Red-green color blindness is inherited through an X-linked, recessive allele (b). Two parents, Fred and Ginger, have normal vision. They have two daughters, Takiyah and Kelly, who also have normal vision, and a color-blind son, David. Daughter Kelly has a color-blind son, Kevin. Daughter Takiyah has five sons, all with normal vision. What are the genotypes of all the individuals? Show all your work! Fred Ginger David Takiyah Kelly Kevin Takiyah’s five sons If Kelly marries a man with normal vision, what is the probability that she’ll have a color-blind son? a color-blind daughter?
The Correct Answer and Explanation is :
To determine the genotypes of the individuals and answer the probability questions, let’s break it down step by step.
Key Facts about X-Linked Recessive Inheritance:
- The gene for red-green color blindness is located on the X chromosome.
- Males have one X and one Y chromosome (XY). If their X chromosome carries the recessive allele bb, they will be color-blind.
- Females have two X chromosomes (XX). They must inherit two recessive alleles bb to be color-blind; if they inherit only one, they are carriers (heterozygous) but not color-blind.
Genotypes of Individuals:
- Fred (Father): Male with normal vision. His genotype is XYX^Y, where XX is the normal allele.
- Ginger (Mother): Female with normal vision, but since she has a color-blind son (David), she must be a carrier. Her genotype is X^X^b.
- David (Son): Male and color-blind. His genotype is XbYX^bY (he inherits XbX^b from his mother and YY from his father).
- Takiyah (Daughter): Female with normal vision. She inherits one XX from her father and XbX^b from her mother. Her genotype is X^X^b (carrier).
- Kelly (Daughter): Female with normal vision. Like Takiyah, her genotype is X^X^b (carrier).
- Kevin (Kelly’s Son): Male and color-blind. His genotype is XbYX^bY.
- Takiyah’s five sons: All have normal vision. Their genotype is XYXY (inheriting XX from Takiyah’s normal allele X^).
Probabilities if Kelly marries a man with normal vision:
- Color-Blind Son: Kelly’s genotype is X^X^b. The probability she passes XbX^b is 12\frac{1}{2}. A normal father contributes YY. Thus, the probability of a color-blind son is: 12×12=14.\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.
- Color-Blind Daughter: For a daughter to be color-blind, Kelly must pass XbX^b (12\frac{1}{2}), and the father must also contribute XbX^b (00, as he only has XX). Thus, the probability is: 0.0.
Explanation (300 Words):
Red-green color blindness is inherited through an X-linked recessive allele, which affects males more frequently because they only need one copy of the defective XbX^b allele. Females need two copies to express the condition. In this problem, both parents, Fred and Ginger, have normal vision, but Ginger is a carrier (X^X^b), as evidenced by her color-blind son, David. Their daughters, Takiyah and Kelly, inherited XbX^b from Ginger but XX from Fred, making them carriers.
Kelly’s son Kevin inherited XbX^b from her and is color-blind. Takiyah’s sons inherited only her normal XX allele, resulting in normal vision.
If Kelly marries a man with normal vision (XYXY), their sons will inherit YY from the father and either XX or XbX^b from Kelly, giving a 25% chance of a color-blind son. Their daughters will inherit XX from the father and either XX or XbX^b from Kelly, but none will be color-blind because the father’s normal XX protects them.
This inheritance pattern highlights how X-linked traits like red-green color blindness are passed through carrier females and predominantly affect males.