Eye color is determined by genetic combination. Let R represent the gene for brown eyes and L represent the gene for blue eyes. Any gene combination including R results in brown eyes. Consider the offspring of a parent with homozygous brown-eyed,RL,gene combination.

What percent of the possible genetic combinations are carriers for the blue-eyed gene?

Thank you for your time and help in solving this problem.

biologists use a Punnett square to show the geneotypes

if both parents are RL, the following outcomes are equally probable
... RR, RL, LR, LL

To determine the percent of possible genetic combinations that are carriers for the blue-eyed gene, we need to analyze the given information.

In this case, the parent has a homozygous brown-eyed gene combination, RL. This means that both copies of their eye color gene are for brown eyes.

Let's break down the genetic combinations of the offspring:

1. RR: This combination occurs when the offspring inherits the brown eye gene (R) from both parents. In this case, none of the copies of the eye color gene are carrying the blue-eyed gene. The offspring will have brown eyes.

2. RL: This combination occurs when the offspring inherits one copy of the brown eye gene (R) and one copy of the blue-eye gene (L) from the parent. In this case, one copy of the eye color gene is carrying the blue-eyed gene. The offspring will have brown eyes.

Since both combinations result in brown eyes, none of the possible genetic combinations of the offspring are carriers for the blue-eyed gene.

Thus, the percentage of the possible genetic combinations that are carriers for the blue-eyed gene is 0%.

Note: It's essential to remember that this specific scenario assumes that the brown eye gene is dominant over the blue eye gene. The combination RL results in brown eye color. If the blue eye gene were dominant, the answer would be different.