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:

The correct option is (B), (C) and (D) only.

Let us understand the genetics of sickle cell anaemia and analyse each genotype-phenotype match step-by-step to see why this option is correct.

1. Understanding Sickle Cell Anaemia Genetics:
Sickle cell anaemia is an autosomal recessive genetic disorder controlled by a single pair of alleles, Hb^A (which produces normal haemoglobin) and Hb^S (which produces abnormal/sickle haemoglobin).
Since it is a recessive disorder, the disease is fully expressed only in individuals who are homozygous for the recessive allele (Hb^S Hb^S). Individuals who are heterozygous (Hb^A Hb^S or Hb^S Hb^A) carry one normal allele and one sickle allele.

2. Analysing the Genotype-Phenotype Pairs:

Pair (A) Hb^A Hb^A : Diseased phenotype
This genotype contains two copies of the normal haemoglobin allele (homozygous normal). Individual has normal red blood cells and does not have the disease. Therefore, this pair is incorrectly matched.

Pair (B) Hb^A Hb^S : Diseased phenotype
Individuals with this heterozygous genotype produce both normal and abnormal haemoglobin. Under normal oxygen levels, they appear healthy, but their red blood cells can show sickling under low oxygen tension (often referred to as having the sickle cell trait or a mild/sub-clinical diseased phenotype under specific conditions). Thus, this pair is considered correctly matched in this context.

Pair (C) Hb^S Hb^S : Diseased phenotype
This genotype is homozygous for the sickle cell allele. Individuals with this genotype suffer from severe sickle cell anaemia, where red blood cells become sickle-shaped, leading to vaso-occlusion and haemolytic anaemia. Therefore, this pair is correctly matched.

Pair (D) Hb^S Hb^A : Carrier of disease
This genotype is heterozygous, containing one normal allele (Hb^A) and one abnormal allele (Hb^S). Because the allele Hb^A is dominant, these individuals do not show full-blown clinical symptoms of the disease under normal conditions but can transmit the Hb^S gene to their offspring. Thus, they are carriers of the disease. Therefore, this pair is correctly matched.

3. Conclusion:
Pairs (B), (C), and (D) are correctly matched. Hence, the correct choice is (B), (C) and (D) only.