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B1 ILTS – 09

Molecular Basis of Inheritance

Botany · Class 12 Chapter 6 · error and blank review

Test date 06-09-2026Marks obtained 526NEET 2027 · Gr 12 Live Full Course
6questions in this chapter
1answered wrong
5left blank

Weak areas from this paper

Reading past the operative wordrecurring

Q92

“Sugar, base and phosphate” is a correct sentence — about a nucleotide. The question asked about the backbone. The chemistry was known; the noun was not read closely enough.

Q120 on the same paper tests the neighbouring distinction (nucleoside versus nucleotide) and was left blank, so this vocabulary layer is the thing to shore up, not the structural biology underneath it.

Fix to drill. Underline the operative noun in every structural question before looking at the options: backbone, nucleotide, nucleoside, base pair, monomer. Each has a different answer, and papers exploit the overlap.

Five of six left blank in a high-yield chapterattempt rate

Q101, 118, 119, 120, 132

Q120 asks for the monomer of a nucleic acid. Q119 asks which histone lies outside the core. Q132 restates the NCERT description of the double helix. These are single-sentence recall.

Q118 is the only one needing any working, and it is two multiplications.

Molecular Basis of Inheritance is one of the most heavily weighted chapters in NEET Biology, so a near-total blank here is expensive out of proportion to its difficulty.

Fix to drill. Re-attempt these five closed-book with no time limit first, to separate “did not know” from “did not attempt”. Whatever you answer correctly on that untimed pass was a confidence problem, not a knowledge one, and should be drilled for speed rather than re-taught.

Assertion-reason, the same gap as Physics and Chemistryattempt rate

Q101 blank here; Q15, Q44 wrong in Physics; Q54, Q72 wrong in Chemistry

Q101 was not attempted, so it adds no new diagnostic information on its own. But taken with the four assertion-reason errors elsewhere on this paper, the pattern is consistent: this format is being avoided when it is not being answered in the wrong direction.

Q101 is worth revisiting because it is the easiest of the five. Both statements are false, and the date alone settles it.

Fix to drill. Practise the format across subjects in one sitting rather than chapter by chapter. The three-step written defence — verdict on A, verdict on R, then options — is subject-independent, and this paper shows it is the single highest-value habit available right now.

Bases versus base pairswatch

Q118

Q118 was blank, so there is no evidence yet of an error here. It is flagged because the conversion it hinges on — 100 bases means 50 pairs — is the one that generates every wrong option in the question, and it recurs throughout this chapter in E. coli length calculations and in Chargaff problems.

Related note already on file: the NCERT E. coli DNA length figure does not reconcile with its stated base-pair count. Always compute from the base-pair number given in the question, never from the remembered length.

Fix to drill. In any nucleic-acid calculation, write down the number of base pairs as your first line, even when the question gives bases. Every subsequent step uses pairs.
Question by question
Q92Marked wrongStructure of the polynucleotide chain

Backbone of polynucleotide chain of DNA is made up of

  • sugar and base
  • base and phosphate
  • sugar and phosphate
  • sugar, base and phosphate
Given
A DNA polynucleotide chain.
Asked
What its backbone is made of.
Concepts

A nucleotide has three parts: phosphate, pentose sugar, and nitrogenous base. But only two of them are in the chain.

The chain is built by phosphodiester bonds, which run from the 3' carbon of one sugar, through a phosphate, to the 5' carbon of the next sugar. Sugar, phosphate, sugar, phosphate — and never a base.

Each base hangs off its own sugar by an N-glycosidic bond, projecting sideways into the middle of the helix. It is a side group, not a link in the chain.

Formula
backbone: sugar – phosphate – sugar – phosphate
linkage: 3',5'-phosphodiester bond

base attaches to sugar by N-glycosidic bond
and projects inward
Baby steps
  1. Ask what is physically joined end to end: the phosphodiester bond connects sugars via phosphates.
  2. The bases branch off the sugars sideways; remove all of them and the chain still holds together.
  3. So the backbone is sugar and phosphate only.
  4. The word backbone is doing all the work in this question — the distractor exists purely for anyone who reads it as “composition”.
Answer — sugar and phosphate
You marked: “sugar, base and phosphate” — all three components of a nucleotide were listed. But the question asks what forms the backbone, not what a nucleotide contains. Bases are attached to the backbone; they are not part of it.
the backbone is the outer rail only — bases hang off it, inwardsPSPSPSPSPSPSPSPSPSPSATGCTACGAT5'3'3'5'backbone = sugar + phosphate onlythe bases are attached to it, not part of itphosphodiester bonds join sugar to phosphate — no base is ever in the chain

Sugar and phosphate form the rails; bases project inward as rungs.

