Weighted to match the chapter's real NEET profile: transcription and replication heaviest, all three numerical types covered, and every current question format represented.
Every numerical answer verified by computation before being written into the question.
How to use this file. Cover the solution and attempt each question closed-book first, giving yourself
about 60 seconds per question — 20 minutes for the set. Then read the card. The last two boxes of
every card, Shortcut and Where it goes wrong, are the parts worth re-reading later; the working
above them is there for the first pass only.
A DNA molecule contains 6.6 × 10⁹ base pairs. Its approximate length is:
2.2 mm
2.2 m
22 m
0.22 m
GivenA double-stranded DNA molecule of 6.6 × 10⁹ base pairs.
AskedThe length of the molecule.
Concept to useIn B-form DNA consecutive base pairs are stacked a fixed distance apart, so length is nothing more than a count multiplied by a spacing.
Formula or rulelength = number of base pairs × 0.34 nm
Baby steps
Write the spacing in metres first: 0.34 nm = 3.4 × 10⁻¹⁰ m.
Set up the product: 6.6 × 10⁹ × 3.4 × 10⁻¹⁰ m.
Powers of ten: 10⁹ × 10⁻¹⁰ = 10⁻¹.
Digits: 6.6 × 3.4 = 22.44.
Combine: 22.44 × 10⁻¹ = 2.244 m.
Answer(B) About 2.2 m — which is exactly the figure NCERT quotes for the DNA in one human cell.
Why not the others
2.2 mm comes from treating a nanometre as 10⁻⁶ m instead of 10⁻⁹ m.
22 m and 0.22 m are each one power of ten out — the classic slip when the exponent is done last.
ShortcutWhen the base-pair count is quoted as “× 10⁹”, the answer comes out directly in metres: just multiply the coefficient by 0.34. Here 6.6 × 0.34 = 2.24, so 2.24 m. No conversion needed at all.
Where it goes wrongThis is a powers-of-ten question wearing a Biology costume. Convert 0.34 nm into metres before multiplying, never after. Every wrong option here is the right digits with the wrong exponent.
Q2. DNA structure
A double-stranded DNA segment has 23,000 base pairs. The number of complete helical turns and the total length are:
2300 turns and 7.82 μm
23,000 turns and 7.82 μm
2300 turns and 0.782 μm
230 turns and 7.82 μm
GivenA double-stranded DNA of 23,000 base pairs.
AskedNumber of complete turns, and the length.
Concept to useB-form DNA has two fixed geometric constants: ten base pairs make one turn, and each base pair adds 0.34 nm of length.
Formula or ruleturns = bp ÷ 10 and length = bp × 0.34 nm
Baby steps
Turns: 23,000 ÷ 10 = 2300.
Length: 23,000 × 0.34 nm = 7820 nm.
Convert: 1 μm = 1000 nm, so 7820 nm = 7.82 μm.
Cross-check by the other route: 2300 turns × 3.4 nm per turn = 7820 nm. Same answer.
Answer(A) 2300 turns and 7.82 μm.
Why not the others
23,000 turns treats every base pair as a turn.
0.782 μm divides the length by ten as well as the turns.
230 turns divides by 100 instead of 10.
ShortcutTwo routes to the same length — bp × 0.34 or turns × 3.4. Doing both takes five seconds and catches any slip immediately.
Where it goes wrongMixing the two spacings. 0.34 nm belongs to one base pair; 3.4 nm belongs to one turn. Write the unit beside the number every single time and they cannot be swapped.
Q3. DNA structurePYQ shape: AIPMT 2015
If a double-stranded DNA has 20 per cent cytosine, the percentage of adenine in it is:
20%
40%
30%
60%
GivenDouble-stranded DNA in which cytosine is 20% of the bases.
AskedThe percentage of adenine.
Concept to useChargaff's rules. In double-stranded DNA the two members of each pair are always present in equal amounts.
Formula or ruleA = T, G = C, and A + T + G + C = 100%
Baby steps
C = 20%, so G = 20%.
Together G + C = 40%.
