Two-thirds of the loss sits in one block, and most of that block traces to a single sentence — see section 3.
2. What went right
All six match-the-following questions correct (Q9, 24, 32, 44, 53, 77). This was her worst format in the 20 Sep ILTS; working the rows in the margin has fixed it.
Statement-count questions 5 / 5 (Q12, 18, 35, 41, 70) — and none left blank, unlike the ILTS where eight statement questions were skipped.
Assertion–Reason 5 / 6 (only Q62 missed).
Meselson–Stahl and Taylor, Q71–80: 10 / 10, including the hybrid-DNA proportion and the band figure.
RNA world, Q37–44: 8 / 8.
She circled the word NOT in Q8 and the word 'no' in Q40 — exactly the habit that stops planted-word errors.
3. One sentence fixes four questions
A new nucleotide can only be joined to the free 3′-OH end of the growing chain, so every new strand grows 5′→3′. Q49 asked it directly; Q50, Q58 and Q62 are all consequences of it. That is 20 marks from one sentence.
Method for every fork figure: 1. Draw the arrow showing where the fork is moving. 2. The strand that can keep growing 5′→3′ in that direction is the leading strand — its template runs 3′→5′ towards the fork. 3. The other template gives Okazaki fragments, whose arrows point away from the fork.
The Search for the Genetic Material — 3 wrong · 0 blank
Q2WrongNNCERT definitionGriffith's definition of transformation
In Griffith's terms, transformation is?
Her answerThe recovery of heat-killed S bacteria in the body of the mouse
Correct answerThe change of a living R bacterium into an S bacterium by a substance from dead S cells
WhyTransformation is the conversion of the harmless R strain into the virulent S strain by some substance passing from the dead S cells. Her option describes something that never happened — heat-killed cells are not 'recovered'; what Griffith recovered from the dead mice were LIVING S bacteria.
Where it went wrong · fixDefinition question. Her tick sat on the option below the right one — see the note on tick placement at the end.
Q19WrongSArrangement / orderOrder of the three discoveries
Arrange in the order established: P DNA enters the bacterium during phage infection · Q a transforming principle exists · R the transforming principle is DNA.
Her answerP → Q → R
Correct answerQ → R → P
WhyGriffith 1928 showed a transforming principle exists (Q). Avery, MacLeod and McCarty 1944 showed it is DNA (R). Hershey and Chase 1952 showed DNA enters the bacterium (P).
Where it went wrong · fixSequence question — fix the first and last items by their years before choosing. The dates were answered correctly in Q1 and Q11.
The bacterium used by Hershey and Chase as the host was?
Her answerStreptococcus pneumoniae
Correct answerEscherichia coli
WhyHershey and Chase infected E. coli with bacteriophage T2. Streptococcus pneumoniae is Griffith's and Avery's organism.
Where it went wrong · fixAttribution swap between the two classic experiments. Card: Griffith and Avery → pneumococcus · Hershey–Chase → E. coli + T2 phage.
Properties of the Genetic Material — 1 wrong · 0 blank
Q34WrongSSilly / recall slipBase that replaces uracil
In DNA, the base that replaces uracil is?
Her answerAdenine
Correct answerThymine
WhyDNA uses thymine (5-methyl uracil) in place of RNA's uracil, and that methyl group is part of what makes DNA more stable.
Where it went wrong · fixA one-second recall answered correctly elsewhere in the paper (Q32 links thymine with extra stability). Read the four options before ticking.
DNA Replication — 6 wrong · 3 blank
Q45WrongCConcept gapWhat semiconservative means
In semiconservative replication, each daughter DNA molecule contains?
Her answerStrands that are a patchwork of old and new
Correct answerOne parental strand and one newly made strand
WhySemiconservative = one strand conserved from the parent, one newly made. The 'patchwork' option describes the DISPERSIVE model, which Meselson and Stahl ruled out.
Where it went wrong · fixShe named the dispersive model. Three models card: conservative (both old / both new) · semiconservative (one each) · dispersive (patchwork).
Q49WrongCConcept gapWhere a new nucleotide is added
A new nucleotide can be added only to the?
Her answer5′-phosphate end of the growing chain
Correct answerFree 3′-OH end of the growing chain
WhyDNA polymerase joins the incoming nucleotide's phosphate to the free 3′-OH of the growing chain, which is why synthesis runs 5′→3′ only.
Where it went wrong · fixThis is the root error of the whole replication block. Q50, Q58 and Q62 all follow from it. Learn it as one sentence: new nucleotides go on the free 3′-OH, so the new strand grows 5′→3′.
Q50WrongCDirection reversalTemplate for the continuous strand
In the fork shown, P is made continuously. The template on which it is built runs?
