Error Notes — ILTS AITS, 4 October

The headline. 554 against 492 on Mock #2 four days earlier — +62, and 14 marks above the 540 danger line. Accuracy on what she attempted is 86.7% (144 of 166). Chemistry has gone from the worst section on the paper to the best. The whole of the remaining gap to 600 now sits in one section.

Section by section, against Mock #2

Section	Correct	Wrong	Blank	Marks	Mock #2	Change
Physics	28	9	8	103	104	-1
Chemistry	41	3	1	161	111	+50
Botany	37	5	3	143	135	+8
Zoology	38	5	2	147	142	+5
Total	144	22	14	554	492	+62

Physics is now the entire problem. 103 marks against 161 in Chemistry — and 17 of the 36 lost items are in Physics alone: 9 wrong and 8 left blank. Those 8 blanks are 32 marks of opportunity, and six of them are pure Lenz's-law direction questions that need no calculation at all. Chemistry (+50), Zoology (+5) and Botany (+8) all moved the right way.

Two patterns worth naming. First, every Physics blank is a reasoning question, not an arithmetic one — Q27, 29, 32, 33, 34, 35 are all Lenz's law in words. She is skipping the questions she could answer fastest. Second, the two-statement and assertion–reason items (Q2, Q34, Q180) still cost marks, and in Q180 the planted word again sits at the end of the sentence.

Lost marks by cause

Cause	Questions
Left blank	Q27, Q29, Q32, Q33, Q34, Q35, Q41, Q44, Q72, Q94, Q99, Q110, Q136, Q180
Assertion–Reason: a false Reason accepted	Q2
Definition confused with Faraday's law	Q5
Vector product mishandled	Q9
Opposition to motion ignored	Q10
Steady versus changing field	Q21
Angle factor dropped	Q25
Charge formula replaced by current	Q26
Direction and sign	Q31
Missing factor of 2	Q43
Statement count short by one	Q54
Group placement	Q56
Reagent behaviour reversed	Q74
Base pairing arithmetic	Q95
Attribution swapped	Q107
Rate of mutation reversed	Q116
Hierarchy order	Q130
Scope of the statement	Q134
Sequence order	Q144
Enzyme class confused	Q146
Insertional inactivation reversed	Q153
Colour logic inverted	Q154
Palindrome construction	Q155

Physics — 103 marks 9 wrong · 8 blank

Wrong answers

Q2Lenz's law / ARwrong

Her answer: Both A and R true, R explains A · Correct: A is true but R is false

Why: A is true: rotating a cylindrical magnet about its own axis changes nothing about the flux through the coil, so no current is induced. R is false: the direction of an induced current is found from Lenz's law (or Fleming's RIGHT hand rule). Fleming's LEFT hand rule gives the force on a current-carrying conductor.
The fix: Judge R on its own before pairing it with A. Left hand = force on a current; right hand = induced current.

Q5Lenz's lawwrong

Her answer: Magnitude of induced emf · Correct: Direction of induced current

Why: Lenz's law supplies only the DIRECTION (the minus sign). The magnitude of the induced emf comes from Faraday's law, ε = −N dΦ/dt.
The fix: One sentence to hold: Faraday gives how much, Lenz gives which way.

Q9Motional emf, vector formwrong

Her answer: 20 volt · Correct: 25 volt

Why: ε = (v × B)·L. With v = 1î and B = 3î + 4ĵ + 5k̂, v × B = 4k̂ − 5ĵ. The rod is L = 5ĵ, so ε = (4k̂ − 5ĵ)·5ĵ = −25, magnitude 25 V. Only the component of B perpendicular to BOTH v and the rod contributes — here that is B_z = 5, giving ε = B_z·l·v = 5 × 5 × 1 = 25 V.
The fix: In a vector field problem, pick the component of B perpendicular to both v and the rod. Choosing B_y = 4 gives the 20 V distractor.

Q10Eddy currentswrong

Her answer: Acceleration equal to g · Correct: Almost constant speed

Why: Falling through a copper tube, the magnet's changing flux drives eddy currents in the tube whose magnetic effect opposes the fall. The retarding force grows with speed until it balances the weight and the magnet descends at an almost constant (terminal) speed.
The fix: Any conductor + relative motion = opposition. Acceleration g is the answer only for a non-conducting tube.

Q21Eddy currentswrong

Her answer: Metal kept in a steady magnetic field · Correct: Metal kept in a varying magnetic field

Why: Eddy currents need a CHANGING flux through the bulk of the conductor. A steady field through a stationary metal changes no flux and induces nothing.
The fix: Steady field + stationary metal = no induction, ever.

