ERROR NOTESMBI · THE DNA · SET 1
Molecular Basis of Inheritance — The DNA
From the 40-question PYQ paper attempted 02/09/26 · NCERT §5.1, structure of the polynucleotide chain and packaging of the DNA helix · Aamirah Fathima
126 / 16033 correct · 6 wrong · 1 blank78.75%
20/20Section A recall
8/10Assertion–reason
2/6Match the column
3/4Numerical / multi-stmt
39/40Attempt rate
0Marks lost to unknown content
DIAGNOSIS — WHERE THE 34 MARKS WENTEvery mark lost was procedural, not a content gap
| Error family | Questions | Marks | What actually happened |
|---|
Right content, wrong arrangement
Recurring family — confirmed across all four subjects | 31, 32, 35 | 12 lost
(3 × −4) | In all three, the four items were known and exactly two were transposed into each other's rows. Never three or four — always a clean pair swap. Both numeric match questions (33, 34) were correct, so the transposition is specific to word-to-word matching, where nothing internal contradicts a wrong placement. |
Assertion–reason causality
Recurring family — unsettled procedure | 22, 27 | 8 lost
(2 × −4) | Both were the “both true, R is not the explanation” verdict and both were answered as “R is the correct explanation”. Both statements were true in each case, so the truth check passed and the reasoning stopped there. The causal question was never asked. The eight A–R questions she got right were all ones where a statement was false — i.e. the truth check alone was sufficient. |
| Known fact not carried into the calculation | 39 | 4 lost | Q7 (A=T is two bonds, G≡C is three) was answered correctly. Q39 then applied two bonds to every pair. The fact was available; the numerical step that needed it did not reach for it. |
| Blank | 36 | 0 lost | The only blank on the paper, and it sits immediately after three match questions that went wrong. Reads as a confidence drop inside Section C rather than a gap — the content (H1, octamer, NHC proteins, heterochromatin) was answered correctly elsewhere on this same paper. |
The headline: Section A was 20/20. Not one mark was lost to unknown content anywhere on this paper. All 34 marks went to two checking procedures that are not yet automatic — verifying a match by reading it backwards, and testing whether a reason actually explains an assertion. Both are trainable in a single sitting, which makes this a much better result than 78.75% suggests.
CARD 1 · Q31 — MATCH THE COLUMNRight content, wrong arrangement
Q31 · Types of bondStructure of the polynucleotide chain
Her answer — DI-C · II-D · III-A · IV-B
Correct — BI-B · II-D · III-A · IV-C
| List I | Should join | She put | Verdict |
|---|
| I · N-glycosidic bond | Nitrogenous base and pentose sugar | Two complementary bases | Swapped |
| II · Phosphoester bond | Pentose sugar and phosphate group | Same | Correct |
| III · Phosphodiester bond | Two adjacent nucleotides | Same | Correct |
| IV · Hydrogen bond | Two complementary bases of opposite strands | Base and pentose sugar | Swapped |
RuleThe four bonds sit in a fixed chain of assembly: base —N-glycosidic— sugar —phosphoester— phosphate —3′–5′ phosphodiester— next nucleotide, and only then —hydrogen bond— across to the other strand.
Why D failsHydrogen bonds act between the two strands, never inside a single nucleotide. The moment a hydrogen bond is placed on “base and pentose sugar”, it has been made an intra-nucleotide bond, which contradicts the entire double-helix model.
ShortcutThree of these four bonds are within one strand; only the hydrogen bond crosses between strands. Place the hydrogen bond first — it is the only one with a partner on the opposite strand — and the other three fall into the assembly order automatically.
Where it went wrongThe two ends of the list, I and IV, were exchanged while the middle two stayed correct. This is the signature of matching down the list in one pass without a reverse check.
CARD 2 · Q32 — MATCH THE COLUMNRight content, wrong arrangement · attribution
Q32 · Scientists and contributionsHistory of the double helix
Her answer — CI-B · II-A · III-D · IV-C
Correct — DI-C · II-A · III-D · IV-B
| List I | Contribution | She put | Verdict |
|---|
| I · Friedrich Miescher | Isolated DNA and named it ‘nuclein’ | Base equivalence A=T, G=C | Swapped |
| II · Watson and Crick | Proposed the double helix model | Same | Correct |
| III · Wilkins and Franklin | Produced the X-ray diffraction data | Same | Correct |
| IV · Erwin Chargaff | Showed equivalence of A with T, G with C | Isolated and named ‘nuclein’ | Swapped |
RuleFix the four by date and by type of work: Miescher isolated the substance (1869, chemistry); Chargaff measured base ratios (analysis); Wilkins and Franklin photographed it (physics); Watson and Crick assembled the model from the other three (1953, synthesis).