The rest is A and T: 100 − 40 = 60%.
A and T are equal, so A = 60 ÷ 2 = 30%.
Answer(C) 30%.
Why not the others
20% sets A equal to C, which Chargaff never says.
40% is G + C.
60% is A + T — the answer you get if you stop one step early.
ShortcutOne line: A = (100 − 2 × given) ÷ 2. Here (100 − 40) ÷ 2 = 30.
Where it goes wrongForgetting the final halving. That produces 60%, and 60% is sitting in the option list precisely because it is the commonest stopping point.
Q4. DNA structure
In a sample of double-stranded DNA, the ratio (A + T)/(G + C) is 1.5. The percentage of guanine is:
30%
40%
60%
20%
GivenDouble-stranded DNA with (A + T)/(G + C) = 1.5.
AskedThe percentage of guanine alone.
Concept to useThe ratio tells you how the 100% splits between the two pairs; Chargaff then splits each pair down the middle.
Formula or rule(A + T) + (G + C) = 100, with A = T and G = C
Baby steps
Say which group is bigger, in words, before any arithmetic. A + T is on top of the fraction, so A + T is the larger group.
Let G + C = y. Then A + T = 1.5y.
1.5y + y = 100 → 2.5y = 100 → y = 40.
So G + C = 40%.
G and C are equal, so G = 20%.
Answer(D) 20%.
Why not the others
30% is what you get by flipping the ratio the wrong way up — and it is also the correct value for adenine, so it feels right twice over. It is the trap option.
40% is G + C, not G.
60% is A + T.
ShortcutWith r = (A+T)/(G+C): G + C = 100/(r + 1). Here 100/2.5 = 40, so G = 20. Works for any ratio given in this form.
Where it goes wrongThis is the ratio-reversal family, in its Biology form. The fix is mechanical: before substituting anything, write the sentence “the group on top is the bigger group” on the paper. The exam has deliberately placed the flipped answer among the options.
Q5. ReplicationPYQ shape: NEET 2019
E. coli grown for many generations on ⁵N medium is transferred to ¹⁴N medium. After two generations, the proportion of DNA molecules that are hybrid is:
100%
50%
25%
0%
GivenAll DNA initially heavy (⁵N); growth continues in light (¹⁴N) medium for two generations.
AskedPercentage of hybrid (one heavy strand, one light strand) molecules.
Concept to useReplication is semiconservative, so the two original strands are never destroyed. They simply end up in two different molecules and stay there forever.
Formula or rulehybrid % = (2 ÷ 2ⁿ) × 100, where n is the number of generations
Baby steps
After n = 2 generations the total number of molecules is 2² = 4.
Exactly two of the original heavy strands exist, and each sits in one molecule.
So hybrid molecules = 2.
2 ÷ 4 = 0.5, that is 50%.
Answer(B) 50%.
Why not the others
100% is the picture after one generation, not two.
25% is generation three.
0% would require the conservative model, which this experiment ruled out.
ShortcutThe hybrid count is frozen at 2 forever; only the total doubles. So the percentage just halves each generation: 100, 50, 25, 12.5. Count the generations and halve that many times from 100.
Where it goes wrongTwo counting errors here. First, generation 0 is the parent — “two generations” means n = 2, giving four molecules. Second, hybrid molecules never disappear; if an option says 0%, it is testing whether you believed the old strands got used up.
Q6. Replication
Assertion (A): On one of the two templates at a replication fork, the new DNA strand is made in short, separate pieces. Reason (R): DNA polymerase can add a nucleotide only to a free 3′-OH group.
Both A and R are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A
A is true but R is false
A is false but R is true
GivenAn assertion about the lagging strand and a reason about the polymerase's chemistry.
AskedWhether each is true, and whether the reason explains the assertion.
Concept to useAssertion–reason is never read as one sentence. It is three separate judgements made in a fixed order.
Formula or ruleStep 1: is A true? Step 2: is R true? Step 3: does R cause A?
Baby steps
A alone. True. That is the lagging strand, built as Okazaki fragments.
