Q.If haploid number in a cell is 18, the double monosomic and trisomic number will be
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Start your 14-day free trial to unlock the full solution →Step 1. Find the diploid number. Haploid number n = 18, so diploid 2n = 36.
Step 2. Compute the trisomic number. A trisomic individual carries one extra chromosome for one homologous pair: trisomic = 2n + 1 = 36 + 1 = 37.
Step 3. Compute the double-monosomic number. 'Double monosomic' means the individual is simultaneously monosomic (missing one homolog) for TWO DIFFERENT chromosome pairs, so two chromosomes are missing in total: double monosomic = 2n − 1 − 1 = 2n − 2 = 36 − 2 = 34. (This matches this same chapter's own aneuploidy classification table, where 'two individual chromosomes lose from diploid' is defined as double monosomy.)
Step 4. Compare against the printed options. The mathematically correct pair is 34 (double monosomic) and 37 (trisomic). (a) 35 and 37 — 35 is a simple monosomic number (2n−1), not the double-monosomic number, so its first value is wrong. (b) 34 and 35 — its first value (34) is exactly the correct double-monosomic number; its second value (35) is a simple monosomic number, not the trisomic number (which should be 37). (c) 37 and 35 — correctly has 37 (trisomic) but pairs it with 35 (monosomic) rather than with the double-monosomic figure, and lists 37 first rather than matching the question's 'double monosomic and trisomic' order. (d) 17 and 19 — far too small (close to the haploid number, …
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