Why C failsIt credits Miescher with a base-ratio measurement that could not exist in 1869 — the bases had not been quantified. And it gives Chargaff the isolation, which was already eighty years old by the time he worked.
ShortcutOrder them on a timeline before matching: 1869 Miescher → 1950 Chargaff → 1952 Wilkins & Franklin → 1953 Watson & Crick. Miescher is always first and Watson–Crick always last, so the two you actually have to decide between are the middle pair.
Where it went wrongSame shape as Q31 — the first and last rows exchanged, middle two intact. This is also the attribution error family already logged for Mendel/Morgan and Sturtevant/Morgan in genetics: the names are known, the pairing is not stable.
CARD 3 · Q35 — MATCH THE COLUMNRight content, wrong arrangement · near-identical terms
Q35 · Nucleoside / nucleotide / nucleosome / nucleoidTerminology
Her answer — CI-C · II-B · III-A · IV-D
Correct — BI-C · II-A · III-B · IV-D
| List I | Composition | She put | Verdict |
|---|
| I · Nucleoside | Base + pentose sugar | Same | Correct |
| II · Nucleotide | Base + pentose sugar + phosphate | 200 bp on a histone octamer | Swapped |
| III · Nucleosome | 200 bp of DNA wrapped on a histone octamer | Base + sugar + phosphate | Swapped |
| IV · Nucleoid | Bacterial DNA with non-histone proteins | Same | Correct |
RuleTwo of these are chemical units (nucleoside, nucleotide — differing only by a phosphate) and two are whole structures (nucleosome, a DNA–protein particle; nucleoid, a region of a bacterial cell). A chemical unit can never be measured in base pairs.
Why C failsIt assigns 200 bp to a nucleotide. A nucleotide is one base with one sugar and one phosphate — it cannot contain 200 base pairs of anything. That single mismatch of scale invalidates the option without needing to check the other three rows.
ShortcutSort by the ending before matching: -side and -tide are molecules (tide has the phosphate — “tide = triphosphate side”); -some is a body/particle; -oid is a region. Then check the units: anything quoted in base pairs must be a structure, never a unit.
Where it went wrongHere the swap was the middle pair rather than the ends, so it is not a positional habit — it is genuinely the two most similar-sounding words being exchanged. Nucleotide and nucleosome differ by three letters and by six orders of magnitude in size.
CARD 4 · Q22 — ASSERTION & REASONCausality not tested
Q22 · DNA as a polymerDefinition
Assertion (A)DNA is a long polymer of deoxyribonucleotides.
Reason (R)The length of a DNA molecule is usually defined by the number of nucleotides or base pairs present in it.
Her answer — ABoth true and R is the correct explanation of A
Correct — BBoth true but R is not the correct explanation of A
RuleR must answer the question “why is A true?” Here, A is a statement about what DNA is made of; R is a statement about how we choose to measure it. A measuring convention cannot be the cause of a chemical composition.
The testReverse the logic: if scientists measured DNA in micrometres instead of base pairs, would DNA stop being a polymer of deoxyribonucleotides? No. So R is not the cause of A — it is a consequence of A, which is a different relationship.
ShortcutWhen both statements are true, ask which direction the arrow runs. Here it runs A → R (because DNA is a polymer of nucleotides, we can count nucleotides). The option demands R → A. A reversed arrow always means answer (b).
Where it went wrongBoth statements were true and both were about DNA, so the pair felt connected. Topical relatedness was accepted in place of causation.
CARD 5 · Q27 — ASSERTION & REASONCausality not tested
Q27 · Thymine and uracilNitrogenous bases
Assertion (A)Thymine is present in DNA but is replaced by uracil in RNA.
Reason (R)DNA is normally a double-stranded molecule.
Her answer — ABoth true and R is the correct explanation of A
Correct — BBoth true but R is not the correct explanation of A
RuleWhich pyrimidine a nucleic acid carries is a matter of chemistry — thymine is 5-methyl uracil, and the methyl group confers stability and protects against a specific mutation. How many strands the molecule has is a separate property. Neither determines the other.
The testCounter-example: single-stranded DNA (as in bacteriophage φ×174, which appeared in Q34 of this same paper) still contains thymine, not uracil. Being double-stranded is therefore not what puts thymine in DNA.
ShortcutIf you can name a single case where R is true and A is false — or R false and A true — then R cannot be the explanation of A. One counter-example settles it.
Where it went wrongIdentical to Q22: two true statements about DNA accepted as causally linked. Both of the A–R errors on this paper were the same verdict in the same direction — that consistency is useful, because it means one procedure fixes both.
