In Drosophila, gray body color is dominant to ebony body color, while long wings are dominant to vestigial wings. Assuming that the P₁ individuals are homozygous, work the following crosses through the F₂ generation, and determine the genotypic and phenotypic ratios for each generation.
gray, vestigial x ebony, long

Question

In Drosophila, gray body color is dominant to ebony body color, while long wings are dominant to vestigial wings. Assuming that the P₁ individuals are homozygous, work the following crosses through the F₂ generation, and determine the genotypic and phenotypic ratios for each generation.
gray, vestigial x ebony, long

gray, vestigial x ebony, long

Accepted Answer

Hi everyone. Let's look at our next problem. It says a tall pea plant with purple flowers. Big B. Little B. Big C. Little C. Is crossed with another tall pleat pant with purple flowers of the same genotype identify which genotype has a 1/16 chance of appearing in the F. One generation. So we know that the genotype of our two parents is identical. They are both heros I guess for both genes. So when we look to set up our punnett square, put that on here, We fill in the four types of gametes from the two parents. So that big C. Big C. Big B. Little C. Little B. Big C. Little B. Little C. And then of course the same four combinations from the other parent. So let's look at our filled out pendant square. I'll pace that up here just a second and we can walk through our answer choices. Um If you've solved problems like this before, uh if you're thinking of the usual fanatic ratio of a di hybrid cross like this one, we know that the fanatic ratio is 9 to 3 to 3 to 1 where this one is homos, I guess recessive. So that can lead us to guess that the genotype that has a 1/ chance of appearing would be the genotype that leads to the homos, I guess recessive Vienna type. So that would be a little B. Little B. Little C. Little C. And indeed we can check that on upon it square. And we see that this only occurs in one box. So indeed this genotype only has a one and 16 chance of appearing in the F. One generation. So our answer here is going to be D. The homicide is recessive. Little B. Little B. Little C. Little C. If we want to confirm that, we can just look at our other answer choices and check them against the pennant square. Uh We've got choice A. Because again this 9 to 3 to 3 to 1 is a fanatic ratio. But of course different genotype can lead the same phenotype, which is why you have nine out of 16 offspring showing the dominant phenotype since several different Jenna types can lead to that outcome. But if we wanted to double check for instance, Choice A. Says Big B. Big B. Big C. Little C. When we look here, we can see that we've got this occurring in more than one place. So again here we've got our big B. And Big B. And then we need big C. Little C. Up in here. So we see that occurring at least two boxes. And therefore not our correct answer. When we look at Choice B. It's big B. Little B. Big C. Big C. Well, where do we find big C. Big C occurring? We see it here in this corner and we see it up here in this corner, do we find that particular genotype here in this square. And then if we trace down here a second square. So again occurring more than one out of 16th times. And finally, Choice C. Is hetero sickos at both genes, just like the parents. We're going to see that happening even more often, and indeed we can look for that And we would find that happening one for obviously more than 1/16 of the time. So Choice C. Also not our answer. So again, if we have these two hetero parents, Big B. Little B. Big C. Little C crossed with each other, which genotype has a 1/16 chance of appearing in the F one generation. And that's Choice D. Little B. Little B. Little C, little C. See you in the next video.

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Key Concepts

Mendelian Inheritance and Dominance

Genotypic and Phenotypic Ratios

Dihybrid Cross and Independent Assortment