R alone. True. This is the fundamental constraint on every DNA polymerase.
Does R cause A? Yes. Because the enzyme can only extend a 3′-OH, it can only build 5′→3′. The two templates are antiparallel, so only one of them can be read smoothly toward the opening fork; on the other the enzyme must keep restarting as fresh template appears, and that produces fragments.
Both true, and R explains A.
Answer(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
Option B would need the two statements to be unrelated, but the polymerase constraint is precisely the cause.
Options C and D each declare one of two true statements false.
ShortcutIn this chapter, whenever the reason is the “5′→3′ only” rule and the assertion is anything about the lagging strand, Okazaki fragments or ligase, the answer is (A). It is the chapter's most repeated assertion–reason shape.
Where it goes wrongAssertion–reason instability. The failure is almost never a knowledge gap; it is reading the pair as a single flowing sentence and answering on impression. Write a small T or F beside each of the two lines on the question paper before looking at the four options. That single habit stabilises the format.
Q7. Replication
Consider the following statements about DNA replication in E. coli. I. Deoxyribonucleoside triphosphates act both as substrate and as a source of energy. II. The entire DNA molecule unwinds before synthesis begins. III. Replication begins at a single origin of replication. IV. DNA ligase joins the fragments made on the lagging strand. How many of these are correct?
One
Two
Three
Four
GivenFour statements about bacterial replication.
AskedHow many are correct.
Concept to useMulti-statement items are four true/false questions charging one mark. Each clause is judged on its own.
Formula or ruleMark each statement T or F on the paper, then count the Ts.
Baby steps
I — True. The two terminal phosphates released from a dNTP supply the energy for the join, and the remaining nucleotide is the building block.
II — False. Opening the whole molecule would cost an impossible amount of energy. A small opening, the replication fork, travels along instead.
III — True. Bacteria have one origin; eukaryotic chromosomes have many.
IV — True. Ligase seals the nicks between Okazaki fragments.
Three true.
Answer(C) Three.
Why not the others
One and Two undercount — usually the result of doubting statement I.
Four accepts statement II, which is the fixed distractor of this topic.
ShortcutAny statement in this chapter claiming that the whole DNA molecule unwinds is false. It is a stock false clause and appears again and again.
Where it goes wrongCounting in your head. Put a tick or cross next to each Roman numeral on the paper. Losing track of one statement in four is the standard way this format is failed, and it is entirely mechanical.
Q8. TranscriptionPYQ shape: NEET 2023
The coding strand of a transcription unit reads 5′-ATGCATGCATGC-3′. The mRNA transcribed from this unit is:
5′-AUGCAUGCAUGC-3′
5′-UACGUACGUACG-3′
3′-AUGCAUGCAUGC-5′
5′-CGUACGUACGUA-3′
GivenThe coding strand, written 5′→3′.
AskedThe mRNA sequence, with its polarity.
Concept to useThe template strand is the one copied, but the RNA that results is identical to the coding strand except that uracil replaces thymine. So the coding strand is a direct read-off.
Formula or rulemRNA = coding strand, with every T replaced by U, and the same 5′→3′ polarity
Baby steps
Check the label: this is the coding strand, so no complementing is needed.
Copy it letter for letter: A T G C A T G C A T G C.
Swap every T for U: A U G C A U G C A U G C.
Keep the polarity as it was: 5′-AUGCAUGCAUGC-3′.
Answer(A) 5′-AUGCAUGCAUGC-3′
Why not the others
Option B is what you get by complementing when you should not have — it is the template strand written with U.
Option C has exactly the right bases but the ends labelled the wrong way round.
Option D is the sequence read backwards.
ShortcutDecide first which strand you have been handed. Coding → swap T for U, one step.Template → complement first, then swap, two steps. A sequence written 5′→3′ and called coding needs one step only.
Where it goes wrongRight-content-wrong-arrangement. Option C carries the correct bases in the correct order with the polarity reversed — the exact shape of the error family that has cost marks across all four subjects. Write the 5′ and 3′ labels on your own answer before you look at the options, and compare labels first.
