Match List I with List II
| List I | List II |
|---|---|
| A. Phenylketonuria | I. Sex linked recessive disease |
| B. Down’s syndrome | II. Autosomal Recessive disease |
| C. Turner’s syndrome | III. Trisomy of chromosome number 21 |
| D. Haemophilia | IV. Monosomy of X chromosome |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-III, C-IV, D-I
(A) Phenylketonuria — this matches (II) Autosomal Recessive disease. The person lacks the enzyme that converts phenylalanine into tyrosine. The gene is on an autosome, and the disease shows only when both alleles are defective.
(B) Down’s syndrome — this matches (III) Trisomy of chromosome number 21. The person has an extra copy of chromosome 21, so 47 chromosomes in all.
(C) Turner’s syndrome — this matches (IV) Monosomy of X chromosome. The female has only one X chromosome (45 with X0) and has sterile ovaries.
(D) Haemophilia — this matches (I) Sex linked recessive disease. The gene for a blood-clotting protein lies on the X chromosome. A carrier mother passes it to half her sons, who then bleed for a long time even from a small cut.
Note: A similar question was also asked in CUET (UG) 2026 on 07-06-2026 (Slot 2), and in CUET (UG) 2025 on 29-05-2025 (Slot 1).