Answer:
1. XaY,BB and XaY,Bb
2. Father: XaY,Bb
Mother: XAXa,bb
Son: XaY,Bb
Daughter: XaXa,bb
Explanation:
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
