examined the genotypes several individuals for this locus and obtained the following numbers B/B: 320individuals B/b: 60individuals b/b: 12individuals If the B locus is at Hardy-Weinberg equilibrium, what would the expected number of individuals with the Bb genotype? Round your answer to the closest full number.
The following data were obtained from a Ghanaian population of 152 people. Blood Type Number of Individuals M 61 MN 64 N 27 Calculate the frequencies of the M and N alleles in this population. What are the expected numbers of individuals of each genotypic class in this population? Assume Hardy-Weinberg conditions. Use the chi-squared test to determine if these data fit the Hardy-Weinberg equilibrium model. The degrees of freedom for this test should be 1. Why is this appropriate?
Pretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So your hypothesis is that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out a chisquare goodness of fit test and calculate a total chisquare value of 0.03 with 1 degree of freedom (see table), what does this mean? (select all true statements)a) The data do NOT fit the hypothesized distribution.b) The data do fit the hypothesized distribution well enough, so we accept the hypothesis at this time (i.e. we cannot reject the hypothesis). c) The probability that the data came from a population in Hardy-Weinberg equilibrium is too small, so we reject the hypothesis.d) The probability that the data came from a population in Hardy-Weinberg equilibrium is too big, so we reject the hypothesis.e) The data support Hardy-Weinberg expectations – there is no…
A sample of 100 individuals from a population that is dimorphic at the A locus has genotype counts as follows. AA: 30 Aa: 60 aa: 10 a) What are the allele frequencies in the population? b) What are the expected genotype frequencies, if the population were at HardyWeinberg equilibrium? c) Is the proportion of heterozygotes lower or higher than expected at Hardy-Weinberg equilibrium? What deviations from the assumptions of the model would best explain the observed difference?
If you survey the genotype frequencies at locus E, which has alleles E (frequency =p) and e (frequency=q), and find that p2 + 2pq + q2 is much greater than 1, you can confidently conclude that
In corn, kernel color is governed by a dominant allele for white color (W) and by a recessive allele for yellow (w). A random sample of 152 kernels from a population that is in Hardy- Weinberg equilibrium reveals that 35 kernels are yellow and 117 kernels are white. What is the frequency of individuals with the homozygous dominant genotype?
The B locus has two alleles B and b with frequencies of0.95 and 0.05, respectively, in a population in the currentgeneration. The genotypic fitnesses at this locus are wB/B= 1.0, wB/ b = 1.0, and wb / b = 0.0a. What will the frequency of the b allele be in twogenerations?b. What will the frequency of the b allele be in twogenerations if the fitnesses were wB/B = 1.0, wB/ b = 0.0,and wb / b = 0.0?c. Explain why there is a difference in the rate ofchange for the frequency of the b allele under parts (a)and (b) of this problem
****Note for question 1 - you have been given the frequency of the two alleles (dominant and recessive = p + q=1) and you are using those allele frequencies to calculate the genotype frequencies. p2 + 2pq + p2