What is the probability that a phenotypically normal sixth child will be a carrier in a family where both parents are carriers?

What is the probability that a phenotypically normal sixth child will be a carrier in a family where both parents are carriers?

• A. 1/4
• B. 2/4
• C. 0
• D. 3/4

Answer: (C)

In this scenario, both parents are carriers of a recessive trait, typically denoted as heterozygous (Aa, where "A" represents the normal allele and "a" represents the recessive allele). When determining the probability of their children being carriers or expressing the recessive phenotype, we can use a Punnett square to visualize the potential outcomes. When two heterozygous parents (Aa x Aa) have a child, the possible genotypes of their offspring are as follows: - AA (normal, not a carrier) - Aa (normal, carrier) - Aa (normal, carrier) - aa (affected, with the recessive phenotype) From this cross, we can see there are four possible combinations: one homozygous dominant (AA), two heterozygous (Aa), and one homozygous recessive (aa). This results in a 1:2:1 ratio, meaning there is: - 1 chance of being AA (not a carrier) - 2 chances of being Aa (a carrier) - 1 chance of being aa (affected) Thus, the probability that a child will be phenotypically normal but a carrier (Aa) is 2 out of 4 possible outcomes,

In this scenario, both parents are carriers of a recessive trait, typically denoted as heterozygous (Aa, where "A" represents the normal allele and "a" represents the recessive allele). When determining the probability of their children being carriers or expressing the recessive phenotype, we can use a Punnett square to visualize the potential outcomes.

When two heterozygous parents (Aa x Aa) have a child, the possible genotypes of their offspring are as follows:

• AA (normal, not a carrier)

• Aa (normal, carrier)

• Aa (normal, carrier)

• aa (affected, with the recessive phenotype)

From this cross, we can see there are four possible combinations: one homozygous dominant (AA), two heterozygous (Aa), and one homozygous recessive (aa). This results in a 1:2:1 ratio, meaning there is:

• 1 chance of being AA (not a carrier)

• 2 chances of being Aa (a carrier)

• 1 chance of being aa (affected)

Thus, the probability that a child will be phenotypically normal but a carrier (Aa) is 2 out of 4 possible outcomes,