Question

colorblindness pedigree
remember the following when
working pedigree charts:

1. circles are females and

squares are males

1. a shaded circle or square

indicates that a person has
the trait.
use the pedigree above to answer the questions below. shaded individuals are color blind.

1. what is the possible genotype of the mother, number 3? how did you determine your answer?
2. number 8 was color blind just like his father. where did the son get his allele for colorblindness?
3. neither numbers 1 nor 2 were color blind. how did they have a color blind son (6)?
4. what must be the genotypes of the parents of a color blind daughter? explain.
5. if number 13 marries a normal man, what is the probability that their sons will be color blind?

Explanation:

1. Colorblindness is an X-linked recessive trait. Males have an XY genotype, so a single recessive X allele ($X^c$) causes colorblindness. Females have XX genotypes, so they need two recessive alleles ($X^cX^c$) to be colorblind, or can be carriers with one recessive allele ($X^CX^c$). Individual 3 is not colorblind, so she has at least one dominant $X^C$ allele. She has a colorblind son (individual 8), who must have inherited his $X^c$ allele from his mother (since males get their X chromosome from their mother and Y from their father). Thus, individual 3 must carry the recessive allele.
2. Males inherit their only X chromosome from their mother, and their Y chromosome from their father. Since colorblindness is carried on the X chromosome, the son's colorblind allele comes from his mother.
3. Individual 6 is a colorblind male, so his genotype is $X^cY$. He inherited his Y chromosome from his father (individual 2) and his $X^c$ allele from his mother (individual 1). This means individual 1 is a carrier ($X^CX^c$), so she can pass the recessive allele to her son even though she is not colorblind herself.
4. A colorblind daughter has the genotype $X^cX^c$. She must inherit one $X^c$ allele from each parent. Her father, being male, only has one X chromosome, so it must be the recessive $X^c$ (meaning he is colorblind, $X^cY$). Her mother must contribute the second $X^c$ allele, so she can either be a carrier ($X^CX^c$) or colorblind ($X^cX^c$).
5. Individual 13 is a female carrier ($X^CX^c$): she is not colorblind, but her father (individual 8) was colorblind, so she inherited his $X^c$ allele. A normal male has the genotype $X^CY$. When they have children, sons will inherit their Y chromosome from the father and either the $X^C$ or $X^c$ allele from the mother. There is a 50% chance of inheriting the $X^c$ allele (which causes colorblindness in males) and a 50% chance of inheriting the $X^C$ allele (which results in normal vision).

Answer:

1. The possible genotype of individual 3 is $X^CX^c$. She is not colorblind, so she has one dominant $X^C$ allele, but she passed an $X^c$ allele to her colorblind son (individual 8), proving she is a carrier.
2. Number 8 got his colorblind allele from his mother (individual 7).
3. Individual 1 (the mother) is a carrier of the colorblind allele ($X^CX^c$). She passed the recessive $X^c$ allele to her son (individual 6), who inherited a Y chromosome from his father, resulting in his colorblindness.
4. The father must be colorblind ($X^cY$), and the mother must be either a carrier ($X^CX^c$) or colorblind ($X^cX^c$). The daughter needs two $X^c$ alleles: one from her father (who only has an $X^c$ to pass to daughters) and one from her mother.
5. The probability that their sons will be colorblind is 50% or $\frac{1}{2}$.