Cytoplasmic male sterility (CMS) in plants has been expl...

Question

Cytoplasmic male sterility (CMS) in plants has been exploited to produce hybrid seeds (see Experimental Insight 17.1). Specific CMS alleles in the mitochondrial genome can be suppressed by specific dominant alleles in the nuclear genome, called Restorer of fertility alleles, RF. Consider the following cross:

♀CMS 1Rf 1/Rf1 rf2/rf2 × ♂CMS2rf 1/rf1 Rf2/Rf2

What genotypes and phenotypes do you expect in the F₁? If some of the F₁ plants are male fertile, what genotypes and phenotypes do you expect in the F₂?

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next question. It says if a true breeding, smooth and yellow seeded pea big R, big R big. Why big? Why is crossed with another true breeding? Wrinkled and green. Seated pea little R, little r little. Why little Y Which of the following phenotypes would be expected in 16 F two progeny. And our answer choices are a nine, smooth and yellow seated peas. B, five wrinkled and green seated peas, C two smooth and green seated peas or choice D all options are correct. So keep in mind we do have that d selection there that all options are correct. So we'll think about the fact that we have an original cross, which was the homozygous dominant in both traits with homozygous recessive in both traits. So big R big R, big, Y big, Y crossed with little R, little, R, little by little Y. So the F one generation will all be the same since each parent can only contribute one type of game. And the F one generation, all of them will be heterozygous for both traits. Big R little R, big, Y little Y And if we cross two members of this generation with each other for our F two generation. That's going to be what's called a die hybrid, cross Two parents heterozygous and two different traits. And we can recall from our content video that the phenotype ratio of a dye hybrid cost is 9-3 to three to one where those numbers stand for the nine being um dominant in both traits. And then the two middle threes showing one dominant, one recessive trait. And then the one at the end being recessive in both traits. Well, we kind of have a shortcut to our answer here because we have 16 F two progeny. We're talking about our ratio is all if expressed as fractions are fractions out of 16 because add up 933 and one and you get 16. So any possible correct answer must be one of these numbers 93 or one. Since those are the ratios we expect we only have one answer choice that has one of those numbers. And that's choice A which is nine smooth and yellow seated PS. And we can double check by looking at the phenotypes here. We see that smooth is dominant and yellow is also dominant. So as expected nine out of 16 showing both dominant traits. So that takes us right to saying that choice A is our correct answer and we can eliminate all the others because they are not. Choice B has five wrinkled and green seated P, we don't even need to evaluate the traits. Five isn't an expected uh phenotype amount. Same with choice C two, smooth and green, seated P two is not going to be an expected number. And then of course choice D all options are correct has been ruled out. So again, we have our um di hybrid cross, we have heterozygous P for big R little R, big Y little Y which of the following phenotypes expected in 16 F two progeny. And that's choice. A nine, smooth and yellow. Seated ps. See you in the next video.

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