Q9. TranscriptionPYQ shape: NEET 2024
Match the eukaryotic RNA polymerases with their products. (i) RNA polymerase I (ii) RNA polymerase II (iii) RNA polymerase III (p) tRNA, 5S rRNA and snRNAs (q) 28S, 18S and 5.8S rRNA (r) hnRNA
i–r, ii–q, iii–p
i–q, ii–r, iii–p
i–p, ii–r, iii–q
i–q, ii–p, iii–r
GivenThree eukaryotic RNA polymerases and three product sets.
AskedThe correct pairing.
Concept to useEach of the three eukaryotic polymerases has one fixed job, and the split is tested for precision rather than understanding.
Formula or ruleI → the large rRNAs | II → hnRNA | III → tRNA plus the small RNAs
Baby steps
Pol I makes the large ribosomal RNAs: 28S, 18S and 5.8S. So i–q.
Pol II makes hnRNA, the precursor of mRNA. So ii–r.
Pol III makes tRNA, 5S rRNA and the snRNAs. So iii–p.
That gives i–q, ii–r, iii–p.
Answer(B) i–q, ii–r, iii–p
Why not the others
Option A swaps I and II.
Option C gives the rRNAs to Pol III and the small RNAs to Pol I — the pairing fully inverted.
Option D swaps II and III.
ShortcutCount I, II, III and say r, m, t — ribosomal, messenger, transfer. Then attach the two strays: 5S rRNA and snRNA go with III, not with I.
Where it goes wrongThe question is really testing one fact: 5S rRNA is a ribosomal RNA but it is made by polymerase III, not polymerase I. Anyone who sorts by the words “rRNA goes with I” will pick a wrong option. Sort by the full product list, not by the name.
Q10. TranscriptionPYQ shape: NEET 2025
Consider the following statements about the processing of hnRNA. I. Capping adds methyl guanosine triphosphate to the 5′ end. II. Tailing adds 200–300 adenylate residues at the 3′ end in a template-dependent manner. III. Splicing removes introns and joins exons. IV. The fully processed hnRNA is then called mRNA and is transported out of the nucleus. How many are correct?
One
Two
Three
Four
GivenFour statements on post-transcriptional modification.
AskedHow many are correct.
Concept to useAll three modifications are correct as facts; the question turns on one qualifying word inside statement II.
Formula or ruleMark T or F beside each numeral, then count.
Baby steps
I — True. The 5′ cap is methyl guanosine triphosphate, an unusual nucleotide.
II — False. The number 200–300 and the 3′ position are both right, but tailing is template-independent. There is no run of Ts in the DNA to copy.
III — True.
IV — True. Only the fully processed molecule is called mRNA, and only then does it leave the nucleus.
Three true.
Answer(C) Three.
Why not the others
Four is the answer for anyone who read statement II for its numbers and skipped the qualifier.
One and Two reject statements that are correct as written.
ShortcutUnderline the qualifying word in each statement as you read: template-dependent, 5′, 3′, introns. Multi-statement items in this chapter are almost always decided by a single qualifier, not by the main claim.
Where it goes wrongJudging a statement true because most of it is true. Statement II is correct in three respects and wrong in one word, and one wrong word makes the whole statement false. Read to the full stop before deciding.
Q11. Transcription
In bacteria, the factor that enables RNA polymerase to terminate transcription is:
sigma factor
DNA ligase
primase
rho factor
GivenBacterial transcription.
AskedThe termination factor.
Concept to useBacteria have a single RNA polymerase that needs a different helper at each end of the job.
Formula or rulesigma (σ) → initiationrho (ρ) → termination
Baby steps
Bacterial RNA polymerase makes all three kinds of RNA on its own.
To recognise the promoter and begin, it associates with the sigma factor.
To let go at the terminator, it needs the rho factor.
The question asks for termination, so the answer is rho.
Answer(D) Rho factor.
Why not the others
Sigma is initiation, not termination — the intended trap.
DNA ligase and primase belong to replication, not transcription.