Q25Magnetic fluxwrong

Her answer: 0.08 Wb · Correct: 0.02 Wb

Why: Φ = BA cos θ with B = 1/π Wb m⁻², A = πr² = π(0.2)² = 0.04π m², so BA = 0.04 Wb. The axis makes 60° with B, and the axis IS the normal, so Φ = 0.04 cos 60° = 0.02 Wb.
The fix: Write BA first, then apply cos θ. Here the π cancels, which is the point of the numbers chosen.

Q26Induced chargewrong

Her answer: 9 C · Correct: 7.5 C

Why: q = ΔΦ/R, and ΔΦ = φ(3) − φ(0) = (9 + 6) − 0 = 15 Wb. So q = 15/2 = 7.5 C. The time does not appear anywhere in the charge.
The fix: Charge needs only the change of flux and R. 9 C comes from using the emf at t = 3 s (dφ/dt = 8 V, I = 4 A) and multiplying by something.

Q31Lenz's law, two loopswrong

Her answer: Clockwise in both A and B · Correct: Clockwise in A, anticlockwise in B

Why: The two loops sit on opposite sides of the wire, so the field of the wire threads them in OPPOSITE senses. As I increases, each loop opposes its own growing flux — and because the fluxes point opposite ways, the induced currents must circulate in opposite senses.
The fix: When two loops straddle a wire, settle the direction of B through each loop separately before applying Lenz's law.

Q43Faraday's law, reversalwrong

Her answer: 5.0 V · Correct: 10.0 V

Why: Reversing the flux changes it from +Φ to −Φ, so ΔΦ = 2Φ = 2 × 500 µWb = 10⁻³ Wb. Then ε = N ΔΦ/Δt = 200 × 10⁻³ / (20 × 10⁻³) = 10 V.
The fix: The word 'reversed' always doubles ΔΦ. 5.0 V is exactly the answer without that factor of two.

Left blank

Q27Lenz's law, directionblank

Left blank · Correct: B to A and D to C

Why: The field into the page is increasing, so the induced current opposes it and flows anticlockwise round the frame as drawn. Traced round the frame, that is B to A in one wire and D to C in the other.
The fix: Flux into page and growing → anticlockwise → read the lettering off the figure.

Q29Lenz's law and energyblank

Left blank · Correct: Deflection both sides, amplitude steadily decreasing

Why: The magnet moves in and out, so the flux rises and falls alternately and the galvanometer swings both ways. The induced currents oppose the motion and dissipate energy, so the oscillation — and with it the deflection — dies away.
The fix: Alternating motion → alternating deflection; opposition → decaying amplitude.

Q32Lenz's law, switch on and offblank

Left blank · Correct: Clockwise pulse, then anticlockwise pulse

Why: Closing the switch makes the flux through the loop grow: one pulse opposing the growth. Opening it makes the flux die: a second pulse in the opposite sense, supporting the dying flux.
The fix: Switch ON and switch OFF are mirror images — one pulse each way.

Q33Lenz's law, approaching loopsblank

Left blank · Correct: The current in each loop will decrease

Why: The loops carry current the same way, so as they approach, the flux each one links from the other increases. Each loop opposes that increase, and the opposing induced emf reduces the current in it.
The fix: Approach → flux grows → induced effect opposes → current falls.

Q34Faraday and Lenz, two-statementblank

Left blank · Correct: Statement I false, Statement II true

Why: Statement I is false: the emf depends on the RATE of change of flux, not on how large the flux is. Statement II is true: Lenz's law is a direct consequence of the conservation of energy.
The fix: The flux value versus its rate of change — the single commonest false statement in this chapter.

Q35Eddy currents, dampingblank

Left blank · Correct: Come to rest very soon

Why: As the metallic ring swings through the field, the flux through it changes and eddy currents are induced. Their magnetic effect opposes the motion, so the pendulum is damped and stops quickly.
The fix: Metal + field + motion = electromagnetic damping. Same physics as Q10 and Q21 on this paper.

Q41Motional emfblank

Left blank · Correct: 10⁻³ volt

Why: ε = B_V l v with B_V = 0.2 × 10⁻⁴ Wb m⁻², l = 1 m and v = 180 km/h = 50 m s⁻¹. So ε = 0.2 × 10⁻⁴ × 1 × 50 = 10⁻³ V.
The fix: Convert km/h to m/s first (÷3.6). Everything else is one multiplication.

Q44Motional emf, free fallblank

Left blank · Correct: 1.20 mV, with P at higher potential

Why: After 2 s of free fall v = gt = 20 m s⁻¹. ε = B_H l v = 0.3 × 10⁻⁴ × 2 × 20 = 1.2 × 10⁻³ V = 1.20 mV. The force on the positive carriers, qv × B, drives them towards P, so P is at the higher potential.
The fix: Two steps: v = gt first, then Blv. The potential end comes from v × B, not from guesswork.

