Q.If the sequence of one strand of DNA is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Write down the sequence of complementary strand in 5' → 3' direction.

Step 1 — The five processes, in the order they must happen.

1. Fragmentation. Detritivores (e.g. earthworms) physically break the detritus into smaller particles. This must be first: it hugely increases the surface area on which everything else acts.
2. Leaching. Water percolating through the fragmented litter washes water-soluble inorganic nutrients down into the soil, where they get precipitated as unavailable salts.
3. Catabolism. Bacterial and fungal enzymes degrade the detritus into simpler inorganic substances. (This is the chemical breakdown, and it needs the surface area created in step 1.)
4. Humification. The three steps above operate simultaneously on the litter, and they leave behind a dark-coloured, amorphous, highly resistant substance called humus — which is highly resistant to microbial action and decomposes very slowly, acting as a nutrient reservoir.
5. Mineralisation. Some microbes finally degrade the humus, releasing inorganic nutrients into the soil. This is the release step, so it can only be last.

Step 2 — Use the logical constraint to pick the option.

The key ordering rule: humus must exist before it can be mineralised, so humification always precedes mineralisation. That single test settles the question:

• (A) …Mineralisation → Humification ✗ (wrong order at the end)
• (B) Fragmentation → Mineralisation → … ✗ (mineralisation second — impossible)
• (D) Fragmentation → Catabolism → Leaching ✗ (leaching acts on the freshly fragmented litter, before catabolism)
• (C) Fragmentation → Leaching → Catabolism → Humification → Mineralisation ✓

Step 3 — Note on rate.

Decomposition is largely an oxygen-requiring process; its rate is controlled by the chemical quality of detritus (rich in lignin/chitin ⇒ slow; rich in nitrogen and water-soluble sugars ⇒ fast) and by the climatic factors (warm and moist ⇒ fast).

ANSWER: C

Step 1 — Recall Griffith's 1928 experiment with Streptococcus pneumoniae

Griffith worked with two strains:

• S strain — smooth colonies, has a mucous (polysaccharide) coat, is virulent.
• R strain — rough colonies, no coat, is non-virulent.

Step 2 — The four injections and their outcomes

Experiment 4 is the famous result: some "transforming principle" passed from the dead S bacteria to the living R bacteria, converting them into virulent, coat-bearing S bacteria. (Avery, MacLeod and McCarty later showed that transforming principle is DNA.)

Step 3 — Check each printed statement against the table

• P. "R strain → inject into mice → mice die" — FALSE. Row 2: the R strain is non-virulent, so the mice live.
• Q. "S strain (heat killed) → inject into mice → mice die" — FALSE. Row 3: heat killing destroys virulence, so the mice live. (If heat-killed S alone had killed the mice, experiment 4 would have proved nothing.)
• R. "R strain → inject into mice → mice live" — TRUE. This is exactly row 2.

Step 4 — Select

Only statement R is correct.

• (A) "P only" — P is false.
• (C) "P and R" — includes the false P; also P and R contradict each other (the same injection cannot both kill and spare the mice), which is a quick internal check that they cannot both be right.
• (D) "Q and R" — includes the false Q.

ANSWER: B

Step 1 — Recall the structure of a structural gene.

A structural gene is double-stranded. One strand is the template (antisense) strand, which has $3'\rightarrow 5'$ polarity and is actually copied by RNA polymerase. The other is the coding (sense) strand, which has $5'\rightarrow 3'$ polarity; it is not transcribed, but its sequence is identical to the mRNA (with T in place of U) — which is exactly why it is called the coding strand.

Step 2 — Apply the two rules that fix the answer.

1. Complementarity (Chargaff / Watson–Crick): $A$ pairs with $T$, $G$ pairs with $C$.
2. Antiparallel orientation: if the template reads $3'\rightarrow 5'$ left-to-right, the coding strand written under it reads $5'\rightarrow 3'$ left-to-right.

Step 3 — Build the complement, base by base.

Checking the first four: $A\rightarrow T$, $T\rightarrow A$, $G\rightarrow C$, $C\rightarrow G$, giving the repeating unit TACG, repeated four times over the 16 bases.

So the coding strand is $5'\text{-TACGTACGTACGTACG-}3'$.

Step 4 — Screen the options (the polarity labels are half the question).

