A three-gene system of additive genes (A, B, and C) controls plan...

Question

A three-gene system of additive genes (A, B, and C) controls plant height. Each gene has two alleles (A and a, B and b, and C and c). There is dominance among the alleles of each gene, with alleles A, B, and C dominant over a, b, and c. Under this scheme, the dominant genotype for a gene contributes 10 cm to height potential, and the recessive genotype contributes 4 cm.

What is the height potential of the F₁ progeny of the homozygous plants identified in (a) and (b) of this problem?

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. In a pea plant. The gene for flower color is controlled by two allys dominant and recessive are the gene for seed color is also controlled by two alleys dominant and recessive. Y if a plant with a genotype heterozygous for both genes is crossed to the plant. Also with the genotype heterozygous. For both all, what is the probability of getting offspring with a genotype homozygous dominant r heterozygous Y choice A 1/16 b, 1/8 C 1/4 or D three eights. So we have two independently assorting genes. So in that case, when we want to look at the probability of two events happening, that equals the product or the probability of the individual events. In this case, we can calculate the probability of our offspring having these two genotypes from the two different genes, homozygous dominant R, head of Z Y by multiplying the probability of each of those things happening. So the probability of being homozygous dominant for R, we have two heterozygote crossed. So we have a very simple heterozygous punnet square for one treat. So it's a little four square pune square with dominant R, recessive R on the top, dominant R, recessive R on the left. And we end up with homozygous dominant, two heterozygous in the bottom, left and top right. And Homozygous recessive on the right. So that means the probability of being homozygous dominant is 1/ equals 1/4. Now, the probability of being hetero, I guess for why? Well, you have the same double heterozygous cross. We don't even need to draw out another planet square because we can see it'll be the same kind of cross as with gene R. So in that case, the probability of getting a heterozygous outcome and I'm going to highlight it on my R panne square. Just to remind me, we see we get two out of four that are heterozygous. I'm gonna highlight why here in blue. So I remember that that corresponds. So the probability of getting he Torey it or why is one half, two out of four or one half. So our overall probability here for obtaining offspring with the genotype homozygote was dominant R heterozygous Y will equal 1/ multiplied by one half, which will equal 1/8. And when we look at our answer choices, we see that choice B is 1/8. So that is our answer to this question. See you in the next video.

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