Her answer3′→5′ away from the fork
Correct answer3′→5′ towards the fork
WhyThe leading strand grows 5′→3′ following the fork, so its template must run 3′→5′ in the same direction — towards the fork. The template that runs 5′→3′ towards the fork gives the lagging strand (Okazaki fragments), which she answered correctly in Q51.
Where it went wrong · fixShe had the polarity right (3′→5′) and the direction wrong. On any fork figure, mark the fork's direction of travel first, then label each template.
Q55WrongNNCERT definitionReplication fork vs origin
The small opening of the helix in which replication takes place is called the?
Her answerOrigin of replication
Correct answerReplication fork
WhyThe origin is the fixed SITE where replication starts; the fork is the moving Y-shaped opening where synthesis happens.
Where it went wrong · fixVocabulary pair — both appeared in the same question set (Q53 matched the origin correctly). Card: origin = where it starts · fork = where it is happening now.
Q56BlankTBlank — not attemptedWhy only a small region opens
If the two strands of a long DNA molecule were separated completely before replication began, then?
Her answerLeft blank
Correct answerAn enormous amount of energy would be needed, which is why only a small region is opened at a time
WhyUnwinding a whole long molecule would demand a huge amount of energy and the separated strands would also supercoil, so the helix is opened only within the fork.
Where it went wrong · fixBlank. This is the NCERT line that explains why replication happens inside a small fork — it links directly to Q55.
Q58WrongCDirection reversalDirection of fork travel
The overall direction in which a replication fork moves along the DNA is?
Her answerOpposite to the direction of continuous synthesis
Correct answerThe same as the direction of continuous synthesis
WhyThe leading strand is made continuously as it chases the fork, so the fork and the continuous synthesis travel the same way. Okazaki fragments are the ones made away from the fork (answered correctly in Q68).
Where it went wrong · fixSame root as Q49 and Q50. Draw the fork, mark its direction with an arrow, then ask which strand can keep following it.
Q61BlankTBlank — flagged, never revisitedStatement count on polymerase and ligase
A: DNA polymerase uses deoxyribonucleoside triphosphates as substrates. B: Replication is semiconservative. C: DNA ligase joins two fragments by a phosphodiester bond. D: DNA polymerase can start a new chain from a bare template with no free 3′-OH.
Her answerLeft blank
Correct answerOnly three (A, B and C)
WhyD is false: polymerase cannot start from nothing — it needs a free 3′-OH, which is why a primer is required. A, B and C are all NCERT.
Where it went wrong · fixShe circled this question to come back to it and never did. Same statement D as Q49 — one concept, two lost questions.
Q62WrongCConcept gapA–R: why not 3′→5′
A: Replication cannot proceed in the 3′→5′ direction. R: DNA polymerase adds a nucleotide only to the free 3′-OH end of the growing chain.
Her answerA is false but R is true
Correct answerBoth true, R is the correct explanation of A
WhyA is true — no polymerase makes a strand 3′→5′. R is true and is exactly the reason.
Where it went wrong · fixShe marked A false, which contradicts what she answered in Q72. Judge A on its own first: could a new strand ever grow 3′→5′? No.
Q65BlankTBlank — flagged, never revisitedDNA amount after replication
In a cell that has just completed replication but has not divided, the amount of DNA is?
Her answerLeft blank
Correct answerTwice that of the cell before replication
WhyReplication doubles the DNA; division then halves it back. Between the two, the cell holds double.
Where it went wrong · fixAlso circled and never returned to. Flagged questions must be answered before the paper ends — an unanswered flag is a guaranteed zero.
Fix list
Replication direction block (Q45, 49, 50, 55, 58, 62): redraw the fork three times from memory with every label, then re-solve these six cold.
Answer every flagged question before the paper ends. Q61 and Q65 were circled to return to and never answered — 8 marks for two questions she could do. Rule: at the 70-minute mark, stop and clear all flags.
Vocabulary card: origin = where replication starts · fork = the moving opening · leading = continuous · lagging = Okazaki · ligase = seals · polymerase = builds.
Three models card: conservative (both old / both new) · semiconservative (one each) · dispersive (patchwork).
Experiment card: Griffith and Avery → pneumococcus · Hershey–Chase → E. coli and T2 phage · 1928 → 1944 → 1952.
Read all four options before ticking (Q34, thymine).
Note on tick placement. On three questions (Q2, Q19, Q21) her tick sits on the letter of the option immediately BELOW the correct one, and in each case the option above is the right answer. Where an option wraps onto two lines, the tick can drift to the next letter. These were graded as marked; if she confirms she meant the option above in all three, the score becomes 273. Ask her to place the tick beside the option letter, not in the gap between lines.
Error tags: C concept · N NCERT/definition · S silly or recall slip · T not attempted. Scope: search for the genetic material, properties of genetic material, the RNA world, DNA replication and the Meselson–Stahl and Taylor experiments.