• (A) $5'$-ATGCATGCATGCATGC-$3'$ — this is the template's own base sequence, not its complement. ✗
• (B) $3'$-GCATGCATGCATGCAT-$5'$ — wrong bases and wrong polarity. ✗
• (C) $5'$-TACGTACGTACGTACG-$3'$ — correct complement and correct $5'\rightarrow 3'$ polarity for a coding strand. ✓
• (D) $3'$-TACGTACGTACGTACG-$5'$ — right bases, but labelled $3'\rightarrow 5'$, which is the template's polarity, not the coding strand's. ✗

Cross-check with the mRNA: RNA polymerase reads the $3'\rightarrow 5'$ template and synthesises mRNA $5'\rightarrow 3'$, giving $5'\text{-UACGUACGUACGUACG-}3'$. Replacing U with T reproduces option (C) exactly — confirming (C) is the coding strand.

ANSWER: C

Step 1 — What a DNA palindrome means.

A palindrome in DNA is not simply a sequence that reads the same backwards. It is a sequence in which both strands read identically when each is read in the $5'\to3'$ direction. This is the definition used in the NCERT Biotechnology chapter.

Step 2 — Apply the test to the given sample.

• Top strand, read $5'\to3'$ (left to right): GAATTC
• Bottom strand, read $5'\to3'$: the $5'$ end of the bottom strand is on the right, so read it right-to-left: G, A, A, T, T, C $\Rightarrow$ GAATTC

Both strands read GAATTC. The palindrome test is satisfied. $\checkmark$

(Note the strands are also properly complementary: G–C, A–T, A–T, T–A, T–A, C–G. $\checkmark$)

Step 3 — Recognise the sequence.

GAATTC is the recognition site of EcoRI, the best-known restriction endonuclease. Restriction enzymes recognise palindromic sites and cut between G and A on both strands, producing sticky ends — which is exactly why palindromes matter in recombinant DNA technology.

Step 4 — Eliminate the others.

• (A) Completion of replication — nothing here indicates a replication terminus.
• (B) Initiator codon at 5' end — the initiator codon is AUG (in mRNA) / ATG (in DNA), and it is a translation feature. GAATTC is not ATG.
• (D) Deletion mutation — no reference sequence is given to compare against, and both strands are perfectly complementary with no gap.

ANSWER: C

The citric acid cycle (Krebs cycle) is a central metabolic pathway where acetyl‑CoA is completely oxidised to CO₂. A key feature is that the cycle intermediates are organic acids with varying carbon chain lengths — 4C, 5C, and 6C compounds appear at different steps. The question tests your ability to match each acid to its carbon count.

Let’s trace the carbon numbers of the major intermediates in order:

1. Citric acid (6C) — The cycle begins when oxaloacetate (4C) condenses with acetyl‑CoA (2C) to form citrate, a 6‑carbon tricarboxylic acid.
2. α‑Ketoglutaric acid (5C) — After two decarboxylation steps (isocitrate → α‑ketoglutarate), one carbon is lost as CO₂, leaving a 5‑carbon keto acid.
3. Succinic acid (4C) — α‑Ketoglutarate undergoes another decarboxylation (oxidative decarboxylation) to form succinyl‑CoA, which is then converted to succinate — a 4‑carbon dicarboxylic acid.
4. Malic acid (4C) — Later in the cycle, fumarate (4C) is hydrated to malate (also 4C).
5. Oxaloacetic acid (4C) — Finally, malate is oxidised back to oxaloacetate (4C), completing the turn.

Now check the options:

• (A) Oxaloacetic acid (4C), Citric acid (6C), Succinic acid (4C) — gives 4C, 6C, 4C, not the required 4C, 5C, 6C.
• (B) Succinic acid (4C), α‑Ketoglutaric acid (5C), Citric acid (6C) — exactly matches the order.
• (C) Pyruvic acid (3C, not in cycle), Malic acid (4C), α‑Ketoglutaric acid (5C) — wrong carbon counts and includes pyruvate.
• (D) Pyruvic acid (3C), α‑Ketoglutaric acid (5C), Citric acid (6C) — again includes pyruvate.

Step 1 — What each RNA does in translation.

Step 2 — Why the adaptor must exist.

Francis Crick's adaptor hypothesis pointed out that there is no chemical affinity between a nucleotide triplet and an amino acid. Something must physically bridge the two. That bridge is tRNA: aminoacyl-tRNA synthetase charges the tRNA with its cognate amino acid (aminoacylation), and the charged tRNA then delivers it to the A-site of the ribosome, where its anticodon pairs with the codon.

