Question

a gene is composed of two alleles. an allele can be either dominant or recessive. suppose that a couple, who are both carriers of the sickle - cell anemia allele but do not have the disease, decide to have a child. because both parents are carriers of the disease, each has one dominant normal - cell allele (s) and one recessive sickle - cell allele (s). therefore, the genotype of each parent is ss. each parent contributes one allele to their offspring, with each allele being equally likely. complete parts (a) through (c).

vice versa). list the other two possible genotypes of the offspring.
ss, ss
(use a comma to separate answers as needed.)

(b) what is the probability that the offspring will have sickle - cell anemia? in other words, what is the probability that the offspring will have genotype ss? interpret this probability.
the probability is \\( \frac { 1 } { 4 } \\). in 100 instances where both parents have one dominant normal - cell allele (s) and one recessive sickle - cell allele (s), we would expect approximately 25 of the offspring to have genotype ss.
(simplify your answers. do not round.)

(c) what is the probability that the offspring will not have sickle - cell anemia but will be a carrier? in other words, what is the probability that the offspring will have one dominant normal - cell allele and one recessive sickle - cell allele? interpret this probability.
the probability is \\( \square \\). in 100 instances where both parents have one dominant normal - cell allele (s) and one recessive sickle - cell allele (s), we would expect approximately \\( \square \\) of the offspring will be a carrier of, but will not have, sickle - cell anemia.
(simplify your answers. do not round.)

Explanation:

Step1: Determine the possible genotypes

When both parents have genotype \( Ss \), we can use a Punnett square. The possible combinations are \( SS \), \( Ss \), \( Ss \), \( ss \).

Step2: Calculate the probability of being a carrier (\( Ss \))

There are 4 possible genotypes. The number of \( Ss \) genotypes is 2. So the probability \( P=\frac{2}{4}=\frac{1}{2}\).

Step3: Calculate the expected number in 100 instances

If the probability is \(\frac{1}{2}\), then in 100 instances, the expected number is \(100\times\frac{1}{2} = 50\).

Answer:

The probability is \(\frac{1}{2}\). In 100 instances where both parents have one dominant normal - cell allele (\( S \)) and one recessive sickle - cell allele (\( s \)), we would expect approximately \(50\) of the offspring will be a carrier of, but will not have, sickle - cell anemia.