Question

If a man and a woman are heterozygous for a gene, and if they have three children, what is the chance that all three will also be heterozygous

• A. 1/8
• B. 2/8
• C. 3/8
• D. 3/4

Hint:

Always use the product rule for independent "and" events, and the sum rule for mutually exclusive "or" events.
Here, the birth of each child is independent, so we multiply: \( (1/2)^3 = 1/8 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When studying single-gene inheritance (Mendelian genetics), we can determine the genotypic probabilities of the offspring using a Punnett square.
Since the birth of each child is an independent event, we apply the product rule of probability to calculate the joint probability of multiple independent outcomes.
Key Formula or Approach:
For independent events \( E_1, E_2, \dots, E_n \), the joint probability is:
\[ P(E_1 \text{ and } E_2 \text{ and } \dots \text{ and } E_n) = P(E_1) \times P(E_2) \times \dots \times P(E_n) \]

Step 2: Detailed Explanation:
Let the heterozygous parents have the genotype \( Aa \).
The cross between them is:
\[ Aa \times Aa \] The expected genotypic ratio of their offspring is:
- \( 1/4 \) Homozygous Dominant (\( AA \))
- \( 1/2 \) Heterozygous (\( Aa \))
- \( 1/4 \) Homozygous Recessive (\( aa \))
Thus, the probability of any single child being heterozygous is:
\[ P(\text{Heterozygous}) = \frac{1}{2} \] Since the genotypes of the three children are independent events, the probability that all three will be heterozygous is:
\[ P(\text{All 3 heterozygous}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{8} \]

Step 3: Final Answer:
The probability that all three children will be heterozygous is 1/8.