Chemistry — 161 marks 3 wrong · 1 blank

Wrong answers

Q54Anion analysis, chromyl chloridewrong

Her answer: 3 · Correct: 4

Why: All four are correct. NaCl with K₂Cr₂O₇ and conc. H₂SO₄ gives chromyl chloride CrO₂Cl₂, a deep red vapour; passed into NaOH it gives yellow Na₂CrO₄; and chlorine is evolved as well. Every line in the list is a standard observation of the chromyl chloride test.
The fix: In a count question, verify each line separately — stopping at three means one true line was rejected on suspicion.

Q56Cation analysis, group orderwrong

Her answer: B, C, A, D, E · Correct: B, A, D, C, E

Why: By analysis group: Cu²⁺ is Group II, Al³⁺ Group III, Co²⁺ Group IV, Ba²⁺ Group V, Mg²⁺ Group VI. In increasing group number that is B, A, D, C, E.
The fix: Learn the groups as a ladder: I Pb · II Cu · III Al, Fe · IV Zn, Mn, Ni, Co · V Ba, Sr, Ca · VI Mg.

Q74Anion analysis, sulphate vs carbonatewrong

Her answer: Na₂SO₄ · Correct: Na₂CO₃

Why: Barium carbonate is white but DISSOLVES in dilute HCl, so in the presence of the acid no precipitate appears — which is what the question asks for. Barium sulphate is white and insoluble in acid, so Na₂SO₄ gives the precipitate either way.
The fix: Read the 'but not in presence of dil. HCl' clause: it is asking which precipitate the acid destroys.

Left blank

Q72Cation analysis, Group Vblank

Left blank · Correct: Ammonium oxalate

Why: Calcium is confirmed by a white precipitate of calcium oxalate with ammonium oxalate, insoluble in acetic acid. Potassium chromate is barium's test.
The fix: Group V confirmations: Ba → yellow chromate, Ca → white oxalate, Sr → crimson flame.

Botany — 143 marks 5 wrong · 3 blank

Wrong answers

Q95DNA structure, hydrogen bondswrong

Her answer: 240 · Correct: 130

Why: 20 adenines means 20 thymines, so 20 A–T pairs with 2 bonds each = 40. The remaining 100 − 40 = 60 bases are G and C, which is 30 G–C pairs with 3 bonds each = 90. Total 40 + 90 = 130.
The fix: Count PAIRS, not bases: 100 bases = 50 pairs. The 240 distractor comes from treating all 100 bases as pairs.

Q107History of DNA, matchwrong

Her answer: A–IV, B–III, C–II, D–I · Correct: A–III, B–IV, C–II, D–I

Why: Crick is the central dogma (III); Wilkins is the X-ray diffraction data (IV). She exchanged the two. Chargaff (A = T ratios) and Miescher (DNA as an acidic substance) were correct.
The fix: Fix the pair that is always swapped: Wilkins and Franklin took the X-ray pictures; Crick wrote the central dogma.

Q116Properties of genetic materialwrong

Her answer: I and II only · Correct: II, III and V only

Why: Statement I is FALSE: RNA, being unstable, mutates at a FASTER rate, not a slower one. II (thymine confers stability), III (DNA is the genetic material in all cellular organisms and some acellular ones) and V (DNA is acidic) are true. IV is false — DNA is digested by nucleases, not ligases.
The fix: Two reversals in one question: RNA mutates faster, and ligase joins while nuclease digests.

Q130Organisation of genetic materialwrong

Her answer: Chromosome, gene, genome, nucleotide · Correct: Genome, chromosome, gene, nucleotide

Why: The genome is the whole genetic complement, which contains the chromosomes; a chromosome carries many genes; a gene is made of nucleotides.
The fix: Largest to smallest: genome ⊃ chromosome ⊃ gene ⊃ nucleotide.

Q134Genetic materialwrong

Her answer: Most of the prokaryotes · Correct: All organisms

Why: Nucleic acid — DNA or RNA — is the genetic material in ALL organisms. Restricting it to prokaryotes or to 'most' organisms is the error; the exceptions people think of (RNA viruses) are still nucleic acid.
The fix: The word is nucleic ACID, not DNA. That makes the statement universal.

Left blank

Q94DNA structure, matchblank

Left blank · Correct: A–II, B–III, C–IV, D–I

Why: Length of one coil (pitch) = 3.4 nm; bases per turn = 20… the paper's key pairs one coil with 3.4 nm (II), bases in one turn with 20 (III), diameter with 20 Å (IV) and adenosine with nucleoside (I).
The fix: Four numbers worth memorising together: 0.34 nm per base pair, 3.4 nm per turn, 20 Å diameter, 10 bp per turn.

Q99Nucleotide chemistry, matchblank

Left blank · Correct: A–II, B–III, C–I

Why: The N-glycosidic bond joins the base to the pentose sugar (II); the phosphoester linkage is found in adenylic acid, a nucleotide (III); the phosphodiester linkage is what joins nucleotides in NAD and in a polynucleotide chain (I).
The fix: Three bonds, three places: base–sugar, sugar–phosphate, nucleotide–nucleotide.

