Question

Select the correctly matched pair about sickle cell anaemia: Genotype : Phenotype
• [(A)] \(Hb^A Hb^A\) : Diseased phenotype
• [(B)] \(Hb^A Hb^S\) : Diseased phenotype
• [(C)] \(Hb^S Hb^S\) : Diseased phenotype
• [(D)] \(Hb^S Hb^A\) : Carrier of disease Choose the correct answer from the options given below:

• A. \((C)\ \text{and}\ (D)\ \text{only}\)
• B. \((A)\ \text{and}\ (C)\ \text{only}\)
• C. \((B),\ (C)\ \text{and}\ (D)\ \text{only}\)
• D. \((A),\ (B)\ \text{and}\ (C)\ \text{only}\)

Hint:

\[ Hb^S Hb^S \Rightarrow \text{Diseased} \] \[ Hb^A Hb^S \Rightarrow \text{Carrier} \]

Answer 1

The Correct Option is A

Sickle cell anaemia is an autosomal recessive genetic disorder caused by abnormal haemoglobin: \[ Hb^S \]
Step 1: Analyze genotype \(Hb^A Hb^A\)
• Both alleles are normal.
• Individual is healthy. Thus: \[ (A)\ \text{is incorrect} \]
Step 2: Analyze genotype \(Hb^A Hb^S\)
• Individual is heterozygous.
• Such individuals are carriers of disease.
• They generally do not show diseased phenotype. Thus: \[ (B)\ \text{is incorrect} \]
Step 3: Analyze genotype \(Hb^S Hb^S\)
• Both alleles are sickle-cell alleles.
• Individual shows diseased phenotype. Thus: \[ (C)\ \text{is correct} \]
Step 4: Analyze genotype \(Hb^S Hb^A\)
• This is heterozygous condition.
• Individual acts as carrier of disease. Thus: \[ (D)\ \text{is correct} \] Therefore, the correctly matched pairs are: \[ (C)\ \text{and}\ (D) \] Hence: \[ \boxed{\text{(A)}} \]