In sickle-cell anaemia, what proportion of the children of two carrie…

The correct answer is Option (2) → One-fourth

Sickle-cell anaemia is an autosome-linked recessive disorder controlled by one pair of alleles, Hb^A (normal) and Hb^S (mutant). Only Hb^SHb^S shows the disease. A carrier (Hb^AHb^S) looks unaffected but passes the mutant allele to half of its gametes.

• Parents: Hb^AHb^S × Hb^AHb^S
• Gametes of each parent: Hb^A and Hb^S, in equal numbers

	HbA	HbS
HbA	HbAHbA	HbAHbS
HbS	HbAHbS	HbSHbS

• Hb^AHb^A (normal) : Hb^AHb^S (carrier) : Hb^SHb^S (diseased) = 1 : 2 : 1
• Children with the diseased phenotype = 1 out of 4 = one-fourth

Option (1) One-half — Incorrect. One-half is the share of children who are carriers (Hb^AHb^S). They appear unaffected, so they do not show the diseased phenotype.

Option (2) One-fourth — Correct. Only the Hb^SHb^S children, one in four, receive the mutant allele from both parents and show the disease.

Option (3) Three-fourths — Incorrect. Three-fourths is the share of children who appear unaffected (Hb^AHb^A and Hb^AHb^S together), not of those who show the disease.

Option (4) One-third — Incorrect. A one-in-three share does not arise from a single-gene cross of two heterozygotes; the cross gives quarters (1 : 2 : 1).

NCERT Extract:
Mendelian Disorders – “… Sickle-cell anaemia : This is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene (or heterozygous). The disease is controlled by a single pair of allele, Hb^A and Hb^S. Out of the three possible genotypes only homozygous individuals for Hb^S (Hb^SHb^S) show the diseased phenotype. Heterozygous (Hb^AHb^S) individuals appear apparently unaffected but they are carrier of the disease as there is 50 per cent probability of transmission of the mutant gene to the progeny, thus exhibiting sickle-cell trait (Figure 4.15).”