If both parent are bald and their first female child is normal, then the chances of baldness in their second male child is :-

Options

A. 0.5
B. 1
C. 0.25
D. 0

Answer

To solve the problem, we need to analyze the genetics of baldness in the given scenario. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Genetics of Baldness
Baldness is a trait that is dominant in males and recessive in females. This means:
- Males can be bald if they have at least one dominant allele (B).
- Females require two recessive alleles (bb) to be bald.

### Step 2: Determine the Genotypes of the Parents
According to the question:
- Both parents are bald.
- The mother must have the genotype bb (homozygous recessive) since she is bald.
- The father can either be Bb (heterozygous) or BB (homozygous dominant) since he is bald.

### Step 3: Analyze the First Female Child
The first female child is normal, which means she cannot have the genotype bb. The only possible genotype for her is Bb (heterozygous) since she must inherit one b allele from her mother and one B allele from her father.

### Step 4: Determine the Father's Genotype
Since the first female child is Bb, the father must be Bb (heterozygous). If he were BB, all female children would be BB and thus bald.

### Step 5: Set Up a Punnett Square
Now we can set up a Punnett square to determine the possible genotypes of their male children:

| | B | b |
|-----|-----|-----|
| b | Bb | bb |
| b | Bb | bb |

### Step 6: Analyze the Punnett Square Results
From the Punnett square:
- 50% of the offspring (Bb) will be bald (since B is dominant in males).
- 50% of the offspring (bb) will not be bald.

### Step 7: Determine the Chances for the Second Male Child
Since we are interested in the second male child, we look at the results:
- There is a 100% chance that a male child will be bald if he inherits the B allele (Bb).
- Therefore, the chances of baldness in their second male child is 1 (or 100%).

### Conclusion
The chances of baldness in their second male child is 1.

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