Exam 5: Linkage, Recombination, and Eukaryotic Gene Mapping

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Genetic distances within a given linkage group:

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Geneticists often assume that map distances less than 7 to 8 map units or cM are quite accurate.Map distances that exceed this threshold significantly are assumed to be less accurate and the level of accuracy decline increases as map distances increase.Briefly explain this observation.

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The farther apart two genes are the more likely it is that multiple crossovers will occur.Double crossovers (or an even number of crossovers)between two genes may lead to parental genotypes in the offspring that will not be counted as recombinants, even though crossing over has taken place.Since genetic maps are created by counting observable recombinants in the offspring, double crossovers will lead to an underestimation of the true map distance.

What is the parental genotype? What is the parental genotype?

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Crossing over occurs during:

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Assume that an individual of AB/ab genotype is testcrossed and four classes of testcross progeny are found in equal frequencies.Which of the following statements is TRUE?

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Two genes, A and B, are located 30 map units apart.The dihybrid shown below is mated to a tester aa bb.What proportion of the offspring is expected to be dominant for both traits? Two genes, A and B, are located 30 map units apart.The dihybrid shown below is mated to a tester aa bb.What proportion of the offspring is expected to be dominant for both traits?

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A physical map often measures _____, whereas a genetic map measures _____.

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In maize (corn), assume that the genes A and B are linked and 30 map units apart.If a plant of Ab/aB is selfed, what proportion of the progeny would be expected to be of ab/ab genotype?

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Assume that you discover a new human gene that you believe is located on the Y chromosome although not in the region (pseudoautosomal)of the Y that is homologous with part of the X chromosome.How would you map this gene with respect to the other genes on the Y chromosome?

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In addition to determining genotypes, two- and three-factor testcrosses can be used to:

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If the recombination frequency between genes (A)and (B)is 5.3%, what is the distance between the genes in map units on the linkage map?

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You are examining the following human pedigree and want to determine if the rare dominant disease allele (D)is linked to a specific DNA sequence location, referred to as a "molecular marker." You are using two molecular markers, R1 and R2.Parental and progeny genotypes and phenotypes are indicated.Note that the father is a dihybrid at both loci, but the mother is homozygous recessive at both loci and linked to R1 (d-R1/d-R1).There is complete penetrance of the trait and a linkage phase of D-R1/d-R2 in the father.Assuming that the marker and the gene are linked, what is the best estimate of the map distance between the two loci? You are examining the following human pedigree and want to determine if the rare dominant disease allele (D)is linked to a specific DNA sequence location, referred to as a molecular marker. You are using two molecular markers, R1 and R2.Parental and progeny genotypes and phenotypes are indicated.Note that the father is a dihybrid at both loci, but the mother is homozygous recessive at both loci and linked to R1 (d-R1/d-R1).There is complete penetrance of the trait and a linkage phase of D-R1/d-R2 in the father.Assuming that the marker and the gene are linked, what is the best estimate of the map distance between the two loci?

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In flower beetles, pygmy (py)is recessive to normal size (py+), and red color (r)is recessive to brown (r+).A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red.The following are progeny phenotypes from this testcross: In flower beetles, pygmy (py)is recessive to normal size (py<sup>+</sup>), and red color (r)is recessive to brown (r<sup>+</sup>).A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red.The following are progeny phenotypes from this testcross:   Carry out a series of chi-square tests to determine if the two loci are assorting independently.What is the CORRECT chi-square value and how many degree(s)of freedom should be used in its interpretation? Carry out a series of chi-square tests to determine if the two loci are assorting independently.What is the CORRECT chi-square value and how many degree(s)of freedom should be used in its interpretation?

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Dr.Disney has been raising exotic fruit flies for decades.Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb.He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw.He is able to establish flies that are homozygous for both mutant traits.He mates these two strains with each other.Dr.Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males.In the resulting testcross progeny he observes 500 flies that are of the following make-up: Dr.Disney has been raising exotic fruit flies for decades.Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb.He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw.He is able to establish flies that are homozygous for both mutant traits.He mates these two strains with each other.Dr.Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males.In the resulting testcross progeny he observes 500 flies that are of the following make-up:   What is the relationship with respect to location between the two genes? What is the relationship with respect to location between the two genes?

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A testcross is performed on an individual to examine three linked genes.The most frequent phenotypes of the progeny were Abc and aBC, and the least frequent phenotypes were abc and ABC.What was the genotype of the heterozygous individual that is testcrossed with the correct order of the three genes?

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In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol.Another dominant allele, M, is responsible for the ability of soreflies to sing like birds.A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-breeding wild-type (i.e., mute, Loritol-sensitive)males.Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive.What will be the results of a chi-square test for independent assortment?

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In Drosophila melanogaster, cut wings (ct)is recessive to normal wings (ct+), sable body (s)is recessive to gray body (s+), and vermilion eyes (v)is recessive to red eyes (v+).All three recessive mutations are X-linked.A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross.The following are the progeny resulting from the testcross: In Drosophila melanogaster, cut wings (ct)is recessive to normal wings (ct<sup>+</sup>), sable body (s)is recessive to gray body (s<sup>+</sup>), and vermilion eyes (v)is recessive to red eyes (v<sup>+</sup>).All three recessive mutations are X-linked.A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.The F<sub>1</sub> females produced by this cross were mated with cut, sable, vermilion males in a testcross.The following are the progeny resulting from the testcross:   What the interference value shown by this cross? What the interference value shown by this cross?

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Is it possible for two different genes located on the same chromosome to assort independently?

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In Drosophila melanogaster, cut wings (ct)is recessive to normal wings (ct+), sable body (s)is recessive to gray body (s+), and vermilion eyes (v)is recessive to red eyes (v+).All three recessive mutations are X-linked.A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross.The following are the progeny resulting from the testcross. In Drosophila melanogaster, cut wings (ct)is recessive to normal wings (ct<sup>+</sup>), sable body (s)is recessive to gray body (s<sup>+</sup>), and vermilion eyes (v)is recessive to red eyes (v<sup>+</sup>).All three recessive mutations are X-linked.A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.The F<sub>1 </sub>females produced by this cross were mated with cut, sable, vermilion males in a testcross.The following are the progeny resulting from the testcross.   What is the CORRECT genetic map with respect to gene order and distances (in cM)for these three genes? What is the CORRECT genetic map with respect to gene order and distances (in cM)for these three genes?

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Assume that A and B are two linked genes on an autosome in Drosophila.A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below.However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted.What is the most precise map distance that can be calculated from these data? Aa Bb = 235 Aa bb = 225 Aa Bb = 20

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