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While studying an autosomal recessive di...

While studying an autosomal recessive disorder in a family . It was noted that husband and wife both did not suffer from disease but their fathers had the disease. What is the probability that the couple will have a girl child without the disorder ?

A

`1/2`

B

`3/4`

C

`3/8`

D

`1/8`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining the probability that a couple will have a girl child without an autosomal recessive disorder, we can break down the solution step by step: ### Step 1: Understand the Genotypes The disorder in question is autosomal recessive. This means: - Individuals with the genotype **AA** (homozygous dominant) are normal. - Individuals with the genotype **Aa** (heterozygous) are carriers and also normal. - Individuals with the genotype **aa** (homozygous recessive) are affected by the disorder. ### Step 2: Determine the Genotypes of the Parents The problem states that both the husband and wife do not suffer from the disease, but their fathers do. This implies: - The fathers of both the husband and wife have the genotype **aa** (affected). - Since the fathers are **aa**, they can only pass on the **a** allele to their children. - Therefore, both the husband and wife must be carriers, having the genotype **Aa** (heterozygous). ### Step 3: Set Up a Punnett Square To find the probability of their offspring's genotypes, we can use a Punnett square. The gametes from each parent (both **Aa**) will be: - Male gametes: A and a - Female gametes: A and a The Punnett square will look like this: | | A | a | |-------|-------|-------| | **A** | AA | Aa | | **a** | Aa | aa | ### Step 4: Analyze the Outcomes From the Punnett square, we can see the possible genotypes of the offspring: - **1 AA** (normal) - **2 Aa** (carriers, normal) - **1 aa** (affected) This gives us a total of 4 outcomes: - 3 out of 4 (AA + Aa + Aa) are normal (no disorder). - 1 out of 4 (aa) is affected. ### Step 5: Calculate the Probability of Having a Girl Child The question asks for the probability of having a girl child without the disorder. We know: - The probability of having a normal child is **3/4**. - The probability of having a girl child is **1/2**. ### Step 6: Combine the Probabilities To find the probability of having a girl child without the disorder, we multiply the two probabilities: \[ \text{Probability of normal child} \times \text{Probability of girl child} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \] ### Final Answer The probability that the couple will have a girl child without the disorder is **3/8**. ---
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