Nucleic acids — DNA, RNA; base pairing, double helix, replication, transcription, translation (overview): Worked examples

Before we count anything, one reminder that we will lean on constantly. Base pairing is held together by hydrogen bonds, and the pairing is specific: A only meets T (or U in RNA), G only meets C. That single fact is the engine behind every example below.

The scenario matrix

Here is the full landscape of cases this topic can throw at you. Each later example is labelled with the cell it fills.

#	Case class	What makes it tricky	Example
1	Standard base counting (one % given)	just apply Chargaff	Ex 1
2	Degenerate 50/50 input	%A = %G — is that even allowed?	Ex 2
3	Zero / extreme input	%A = 0, or %GC = 100%	Ex 3
4	H-bond total counting	mixes pairing rules with bond counts	Ex 4
5	Build the complementary strand	direction (5′/3′) and antiparallel trap	Ex 5
6	Transcription (DNA → mRNA)	U-for-T swap + reading direction	Ex 6
7	Translation (codon reading)	grouping in 3s, start/stop	Ex 7
8	Real-world word problem	melting temperature vs GC content	Ex 8
9	Exam twist (impossible data)	data that violates Chargaff → spot the trap	Ex 9

The tools we will re-use:

Example 1 — Standard base counting (Cell 1)

Example 2 — The degenerate 50/50 case (Cell 2)

Example 3 — Zero and extreme inputs (Cell 3)

Worked example Two edge cases at once

(a) A hypothetical DNA has %A = 0%. What are the others, and is it physically sane? (b) Another DNA has GC content = 100%. Find all four bases.

Forecast: In (a), if there is no A, can there be any T? In (b), how much A?

Step 1 — Part (a): use A–T pairing. Why this step? $% T = % A = 0$ . With no A, there can be no T (each T needs an A partner across the helix).

Step 2 — Part (a): fill the rest with G,C. Why this step? Closure: $% G + % C = 100$ . Chargaff splits them equally: $% G = % C = 50%$ . This DNA is all G–C — extreme but internally consistent (real GC-rich organisms approach this).

Step 3 — Part (b): GC = 100 forces AT = 0. Why this step? $% A + % T = 100 - 100 = 0$ , so $% A = % T = 0$ , and $% G = % C = 50%$ .

Answer: (a) A=0, T=0, G=50, C=50. (b) identical: A=0, T=0, G=50, C=50.

Verify: Both give $0 + 0 + 50 + 50 = 100 ✓$ . Notice the two edge phrasings collapse to the same DNA — a nice consistency check.

Example 4 — Counting hydrogen bonds (Cell 4)

Example 5 — Build the complementary strand (Cell 5)

Worked example Watch the direction!

Given one DNA strand $5^{'} - A T G C C A - 3^{'}$ , write its complementary strand with correct ends labelled.

Forecast: Which base sits opposite each? And which end (5′ or 3′) does the new strand start on?

Step 1 — Pair each base. Why this step? Complementary base pairing: A↔T, G↔C. Reading the given strand left to right ( $A, T, G, C, C, A$ ) the partners are $T, A, C, G, G, T$ .

Step 2 — Flip the direction. Why this step? Strands are antiparallel (parent note, mistake #4). The partner opposite the $5^{'}$ end of the top strand is the $3^{'}$ end of the bottom strand. Look at the figure: the arrows point in opposite directions.

Step 3 — Write it correctly. Base-paired string (aligned under the original): $3^{'} - T A C G G T - 5^{'}$ . Conventionally we write $5^{'} \to 3^{'}$ , so rewrite it reversed: $5^{'} - T G G C A T - 3^{'}$ .

Answer: complementary strand $= 5^{'} - T G G C A T - 3^{'}$ .

Verify: Re-pair $5^{'} - T GGC A T - 3^{'}$ back against the original and check base counts: original has A=2,T=1,G=1,C=2; complement should have A=1,T=2,G=2,C=1 — and it does ( $T, G, G, C, A, T$ → T=2,G=2,C=1,A=1) $✓$ .

Example 6 — Transcription: DNA → mRNA (Cell 6)

Worked example The U-for-T swap

A DNA template strand reads $3^{'} - T A C G T A - 5^{'}$ . Build the mRNA.

Forecast: Which DNA base becomes which RNA base? Watch the T's.

Step 1 — Read the template. Why this step? RNA polymerase reads the template $3^{'} \to 5^{'}$ and builds mRNA $5^{'} \to 3^{'}$ , laying down the complementary base to each template base.

Step 2 — Pair, but U replaces T. Why this step? RNA has no thymine. So opposite a template A we place U (not T); all other pairings are normal.

Template	mRNA base
T	A
A	U
C	G
G	C
T	A
A	U

Step 3 — Assemble. mRNA ( $5^{'} \to 3^{'}$ ) $= A U G C A U$ .

Answer: $5^{'} - A U G C A U - 3^{'}$ .

