Two pure-breeding strains of summer squash producing yellow fruit, Y₁ and Y₂, are each crossed to a pure-breeding strain of summer squash producing green fruit, G₁, and to one another. The following results are obtained:
If the F₁ of Crosses I and II are mated, predict the phenotype ratio of the progeny.

Question

Two pure-breeding strains of summer squash producing yellow fruit, Y₁ and Y₂, are each crossed to a pure-breeding strain of summer squash producing green fruit, G₁, and to one another. The following results are obtained:

If the F₁ of Crosses I and II are mated, predict the phenotype ratio of the progeny.

Table showing crosses of summer squash with phenotypic ratios for yellow and green fruit.

If the F₁ of Crosses I and II are mated, predict the phenotype ratio of the progeny.

Accepted Answer

Hi, everyone. Welcome back. This is our next problem. It says two pure breeding strains of a plant designated P one and P two produce flowers of the same color, which is purple. Each of these strains when crossed with a pure breeding strain, ax produces F1 progeny plants that have pink flowers determine the genotype of the progeny in the F one generation. And our answer choices are a big, a little, a big B, little B and big, a little, a big b, little bee choice. B big A big, a big b, big B and big, a little A big b, little B choice C big, a little, a big B, big B and big A big, a big b, little B and choice D little, a little, a little B, little B and big, a little, a big b, little B. So this is a lot to think through. It's a rather complex problem, but we'll start with the fact that we have two genes A and B determining a single color. So we've got two genes determining a single being a typical trait. So we can recall from our content videos that this means we must have an epis static interaction between those genes that the interaction of the expression of one gene will affect or mask the expression of the other. We know we have strains P one and P two, they're true breeding. So both must be homozygous for whatever their genotype is and each of them in our true breathing state gives purple flowers. So despite the fact that there are two different strains with two different genotypes, they result in the same phenotypic trait. Then we have strain AX, which also must be homozygous. As we know. It's also true breeding though, we don't know what color of flowers it has. But we do know that when we cross P one with x or P two with X, that we get pink flowers. So this leads us to expect that let's, we can posit that strain X has a dominant trait in both genes A and B. Therefore allowing it to override what's ever going on in P two, P one and P two. So let's pause it. That strain X has the genotype, big A big A big B, big B, this satisfies the condition that it must be homozygous. Now, let's see if that makes sense of our pink versus purple. So perhaps we'd say that the purple trait is then overridden by the condition of being dominant in both genes. So a dominant ally A. So A blank B blank leads the phenotypic trait of pink flowers So then how would we have these two other strains that both generate the same color purple? But are then overridden by X? Well, that would be the case if only one of A or B is present as a dominant allele, the flowers are purple. And let's see. Does this work? Well, if we say that P one is big, a big, a little B, little B remembering that it has to be homozygous and there it has only one dominant ally. A B is homozygous recessive. That would lead it to be purple. Then P two, we'd say we have little, a little A big B big B satisfies the conditions of being homozygous. It has only one dominant gene B A is homozygous recessive. So that would lead to purple flowers. Now, does that make sense of having pink flowers in the F1 generation if either of these is crossed with strain X? Well, yes, it does because strain X will always give a dominant, a ally and a dominant B ally. So in the F two generation, excuse me, in the F one generation, anything that's been um since it's been crossed with strain X, every member of the F one generation is going to have at least one A alley and one B alley. So I'm gonna highlight that a big A blank, big B blank leading to the pink phenotype. That's going to be the case for 100 Of my F one generation. Now we saw in our answer choices that they all have two different possibilities for genotypes. And that makes sense because P1 and P two have two different genotypes. So let's think about what those would be based on what we figured out. So we're gonna pause it. Let's say P1 is big A big B, I'm sorry, big A big, a little B, little B now it's crossed with strain X. So it's crossed with big A big A big B, big B. So the F one progeny of this strain P one, they can only inherit a dominant A alley. So there'll be big a big A and then they're going to inherit a small B alley from P one and a big B all from X. So they'll all be big B little B. So that will be the first possibility. Big A big A big B, little B, they will be pink as they have a dominant A and a dominant B. Now, let's look at the possibility for the cross of P two which we've said is little, a little, a big B, big B cross with strain X. Big A big A big B big B. Then the F one progeny will inherit a little A A all from P two and a big A all from X. So there'll be big a little A and then they can only inherit a dominant B Allal from both strained parents. So big B big B. So the F one progeny of that cross are big, a little A big B, big B and that will also be pink as they are dominant. Alls of both A and B. So that will lead us to our answer choice, which is going to be choice C and that is big, a little A big B, big B and big A big A big B, little B. So that each will be homozygous dominant for one of those genes and heterozygous for the other. See you in the next video.

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Key Concepts

Mendelian Genetics

Phenotypic Ratios

Test Cross