CARD 6 · Q39 — NUMERICALKnown fact not carried into the calculation
Q39 · Hydrogen bond countBase pairing · numerical
QuestionA double-stranded DNA segment of 100 base pairs contains 60 adenine bases. Find the total number of hydrogen bonds.
Her answer — D200 (= 100 × 2)
Correct — B240
RuleTotal H-bonds = 2 × (number of A–T pairs) + 3 × (number of G–C pairs). The two pair types must be counted separately — this is the whole question.
Baby steps
- 100 bp means 200 bases in total across both strands.
- A = 60, so T = 60 (Chargaff). Adenine bases and A–T pairs are equal in number ⇒ 60 A–T pairs.
- Remaining pairs = 100 − 60 = 40 G–C pairs.
- H-bonds = (60 × 2) + (40 × 3) = 120 + 120 = 240.
Why 200 fails200 is what you get from 100 pairs × 2 bonds — treating every pair as A–T. It ignores the 40 G–C pairs entirely, and it makes the “60 adenine” figure in the question redundant. If a number given in the stem never enters your working, the method is wrong.
ShortcutThe answer must lie strictly between 2 × 100 = 200 (all A–T) and 3 × 100 = 300 (all G–C). Any option equal to 200 or 300 is an endpoint and cannot be right for a mixed molecule. That eliminates the chosen option before any arithmetic.
Where it went wrongThe fact was secure — Q7 on this same paper (A=T two bonds, G≡C three) was answered correctly. It simply was not reached for when the question was numerical rather than verbal.
CARD 7 · Q36 — THE BLANKNot a content gap
Q36 · Chromatin componentsPackaging of the DNA helix
Her answerLeft blank — the only blank on the paper
Correct — AI-B · II-D · III-C · IV-A
| List I | Role or property |
|---|
| I · Histone H1 | Lies outside the core, on the linker DNA |
| II · H2A, H2B, H3 and H4 | Together form the histone octamer (two copies each) |
| III · Non-histone chromosomal proteins | Required for packaging chromatin at higher levels |
| IV · Heterochromatin | Densely packed and transcriptionally inactive |
Evidence it was knownOn this same paper she answered Q17 correctly (octamer = two copies each of H2A, H2B, H3, H4), Q20 correctly (euchromatin loose, light, active), Q26 correctly (H1 is not part of the core) and Q29 correctly (the octamer does not contain H1). Every one of the four rows above was independently answered right elsewhere.
ShortcutTwo of the four rows can be placed on sight — H1 goes with the linker and heterochromatin goes with “dense and inactive”. With two rows fixed, the remaining two are forced. Even a partial match usually resolves the whole question.
Where it went wrongPosition, not content. It is the fourth match question in a row, and it follows Q31, Q32 and Q35, all of which went wrong. The blank is a confidence effect inside a section that had started to feel unreliable — the recorded blank-rate pattern of clustering in the weakest tier, appearing here at the scale of a single section.
THE TWO PROCEDURES TO INSTALLThese two habits would have converted 126 into 154
Procedure 1 — the reverse check for every match question. Match forward down List I as usual, then read your answer backwards: take each item of List II in turn and name which List I item you have given it. The pair swap that cost 12 marks on this paper is invisible reading forwards (each row looks individually plausible) but obvious reading backwards — “hydrogen bond joins base to sugar” fails instantly when read from that direction. This takes about fifteen seconds per question.
Then check the scale. Before finalising, confirm every row is dimensionally sensible: a chemical unit cannot be measured in base pairs, a scientist cannot make a measurement that pre-dates the technique. Q31, Q32 and Q35 each contained one row that fails this test on its own.
Procedure 2 — the three-step written defence for assertion–reason. This is already prescribed and it is not yet being used. Write, in the margin, three separate verdicts before touching the options:
- Is A true? Decide with the reason covered.
- Is R true? Decide independently, without looking back at A.
- Does R answer “why is A true?” — and specifically, could A still be true if R were false? If yes, R is not the explanation.
Step 3 is the one being skipped. All eight A–R questions answered correctly on this paper were decided at steps 1–2, because one statement was false; both errors were questions that survived to step 3. When both statements are true, the question has not yet been answered.
Re-test schedule. Sections B and C are the only ones that need retesting — Section A is secure at 20/20 and does not need repeating. Attempt a targeted set of 12 match-the-column questions built with two-row transposition distractors and 8 assertion–reason items where R is true but non-explanatory, with the three-step defence written out for every item. Then re-attempt this same paper cold in a fortnight; the target is not a higher score but zero errors of these two families, since the content is already in place.