ERROR NOTESNCERT FIGURE PAPER · PAPER 3
Molecular Basis of Inheritance — Diagram Paper
36 questions set on 13 NCERT figures, attempted 15/09/2026 · 34 answered, 2 left blank · Aamirah Fathima
0Identification errors
2Swap errors (S)
3Interpretation errors (I)
6/6Transcription figures
13/13Structure & packaging
8Marks left on the table
WHERE THIS PAPER SITSThree papers, three formats
| Paper | Score | Accuracy on attempted | Reading |
|---|---|---|---|
| Paper 1 — §5.1 only, 40 Q | 126/160 78.8% | 84.6% | Match-column transpositions and assertion–reason causality. |
| Paper 2 — §5.1–5.6, 50 Q | 170/200 85.0% | 88.0% | Best of the three. Reversal became the dominant family. |
| Paper 3 — figures, 36 Q | 111/144 77.1% | 85.3% | The lowest percentage, but not a regression — figure interpretation is a harder format, and accuracy on what she actually answered held at 85%. The gap between 77.1% and 85.3% is entirely the two blanks. |
DIAGNOSIS — 33 MARKS, AND WHAT THEY WERENothing was lost to failing to recognise a figure
| Cause code | Questions | Marks | What it means |
|---|---|---|---|
| L — could not identify the labelled part | none | 0 | Across all 36 questions and 13 figures, every labelled part was named correctly. Every letter marker — a, b, c, d on the polynucleotide chain, the nucleosome, the fork, both transcription figures — was read right. The diagrams themselves are known. |
| S — identified it, then swapped it | 3, 22 | 10 | A polarity written backwards, and the two new strands given the names belonging to a different process. The content was right; the label attached to it was not. |
| I — read it, misinterpreted it | 17, 19, 31 | 15 | Two of these are the same misreading twice (Q17 and Q19), and the third is a miscount in the codon table. |
| Blank | 35, 36 | 8 | The last two questions on the paper, on the easiest figure of the set. This reads as running out of time, not as a gap. |
The single most useful line in this report: zero L errors. On a paper built entirely out of diagrams, she identified every labelled structure correctly and lost marks only on what the figures meant. That is a much better position to be in than the reverse, and it means the fix is reasoning practice on a handful of figures, not redrawing them.
CARD 1 · Q3 — POLARITY OF A WRITTEN SEQUENCES · reversal, third paper running
Q3 · Polynucleotide chainFigure 1
QuestionReading the bases of the chain shown in the correct polarity gives the sequence …
Her answer — B3′-ATGC-5′
Correct — C5′-ATGC-3′
RuleA nucleic acid sequence is always written starting from the 5′ end. In the figure, a is the free phosphate, so the left-hand end is 5′ and the bases read A, T, G, C in that direction.
NoteShe read the bases correctly — her answer has ATGC in the right order. Only the two end labels were exchanged, so this was one character away from four marks.
ShortcutIf the sequence starts with 3′, rewrite it before you compare it with the options. There is no situation in NEET where a sequence should be left written 3′-first.
Where it went wrongThe reversal family, now present on all three papers: 5′/3′ ends on Paper 1, template-strand polarity and promoter/terminator on Paper 2, and sequence polarity here.
CARDS 2 AND 3 · Q17 AND Q19 — THE FIRST GENERATIONI · one misreading, made twice on two different figures
Q17 · Watson–Crick semiconservative modelFigure 7
QuestionEach of the two DNA molecules produced after replication contains …
Her answer — Ctwo newly synthesised strands
Correct — Aone parental strand and one newly synthesised strand
Reading the figureIn the NCERT drawing the parental strands are shown in one shade and the new strands in another. Follow either daughter molecule down the page: one of its two strands runs all the way back into the parent. That strand is old.
She knows thisOn the very next question, Q18, she correctly described what conservative replication would have looked like — one all-old molecule and one all-new. The contrast is understood; the product of the first round is being read wrongly.
Q19 · Meselson and StahlFigure 8
QuestionGeneration I shows a single band of hybrid density because …
Her answer — Ball the molecules still contain two 15N strands
Correct — Cevery molecule contains one 15N strand and one 14N strand
Why B cannot be rightIf the molecules still had two 15N strands they would be heavy, and the band would sit exactly where it did at time zero. The figure shows it has moved, and the tube is labelled Hybrid.
She knows this tooQ20 (Generation II is 1:1) and Q21 (15N is heavy, not radioactive) were both correct. Generation II was read properly; Generation I was not.
Shortcut for bothBefore answering anything about a replication figure, say the sentence: “After one round, every molecule is half old and half new.” Both Q17 and Q19 are answered by it. The parental strands are never destroyed and never end up together again — after the first round there is no heavy molecule left anywhere.
