Question

search the nih gene database for the term colorblindness. use the results of the database search to explain how a father and mother who are not colorblind could have a son who is colorblind. your model can be a pedigree chart, a punnett square, or a diagram of chromosomes.

Explanation:

1. NIH Gene Database Search: Searching the NIH gene database for "colorblindness" reveals that color blindness (e.g., red - green color blindness) is an X - linked recessive trait. The relevant genes (like OPN1LW, OPN1MW for red - green color blindness) are located on the X chromosome.
2. Genotype of Parents:

• A father has one X and one Y chromosome. Since he is not colorblind, his genotype is $X^{B}Y$ (where $X^{B}$ represents the normal allele on the X chromosome and Y is the Y chromosome).
• A mother who is not colorblind but can pass on the color - blindness allele must be a carrier. Her genotype is $X^{B}X^{b}$ (where $X^{b}$ is the allele for color blindness).

1. Punnett Square Explanation:

• When we create a Punnett square for the cross between $X^{B}Y$ (father) and $X^{B}X^{b}$ (mother), we consider the possible gametes. The father can produce two types of gametes: $X^{B}$ and $Y$. The mother can produce two types of gametes: $X^{B}$ and $X^{b}$.
• The possible offspring genotypes are:
• $X^{B}X^{B}$ (female, normal), $X^{B}X^{b}$ (female, carrier), $X^{B}Y$ (male, normal), and $X^{b}Y$ (male, colorblind). So, there is a 25% chance (1 out of 4) of having a colorblind son.
• In terms of a pedigree chart, we would represent the father as a square (male) with a normal phenotype, the mother as a circle (female) with a normal phenotype (but being a carrier). Their son would be a square with the colorblind phenotype, inheriting the $X^{b}$ from the mother and the Y from the father.
• For a chromosome diagram, the father's sex chromosomes are X (with $X^{B}$) and Y. The mother's sex chromosomes are two Xs, one with $X^{B}$ and one with $X^{b}$. During meiosis, the mother can pass on the $X^{b}$ chromosome, and the father passes on the Y chromosome, resulting in a son with $X^{b}Y$ genotype, which leads to color blindness.

Answer:

To explain how a non - colorblind father ($X^{B}Y$) and a non - colorblind (carrier, $X^{B}X^{b}$) mother can have a colorblind son:

1. Gene Basis (from NIH Database)

Color blindness (e.g., red - green) is X - linked recessive. Genes like OPN1LW/OPN1MW (for red - green) are on the X chromosome. A male with the recessive allele on his single X (from a carrier mother) and Y (from father) is colorblind.

2. Parental Genotypes

• Father: $X^{B}Y$ (normal, as Y has no color - blindness allele, and X has normal allele).
• Mother: $X^{B}X^{b}$ (carrier, has one normal $X^{B}$ and one recessive $X^{b}$ allele, so phenotypically normal).

3. Offspring Inheritance (Punnett Square Example)

Father's Gametes\Mother's Gametes	$X^{B}$	$X^{b}$
$Y$	$X^{B}Y$ (normal male)	$X^{b}Y$ (colorblind male)

There is a 25% chance of a $X^{b}Y$ (colorblind son) as he gets $X^{b}$ from mother and $Y$ from father. A pedigree would show the father (normal male symbol), mother (normal female symbol, carrier), and a colorblind son (filled male symbol). A chromosome diagram would show the mother passing $X^{b}$ and father passing $Y$ to the son, resulting in $X^{b}Y$ genotype and color blindness.