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 a homozygous brown-eyed, RR, and a parent with a heterozygous brown-eyed, Rl, gene combination.
Show how you could use a polynomial to model the possible genetic combinations of the offspring.
What percent of the possible genetic combinations result in brown eyed offspring?******Based on a Punnett square, I think this is 75%, but not sure
What percent of the possible genetic combinations are carriers for the blue eyed gene?****** Also based on the square i think this is 50%, but still not sure
All offspring will have the dominant R gene and brown phenotype, since that is all the RR parent can give. Did you actually use a Punnett square?
But you are right with half as carriers.
She needs the polynomial... it isn't hard to figure out the 50%
To model the possible genetic combinations of the offspring, we can use a polynomial expression. Let's break down the parental gene combinations first:
Parent 1: RR (homozygous brown-eyed)
Parent 2: Rl (heterozygous brown-eyed)
To determine the possible genotypes, we can multiply (R + R) with (R + l) using the distributive property of mathematics:
(R + R) * (R + l)
= (R * R + R * l) + (R * R + R * l)
= R^2 + Rl + Rl + l^2
= R^2 + 2Rl + l^2
The polynomial expression R^2 + 2Rl + l^2 represents the possible genetic combinations of the offspring, where R^2 represents the homozygous brown-eyed gene combination (RR), 2Rl represents the heterozygous brown-eyed gene combination (Rl), and l^2 represents the homozygous blue-eyed gene combination (ll).
Now, let's determine the percentage of possible genetic combinations resulting in brown-eyed offspring:
Out of the three possible genetic combinations (R^2 + 2Rl + l^2), R^2 represents the homozygous brown-eyed gene combination (RR). Therefore, there is only one genotype for brown eyes.
Since there are a total of three possible genetic combinations, and only one results in brown eyes, we can calculate the percentage by dividing 1 by 3 and multiplying by 100:
Percentage of brown-eyed offspring:
= (1 / 3) * 100
= 33.33%
So, approximately 33.33% of the possible genetic combinations result in brown-eyed offspring.
Next, let's determine the percentage of possible genetic combinations that are carriers for the blue-eyed gene:
Out of the three possible genetic combinations (R^2 + 2Rl + l^2), l^2 represents the homozygous blue-eyed gene combination (ll). Therefore, there is only one genotype for blue eyes.
Since there are a total of three possible genetic combinations, and only one results in blue-eyed individuals, we can calculate the percentage by dividing 1 by 3 and multiplying by 100:
Percentage of carriers for the blue-eyed gene:
= (1 / 3) * 100
= 33.33%
So, approximately 33.33% of the possible genetic combinations are carriers for the blue-eyed gene.
Note: The remaining percentage, 33.33%, represents the heterozygous brown-eyed gene combinations (Rl), which do not carry the blue-eyed gene.
Therefore, the percentage of brown-eyed offspring is approximately 33.33%, and the percentage of carriers for the blue-eyed gene is also approximately 33.33%.