Question

1. sickle cell anemia is a codominant gene. if two parents who are heterozygous for the sickle cell trait have children, what percentage would be homozygous for the disease?

Explanation:

Step1: Define the alleles

Let's denote the normal allele as \( A \) and the sickle cell allele as \( S \). Since the parents are heterozygous for the sickle cell trait, their genotype is \( AS \).

Step2: Set up the Punnett square

When we cross two \( AS \) parents, the possible gametes from each parent are \( A \) and \( S \). The Punnett square will be:

	\( A \)	\( S \)
\( S \)	\( AS \)	\( SS \)

Step3: Determine the genotypes

From the Punnett square, the possible genotypes of the offspring are \( AA \) (homozygous normal), \( AS \) (heterozygous), and \( SS \) (homozygous for the sickle cell disease).

Step4: Calculate the percentage

There are 4 possible offspring. The number of offspring with the \( SS \) genotype is 1. To find the percentage, we use the formula: \( \text{Percentage} = \frac{\text{Number of } SS \text{ offspring}}{\text{Total number of offspring}} \times 100 \)
So, \( \text{Percentage} = \frac{1}{4} \times 100 = 25\% \)

Answer:

\( 25\% \)