Three independently assorting genes (A, B, and C) are known to control the following biochemical pathway that provides the basis for flower color in a hypothetical plant:
Three homozygous recessive mutations are also known, each of which interrupts a different one of these steps. Determine the phenotypic results in the F₁ and F₂ generations resulting from the P₁ crosses of true-breeding plants listed here:
speckled (AABBCC) × yellow (AAbbCC)

Question

Three independently assorting genes (A, B, and C) are known to control the following biochemical pathway that provides the basis for flower color in a hypothetical plant:

Biochemical pathway showing gene A controls colorless to yellow, gene B yellow to green, and gene C green to speckled.

Three homozygous recessive mutations are also known, each of which interrupts a different one of these steps. Determine the phenotypic results in the F₁ and F₂ generations resulting from the P₁ crosses of true-breeding plants listed here:

speckled (AABBCC) × yellow (AAbbCC)

Accepted Answer

Hi everyone, Let's take a look at this practice problem together. If we cross a red and tall flower big a little a big B. Little B. With a red and short flower big a big a little B. Little B. What is the probability of producing offspring with genotype? Big A big a little B. Little B. The problem is asking us to do a cross between these two parent types and to determine how many of their offspring will look like big A big a little B, little B. So we'll have to do a punnett square to determine that. But in order to do that punnett square, we need to determine the types of gametes that each parent can produce. So for parent one, which is big, a little a big B. Little B. We can do a quick punnett square to determine the gammy gina types. So on the top we'll have the A's along the side, the bees and we do a combination and we'll see that we get four gametes that each look different. Now we need to do the same for parents too. And parent too is big. A big a little b little B. So let's do the punnett square for parent too. And we can quickly see by doing this punnett square, that parent to produces all four gametes that look identical. They are all big a little B. Now we can use these possible gametes to produce our punnett square and cross the parents. So when we do that, it's going to look like the following? And we will have parent one gametes Along the top and parent two gametes along the side. And we see all of the offspring. Now, recall that the practice problem asked us for the genotype. Big A big a little B little B. So how many of the offspring look like that? When we look at the results, we see that there are four that are big, a big a little B little B. So four total have that genotype and there are 16 offspring total that are produced. And so we can reduce that down to one out of four and we can see from our answer choices that be one out of four will have the genotype Big A big a little B. Little B. Alright, everyone. I hope you found this helpful. And I'll see you for the next practice problem.

Verified video answer for a similar problem:

Key Concepts

Independent Assortment of Genes

Biochemical Pathways and Gene Function

Genotypic and Phenotypic Ratios in Crosses