Shortcutσ for start (sigma, start — both begin with s), ρ for release (rho, release — both begin with r). Two words, and the pair can never be swapped again.
Where it goes wrongSigma and rho being interchanged. They appear together in every question on this topic precisely because the two names are easy to hold but hard to assign. Fix them with the s/r anchor rather than trying to remember them separately.
Q12. Translation
Which one of the following statements about tRNA is incorrect?
Francis Crick postulated it as an adapter molecule
The amino acid attaches at its 3′ end
Its anticodon loop pairs with the codon on mRNA
A specific tRNA exists for each of the three stop codons
GivenFour statements about tRNA, three of them true.
AskedThe one that is incorrect.
Concept to usetRNA is the adapter between a three-letter code and an amino acid. It has an anticodon at one end and an amino acid at the other — but only for codons that specify an amino acid.
Formula or rule61 codons have tRNAs; the 3 stop codons have none.
Baby steps
Statement 1: true. Crick proposed the adapter before any tRNA had been seen.
Statement 2: true. The amino acid acceptor end is the 3′ end.
Statement 3: true. That is the whole function of the anticodon loop.
Statement 4: false. There is no tRNA for UAA, UAG or UGA. Termination is done by a release factor, a protein.
The incorrect statement is the fourth.
Answer(D) A specific tRNA exists for each of the three stop codons — this is the false statement.
Why not the others
The other three are all accurate descriptions of tRNA and are stated exactly as NCERT gives them.
ShortcutThe absence of stop-codon tRNAs is not an incidental fact; it is the mechanism of termination. If a tRNA existed for UAA, the chain would never stop. Reasoning from the mechanism gives the answer without recall.
Where it goes wrongAnswering a negative-stem question as though it were positive. Circle the word incorrect in the stem before reading a single option. Three of these four statements are true, so a candidate who reads for truth will confidently pick a wrong answer.
Q13. TranslationPYQ shape: NEET 2021
In a bacterial ribosome, the peptide bond between two amino acids is formed by:
a protein enzyme of the small subunit
23S rRNA of the large subunit
16S rRNA of the small subunit
the release factor
GivenPeptide bond formation during bacterial translation.
AskedWhat catalyses it.
Concept to useThe catalytic part of the bacterial ribosome is RNA, not protein. An RNA that acts as an enzyme is a ribozyme.
Formula or rulepeptidyl transferase activity = 23S rRNA of the 50S (large) subunit
Baby steps
The large subunit is the one that holds two tRNAs side by side, so the bond must form there.
In bacteria the large subunit is 50S and contains the 23S rRNA.
That rRNA carries the peptidyl transferase activity — it is a ribozyme.
So the answer is 23S rRNA of the large subunit.
Answer(B) 23S rRNA of the large subunit.
Why not the others
A protein enzyme is the intuitive answer and the wrong one — the whole point of this fact is that no protein is involved.
16S rRNA is real, but it belongs to the small (30S) subunit and does not form peptide bonds.
The release factor ends translation; it does not build bonds.
ShortcutTwo tRNAs sit in the large subunit, so the bond forms in the large subunit. Large subunit in bacteria = 50S = contains 23S. The location settles the answer before any recall of the number.
Where it goes wrongReaching for “an enzyme” and assuming enzyme means protein. This is also the single fact that links this chapter to Evolution, since the ribozyme is the standard molecular evidence for the RNA world — worth flagging in both places.
Q14. Genetic code
Assertion (A): The genetic code is degenerate. Reason (R): A single codon can specify more than one amino acid.
Both A and R are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A
A is true but R is false
A is false but R is true
GivenAn assertion about degeneracy and a reason that describes ambiguity.
AskedWhether each is true, and whether R explains A.
Concept to useDegeneracy and ambiguity are opposite directions of the same relationship, and the code has one but not the other.
Formula or ruleDegenerate: many codons → one amino acid. Ambiguous: one codon → many amino acids. The code is degenerate and unambiguous.
Baby steps
A alone. True. Most amino acids are specified by more than one codon; 61 codons cover 20 amino acids.
