Question

question 1-14
a man with red - green color blindness ((x^{r}y)), a recessive sex - linked trait, has a child with a woman that is homozygous dominant ((x^{r}x^{r})). what is the probability that a child produced from this couple will have red - gre

• (0%) for male children and (0%) for female children
• (0%) for male children and (100%) for female children
• (50%) for male children and (50%) for female children
• (100%) for male children and (0%) for female children

Explanation:

To solve this, we analyze the inheritance of the sex - linked trait:

Step 1: Determine the gametes

• The man with color blindness has the genotype \(X^{r}Y\). So, he can produce two types of gametes: \(X^{r}\) and \(Y\).
• The woman, who is homozygous dominant (\(X^{R}X^{R}\)), can only produce one type of gamete: \(X^{R}\).

Step 2: Analyze the offspring genotypes

• For male children: Males inherit the \(Y\) chromosome from their father and an \(X\) chromosome from their mother. Since the mother only gives \(X^{R}\), the genotype of male children will be \(X^{R}Y\). As the dominant allele (\(R\)) is present, they will not have color blindness. So the probability of male children having color blindness is \(0\%\).
• For female children: Females inherit one \(X\) chromosome from each parent. The father gives \(X^{r}\) and the mother gives \(X^{R}\), so the genotype of female children will be \(X^{R}X^{r}\). Since the dominant allele (\(R\)) is present, they will not have color blindness. So the probability of female children having color blindness is \(0\%\).

The correct option is: 0% for male children and 0% for female children

Answer:

To solve this, we analyze the inheritance of the sex - linked trait:

Step 1: Determine the gametes

• The man with color blindness has the genotype \(X^{r}Y\). So, he can produce two types of gametes: \(X^{r}\) and \(Y\).
• The woman, who is homozygous dominant (\(X^{R}X^{R}\)), can only produce one type of gamete: \(X^{R}\).

Step 2: Analyze the offspring genotypes

• For male children: Males inherit the \(Y\) chromosome from their father and an \(X\) chromosome from their mother. Since the mother only gives \(X^{R}\), the genotype of male children will be \(X^{R}Y\). As the dominant allele (\(R\)) is present, they will not have color blindness. So the probability of male children having color blindness is \(0\%\).
• For female children: Females inherit one \(X\) chromosome from each parent. The father gives \(X^{r}\) and the mother gives \(X^{R}\), so the genotype of female children will be \(X^{R}X^{r}\). Since the dominant allele (\(R\)) is present, they will not have color blindness. So the probability of female children having color blindness is \(0\%\).

The correct option is: 0% for male children and 0% for female children