Verify: No thymine appears in the RNA product $✓$ . First triplet is $A U G$ = the start codon (Methionine) $✓$ — a good sign the reading frame is set correctly.

Example 7 — Translation: reading codons (Cell 7)

Worked example Group in threes, spot start and stop

An mRNA reads $5^{'} - A U G U U U U A A - 3^{'}$ . How many codons? What does each mean structurally?

Forecast: How many amino acids will this actually add to the protein?

Step 1 — Count and group. Why this step? Codons are triplets ( $4^{3} = 64$ possibilities, enough for 20 amino acids). Total bases $= 9$ , so $9/3 = 3$ codons: AUG | UUU | UAA.

Step 2 — Read each codon's role. Why this step? Ribosomes read $5^{'} \to 3^{'}$ . AUG = start (Methionine). UUU = Phenylalanine. UAA = stop (a genetic-code signal, codes for no amino acid — it ends the chain).

Step 3 — Count amino acids added. Why this step? Start and one coding codon add amino acids; the stop codon adds none. So the peptide has 2 amino acids (Met–Phe), then translation halts.

Answer: 3 codons, 2 amino acids in the finished (tiny) peptide.

Verify: Bases divide evenly into triplets ( $9 = 3 \times 3$ , no leftover) $✓$ . Amino-acid count $= (codons) - (stop codons) = 3 - 1 = 2 ✓$ .

Example 8 — Real-world word problem: melting temperature (Cell 8)

Worked example Which DNA melts first?

Two DNA samples: Sample X has 70% GC, Sample Y has 30% GC. If you heat both, which "melts" (strands separate) at the lower temperature? Estimate the difference in H-bonds per 10 base pairs.

Forecast: More GC = more or fewer H-bonds = easier or harder to pull apart?

Step 1 — Link GC content to bonds. Why this step? Each GC pair = 3 H-bonds, each AT pair = 2 H-bonds. More GC ⇒ more bonds per rung ⇒ more heat needed to break them (an enzyme-free, purely physical melting). So Sample Y (low GC) melts at the lower temperature.

Step 2 — Count bonds in 10 pairs of X. Why this step? 70% of 10 = 7 GC pairs and 3 AT pairs. $7 \times 3 + 3 \times 2 = 21 + 6 = 27 H-bonds .$

Step 3 — Count bonds in 10 pairs of Y. $3 \times 3 + 7 \times 2 = 9 + 14 = 23 H-bonds .$

Answer: Sample Y melts first; X has $27 - 23 = 4$ more H-bonds per 10 base pairs.

Verify: Higher-GC sample has more bonds ( $27 > 23$ ) ⇒ higher melting temperature, matching the mistake-note physics $✓$ .

Example 9 — Exam twist: spot the impossible data (Cell 9)

Worked example Data that breaks the rules

A student reports a double-stranded DNA with %A = 30%, %T = 20%, %G = 25%, %C = 25%. Is this data valid? If not, why?

Forecast: It sums to 100 — does that make it correct?

Step 1 — Check closure. Why this step? $30 + 20 + 25 + 25 = 100$ . Closure passes — but closure alone is not enough.

Step 2 — Check Chargaff pairing. Why this step? In double-stranded DNA we need $% A = % T$ . Here $30 \neq = 20$ . Violation. Every A on one strand must have a T partner on the other, so their totals must match.

Step 3 — Diagnose. Why this step? The G–C pair is fine ( $25 = 25$ ), but A–T is broken. This data cannot describe a normal double-stranded DNA — it could only occur for single-stranded nucleic acid (where Chargaff need not hold).

Answer: Invalid for dsDNA, because $% A \neq = % T$ .

Verify: $% A - % T = 30 - 20 = 10 \neq = 0$ , so the required equality fails $✓$ — the trap is that summing to 100 is necessary but not sufficient.

Recall Self-test across the matrix

Cover the answers and try each. If %C = 32%, what is %A? ::: 18% (since %G = 32, A+T = 36, split equally). Is 25/25/25/25 a legal DNA base composition? ::: Yes — Chargaff holds; it is the degenerate all-equal case. If %A = 0, what is %G? ::: 50% (no A means no T, so G+C fill 100%, split equally). Total H-bonds for 6 A–T + 4 G–C pairs? ::: 24. Complement of 5′-ATGCCA-3′ (written 5′→3′)? ::: 5′-TGGCAT-3′. mRNA from template 3′-TACGTA-5′? ::: 5′-AUGCAU-3′. How many amino acids from AUG-UUU-UAA? ::: 2 (stop codon adds none). Higher GC content ⇒ higher or lower melting temperature? ::: Higher (more H-bonds per pair). Is %A=30, %T=20, %G=25, %C=25 valid for dsDNA? ::: No — %A ≠ %T breaks Chargaff.

See also: the Hinglish overview, and the prerequisite Hydrogen bonding that makes every pairing possible.