NEET (UG) · Botany, Class 12
Molecular Basis of Inheritance: NEET PYQ, 2018–2026
Molecular Basis of Inheritance (Botany, Class 12) has been asked 43 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 6.1 questions a paper, 13% of all Botany questions. It is the most-asked of 17 Botany chapters. NEET 2026 had 2 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 6 |
| 2020 | 4 |
| 2021 | 9 |
| 2023 | 8 |
| 2024 | 6 |
| 2025 | 8 |
| 2026 | 2 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Transcription and translation22
- DNA structure11
- Lac operon5
- Human genome project2
- Replication2
- DNA fingerprinting1
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10 Molecular Basis of Inheritance questions to practise.
Our own questions, written in the NEET pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Botany · Molecular Basis of Inheritance
A tRNA has the anticodon 3′-UAC-5′. Written 5′ to 3′, which codon on the mRNA does it read?
Answer: (D)
Codon and anticodon pair antiparallel. Anticodon 3′-UAC-5′ pairs with codon 5′-AUG-3′: U pairs with A, A with U, C with G. Write the anticodon 3′→5′ alongside the codon 5′→3′ and check the complements. AUG is also the methionine codon.
Q2 · Botany · Molecular Basis of Inheritance
After hybridisation with a radioactive probe, the membrane is exposed to X-ray film to produce an autoradiograph. Why do the dark bands on the film mark the positions of the probe-bound DNA fragments?
Answer: (A)
The probe is radioactively labelled. After hybridisation and washing away unbound probe, only the fragments carrying the complementary repeat remain labelled. Placing the membrane against X-ray film lets the emitted radiation expose the film at those positions, producing dark bands. The band pattern therefore reflects where labelled probe is bound, not the DNA mass or any colour change.
Q3 · Botany · Molecular Basis of Inheritance
A wild-type E. coli culture is grown on glucose and lactose together. Glucose is consumed first, and lac gene transcription stays low until glucose is largely depleted, even though lactose is present throughout. Which explanation best accounts for the low transcription during the glucose phase?
Answer: (C)
Principle: the lac operon needs two signals for high expression. Allolactose (inducer) removes the repressor from the operator, which is the negative control. Catabolite repression is a positive control: cAMP binds CAP, and the cAMP-CAP complex binds upstream of the promoter to help RNA polymerase bind. Glucose lowers cAMP, so CAP is unbound and transcription stays low even with the operator free. Here lactose is present and the repressor is inactivated, so the missing element is the positive signal. Expression rises when glucose is exhausted and cAMP increases. Glucose does not modify the repressor, block lactose uptake (permease is already expressed), or degrade DNA.
Q4 · Botany · Molecular Basis of Inheritance
What does the abbreviation SNP stand for in the study of human genome variation, and what is one major use of SNPs?
Answer: (A)
Principle: a polymorphism is a variation present in a population, and a single nucleotide polymorphism is a change at one base position. Such variations are inherited and can be tracked, so they act as markers near disease genes. Hence the correct expansion and use are single nucleotide polymorphism and disease gene tracing.
Q5 · Botany · Molecular Basis of Inheritance
In prokaryotes, which feature of the nascent RNA causes transcription to terminate without the help of the rho factor?
Answer: (A)
Principle: prokaryotes have two termination mechanisms. In rho-independent termination, a GC-rich inverted repeat in the RNA folds into a hairpin, which stalls RNA polymerase. A string of uracils then gives weak U-A pairing between RNA and DNA, so the transcript is released. Apply this to the stem: the feature must be in the RNA itself and must not involve rho.
Q6 · Botany · Molecular Basis of Inheritance
During DNA replication, the lagging strand is synthesised as short Okazaki fragments, while the leading strand is synthesised as one continuous chain. Which statement correctly explains this difference?
Answer: (A)
DNA polymerase synthesises a new strand only in the 5'→3' direction. At the fork the two parental strands are antiparallel. The leading strand is synthesised continuously towards the fork. The lagging strand's template runs the other way, so its new strand must be made in short segments, each started by a primer and extending away from the fork. These are Okazaki fragments. After DNA polymerase I replaces the primers, DNA ligase seals the remaining nicks into a continuous strand. Students who recall 'lagging = fragments' without the directional reason, or who swap the roles of helicase, topoisomerase and ligase, pick the distractors.
Q7 · Botany · Molecular Basis of Inheritance
Which codon codes for methionine and also acts as the initiator codon?
Answer: (D)
AUG has dual functions: it codes for methionine and it also acts as the initiator codon. UAA, UAG and UGA are the stop (terminator) codons.
Q8 · Botany · Molecular Basis of Inheritance
In the Southern-blot step of DNA fingerprinting, the radioactive probe used for hybridisation is a single-stranded DNA sequence that is complementary to the repeat unit. Why must the probe be single-stranded?
Answer: (A)
After electrophoresis, fragments are denatured and transferred to a membrane, so the target DNA is single-stranded. Hybridisation depends on the probe pairing with its complementary target sequence. A single-stranded probe of the repeat's complementary sequence can anneal to the matching target fragments, and the label then reveals where the probe bound on the autoradiograph.
Q9 · Botany · Molecular Basis of Inheritance
A probe made from a short core repeat sequence of 16 base pairs is hybridised to a Southern blot of genomic DNA digested with a restriction enzyme, and many bands appear in each lane. Which property of the probe most directly explains this?
Answer: (C)
Principle: a probe recognises its complementary sequence wherever it occurs. A single-locus probe targets one unique repeat array, while a multi-locus probe targets a core sequence shared by many arrays. Apply it: a 16-bp core sequence is present at many loci, so the probe hybridises to fragments from many of them, each with a different length. The probe's strandedness and the enzyme's cut sites do not generate multiple bands.
Q10 · Botany · Molecular Basis of Inheritance
During processing of eukaryotic hnRNA to mature mRNA in the nucleus, which segments are removed by splicing?
Answer: (A)
The hnRNA transcript undergoes three modifications before it becomes mature mRNA: capping at the 5′ end, tailing at the 3′ end, and splicing. Splicing removes introns and joins exons. The cap and poly-A tail are added to the mRNA and are not removed.
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All 43 real NEET questions from Molecular Basis of Inheritance.
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About Molecular Basis of Inheritance in NEET.
How many questions come from Molecular Basis of Inheritance in NEET?
Molecular Basis of Inheritance (Botany, Class 12) has been asked 43 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 6.1 questions a paper, 13% of all Botany questions. It is the most-asked of 17 Botany chapters. NEET 2026 had 2 from it.
Which topics of Molecular Basis of Inheritance are asked most in NEET?
Transcription and translation (22), DNA structure (11), Lac operon (5), Human genome project (2), counted over the papers from 2018–2026.
Is Molecular Basis of Inheritance important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 1 of 17 in Botany; use that to decide how much time it gets, not whether to study it.
Where can I solve Molecular Basis of Inheritance NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET question from Molecular Basis of Inheritance with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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