ERROR NOTESMBI 5.1–5.6 · PAPER 2
Molecular Basis of Inheritance — The DNA to the Genetic Code
50-question paper on NCERT topics 5.1 to 5.6, attempted 15/09/2026 · all 50 questions answered, none left blank · Aamirah Fathima
170 / 20044 correct · 6 wrong · 0 blank85.0%
50/50Attempted
88%Accuracy
+6.25Points vs Paper 1
5/6Assertion–reason
4/4Statement I / II
4/5Match the column
3Reversal errors
WHAT CHANGED SINCE PAPER 1Both prescribed procedures were used, and both worked
| Measure | Paper 1
§5.1 only | Paper 2
§5.1–5.6 | Reading |
|---|
| Score | 126 / 160
78.75% | 170 / 200
85.0% | Up on a paper covering six topics instead of one, so the rise is larger than the raw number suggests. |
| Accuracy on attempted | 84.6% | 88.0% | Fewer errors per question answered, with a much wider syllabus in play. |
| Blanks | 1 | 0 | Nothing abandoned. The Section C confidence drop seen last time did not recur. |
| Assertion–reason | 8 / 10 | 5 / 6 | Q45 was the exact item type she failed twice last time — both statements true, reason not the explanation — and she answered it correctly. The three-step defence has taken hold. |
| Statement I / II | — | 4 / 4 | A new format on this paper, handled cleanly. |
| Match the column | 2 / 6 | 4 / 5 | From three transpositions down to one. Improved, but the family is still alive — see Q28. |
THE SIX ERRORS — WHERE 30 MARKS WENTThree of the six are one single habit
| Error family | Questions | Marks | What happened |
|---|
Reversal
Now the dominant family | 30, 32
(and 8) | 12 | Both Q30 and Q32 turn on the polarity and position of the transcription unit. She placed the terminator where the promoter belongs, and accepted "the template strand has 5′→3′ polarity" as true. Q8 is the same habit in a different costume — she inverted both verdicts of an assertion–reason pair after misreading a base-pairing rule. |
| Numerical — wrong rule applied | 5, 22 | 8 | Both are calculations where the correct method is known but the wrong constant or rule was used: dividing by 20 instead of 200, and doubling the hybrid count each generation instead of holding it at two. |
Right content, wrong arrangement
Down from 3 errors to 1 | 28 | 4 | A clean two-row transposition again: ligase and helicase exchanged, the other two rows correct. Same signature as Q31, Q32 and Q35 of Paper 1. |
Topic by topic: 5.1 The DNA 7/9 · 5.2 Search for Genetic Material 8/8 · 5.3 RNA World 3/3 · 5.4 Replication 7/9 · 5.5 Transcription 9/11 · 5.6 Genetic Code 10/10. The genetic code — the newest topic on this paper and the one with the most numerical and table work — was answered perfectly. Nothing was lost to unfamiliar content anywhere; every error was on material she has already demonstrated she knows.
CARD 1 · Q30 — THE TERMINATORReversal · promoter / terminator
Q30 · Transcription unitTopic 5.5
QuestionThe terminator of a transcription unit is …
Her answer — Blocated towards the 5′-end (upstream) of the structural gene
Correct — Dlocated towards the 3′-end (downstream) of the coding strand
RuleA transcription unit runs promoter → structural gene → terminator in the 5′→3′ direction of the coding strand. So the promoter is at the 5′ end (upstream) and the terminator at the 3′ end (downstream).
What she choseHer option describes the promoter's position exactly. The two ends of the unit were exchanged.
ShortcutTranscription ends at the terminator, and the RNA ends at its 3′ end. Terminator and 3′ both mean "the end". Promoter and 5′ both mean "the start". Tie the words together and the pair cannot invert.
Where it went wrongStraight reversal. This is the same family that produced the 5′/3′ and 0.34/3.4 nm errors flagged after Paper 1, now appearing in the transcription unit.
CARD 2 · Q32 — TEMPLATE STRAND POLARITYReversal · 5′ / 3′
Q32 · Transcription unit statementsTopic 5.5
QuestionFive statements A–E about a transcription unit; choose the correct combination. Statement E read: "The template strand has 5′→3′ polarity."
Her answer — AA, B, C, D and E — i.e. statement E accepted as true
Correct — CA, B, C and D only — statement E is false
RuleThe template strand runs 3′→5′. RNA is built 5′→3′ on a template read 3′→5′. The strand with 5′→3′ polarity is the coding strand, which is not copied.
NoteStatements A to D were all judged correctly. The entire question turned on a single reversed arrow in statement E — four right judgements were undone by one.
ShortcutSay the sentence in full, always in the same order: "RNA 5′→3′, template 3′→5′." Any statement that gives the template 5′→3′ is false on sight. In a multi-statement question, the option that includes every statement is almost always the trap.
Where it went wrongSame family as Q30, on the same figure of the transcription unit. Two of the six errors on this paper are the same sentence, misremembered twice.
CARD 3 · Q8 — ASSERTION AND REASON INVERTEDMisread, then both verdicts flipped
Q8 · Base pairing and helix widthTopic 5.1
Assertion (A)The distance between the two strands of a DNA double helix remains uniform at about 2 nm. — TRUE
Reason (R)In DNA, a purine on one strand always pairs with another purine on the opposite strand. — FALSE
Her answer — D(A) is false, but (R) is true
Correct — C(A) is true, but (R) is false
What happenedShe did not get one verdict wrong — she got both wrong, and in exactly opposite directions. That is the signature of reading R as the familiar true statement ("a purine pairs with a pyrimidine") rather than what was actually printed, and then adjusting A to fit.