Where it went wrongThese two are worth 8 of the 33 marks lost and they are a single idea, missed on two adjacent figures. Fixing this one sentence is the highest-value thing on the page.
CARD 4 · Q22 — NAMING THE NEW STRANDSS · vocabulary from the wrong process
Q22 · Replicating forkFigure 9
QuestionThe newly synthesised strands marked a and b are respectively called the …
Her answer — Dcoding strand and the template strand
Correct — Bleading strand and the lagging strand
The mix-upTemplate and coding are transcription words — they describe the two strands of the parent DNA when RNA is being made. Leading and lagging are replication words and describe the two new strands. The question asked about newly synthesised strands, so only the replication pair can apply.
She knows the behaviourQ23 (joined by DNA ligase) and Q24 (discontinuous because polymerase works only 5′→3′) were both correct. Everything about how the strands behave was right; only their names were taken from the wrong chapter section.
ShortcutTwo pairs, two processes: replication → leading / lagging (both new); transcription → template / coding (both old). If the question says “newly synthesised”, the answer must come from the first pair.
CARD 5 · Q31 — COUNTING IN THE CODON TABLEI · read one column, missed the other
Q31 · Codon tableFigure 12
QuestionFrom the table, the number of codons that do not code for any amino acid is …
Her answer — B2
Correct — C3
What she missedUAA and UAG sit next to each other in the U–A column and were both counted. UGA is in the U–G column, a block to the right, and was missed. All three are printed “Stop” in the table.
ShortcutLearn them as a phrase rather than as positions on a grid: “U Are Away, U Are Gone, U Go Away” — UAA, UAG, UGA. All three begin with U, but only two of them share a column. Counting off a table invites exactly this error; reciting a list does not.
The rest of the blockQ32 (degeneracy), Q33 (tryptophan as the other single-codon amino acid) and Q34 (a four-codon mRNA giving three amino acids) were all correct — and Q34 required using UGA as a stop codon, which she did.
CARD 6 · Q35 AND Q36 — THE TWO BLANKS8 marks, and the easiest figure on the paper
Q35 & Q36 · Genome sequencing flowFigure 13
Q35 — answer CThe sequence of steps represents determining the complete base sequence of the human genome. The figure runs person → cell → chromosome → DNA → bases on a screen, which is the logic of the Human Genome Project.
Q36 — answer BThe two rows of bases on the screen are the two strands of DNA, complementary and antiparallel. Read them against each other: one is the complement of the other, written in the opposite direction.
Why this mattersThese were the last two questions on the paper and sit on the only figure that needs no labelling at all. Neither requires recall she does not already have. 8 marks were available for about ninety seconds of work.
FixOn a 36-question paper, spend no more than a minute on any single figure question on the first pass; mark the hard ones and come back. Reaching the end of the paper is worth more than perfecting the middle of it, and the last block is often the easiest because it comes from the shortest topic.
WHAT HAS BEEN FIXED — WORTH READING BEFORE THE NEXT PAPERThree named repairs, all holding
| Previously flagged | This paper | Evidence |
|---|---|---|
| Promoter / terminator reversal Cost 4 marks on Paper 2 (Q30) | Correct | Q25 asked exactly that, on the bacterial transcription figure with the labels masked, and she got it right. |
| Exon / intron and cap / tail reversal | Correct | Q28, Q29 and Q30 on the eukaryotic transcription figure were all correct. The transcription block scored 6/6. |
| Nucleosome and packaging confusion H1 versus the octamer, 200 bp | Correct | Q9 to Q13 all correct, including identifying H1 on the masked figure and the 200 bp value. |
| Match-column transposition | n/a | No match questions on this paper. Still due for a re-test. |
The two sentences to add to the pre-paper list. She already writes three polarity lines at the start of a paper. Add these, because between them they answer four of the five errors on this sheet:
- After one round of replication, every molecule is half old and half new — no heavy molecule survives the first generation. (Q17, Q19)
- Leading and lagging are the new strands; template and coding are the old ones. (Q22)
The existing line “a sequence is always written from the 5′ end” covers Q3, and the stop-codon phrase covers Q31.
Re-test. Three blocks need nothing: structure and packaging (13/13), Hershey–Chase (3/3) and transcription (6/6). Build the next short paper from replication figures only — the semiconservative model, the Meselson–Stahl tubes at every generation, and the replicating fork — plus a codon-table drill. Twelve questions is enough. The target is not a higher percentage but reaching the last question with time in hand: at 85% accuracy on what she attempts, finishing the paper is worth more than any further content work.
Error Notes · Molecular Basis of Inheritance, NCERT figure paper · Paper 3 attempted 15/09/2026 · Aamirah Fathima111/144 · 29 correct · 5 wrong · 2 blank · 0 identification errors