Shortcut
In any DNA question, three words point in three directions: backbone = sugar + phosphate; nucleotide = all three; base pair = the two bases only. Read which word was used before choosing.
Q101Left blankAssertion and reason · history of DNA isolation

Assertion (A): Isolation of intact DNA was feasible during the 19th century.
Reason (R): Isolation techniques were well-developed by the time of Friedrich Meischer.

  • Both (A) and (R) are true and (R) is the correct explanation of (A).
  • Both (A) and (R) are true but (R) is not the correct explanation of (A).
  • (A) is true but (R) is false.
  • Both (A) and (R) are false.
Given
Two historical claims about DNA isolation in the 19th century and about Friedrich Miescher's era.
Asked
Whether each is true.
Concepts

Assertion. Miescher identified nuclein in 1869 — a crude acidic precipitate from the nuclei of pus cells. That was a real discovery, but it was not intact DNA. Isolating DNA whole required tools that did not exist yet: restriction enzymes, gel electrophoresis, gentler lysis chemistry. These are 20th-century developments. So A is false.

Reason. Techniques were not well-developed in Miescher's time — that is precisely why the DNA he obtained was fragmented. R is false too.

Note the internal logic: if R were true, A could plausibly follow. Because R is false, A loses its only support. Both fall together, which is the pattern the question is built around.

Formula
1869 Miescher isolates 'nuclein' (fragmented)
1953 Watson and Crick, double helix
1970s restriction enzymes, intact DNA isolation

intact isolation is a 20th-century technique
Baby steps
  1. Test A: was intact DNA isolated in the 1800s? No — Miescher got nuclein, not intact DNA. False.
  2. Test R: were isolation techniques well developed then? No — they were primitive, which is why the DNA broke up. False.
  3. Both false.
  4. Also check the spelling of the name: the paper writes “Meischer”; NCERT has Miescher. A misspelling never changes the answer, so do not let it slow you down.
Answer — Both A and R are false
1869Miescher isolates'nuclein' from pus cells1953Watson and Crickdouble helix1970srestriction enzymes,intact DNA at lastisolating DNA whole is a 20th-century achievement, not a 19th-century oneboth A and R are false

The gap between naming a substance and isolating it whole is about a century.

Shortcut
For history-of-biology assertion-reason questions, anchor on the date. If the claim puts a modern technique in the 19th century, it is almost always false. 1869 is the only date this chapter needs from that century.
Q118Left blankCounting hydrogen bonds

How many hydrogen bonds are possible in a DNA having 20 adenine bases in a total of 100 bases?

  • 100
  • 130
  • 280
  • 240
Given
Double-stranded DNA; total bases = 100; adenine = 20.
Asked
The total number of hydrogen bonds.
Concepts

Three facts do all the work:

Chargaff. A = T and G = C in double-stranded DNA.

Bond counts. A–T is held by 2 hydrogen bonds; G≡C by 3.

Bases versus pairs. 100 bases is 50 pairs. This is where the question is won or lost — every wrong option comes from getting this conversion wrong or skipping it.

Formula
A = T, G = C (Chargaff)
A–T → 2 H-bonds, G≡C → 3 H-bonds

total bases = 2 × total base pairs
Baby steps
  1. A = 20, so T = 20. Together they account for 40 of the 100 bases, which is 20 A–T pairs.
  2. The remaining bases are 100 − 40 = 60, all G and C. That is 30 G≡C pairs.
  3. Bonds from A–T: 20 × 2 = 40. Bonds from G≡C: 30 × 3 = 90.
  4. Total: 40 + 90 = 130.
Answer — 130
100 bases means 50 base pairs — halve before you count100 bases total÷ 250 base pairsA = 20so T = 2020 A–T pairsA + T = 40100 − 40 = 60G + C = 60G + C = 60÷ 230 G≡C pairs20 A–T pairs × 2 bonds = 4030 G≡C pairs × 3 bonds = 90total = 130

Bases to pairs, then pairs to bonds — two conversions, in that order.

Shortcut
Sanity-check the size of your answer before committing. With 50 pairs, the minimum possible is 50 × 2 = 100 and the maximum is 50 × 3 = 150. Any answer outside 100–150 is wrong by inspection — which kills both 280 and 240 instantly.
Q119Left blankThe linker histone

Type of histone molecule that lies outside the nucleosome core and seals the two turns of DNA by binding at the point where DNA enters and leaves the core is

  • H1
  • H2A, H2B
  • H3
  • H4
Given
A description of a histone that sits outside the core and clamps the DNA at its entry and exit points.
Asked
Which histone this is.
Concepts

The histone octamer is made of two copies each of H2A, H2B, H3 and H4 — eight proteins forming the core. About 200 base pairs of DNA wrap around it in roughly 1.75 turns, and that assembly is the nucleosome.

H1 is not in the octamer. It sits on the outside, binding at the single point where the DNA enters and leaves, and holds the two turns together. That is why it is called the linker histone.

