Question

A color-blind woman marries a man who is not color-blind and they have six children ( 3 boys and 3 girls) All of the sons and none of the daughters are color blind.

a. What does this tell you about the color blind allele?

b. Draw the pedigree of this family, using correct symbols and all of the genotypes you can determine with certainty.

c. One of the daughters marries a man who can see color. State the phenotypic and genotypic ratios you would expect to see for both boys and girls in their offspring.

Answer 1

a.

Since all the sons of the color blind woman are themselves color blind, but none of the daughters are, the color blind allele is expressed as a recessive X linked allele. This is because all the sons receive an X chromosome from their mother. The daughters also receive an X chromosome from their mother, but they also get a second, non-mutated X from their father, who is not color blind.

b.

c.

The daughters for this couple are all carriers for the color blindness allele. Since the man one of the daughters marries does not have color blindness, he has a normal X chromosome, without the allele for color blindness.

The daughter's genotype -> XX^c
Her husband's genotype -> XY

The daughter's gametes will either carry an X or Xc allele and the husband's gametes will carry either a normal X or a Y chromosome.

Punnet's square for the couple

		Maternal Gametes
		X	Xc
Paternal Gametes	X	XX	XXc
	Y	XY	XcY

Of the couple's boys,
Genotypic Ratio --> XY : X^cY :: 1:1
Phenotypic Ratio --> Normal : Affected :: 1:1

Of the couple's girls,
Genotypic Ratio --> XX : XX^c:: 1:1
Phenotypic Ratio --> Normal Non-carriers : Normal Carriers :: 1:1