33 At locus A in mice, the frequency of the A1 allele is 0.3, and the frequency ofthe A2 allele is 0.7.At locus B, the frequency of the B1 allele is 0.1, and the frequency of the B2allele is 0.9.The frequency of all four possible gametes is as follows:p(A1B1) = 0.03p(A1B2) = 0.27p(A2B1) = 0.07p(A2B2) = 0.63Which of the following statements correctly describes linkage equilibriumor disequilibrium between the two loci?abSelected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer.с0.3d0.7A₂The frequencies of the four gametes add up to 1, so the two loci are in linkage equilibrium.B₁B₂0.1 B₁0.90.10.9The frequency of B1 is independent of whether A1 or A2 is present in the gamete, so the two loci are in linkageequilibrium.The frequency of B1 is not equal to the frequency of A1, so the two loci are in linkage disequilibrium.The frequencies of both A1 and B1 are not equal to 0.5, so the two loci are in linkage disequilibrium.A₁B₁0.03A₁B₂ 0.27A₂B₁0.07A₂B₂ 0.63

Linkage equilibrium refers to the random association of alleles at different loci, where the…

At locus A in mice, the frequency of the A1 allele is 0.3, and the frequency of the A2 allele is 0.7. At locus B, the frequency of the B1 allele is 0.1, and the frequency of the B2 allele is 0.9. The frequency of all four possible gametes is as follows: p(A1B1) = 0.03 p(A1B2) = 0.27 p(A2B1) = 0.07 p(A2B2) = 0.63 Which of the following statements correctly describes linkage equilibrium or disequilibrium between the two loci? a b 0.3 d 0.7 Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. A₁ A₂ The frequencies of the four gametes add up to 1, so the two loci are in linkage equilibrium. B₁ 0.9 B₂ B₁ B₂ 0.1 0.9 The frequency of B1 is independent of whether A1 or A2 is present in the gamete, so the two loci are in linkage equilibrium. The frequency of B1 is not equal to the frequency of A1, so the two loci are in linkage disequilibrium. The frequencies of both A1 and B1 are not equal to 0.5, so the two loci are in linkage disequilibrium. A₁B₁ 0.03 A₂B₂ 0.27 A₂B₁ 0.07 A₂B,₂ 0.63

At locus A in mice, the frequency of the A1 allele is 0.3, and the frequency of the A2 allele is 0.7. At locus B, the frequency of the B1 allele is 0.1, and the frequency of the B2 allele is 0.9. The frequency of all four possible gametes is as follows: p(A1B1) = 0.03 p(A1B2) = 0.27 p(A2B1) = 0.07 p(A2B2) = 0.63 Which of the following statements correctly describes linkage equilibrium or disequilibrium between the two loci? a b 0.3 d 0.7 Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. A₁ A₂ The frequencies of the four gametes add up to 1, so the two loci are in linkage equilibrium. B₁ 0.9 B₂ B₁ B₂ 0.1 0.9 The frequency of B1 is independent of whether A1 or A2 is present in the gamete, so the two loci are in linkage equilibrium. The frequency of B1 is not equal to the frequency of A1, so the two loci are in linkage disequilibrium. The frequencies of both A1 and B1 are not equal to 0.5, so the two loci are in linkage disequilibrium. A₁B₁ 0.03 A₂B₂ 0.27 A₂B₁ 0.07 A₂B,₂ 0.63

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 9QP: Using the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic...

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Given the following testcross data for corn in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: PhenotypeNumber Kn + +451 Kn f +134 + + + 97 + f bz436 Kn + bz18 + + bz119 + f +24 Kn f bz86 Total:1,365 Maize geneticists tend to be like Drosophila geneticists utilizing a + to indicate the wild type allele (which also is dominant) and a lower case letter for the mutant allele (which is recessive). The first thing I suggest in attacking this problem is to use the typical convention of upper case letters for dominant alleles and lower case letters for recessive alleles as follows: F = wild type, f = fine stripe; B = wild type, b = bronze aleurone K = wild type, k = knotted leaf Then we can rewrite the phenotypes in a “language” that is more easily understood: PhenotypeNumber k F B451 k f B134 K F B97 K f b436 k F b18 K F b119 K f B24 k f b86 Total:1,365 a) Determine the sequence (order) of the…
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