R alone. False. This describes ambiguity, and the genetic code is specifically unambiguous — each codon has exactly one meaning.
Since R is false and A is true, only one option fits.
Answer: A is true but R is false.
Answer(C) A is true but R is false.
Why not the others
Options A and B both require R to be true, and it is not.
Option D declares the assertion false, but degeneracy is a genuine property of the code.
ShortcutDraw the arrow. Degeneracy points from several codons to one amino acid. If the sentence in front of you points the arrow the other way, it is describing ambiguity and it is false.
Where it goes wrongAssertion–reason instability again, and this is the harder version. Both lines use the same vocabulary, so they feel like they belong together, and the pair reads smoothly as one sentence. That smoothness is the trap. Judge R entirely on its own, ignoring the assertion above it, before deciding anything about the pairing.
Q15. Genetic codePYQ shape: NEET 2017
An mRNA is 999 nucleotides long and is read as 333 codons. If the 901st base is deleted, how many codons at the beginning of the message remain unaffected?
300
333
301
299
GivenAn mRNA of 999 bases read as 333 codons; base number 901 is deleted.
AskedThe number of codons before the deletion that are read normally.
Concept to useThe reading frame is fixed from the start codon. A deletion changes nothing upstream of itself; everything downstream is shifted.
Formula or rulecomplete codons before position p = whole-number part of (p − 1) ÷ 3
Baby steps
The bases before the deleted one are numbers 1 to 900.
(901 − 1) ÷ 3 = 900 ÷ 3 = 300, exactly.
So bases 1–900 form 300 complete, undisturbed codons.
Base 901 would have been the first base of codon 301, and it is gone — so codon 301 onwards is shifted.
Unaffected codons: 300.
Answer(A) 300 codons.
Why not the others
333 is the original codon count, unchanged — correct only if nothing had been deleted.
301 counts the disrupted codon as unaffected.
299 undercounts by one, usually from subtracting an extra codon ‘for safety’.
ShortcutDivide (position − 1) by 3, and take the whole number. Nothing else is required. Worth knowing for a follow-up part: after the deletion 98 bases remain (902–999), giving 32 complete misread codons plus 2 leftover bases.
Where it goes wrongDividing 901 by 3 instead of 900. The minus one is the entire question — the deleted base is not part of any intact upstream codon. Write out “bases 1 to 900” explicitly rather than working from the position number.
Q16. Genetic codePYQ shape: NEET 2018
In sickle-cell anaemia, the change in the β-globin gene is:
GUG → GAG in the sixth codon; valine is replaced by glutamic acid
GAG → GUG in the sixth codon; glutamic acid is replaced by valine
GAG → GUG in the sixth codon; valine is replaced by glutamic acid
a frameshift caused by the deletion of a single base
GivenThe molecular basis of sickle-cell anaemia.
AskedThe exact codon change and the amino acid substitution.
Concept to useSickle-cell anaemia is a point mutation — a single base substitution, not an insertion or deletion.
Formula or ruleGAG (glutamic acid) → GUG (valine), at the sixth codon of the β-globin chain
Baby steps
The mutation is a substitution of one base, so it is a point mutation, not a frameshift.
The normal codon is GAG, which codes for glutamic acid.
The mutant codon is GUG, which codes for valine.
The direction is normal → abnormal, so glutamic acid is replaced by valine.
The position is the sixth codon of the β-globin chain.
Answer(B) GAG → GUG in the sixth codon; glutamic acid is replaced by valine.
Why not the others
Option A reverses the mutation itself — it describes the sickle allele turning back into the normal one.
Option C has the codon change right and the amino acids the wrong way round. It is the deliberate near-miss.
Option D is the wrong class of mutation entirely.
ShortcutA mutation always runs from normal to abnormal, so the normal item goes on the left of every arrow. Glutamic acid is normal, valine is the sickling one. Write your own arrow before reading the options.
