In humans, red-green color blindness is determined by an X-linked recessive allele (a), whereas eye color is determined by an autosomal gene, where brown (B) is dominant over blue (b). a. What gametes can be formed with respect to these genes by a heterozygous, brown-eyed, color-blind male? b. If a blue-eyed mother with normal vision has a brown-eyed, color-blind son and a blue- eyed, color-blind daughter, what are the genotypes of both parents and children? If the dominant allele A is negeceary fos oarino.

1. Gender: Male --> XY, Color-blind male: XaY, Brown-eyed: BB or Bb. Combine the aforementioned, the gametes can be formed with respect to these genes by a heterozygous, brown-eyed, color-blind male are: XaY,BB and XaY,Bb

2. MOTHER: Blue-eyed female with normal vision : XAXA,bb or XAXa,bb

SON: brown-eyed male, color-blind: XaY,BB or XaY,Bb

DAUGHTER: blue- eyed female, color-blind: XaXa,bb

From the genotype of the son, we can assume that the father must has brown eyes: BB or Bb. Besides, the daughter has blue eyes. Thus, the daughter received the autosomal gene "b" from her parents. At the conclusion, the father eyes color genes is Bb. We can assume that Father genre can be XAY,Bb or XaY,Bb.

Both of son and daughter are color-blind. Thus, Mother genotypes should have Xa. Finally, Mother Genotypes is: XAXa,bb.

Because both children are color-blind, the father genre is: XaY,Bb, brown eyes and color-blind male

Mother (XAXa,bb) × Father (XaY,Bb)

Son 1: (XAY,Bb) clear vision and brown eyes
Son 2: (XAY,bb) clear vision and blue eyes
Son 3: (XaY,Bb) color blindness and brown eyes
Son 4: (XaY,bb) color blindness and blue eyes

Daughter 1: (XAXa,Bb) clear vision and brown eyes
Daughter 2: (XAXa,bb) clear vision and blue eyes
Daughter 3: (XaXa,Bb) color blindness and brown eyes
Daughter 4: (XaXa,bb) color blindness and blue eyes