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Genetics A gene is composed of two alleles, either dominant or recessive. Suppose that a husband and wife, 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 his or her offspring, with each allele being equally likely. (a) List the possible genotypes of their offspring. (b) What is the probability that the offspring will have sicklecell anemia? In other words, what is the probability that the offspring will have genotype \(s s ?\) Interpret this probability. (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.

Step by step solution

01

- Determine Possible Genotypes

Each parent has the genotype Ss, meaning they can contribute either S (dominant) or s (recessive) to their offspring. To find all possible combinations, construct a Punnett square. The rows and columns of the square represent the alleles each parent can contribute:| | S | s ||-----|------|------|| S | SS | Ss || s | Ss | ss |Thus, the possible genotypes for the offspring are SS, Ss, and ss.

02

- Calculate the Probability of Genotype ss

The Punnett square shows that there is one ss combination out of four possible combinations (SS, Ss, Ss, ss). Therefore, the probability of an offspring having genotype ss (sickle-cell anemia) is \[\frac{1}{4} = 0.25\].

03

- Interpret the Probability for Genotype ss

The probability of 0.25 means there is a 25% chance that the offspring will have sickle-cell anemia if both parents are carriers.

04

- Calculate the Probability of Being a Carrier

From the Punnett square, there are two Ss combinations out of four possible combinations (SS, Ss, Ss, ss). Therefore, the probability of an offspring being a carrier (genotype Ss) is \[\frac{2}{4} = 0.5\].

05

- Interpret the Probability for Being a Carrier

The probability of 0.5 means there is a 50% chance that the offspring will be a carrier of the sickle-cell allele but will not have sickle-cell anemia.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Punnett square

In genetics, a Punnett square helps us understand the genetic makeup of offspring from two parents by showing the possible combinations of alleles.

Each parent has two alleles for a specific gene, and these alleles can be either dominant or recessive. In the given exercise, both parents are carriers of the sickle-cell anemia allele. This means each parent has one dominant allele \(S\) and one recessive allele \(s\), giving them the genotype \(Ss\).

To construct a Punnett square, we create a grid where we place the alleles from one parent on the top and the alleles from the other parent on the side.

The possible alleles can be:

• Parent 1: S or s
• Parent 2: S or s

We then fill in the grid by combining these alleles:

\ | S | s \
\----|------|------\
\ S | SS | Ss \
\ s | Ss | ss \

So, the possible genotypes of their offspring are \(SS\), \(Ss\), and \(ss\). This visual tool helps us quickly see all the potential genetic outcomes.

genotype probability

Once we've set up the Punnett square, we can calculate the probabilities of each genotype for the offspring.

In the exercise, the possible genotypes are \(SS\) (normal), \(Ss\) (carrier), and \(ss\) (sickle-cell anemia).

We count the number of times each genotype appears in the Punnett square:

• SS: 1 time
• Ss: 2 times
• ss: 1 time

The total number of combinations in the Punnett square is 4.

To find the probability of each genotype, we divide the number of occurrences by the total number of combinations:

• Probability of SS = \frac{1}{4} = 0.25 \ (25%)
• Probability of Ss = \frac{2}{4} = 0.5 \ (50%)
• Probability of ss = \frac{1}{4} = 0.25 \ (25%)

So, there is a 25% chance the child will be unaffected (SS), a 50% chance the child will be a carrier (Ss), and a 25% chance the child will have sickle-cell anemia (ss).

carrier probability

Carrier probability specifically looks at the likelihood that an offspring will inherit one dominant allele \(S\) and one recessive allele \(s\), making them a carrier.

From our earlier Punnett square, the genotype \(Ss\) (carrier) appears twice out of the four possible combinations:

• SS: 1 time
• Ss: 2 times
• ss: 1 time

Thus, the probability of an offspring being a carrier is \frac{2}{4} = 0.5\.

In other words, there is a 50% chance that a child will inherit one normal allele and one sickle-cell allele, making them a carrier. Carriers do not usually show symptoms of sickle-cell anemia because they have one normal allele that typically prevents the disease from manifesting. However, they can still pass the sickle-cell allele on to their own children, which is an important consideration for future family planning and genetic counseling.