Where it goes wrongRight-content-wrong-arrangement, in its purest form. Options B and C contain identical codon information and differ only in which amino acid is assigned to which side. This is exactly the error family that has recurred across papers. The defence is to write the answer out yourself first, then match — never to read the options and choose the familiar-looking one.
Q17. Lac operonPYQ shape: NEET 2022
Match the genes of the lac operon with their products. (i) i (ii) z (iii) y (iv) a (p) permease (q) transacetylase (r) repressor (s) β-galactosidase
i–r, ii–s, iii–p, iv–q
i–s, ii–r, iii–p, iv–q
i–r, ii–p, iii–s, iv–q
i–q, ii–s, iii–p, iv–r
GivenThe four genes of the lac operon and four protein products.
AskedThe correct pairing.
Concept to useOnly three of the four genes are structural genes making enzymes. The fourth, i, is the regulator and makes the repressor.
Formula or rulei → repressorz → β-galactosidasey → permeasea → transacetylase
Baby steps
i is the regulator gene: its product is the repressor protein. So i–r.
z codes for β-galactosidase, the enzyme that splits lactose into galactose and glucose. So ii–s.
y codes for permease, which makes the cell more permeable to β-galactosides. So iii–p.
a codes for transacetylase. So iv–q.
Answer(A) i–r, ii–s, iii–p, iv–q
Why not the others
Option B swaps i and z, making the regulator gene produce an enzyme.
Option C swaps z and y.
Option D assigns the repressor to a and transacetylase to i — both wrong.
ShortcutFix the three structural genes in order and the fourth follows: z, y, a → beta, permease, acetylase. Then remember that i is not a structural gene at all. If an option gives i an enzyme, discard it without checking the rest.
Where it goes wrongAssuming that because i is listed first it must produce the first enzyme. It produces no enzyme. Also worth guarding: the letter i stands for inhibitor, not inducer — the inducer is lactose, and that mix-up is the single most common slip in this topic.
Q18. Lac operon
In E. coli growing in a medium containing lactose, which of the following happens?
The repressor binds the operator and transcription is blocked
Lactose binds the promoter and displaces RNA polymerase
The i gene is switched off, so no repressor is made at all
Lactose binds the repressor, which can then no longer bind the operator, and a polycistronic mRNA is transcribed
GivenE. coli in a lactose-containing medium.
AskedThe state of the operon and what happens at the molecular level.
Concept to useLactose is the inducer. It works by disabling the repressor, not by acting on the DNA itself.
Formula or rulelactose + repressor → repressor changes shape → operator is free → polymerase transcribes z, y and a as one mRNA
Baby steps
Lactose enters the cell and binds the repressor protein.
Binding changes the repressor's shape, so it can no longer hold on to the operator.
The operator is now free, so RNA polymerase can move past the promoter and through the structural genes.
One polycistronic mRNA is made, carrying z, y and a together.
The operon is on.
Answer(D) Lactose binds the repressor, which can then no longer bind the operator, and a polycistronic mRNA is transcribed.
Why not the others
Option A is the correct description of the opposite condition — lactose absent.
Option B has lactose acting on the promoter. It acts on the repressor protein.
Option C is wrong because the i gene is expressed constantly; the repressor is made and then inactivated, not prevented from being made.
ShortcutThe regulation is negative: the default is off, and the signal works by taking a block away, never by adding an activator. Any option where lactose does something to the DNA is wrong.
Where it goes wrongAnswering the other state of the operon. Options A and D are the two states, and both are correct descriptions of something — just not of the same condition. Underline containing lactose in the stem before reading the options.
Q19. Historical experimentsPYQ shape: NEET 2016
Which conclusion follows directly from the Hershey–Chase experiment?
Bacteria can be permanently transformed by material from heat-killed cells
Protein is the genetic material
DNA, and not protein, entered the bacterial cell and directed the production of new phages
RNA can serve as the genetic material in some viruses
GivenThe Hershey–Chase experiment with bacteriophage T2 and E. coli.
AskedThe conclusion that follows from this experiment.
Concept to useThe experiment exploits one chemical asymmetry: DNA contains phosphorus but no sulphur, and protein contains sulphur but no phosphorus.