ShortcutOn the other five assertion–reason questions she was right, so the procedure is working. The one refinement it needs: at step 2, read the reason word by word, not as a shape you recognise. A single substituted word — purine for pyrimidine — is the whole question.
Where it went wrongNot a causality failure this time, which is the genuine improvement. It is a misread of a reason built by swapping one word — the visual cousin of the reversal family.
CARD 4 · Q22 — MESELSON AND STAHL NUMERICALRight method, wrong rule
Q22 · Generations in 14N mediumTopic 5.4
QuestionAn E. coli cell with 15N–15N DNA is moved to 14N medium for 60 minutes. What are the proportions?
Her answer — D4 hybrid and 4 light DNA molecules
Correct — C2 hybrid and 6 light DNA molecules
Baby steps
- One generation of E. coli = 20 minutes, so 60 minutes = 3 generations.
- Total molecules = 23 = 8.
- Only the two original parental strands survive, and each sits in one molecule ⇒ hybrid = 2, always.
- Light = 8 − 2 = 6. Heavy = 0 from the first generation onward.
What she choseHer answer has the right total of 8 but splits it 4 : 4, i.e. she let the hybrid count double along with everything else. It does not — it is pinned at 2 forever.
Shortcuttotal = 2ⁿ · hybrid = 2 · light = 2ⁿ − 2 — the middle term has no n in it. If your hybrid answer changes when the generation number changes, the rule has been misapplied.
Where it went wrongShe converted the time to generations correctly and got the total right. The single failure was treating a constant as a variable.
CARD 5 · Q5 — NUCLEOSOME COUNTOrder-of-magnitude slip
Q5 · Number of nucleosomesTopic 5.1
QuestionA eukaryotic DNA segment contains 4 × 10⁶ base pairs. The approximate number of nucleosomes formed from it is …
Her answer — C2 × 10⁵
Correct — B2 × 10⁴
Working4 × 10⁶ ÷ 200 = 2 × 10⁴. Her answer is 4 × 10⁶ ÷ 20 — the right operation with a factor of ten missing from the divisor.
ShortcutSplit the division in two: first ÷ 2 (4 → 2), then ÷ 100 (10⁶ → 10⁴). Doing the coefficient and the power separately makes a lost zero impossible to miss. Sense check: a human nucleus has about 2 × 10⁷ nucleosomes in total, so a 4 Mb segment must give a number far smaller than that.
Where it went wrongPure arithmetic. The concept — 200 bp per nucleosome — was answered correctly elsewhere on this same paper.
CARD 6 · Q28 — REPLICATION ENZYMESRight content, wrong arrangement
Q28 · Match the columnTopic 5.4
Her answer — CA-III, B-IV, C-I, D-II
Correct — BA-III, B-I, C-IV, D-II
| List I | Correct function | She assigned | Verdict |
|---|
| A · DNA-dependent DNA polymerase | Polymerisation of deoxyribonucleotides | Same | Correct |
| B · DNA ligase | Joining of Okazaki fragments | Relieving supercoiling | Swapped |
| C · Helicase | Unwinding of the DNA duplex | Joining of Okazaki fragments | Swapped |
| D · Topoisomerase | Relieving supercoiling ahead of the fork | Same | Correct |
Why it failsHer version has helicase joining Okazaki fragments. Helicase is the enzyme that opens the duplex; it does not join anything. It also leaves ligase "relieving supercoiling", which no sealing enzyme does.
ShortcutSort the four by when they act before matching: helicase and topoisomerase work ahead of the fork (opening and de-stressing), polymerase and ligase work behind it (building and sealing). Two rows fixed by timing, and the other two follow.
Where it went wrongTwo rows exchanged, the other two correct — the same signature as three questions on Paper 1. The reverse check was either not run or not run on this question; reading "helicase joins Okazaki fragments" backwards would have caught it instantly.
WHAT TO DO NEXTOne family now accounts for half the loss
Keep doing: the three-step assertion–reason defence and the reverse check on matches. Five of six assertion–reason items, four of four Statement I/II items and four of five matches were right, against 8/10 and 2/6 on Paper 1. Neither procedure needs changing — the match check simply needs to be run on every match question, not most of them.
The one thing to add — a polarity sentence said out loud before every transcription question. Three of the six errors (Q8, Q30, Q32) are a reversed relationship, and two of those three are the same sentence about the transcription unit. Before answering anything from topic 5.5, recite:
- Promoter at the 5′ end, upstream. Terminator at the 3′ end, downstream.
- Template strand 3′→5′. Coding strand 5′→3′. RNA is made 5′→3′.
- Purine pairs with pyrimidine — never purine with purine.
All three of those errors are answered by these three lines. Write them at the top of the rough sheet at the start of the paper, before reading question one.
Re-test. Topics 5.2, 5.3 and 5.6 are clean — 21 of 21 — and do not need repeating. Build the next paper from 5.4 and 5.5 only, weighted towards transcription-unit polarity and replication numericals, and re-attempt this paper cold in a fortnight. The target is not a higher score but zero reversal errors: at 88% accuracy the content is in place, and what remains is three sentences that must not be allowed to invert.