Q110DNA replication, polymeraseblank

Left blank · Correct: About 2000 bp per second in E. coli

Why: DNA polymerase uses deoxyribo-nucleoside triphosphates, works only on a template and cannot itself open the helix — so the only correct statement is the polymerisation rate of about 2000 bp per second in E. coli.
The fix: Note the wording trap: 'nucleotide triphosphates' versus 'deoxyribonucleoside triphosphates'.

Zoology — 147 marks 5 wrong · 2 blank

Wrong answers

Q144Principles of biotechnologywrong

Her answer: B, C, A · Correct: C, B, A

Why: The three steps are: identify the DNA with the desirable gene (C), introduce it into the host (B), then maintain it in the host and pass it to the progeny (A).
The fix: You cannot introduce a gene you have not identified. Identification always comes first.

Q146PCRwrong

Her answer: RNA dependent RNA polymerase · Correct: DNA dependent DNA polymerase

Why: Taq polymerase copies a DNA template into DNA, so it is a DNA-dependent DNA polymerase. It is simply a thermostable one, isolated from Thermus aquaticus.
The fix: Name the enzyme by template → product: DNA template, DNA product.

Q153Cloning vectors, pBR322wrong

Her answer: The Pst I statement · Correct: The BamH I / tetracycline statement

Why: BamH I lies inside the tetᴿ gene. Insert alien DNA there and tetracycline resistance is destroyed, so the recombinant will NOT grow on tetracycline — that is the incorrect statement the question asks for. Pst I lies in ampᴿ, so after insertion there ampicillin can no longer distinguish anything, which is a correct statement.
The fix: On the pBR322 map, fix which gene each site sits in: BamH I and Sal I in tetᴿ; Pst I, Pvu I and Sca I in ampᴿ.

Q154Blue–white screeningwrong

Her answer: Desired gene causes insertional inactivation of Tetᴿ · Correct: Desired gene FAILS to inactivate β-galactosidase

Why: Blue colonies have an intact β-galactosidase, which cleaves the chromogenic substrate and makes the colour — so blue means NO insert, that is, a non-recombinant. White (colourless) colonies are the recombinants.
The fix: Blue = no insert = non-recombinant. White = insert = recombinant.

Q155Restriction enzymes, palindromeswrong

Her answer: CTG · Correct: GAC

Why: A palindromic recognition site reads the same on both strands 5′→3′. With the first half 5′-GTC, the full site must be 5′-GTCGAC-3′, whose complement read 5′→3′ is also GTCGAC. So the next three nucleotides are GAC.
The fix: Build it: write the complement of the first half, then reverse it. Complement of GTC is CAG; reversed, GAC.

Left blank

Q136Restriction enzymesblank

Left blank · Correct: 5

Why: Each cut on a LINEAR molecule adds one fragment, so n cuts give n + 1 pieces: 4 recognition sites → 5 fragments. (On a circular molecule, n cuts give n fragments.)
The fix: Linear: cuts + 1. Circular: cuts. One line, two cases.

Q180History of biotechnology, two-statementblank

Left blank · Correct: Both statements are incorrect

Why: Statement I is wrong: Rene Descartes was a French philosopher and mathematician, not a German biologist. Statement II is wrong because Boyer and Cohen constructed the first ARTIFICIAL recombinant DNA — the word 'natural' makes the sentence false.
The fix: The planted word sits at the end again: 'natural' recombinant DNA. Read the last word of every statement.

What to do next

Priority	What	Worth
1	The eight Physics blanks — Q27, 29, 32, 33, 34, 35, 41, 44. Six are Lenz's law in words and need no calculation.	32 marks
2	Re-do the 30-question EMI direction re-test under a clock; the three-line routine (flux → growing or dying → opposing current) covers Q27, 31, 32, 33, 35 and the blanks.	~20 marks
3	Physics factor traps: reversal doubles ΔΦ (Q43), flux needs cos θ (Q25), charge is ΔΦ/R with no time in it (Q26).	12 marks
4	Biotechnology pairs that invert: blue = non-recombinant (Q154), BamH I sits in tetᴿ (Q153), palindrome construction (Q155).	12 marks
5	Botany: hydrogen-bond arithmetic by pairs (Q95), the Crick/Wilkins attribution (Q107), RNA mutates faster (Q116).	12 marks
6	Chemistry — maintain, do not add. 161 with one blank is the model for the other three sections.	—

The arithmetic of the next 46 marks. Attempting the 14 blanks at her current 86.7% accuracy is worth about +44. Fixing nothing else, that is 598. The Physics section alone, brought to the level the other three already sit at, closes the gap to 600.