Question

Red-green color blindness is caused by a recessive sexlinked gene. If a red- green color-blind man marries a woman with normal vision whose father was red- green color blind, what will be the expected phenotypic results in their children?

Short answer

The expected phenotypic results in their children are approximately 25% for each phenotype: daughters with normal vision but carrying the colorblind gene, sons with normal vision, daughters with red-green color blindness, and sons with red-green color blindness.

Step-by-step solution

Determine the genotypes of the parents

Since red-green color blindness is a recessive sex-linked gene, we'll represent the normal vision allele with "X^N" and the color-blind allele with "X^c". Since the father is red-green color blind, his genotype must be X^cY. The mother has normal vision, so her genotype must include an X^N allele. Since her father was color-blind, her genotype must also include an X^c allele, and therefore, her genotype is X^N X^c.

Set up a Punnett square

Now that we have the genotypes of the parents, we can set up a Punnett square to determine the possible genotypes of their children. Label one side of the square with the alleles of the father (X^c and Y) and the other side of the square with the alleles of the mother (X^N and X^c).

Fill in the Punnett square

To fill in the Punnett square, combine the alleles of the parents for each box. You should get the following combinations: 1. X^N X^c (daughter with normal vision, carrier of the colorblind gene) 2. X^N Y (son with normal vision) 3. X^c X^c (daughter with red-green color-blindness) 4. X^c Y (son with red-green color-blindness)

Determine the phenotypic ratios of the children

Now that we have the possible genotypes of the children, we can determine the expected phenotypes. There are four possible phenotypes: 1. Daughter with normal vision, carrier of the colorblind gene: 1/4 (25%) 2. Son with normal vision: 1/4 (25%) 3. Daughter with red-green color-blindness: 1/4 (25%) 4. Son with red-green color-blindness: 1/4 (25%) Thus, the expected phenotypic results in their children are approximately 25% for each phenotype: daughters with normal vision but carrying the colorblind gene, sons with normal vision, daughters with red-green color blindness, and sons with red-green color blindness.

Key concepts

Red-green color blindness

Red-green color blindness is a common type of color vision deficiency, particularly prevalent in males. This condition is caused by a recessive gene located on the X chromosome. Because it is recessive, an individual must inherit two copies of the gene (one from each parent) to express the trait. However, in males, who have one X and one Y chromosome, only one copy of the recessive gene is required to be color blind. For females, who have two X chromosomes, two copies of the gene are necessary.

In the given exercise, a man who is red-green color blind has the genotype X^cY because he has the recessive gene on his singular X chromosome. A woman with normal vision without symptoms of color blindness can still be a carrier of the color blindness gene if her father was color blind, making her genotype X^NX^c. This means she carries one normal vision gene (X^N) and one color blindness gene (X^c). Understanding these genetic foundations is vital for predicting the inheritance patterns of this trait through families.

Punnett square

A Punnett square is a tool used to predict the possible genetic outcomes for offspring based on the genotypes of the parents. This method helps illustrate potential combinations of alleles that the children might inherit from their parents. In the exercise, we set up a Punnett square for a color-blind father (X^cY) and a mother who is a carrier (X^NX^c).

Let's break down the process:

• Write the father's alleles across one side of the square (X^c and Y).
• Write the mother's alleles down the other side (X^N and X^c).
• Fill each box by combining alleles from the father and mother.

This results in four possible genetic combinations for the offspring: X^NX^c (carrier daughter), X^NY (normal vision son), X^cX^c (color-blind daughter), and X^cY (color-blind son). By using a Punnett square, it becomes clearer to visualize how different traits can be inherited.

Genotype and phenotype ratios

Understanding genotype and phenotype ratios is crucial for predicting how traits like red-green color blindness are passed down. The genotype refers to the genetic makeup of an organism, indicating which alleles an individual possesses, while the phenotype represents the observable traits resulting from this genetic makeup.

In our exercise, the genotypes identified through the Punnett square give rise to specific phenotypes:

• The genotype X^NX^c produces a daughter with normal vision who is a carrier of the color blindness gene.
• X^NY results in a son with normal vision.
• X^cX^c gives a daughter with red-green color blindness.
• X^cY results in a son with red-green color blindness.

These combinations result in phenotype ratios of 1:1:1:1 for each type of child: 25% daughters with normal vision who are carriers, 25% sons with normal vision, 25% daughters with red-green color blindness, and 25% sons with red-green color blindness. Understanding these ratios helps predict the likelihood of each phenotype manifesting in offspring.