A Drosophila experiment examining potential genetic linkage of X-...

Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Phenotype Number
1. Echinus     8576
2. Scute       977
3. Crossveinless        716
4. Echinus, scute       681
5. Scute, crossveinless     8808
6. Scute, crossveinless, echinus         4
7. Echinus, crossveinless     1002
8. Wild type         1
20,765

Compare the recombination frequencies and speculate about the source of any apparent discrepancies in the recombination data.

Accepted Answer

Hello, everyone and welcome to today's video. So suppose in a pea garden, the P that is heteros for two traits, which is going to be tall, heterozygosity and purple heteros for P is crossed to a pea plant that is homozygosis. So, between parentheses, we have short homozygote, recessive for tea and white, homos, recessive for P for all the two traits. And they produce the following offspring tall and purple. 134 tall and white. 85 short and white, 56 short and purple. 41. So, what are the genotypes of their recombinant as answer choice? A we have heterozygous T and homozygous dominant P and homozygous dominant T and heteros P as answer choice B. We have fully homozygote recessive T MP as and then we have heterozygote and heterozygous P as answer choice C. We have fully homozygosis T M B and then we have homozygosity and P and then we have D which is going to be heteros for T homozygote, recessive for P and homo cycles, recessive for T and heterozygous four P. Well, all we have to do here is look at the genotypes of the appearance the first plan is going to be heterozygote for both traits T and P. So we're going to be writing that genotype down and then they're going to be crossed with a P plan that is going to be homozygosis for short and white. So we're going to be inputting a fully homo homo Zygo recessive genotype for T and P. So once they're cross, they're going to create recombinant which are going to obtain one set of genes from one parent and another set of genes from another parent. And they're going to create the following two possible recombinant. First recombinant is going to be heterozygous for T and homozygous recessive for P and this second recant is going to be saga 4 p and homo cycles recessive for now, looking over answer choices, the correct answer choice that is going to fit what we've been able to find is answer choice D which is heteros for T homozygote, recessive for P and homozygote recessive for T heterozygous for P. I really hope this video helped you and I hope to see you on the next one.

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