Formula or rule³²P labels DNA → radioactivity in the pellet. ³⁵S labels the coat → radioactivity in the supernatant.
Baby steps
Phages grown with radioactive phosphorus had labelled DNA; those grown with radioactive sulphur had labelled protein coats.
Both were used to infect bacteria, then blended to knock the coats off the bacterial surface, then centrifuged.
Radioactivity from the ³²P batch appeared inside the bacteria, in the pellet.
Radioactivity from the ³⁵S batch stayed outside, in the liquid.
So only DNA entered, and that alone was enough to produce new phages.
Answer(C) DNA, and not protein, entered the bacterial cell and directed the production of new phages.
Why not the others
Option A is a perfectly true statement — but it is Griffith's conclusion, from a different experiment twenty-four years earlier.
Option B is exactly what the experiment ruled out.
Option D is true biology, but no RNA virus was involved here.
ShortcutAsk what was measured, not what is generally true. Hershey and Chase measured where radioactivity ended up. The only conclusion available is about which molecule crossed into the cell.
Where it goes wrongAttribution error — the family that has produced marks lost to crediting the wrong scientist. Option A is a true fact from the wrong experiment, and questions of the form ‘which follows from X’ reliably include one. Check who did it, not only whether it is true: Griffith showed that something transfers, Avery–MacLeod–McCarty showed it was DNA in a test tube, Hershey–Chase showed it in a living infection.
Q20. DNA packagingPYQ shape: NEET 2023
A stretch of chromatin contains 6 × 10⁵ base pairs of DNA. The number of nucleosomes in it, and the composition of the histone core, are:
3000 nucleosomes; core of two each of H1, H2A, H2B and H3
1200 nucleosomes; core of two each of H2A, H2B, H3 and H4
300 nucleosomes; core of one each of H2A, H2B, H3, H4 and H1
3000 nucleosomes; core of two each of H2A, H2B, H3 and H4
GivenA chromatin stretch of 6 × 10⁵ base pairs.
AskedNumber of nucleosomes, and the histone composition of the core particle.
Concept to useOne nucleosome is a fixed length of DNA wrapped around a fixed set of eight histone molecules.
Formula or rulenucleosomes = base pairs ÷ 200; core = two each of H2A, H2B, H3 and H4
The core is a histone octamer: eight molecules made of two copies each of four types.
Those four types are H2A, H2B, H3 and H4. H1 is not part of the core — it sits outside, on the linker DNA.
So 3000 nucleosomes with a core of two each of H2A, H2B, H3 and H4.
Answer(D) 3000 nucleosomes; core of two each of H2A, H2B, H3 and H4.
Why not the others
Option A has the arithmetic right but puts H1 into the core and drops H4.
Option B divides by 500 instead of 200.
Option C is wrong on both counts — divides by 2000, and gives one copy of each histone.
ShortcutDivide the base-pair count by 200 and you have the nucleosome count. For the core, remember it is 8 = 2 × 4, and that the odd one out, H1, is the linker — it links beads, so it cannot be inside one.
Where it goes wrongPutting H1 into the octamer. The two halves of this question are independent: a candidate who gets the arithmetic right can still lose the mark on the histone list, and option A exists to catch exactly that. Answer both halves separately before comparing with the options.
Answer key
#
Topic
Key
Q1
DNA structure
B
Q2
DNA structure
A
Q3
DNA structure
C
Q4
DNA structure
D
Q5
Replication
B
Q6
Replication
A
Q7
Replication
C
Q8
Transcription
A
Q9
Transcription
B
Q10
Transcription
C
#
Topic
Key
Q11
Transcription
D
Q12
Translation
D
Q13
Translation
B
Q14
Genetic code
C
Q15
Genetic code
A
Q16
Genetic code
B
Q17
Lac operon
A
Q18
Lac operon
D
Q19
Historical experiments
C
Q20
DNA packaging
D
Key distribution: five A, five B, five C, five D, with no run of more
than two identical letters. Assertion–reason items keep the standard option order and are not permuted.