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Ch. 5 - Chromosome Mapping in Eukaryotes
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All textbooksKlug 12th EditionCh. 5 - Chromosome Mapping in EukaryotesProblem 17

Chapter 5, Problem 17

Drosophila females homozygous for the third chromosomal genes pink and ebony (the same genes from Problem 16) were crossed with males homozygous for the second chromosomal gene dumpy. Because these genes are recessive, all offspring were wild type (normal). F₁ females were testcrossed to triply recessive males. If we assume that the two linked genes, pink and ebony, are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F₁ males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?

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Hey, everyone. Let's take a look at this practice problem together. What might be the order of the jeans and the linkage map? If the recombination frequency between genes A and B is 6% between A and C is 14% and between B and C is 20%. OK. Recall that recombination frequency is not an exact measurement of distance, but it's an approximation. And that one recombination frequency equals one map unit. And we use map units to help us draw out linkage map. Also recall that genes that are further apart from each other have a higher R F or recombination frequency and genes closer together have a lower recombination frequency. So I'll draw a line to represent a chromosome and the distance of this line is going to be equal to the highest percentage of our recombination frequency because again, that means they'll be the furthest apart. So in our Question, is the furthest apart at 20 map units. And that means that they will be the furthest away from each other. So our gene order will either start or end with B or C. Remember though we can't technically tell which gene goes first because recombination frequency doesn't tell us that. But I'm going to put B here And I'll put, see here. Well, using this information, we can already deduce the answer, but we're going to go ahead and finish the order of jeans for practice. So the distance from B to C is map units. Now, we're told the distance from A to B A 6% or six map units and A two C is 14 map units. So that means that A will be closer to B than it is to see. So I'm gonna put a right here And this distance is going to be six map units And this distance is going to be 14 map units. And so a diagram of our genes and their order would look like the following. So for us, our order of genes is B A C and that is option D. So D would be the answer. However, though, remember, we don't know whether it's B or C that is the start of the genes on the chromosome. And so technically, our genes could also appear in the order C A B, but that's not an option. So D is our answer. Alright, everyone, I hope you found this helpful and I'll see you for the next practice problem.
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Textbook Question
Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. Phenotype Male Offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0 Were any double-crossover offspring expected?
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Textbook Question
In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive mutations on chromosome III. Flies from a Dichaete stock were crossed to homozygous ebony, pink flies, and the F₁ progeny, with a Dichaete phenotype, were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the table, Phenotype Number Dichaete 401 ebony, pink 389 Dichaete, ebony 84 pink 96 Dichaete, pink 2 ebony 3 Dichaete, ebony, pink 12 wild type 13 What is the sequence and interlocus distance between these three genes?
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Textbook Question
In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive mutations on chromosome III. Flies from a Dichaete stock were crossed to homozygous ebony, pink flies, and the F₁ progeny, with a Dichaete phenotype, were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the table, Phenotype Number Dichaete 401 ebony, pink 389 Dichaete, ebony 84 pink 96 Dichaete, pink 2 ebony 3 Dichaete, ebony, pink 12 wild type 13 Diagram this cross, showing the genotypes of the parents and offspring of both crosses.
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Textbook Question
In Drosophila, two mutations, Stubble (Sb) and curled (cu), are linked on chromosome III. Stubble is a dominant gene that is lethal in a homozygous state, and curled is a recessive gene. If a female of the genotype is to be mated to detect recombinants among her offspring, what male genotype would you choose as a mate?
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Textbook Question
If the cross described in Problem 18 were made, and if Sb and cu are 8.2 map units apart on chromosome III, and if 1000 offspring were recovered, what would be the outcome of the cross, assuming that equal numbers of males and females were observed?
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Textbook Question
Are mitotic recombinations and sister chromatid exchanges effective in producing genetic variability in an individual? in the offspring of individuals?
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