The phrase “outside the core” in the question is the whole clue.

Formula
core octamer: 2 × (H2A, H2B, H3, H4) = 8
linker histone: H1, outside the core

nucleosome ≈ 200 bp DNA in 1.75 turns
Baby steps
  1. Recall the four core histones: H2A, H2B, H3, H4, two of each.
  2. H1 is the only one not in that list — and the question says “outside the nucleosome core”.
  3. H1 binds where the DNA enters and exits, sealing the turns.
  4. Answer: H1.
Answer — H1
H1 sits outside the core and clamps the DNA where it enters and leavescoreDNA, 1.75 turns · ~200 bpH1seals the two turnswhich histone does whatH2A, H2B, H3, H4two of each → the octamer coreH1outside the core · the linker histoneH3 and H4 are the mostconserved proteins knownnucleosome = core octamer + ~200 bp DNA; H1 is the clamp, not part of the core

The octamer core, the wrapped DNA, and H1 clamping the entry-exit point.

Shortcut
Only one histone is spoken of alone in NCERT, and it is H1. The other four are always named as a group. If a question describes a single named histone doing something distinctive, it is H1.
Q120Left blankThe monomer of a nucleic acid

The unit of nucleic acid is

  • nucleotide
  • nucleoside
  • pentose sugar
  • monosaccharide
Given
Nucleic acids (DNA and RNA).
Asked
Their repeating structural unit.
Concepts

Build it up one piece at a time, and the names follow:

base → base + sugar = nucleoside → nucleoside + phosphate = nucleotide.

The phosphate is the piece that matters. Only a nucleotide has one, and only a phosphate can form the phosphodiester bond that links units into a chain. A nucleoside is a dead end — nothing to polymerise with.

So the monomer of a nucleic acid is the nucleotide.

Formula
base + sugar = nucleoside
base + sugar + phosphate = nucleotide

nucleotide is the monomer, because only
the phosphate can form the next bond
Baby steps
  1. Recall that the linkage in a nucleic acid is a phosphodiester bond, so the monomer must carry a phosphate.
  2. A nucleoside has no phosphate, so it cannot be the repeating unit.
  3. Pentose sugar and monosaccharide are components, not units — and monosaccharide is the monomer of a polysaccharide, a different polymer entirely.
  4. The unit is the nucleotide.
Answer — nucleotide
add one piece at a time — the names change as you gobasebase+ sugarnucleosidesugarbase+ phosphatenucleotidephosphatesugarbasethe repeating unit of a nucleic acid is the nucleotide — it is the one that can polymerise

Base, then nucleoside, then nucleotide — the phosphate makes the monomer.

Shortcut
Match every biomolecule to its monomer as a set: protein → amino acid; polysaccharide → monosaccharide; nucleic acid → nucleotide; fat → fatty acid and glycerol. Learning them together stops one being borrowed for another, which is exactly what the “monosaccharide” option invites.
Q132Left blankTwo statements on the double helix

Statement-I: DNA is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate, and bases project inside.
Statement-II: The two chains in DNA are coiled in right handed fashion.

  • Both statement I and II are correct
  • Both statement I and II are incorrect
  • Statement I is correct, Statement II is incorrect
  • Statement I is incorrect, Statement II is correct
Given
Two descriptive statements about the Watson–Crick double helix.
Asked
Whether each is correct.
Concepts

Statement I is the standard NCERT description, and it is also the answer to Q92 on this same paper. Sugar-phosphate backbone outside, bases projecting inward and pairing in the middle. Correct.

Statement II. B-DNA, the physiological form, is right-handed. Correct. Left-handed DNA does exist — Z-DNA — but it is a rare alternative form, not the standard structure being described here.

Both correct, and each is independently verifiable without reference to the other.

Formula
B-DNA (the usual form):
  right-handed helix
  pitch 3.4 nm, 10 bp per turn
  rise 0.34 nm per base pair

Z-DNA: left-handed, rare
Baby steps
  1. Check Statement I against the structure: backbone outside, bases inside. Correct.
  2. Check Statement II: B-DNA is right-handed. Correct.
  3. Both correct.
  4. Cross-reference: this paper also asks Q92 on the backbone. The two questions test the same sentence of NCERT, so a firm answer to one settles the other.
Answer — Both statements are correct
right-handed: follow a strand upward and it turns clockwise, away from youright-handed B-DNAStatement Ibackbone is sugar-phosphate,bases project inside✓Statement IIthe two chains coil in aright-handed fashion✓

Right-handed B-DNA with the backbone outside and the bases paired inside.

Shortcut
Keep the four B-DNA numbers together as one block: right-handed, 10 bp per turn, 3.4 nm pitch, 0.34 nm rise. Nearly every structural question on this chapter draws from that block, and remembering them as a set is easier than remembering them one by one.