In pea plants, plant height, seed shape, and seed color ...

Question

In pea plants, plant height, seed shape, and seed color are governed by three independently assorting genes. The three genes have dominant and recessive alleles, with tall (T) dominant to short (t), round (R) dominant to wrinkled (r), and yellow (G) dominant to green (g).

What proportion of the that produce round, green seeds (regardless of the height of the plant) are expected to breed true?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together in pea plants, plant height, flower color and pod color are governed by three independently assorting genes. The three genes have dominant and recessive allys with tall capital T, dominant to short lowercase T axial flower capital, A dominant to terminal lowercase A and green pod capital G, dominant to yellow, lowercase G. If two p plants that are heterozygous for all three genes are crossed, what is the probability that their offspring will have the genotype capital T lowercase T capital, A lowercase A capital G, lowercase G. The answer options are a one out of eight B, one out of 16 C, one out of 64 and D, one out of 128. OK. So let's review the information in the question for this problem. We have two pea plants that are heterozygous for all three of the genes. And we want to calculate the probability that their offspring will also be heterozygous for all three genes. So to calculate this and solve this problem, we need to know the product rule. Now, what is the product rule? Recall that the product rule is the probability of two or more independent events occurring together is calculated by multiplying the individual probabilities together. Now, we know that the probability of each gene in the offspring being heterozygous is independent of each other because we are told in the question that the three independent, independently assorting genes. So let's work through this and apply the product rule now instead of working with all three genes, let's just focus on one for right now. So let's take the height gene as an example. We will do a cross between the two heterozygous parents for the height gene using a Punt square. So our cross will be between parent, one genotype capital T lowercase T with another parent capital T lowercase T. Now, the pundit square results for that cross are on the screen. The ails for the first parent are in the first row and in the first column is the ails for the second parent. So capital T lowercase T, analyzing this Punt square, the genotypic ratios are 1 to 2 to one. So there's one homozygous dominant offspring, two heterozygous offspring and one homozygous recessive offspring. So the probability of having a heterozygous offspring for the height gene is two out of four. Now, we know that the parents are heterozygous for both the flower gene, which is represented by the letter A and the pod color gene which is represented by the letter G. If we did the pundit squares for both of those genes, the results would be the same. Therefore, we know that the probability of the offspring having heterozygous genotype for the flower is two out of four as well. And the probability of the offspring bean headers, I guess for the pod color is also two out of four. So now we know all of the probabilities for each of the genes. And each one is two out of four, we can now use the product rule and multiply each event together. So the probability of having a heterozygous offspring for all three genes is two divided by four, multiplied by two, divided by four, multiplied by two, divided by four, solving this equals eight, divided by 64. And if we reduce that, we get 1/8 or one out of eight offspring will be heterozygous for all three genes. Therefore, the correct answer is a one out of eight. All right, everyone. I hope you found this helpful and I'll see you soon for the next practice problem.

Key Concepts

Independent Assortment

Dominant and Recessive Alleles

Breeding True

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