Previous year questions: Molecular Basis of Inheritance
Molecular Basis of Inheritance28 min readLesson 8 of 8
Previous year questions: molecular basis of inheritance
Seven questions here are printed board questions from 2023, 2024, 2025 and 2026, and three are from official sample papers. The one-mark items test definitions, experiments and vocabulary; the longer items ask for experiments or processes in sequence. Read the question, name the process or experiment, then write the points in order.
Q1 (CBSE Board 2023, 1 mark)
Identify the region X, the factor Y and the enzyme Z in the schematic of transcription in a prokaryote. Printed options: (a) X Terminator, Y Sigma factor, Z RNA polymerase; (b) X Promoter, Y Rho factor, Z RNA polymerase; (c) X Promoter, Y Sigma factor, Z RNA polymerase; (d) X Promoter, Y Sigma factor, Z DNA polymerase.
Answer: Option (c). X is the promoter, the region where RNA polymerase binds to begin transcription. Y is the sigma factor, which directs the polymerase to the promoter, and Z is RNA polymerase, the enzyme that makes RNA. The terminator lies at the other end of the transcription unit, the rho factor helps termination rather than initiation, and DNA polymerase does not synthesise RNA.
Q2 (CBSE Board 2024, 4 marks)
Read the passage. In eukaryotic cells the average transcription unit along DNA is about 8,000 nucleotides, so the RNA product is also that long. But only about 1,200 nucleotides of that RNA are used to translate an average polypeptide of 400 amino acids. (a) Name the RNA product transcribed from the DNA that translates into a polypeptide of 400 amino acids, and name the enzyme that transcribes this type of RNA. (b) Name and explain the process the 8,000-nucleotide RNA undergoes to produce this polypeptide. (c) Write the number of RNA polymerases involved in transcription in a prokaryote and in eukaryotes. OR (c) Mention the difference in the site of transcription in a prokaryote and a eukaryote.
Answer: (a) The product is messenger RNA, formed by processing of the primary transcript (hnRNA), and the enzyme is RNA polymerase II. (b) The primary transcript undergoes splicing, in which the introns are removed and the exons are joined, along with capping at the 5 prime end and tailing by poly-A at the 3 prime end; the mature mRNA is about 1,200 nucleotides long and codes the 400-amino-acid polypeptide. (c) A prokaryote has one RNA polymerase, while a eukaryote has three, RNA polymerase I, II and III. For the OR part, transcription in a prokaryote occurs in the cytoplasm, where translation can begin before transcription ends, while in a eukaryote transcription occurs in the nucleus and translation occurs in the cytoplasm.
Q3 (CBSE Board 2025, 1 mark)
The process of splicing in eukaryotes represents the dominance of the: (A) DNA world, (B) RNA world, (C) protein world, (D) lipid world.
Answer: Option (B). Splicing is carried out by the spliceosome, in which RNA molecules catalyse the reaction. The catalytic role of RNA in a modern cell is evidence for the RNA world hypothesis, in which RNA carried both information and catalysis before DNA and proteins took over.
Q4 (CBSE Board 2025, 1 mark)
Regulation of the lac operon by the repressor is referred to as: (A) inducible regulation, (B) repressible regulation, (C) negative regulation, (D) positive regulation.
Answer: Option (C). The repressor protein binds the operator and prevents transcription, so the control is negative. The operon is also inducible, because lactose removes the repressor, but the specific mechanism asked here is negative regulation.
Q5 (CBSE Board 2025, 1 mark)
SNPs in the Human Genome Project refers to: (A) polymorphism in repetitive sequences, (B) single-base DNA differences, (C) single changes in a nucleotide of mRNA, (D) loss or gain of a gene function.
Answer: Option (B). SNP stands for single nucleotide polymorphism, a position where a single base differs between individuals. The Human Genome Project identified about 1.4 million such positions.
Q6 (CBSE Board 2026, 1 mark)
The smallest part of a DNA molecule that can be changed by a point mutation is: (A) oligonucleotide, (B) codon, (C) gene, (D) nucleotide.
Answer: Option (D). A point mutation changes a single base pair, so the smallest unit affected is one nucleotide. A codon is a group of three nucleotides, and a gene is much larger.
Q7 (CBSE Board 2026, 1 mark)
Assertion (A): Stretches of repeated DNA are believed to have no direct role in coding. Reason (R): They help in understanding chromosome structure, dynamics and evolution. Choose: (A) both true, R explains A; (B) both true, R does not explain A; (C) A true, R false; (D) A false, R true.
Answer: Option (B). Both statements are true: the repeated stretches do not code, and they are useful for studying chromosome structure, dynamics and evolution. The reason gives another role of these sequences rather than explaining why they lack coding function, so it is not the correct explanation of the assertion.
Q8 (CBSE SQP 2024-25, 1 mark)
In Escherichia coli, the lac operon gets switched on when lactose is: (A) present in the medium and it binds to the repressor, (B) not present in the medium and the repressor binds to the operator, (C) not present in the medium and RNA polymerase binds to the operator, (D) active lactose present in the medium binds to RNA polymerase.
Answer: Option (A). In the presence of lactose, allolactose binds the repressor and prevents it from binding the operator. RNA polymerase can then transcribe the operon. In options (B) and (C) the operon stays off, and RNA polymerase does not bind the operator.
Q9 (CBSE SQP 2022-23, 5 marks)
Study the schematic of the genes of the lac operon and answer: (a) the active site of permease in the cell membrane is blocked by an inhibitor, how will it affect the lac operon? (b) the protein produced by the i gene becomes abnormal, explain its impact on lactose metabolism with the reason. (c) if the nutrient medium contains only galactose, will the operon be expressed? Justify.
Answer: (a) Permease brings lactose into the cell. If its active site is blocked, lactose cannot enter, so allolactose is not formed, the repressor stays on the operator, and the operon is not expressed even when lactose is present outside. (b) The i gene codes the repressor. An abnormal repressor that cannot bind the operator leaves the operon switched on all the time, so the three enzymes are made constitutively and lactose is metabolised continuously; if the abnormal repressor cannot bind the inducer, the operon stays permanently off. (c) Galactose is not the inducer; the inducer is allolactose formed from lactose. With only galactose in the medium, the repressor remains bound to the operator, so the operon is not expressed.
Q10 (CBSE SQP 2025-26, 5 marks)
(a) Construct a complete transcription unit with a promoter and a terminator on the basis of the given hypothetical template strand. (b) How is transcription a more complex process in eukaryotic cells? Explain the additional processes a precursor mRNA has to undergo.
Answer: (a) The template strand is read in the 3 prime to 5 prime direction and the RNA is built in the 5 prime to 3 prime direction. The promoter lies upstream, towards the 5 prime side of the coding strand, and the terminator lies downstream, towards the 3 prime side. The transcription unit therefore consists of the promoter, the structural gene and the terminator, and the construct must show the template strand with its 3 prime end facing the start of transcription. (b) Eukaryotes have three RNA polymerases, each transcribing a different class of RNA, and each needs transcription factors to bind the promoter; transcription happens in the nucleus. The precursor mRNA made by RNA polymerase II undergoes capping at the 5 prime end, tailing with poly-A at the 3 prime end, and splicing, in which introns are removed and exons joined. The processed mRNA is then exported to the cytoplasm for translation.
Seven of these are board questions across four board exam years. The pattern is consistent: identify the experiment or process, then write its steps in order and finish with the conclusion. The one-mark items reward exact vocabulary, and the longer items reward a labelled sequence rather than a paragraph of general statements.