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 F₂ are expected to be tall, wrinkled, yellow? ttRRGg?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together. A dye hybrid cross between pea plants with round yellow seed capital, R, lowercase R capital, Y lowercase Y and wrinkled green seed. Lowercase R, lowercase R, lowercase Y lowercase Y is performed white is the probability of obtaining offspring with wrinkled yellow seeds. The answer options are a three out of 16 B one out of 16 C nine out of 16 and D one out of four. So we are calculating how many offspring will have wrinkled yellow seeds. So this is the phenotype. But what is the genotype? Well, we know that parent one is round and yellow. They are heterozygous, round and heterozygous yellow and parent two is wrinkled green. So homozygous recessive for, for both the genes. That means with this information, we can conclude that the round capital R A lil is dominant to the wrinkled, lowercase R a little and yellow capital Y A lil is dominant to green, lower case Y A lil that means a wrinkled yellow seed would have what genotype? Their genotype would be lowercase R lowercase R capital Y lowercase Y. So how do we calculate the probability of that genotype occurring in the F one progeny. We can use the product rule for probability recall. That's where you calculate the probability of each independent event and then multiply them together. So for our question, that formula would be the probability of having wrinkled yellow seeds equals the probability of having wrinkled seeds multiplied by the probability of having yellow seeds. So let's first calculate the probability of having wrinkled seed offspring. The cross is between a heterozygous round parent. So capital R, lowercase R and the homozygous recessive wrinkled parent, lower case R, lower case R. The results of that punt square are on the screen. The parent one alis capital R and lowercase R in the top row and parent two ailes lowercase R, lowercase R are in the first column. The number of wrinkled offspring is two. The homozygous recessive offspring will have the wrinkled phenotype. Therefore, the probability of having wrinkled seed offspring is too out of four. Now, we'll repeat that to solve for the probability of having yellow seed offspring. The cross is between the heterozygous yellow parrot and the homozygous recessive green parent. The pun square results of that cross are on the screen, how many offspring are yellow. So it's the heterozygous offspring that will display the yellow phenotype and there's two. So again, the probability of having yellow seed offspring is two out of four. So now our calculation becomes the probability of having wrinkled and yellow equals two, divided by four, multiplied by two, divided by four. This equals four, divided by 16. And if we reduce that down, it means one out of four offspring will be wrinkled, yellow seeds. Therefore, the correct answer is option D 1/4. All right, everyone. I hope you found this helpful and I'll see you soon for the next practice problem.

Key Concepts

Mendelian Genetics

Genotype and Phenotype

Punnett Square

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