Step 3 — Eliminate the rest.

• (A) rRNA forms the ribosome itself and catalyses peptide-bond formation, but it does not ferry amino acids.
• (B) dsRNA is not part of the translation machinery at all.
• (D) mRNA supplies the codon sequence; it is read, not carried.

ANSWER: C

The concept: Why size determines distance

Agarose gel electrophoresis works like a molecular sieve. DNA is negatively charged, so when you apply an electric field across the gel, all DNA fragments move toward the positive electrode. But the agarose matrix has pores of varying sizes — smaller fragments slip through easily and travel far, while larger fragments get tangled and lag behind.

This means: distance travelled ∝ 1/(size of fragment). The largest fragments stay closest to the well (top of the gel), and the smallest fragments run farthest (bottom).

Step-by-step reasoning

1. Identify what M represents

   In any gel electrophoresis figure, lane M is almost always the molecular weight marker (also called a DNA ladder). This is a mixture of DNA fragments of known, pre-determined sizes — it serves as a ruler to estimate the size of unknown fragments in other lanes.

2. Read the band pattern in lane M

   The marker lane shows multiple distinct bands stacked vertically. The band nearest the well (top) is the largest fragment; the band farthest from the well (bottom) is the smallest. The bands in between correspond to intermediate sizes.

3. Match this to the options

   • Option (A): "Digested DNA bands" — digestion refers to cutting DNA with restriction enzymes, which could be in sample lanes, but M is the marker, not a digested sample.
   • Option (B): "Hybridised DNA bands" — hybridisation (like Southern blotting) happens after electrophoresis, not during it. The gel itself shows only size separation.
   • Option (C): "Largest DNA bands" — correct in the sense that the topmost band in M is the largest, but the statement as a whole means "M shows the largest bands" (i.e., the marker contains the largest fragments). This is true relative to the sample lanes if the sample has smaller fragments.
   • Option (D): "Smallest DNA bands" — the bottom band in M is the smallest, but M contains a range of sizes, not just the smallest.

4. Interpret the figure's intent In typical exam figures, the sample lanes show DNA fragments of various sizes (often from restriction digestion), and lane M shows the marker with the largest fragments at the top. The question asks you to identify what M represents in this context — it is the lane containing the largest DNA bands (the marker), which is option (C).

EcoRI is a restriction endonuclease — a molecular scissors that cuts DNA at a specific recognition site. The key property of these enzymes is that they recognise palindromic sequences: the sequence on one strand, when read 5' to 3', is identical to the sequence on the complementary strand read 5' to 3'. For EcoRI, that sequence is:

The cut is made between the G and the first A on each strand, producing sticky ends. So to identify which option can be cut, we need to find a double-stranded DNA that contains this exact hexamer.

Let's examine each option carefully.

1. Option (A):

   Top strand: 5'ACGAATTCAT3'

   Bottom strand: 3'TGCTTAAGTA5'

   Reading the top strand from 5' to 3', we see GAATTC starting at the third base. The bottom strand, read 5' to 3' (which means reading it from right to left as written), is 5'ATGAATTCGT3' — and that also contains GAATTC. This is the correct palindromic site. So EcoRI will cut here.

2. Option (B):

   Top strand: 3'ACGAATTCAT5' — this is written in the 3' to 5' direction. To check, we must mentally flip it to 5' to 3': it becomes 5'TACTTAAGCA3'. That sequence is TACTTAAGCA — no GAATTC. The bottom strand, when flipped, is 5'ATGAATTCGT3', which does have GAATTC, but the top strand doesn't. For a restriction site to work, both strands must have the recognition sequence in the correct orientation. This fails.

3. Option (C):

   Top strand: 5'TGCTTAAGTA3' — this is TGCTTAAGTA. No GAATTC. The bottom strand, read 5' to 3', is 5'TACTTAAGCA3' — also no GAATTC. So this is not a recognition site.

4. Option (D):

   Top strand: 5'TACTTAAGCA3' — no GAATTC. Bottom strand: 5'ACGAATTCGT3' — this has GAATTC, but again, only one strand. The top strand reads TACTTAAGCA, which is not the complement of GAATTC in the correct orientation. So no cut.

Only option (A) satisfies this condition.

Step 1 — Pair each process with its enzyme.

