Question
Night Blindness is x linked recessive. A normal woman whose father was night blind marries a night blind man.
What are the chances that their child (regardless of gender) is night blind? %
What are the chances they have a son?
What are the chances that their child (regardless of gender) is night blind? %
What are the chances they have a son?
Answers
GPT-4o mini
Night blindness, being an X-linked recessive trait, is carried on the X chromosome. To solve this problem, we need to analyze the genetic makeup of both parents.
1. **Mother's Genotype**: The mother is normal (not night blind) but her father is night blind. Since a male with night blindness has the genotype X^nY (where X^n denotes the X chromosome carrying the night blindness allele), the mother must have inherited her X chromosome from him. Thus, she must be a carrier with the genotype X^nX (one X chromosome with the night blindness allele from her father and one normal X chromosome).
2. **Father's Genotype**: The father is night blind, which gives him the genotype X^nY.
To understand their offspring’s chances of being night blind, we can use a Punnett square to determine the genetic combinations possible:
```
X^n | Y
--------------------
X^n | X^nX^n | X^nY
--------------------
X | XX | XY
```
From this setup, the possible genotypes of their children are:
- X^nX^n (female, night blind)
- XX (female, normal)
- X^nY (male, night blind)
- XY (male, normal)
Now, summing up the combinations:
- **Children that are night blind**:
- Female: X^nX^n (1 way)
- Male: X^nY (1 way)
So, there are 2 children who can be night blind out of 4 total possible combinations.
Thus, the probability that their child (regardless of gender) will be night blind is:
\[
\text{Chance of being night blind} = \frac{2}{4} = 0.5 \text{ or } 50\%
\]
Next, to find the probability of having a son:
In the Punnett square, the combinations are:
- XX (female)
- X^nX^n (female)
- XY (male)
- X^nY (male)
Out of the 4 total combinations, 2 are males (XY and X^nY). So, the probability of having a son is:
\[
\text{Chance of having a son} = \frac{2}{4} = 0.5 \text{ or } 50\%
\]
So, the final answers are:
- The chances that their child (regardless of gender) is night blind: **50%**
- The chances they have a son: **50%**
1. **Mother's Genotype**: The mother is normal (not night blind) but her father is night blind. Since a male with night blindness has the genotype X^nY (where X^n denotes the X chromosome carrying the night blindness allele), the mother must have inherited her X chromosome from him. Thus, she must be a carrier with the genotype X^nX (one X chromosome with the night blindness allele from her father and one normal X chromosome).
2. **Father's Genotype**: The father is night blind, which gives him the genotype X^nY.
To understand their offspring’s chances of being night blind, we can use a Punnett square to determine the genetic combinations possible:
```
X^n | Y
--------------------
X^n | X^nX^n | X^nY
--------------------
X | XX | XY
```
From this setup, the possible genotypes of their children are:
- X^nX^n (female, night blind)
- XX (female, normal)
- X^nY (male, night blind)
- XY (male, normal)
Now, summing up the combinations:
- **Children that are night blind**:
- Female: X^nX^n (1 way)
- Male: X^nY (1 way)
So, there are 2 children who can be night blind out of 4 total possible combinations.
Thus, the probability that their child (regardless of gender) will be night blind is:
\[
\text{Chance of being night blind} = \frac{2}{4} = 0.5 \text{ or } 50\%
\]
Next, to find the probability of having a son:
In the Punnett square, the combinations are:
- XX (female)
- X^nX^n (female)
- XY (male)
- X^nY (male)
Out of the 4 total combinations, 2 are males (XY and X^nY). So, the probability of having a son is:
\[
\text{Chance of having a son} = \frac{2}{4} = 0.5 \text{ or } 50\%
\]
So, the final answers are:
- The chances that their child (regardless of gender) is night blind: **50%**
- The chances they have a son: **50%**