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Ch. 5 - Chromosome Mapping in Eukaryotes

Chapter 5, Problem 13

As in Problem 12, flower color may be red, white, or pink, and flower shape may be personate or peloric. For the following crosses, determine the P₁ and F₁ genotypes:

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Hey everyone, let's take a look at this question together. A fruit fly which is hetero cygnus for gray body normal wings and red eyes is crossed with a fruit fly having the same genotype, what proportion of their offspring will have home as aegis? Gray body normal wings and red eyes assume that no crossing over takes place. So we have an L. L. For body wings and eyes. So let's go ahead and denote B. As the allele for body G. As elio for wings, N. E. Z. L. E. O. For eyes. And so we know that we have a fruit fly that is hetero zegas for gray body, normal wings and red eyes so that fruit flies genotype looks like this and we are crossing it with the fruit fly that has the same genotype so it looks just like the first one and we want to figure out what the proportion of their offspring will be home as agus for gray body, normal wings and red eyes. Which is this. So to solve this question, we're going to go ahead and do the cross and we're going to use the fork line method. So we know that the proportions of crossing the two hetero ziggy's genotype together gives us that 1 to 1, means that we have 1/4 rbb one half are that hetero zegas and then 1/4 are the Houma zegas recessive. And so since we're trying to figure out the home as a guest for the dominant? We only need to focus on the home is dominant. So since we're going to cross the heterosexual or wings, we also end up with that 1 to 1 ratio. So we have 1/4 1 half and 1/4. You have 1/4 Homolka's I guess dominant, one half and then 1/4 home as a guest recessive. And then lastly we're going to cross the heterosexual guys for eye color, which also gives us that 1 to 1 ratio. So we have 1/4 1 half and 1/4 Hamas is dominant, Vegas and Hamas against excessive. And since we're looking at the home as ideas for great body normal wings and red eyes, we're focusing on here here and here. And so we use all three of those proportions. So we have 1/4 times 1/4 times 1/4. Because those are the proportions of getting that home as a guest dominant. Or body wings and eyes when we use the fork line method and we multiply them together because of the product rule and we end up with 1/64 which is answer choice D. The correct answer. Because when we cross that heterosexuals, fruit fly for gray body normal wings and red eyes with another fruit fly that has the same genotype, the proportion of their offspring that will be homesickness for gray body Normal wings and red eyes is 1/64 because each trait is 1/ proportion multiplied by each other to give us that 1/64, which is answer choice d. The correct answer. I hope you found this video to be helpful. Thank you and goodbye.
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Textbook Question
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed: pr–adp 29% pr–vg 13 pr–c 21 pr–b 6 adp–b 35 adp–c 8 adp–vg. 16 vg–b. 19 vg–c 8 c–b. 27 In another set of experiments, a sixth gene, d, was tested against b and pr: d–b 17% d–pr 23% Predict the results of two-point mapping between d and c, d and vg, and d and adp.
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Textbook Question
Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where the color can be red, pink (the heterozygote), or white. The second pair leads to personate (dominant) or peloric (recessive) flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed to those that are white, peloric, and dwarf. Determine the F₁ genotype(s) and phenotype(s). If the F₁ plants are interbred, what proportion of the offspring will exhibit the same phenotype as the F₁ plants?
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In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. Using proper nomenclature, determine the genotypes of the P₁ and F₁ parents.
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Textbook Question
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. Determine the sequence of the three genes and the map distances between them.
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Textbook Question
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. Are there more or fewer double crossovers than expected?

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