1. DNA replication — DNA is copied into DNA. The polymerising enzyme is DNA polymerase → II.
2. Translation — mRNA is decoded into protein on the ribosome. Of the enzymes listed, the one belonging to translation is aminoacyl-tRNA synthetase, which charges each tRNA with its correct amino acid (amino-acylation) → IV.
3. Transcription — DNA is copied into RNA by RNA polymerase → I.
4. Reverse transcription — RNA is copied into DNA (retroviruses, e.g. HIV) by reverse transcriptase (RNA-dependent DNA polymerase) → III.

Step 2 — Assemble the code.

That is the sequence II, IV, I, III.

Step 3 — Eliminate.

Option (A) II, IV, III, I wrongly gives transcription to reverse transcriptase; (C) and (D) misplace DNA/RNA polymerase against translation.

ANSWER: B

Step 1 — The nature of the disease.

Sickle-cell anaemia is an autosomal recessive disorder caused by a single base substitution — a classic point mutation — in the gene for the $\beta$-globin chain of haemoglobin.

Step 2 — The molecular change.

The substitution occurs at the sixth codon of the $\beta$-globin gene, converting

The middle base A is replaced by U (at the DNA level, the corresponding change is an $A\rightarrow T$ substitution).

Step 3 — The consequence.

GAG codes for glutamic acid (Glu) and GUG codes for valine (Val). So the 6th amino acid of the $\beta$-globin chain becomes valine instead of glutamic acid. Valine is non-polar, whereas glutamic acid is charged; this makes the mutant haemoglobin (HbS) polymerise under low oxygen tension, distorting the biconcave RBC into a sickle shape.

Step 4 — Eliminate the others.

(B) GAG $\rightarrow$ GAG is no change at all. (A) and (C) do not correspond to the Glu $\rightarrow$ Val change at the sixth codon as described in the standard account of the mutation.

ANSWER: D

Step 1 — Anatomy of a mature mRNA.

A monocistronic mRNA reads:

The untranslated regions (UTRs) sit at both ends: one before the start codon and one after the stop codon.

Step 2 — Their role.

The UTRs are required for efficient translation — the $5'$ UTR is where the ribosome binds and scans to find the start codon, and the $3'$ UTR carries signals governing mRNA stability and localisation. But since they lie outside the reading frame (start → stop), no amino acids are made from them. They exactly satisfy 'required for translation but not translated'. ✓

Step 3 — Rule out the others.

• (A) Stop codons — these are part of the coding region; they terminate translation. They do not code for an amino acid, but they are not a sequence of the kind described and, more importantly, the standard NCERT answer to this exact phrasing is the UTR. Also, a stop codon is a single triplet, not an untranslated region.
• (B) Anticodons — these are on tRNA, not on mRNA. Immediately disqualified by the wording 'sequences of mRNA'. ✗
• (C) Sense codons — these are precisely the codons that ARE translated into amino acids. ✗

ANSWER: D

Step 1 — The definition (and the trap).

In molecular biology a palindrome is not a word that reads the same backwards letter-by-letter. It is a duplex in which the sequence read $5' \to 3'$ on the top strand is identical to the sequence read $5' \to 3'$ on the bottom strand. Restriction endonucleases recognise exactly such sites — which is why the question matters.

Step 2 — Test option (B).

Read the bottom strand in its own $5'\to3'$ direction, i.e. right to left: $\mathrm{G\,G\,A\,T\,C\,C}$ — identical to the top strand. ✓ Palindrome. (Check complementarity too: G·C, G·C, A·T, T·A, C·G, C·G — all correct base pairs.) This is the $Bam$HI recognition site.

Step 3 — Reject the others.

• (A) Top $5'$-CGATA-$3'$; bottom read $5'\to3'$ = TATCG $\neq$ CGATA. ✗ (Also 5 bp — an odd-length palindrome is impossible.)
• (C) Top $5'$-CCTGC-$3'$; bottom read $5'\to3'$ = GCAGG $\neq$ CCTGC. ✗
• (D) Only one strand is printed ($5'$-GAATTG-$3'$); its complement read $5'\to3'$ would be CAATTC $\neq$ GAATTG, so it is not a palindrome either. (Note $Eco$RI's real site is GAATTC, not GAATTG — this is the near-miss distractor.) ✗

Step 4 — Note the parity rule.

A true palindrome must have an even number of base pairs, which is a fast screen: (A) and (C) with 5 bp cannot be palindromic.

ANSWER: B