Question

1. sickle cell anemia is condition that shows codominance. the genotype for normal blood cells is nn. sickle cell trait is the heterozygous condition (ns) and contains both normal and sickle shaped blood cells. sickle cell disease (ss) is when all of the blood cells are sickle - shaped and has lifelong medical implications. a man with sickle cell trait has a child with a woman with sickle cell trait.

k. complete the punnett square.
l. what is the probability of having a child with sickle cell trait?
m. what is the probability of having a normal offspring?
n. what is the probability of having a child with sickle cell disease?

Explanation:

Sub - question k (already completed as per the given Punnett Square, so we'll focus on sub - questions l, m, n)

Sub - question l:

Step1: Identify genotypes in Punnett Square

The Punnett Square has four cells: \(NN\), \(NS\), \(NS\), \(SS\). Sickle cell trait is \(NS\).

Step2: Count favorable and total outcomes

Number of \(NS\) genotypes (favorable) = 2. Total number of genotypes = 4.

Step3: Calculate probability

Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{2}{4}=\frac{1}{2}\) or 50%.

Step1: Identify normal genotype

Normal genotype is \(NN\).

Step2: Count \(NN\) in Punnett Square

Number of \(NN\) genotypes = 1. Total number of genotypes = 4.

Step3: Calculate probability

Probability = \(\frac{1}{4}\) or 25%.

Step1: Identify sickle cell disease genotype

Sickle cell disease genotype is \(SS\).

Step2: Count \(SS\) in Punnett Square

Number of \(SS\) genotypes = 1. Total number of genotypes = 4.

Step3: Calculate probability

Probability = \(\frac{1}{4}\) or 25%.

Answer:

\(\frac{1}{2}\) (or 50%)

Sub - question m: