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

• Adenine — purine (2 rings) — both DNA and RNA.
• Uracil — pyrimidine (1 ring) — RNA only.
• Cytosine — pyrimidine — both.
• Thymine — pyrimidine — DNA only.
• Guanine — purine — both.

Why: purines carry the two fused rings (mnemonic Pure As Gold → A, G). Thymine is the "stable archive" base of DNA; RNA swaps in the cheaper Uracil.

A phosphate group. So nucleotide $=$ base $+$ sugar $+$ phosphate. Adding phosphate to the nucleoside adenosine gives the nucleotide AMP (adenosine monophosphate).

The $3^{\prime }$ end (the one next to C here). Enzymes extend strands $5^{\prime }\to 3^{\prime }$, meaning they attach each incoming nucleotide's phosphate to the free $3^{\prime }\text{-OH}$ of the growing chain.

Step 1 — %C. $G$ pairs with $C$, so $\%C=\%G=32\%$. Step 2 — remaining for A+T. $\%G+\%C=32+32=64\%$, so $\%A+\%T=100-64=36\%$. Step 3 — split A and T. $A$ pairs with $T$, so $\%A=\%T=36/2=18\%$ each. ✓ Check: $32+32+18+18=100$.

Read the template base-by-base and write the RNA complement (remember $U$ for $A$):

• $T\to A$
• $A\to U$
• $C\to G$
• $A\to U$
• $A\to U$
• $G\to C$

mRNA ($5^{\prime }\to 3^{\prime }$): A U G U U C. The first codon AUG is the start codon → Methionine.

$G\text{–}C$. It holds the two strands with 3 hydrogen bonds vs only 2 for $A\text{–}T$. More bonds per rung → more total glue → higher melting temperature. So DNA rich in $G\text{–}C$ is more thermally stable. (This is pure Hydrogen bonding arithmetic: count the bonds.)

Step 1 — count pairs. 600 base pairs means 600 rungs. Each rung is one $A\text{–}T$ or one $G\text{–}C$ pair. Step 2 — use the given condition. "Number of A = number of G." Every A sits in an $A\text{–}T$ rung and every G sits in a $G\text{–}C$ rung, so the number of AT-rungs $=$ number of GC-rungs $=600/2=300$ each. Step 3 — count H-bonds. AT rungs contribute $2$ each, GC rungs contribute $3$ each: $300\times 2+300\times 3=600+900=1500\text{hydrogen bonds}.$ ✓ The GC half, though equal in count, supplies more bonds (900 vs 600).

Reading the figure, the four rungs are $A\text{–}T$, $G\text{–}C$, $G\text{–}C$, $T\text{–}A$. Bases (both strands, 8 total): A, T, G, C, G, C, T, A → count: $A=2,T=2,G=2,C=2$. Chargaff check: $\%A=\%T=25\%$ ✓ and $\%G=\%C=25\%$ ✓. H-bond total from dashes: two 2-dash rungs $+$ two 3-dash rungs $=2+2+3+3=10$ dashes, matching the drawing.

• It contains U, so it is RNA.
• $\%A\ne \%U$ ($30\ne 10$), and that is allowed: Chargaff's rule ($\%A=\%T$, $\%G=\%C$) only holds for a double-stranded molecule where every base is paired. A single strand has no forced partner, so its base percentages are unconstrained.
• Conclusion: single-stranded RNA — the mismatch is expected, not an error.

(a) Transcribe (complement each template base, $U$ for $A$): template T A C G T A C G A T T → mRNA A U G C A U G C U A A. (b) Codons: AUG | CAU | GCU | AA(?) — the last A A is only 2 bases, an incomplete codon, so it is dropped. The three complete codons are AUG, CAU, GCU. (c) Start codon: the first codon AUG = Methionine (start) ✓. (d) Stop check: the complete codons are AUG, CAU, GCU — none is UAA/UAG/UGA, so no stop appears in the read frame. (Building proteins from these codons is the job of Amino acids and Proteins machinery — the ribosome and tRNA.)

There are 4 bases, and a codon of length $n$ can spell $4^n$ distinct words.

• $n=1$: $4^1=4$ words — far short of 20.
• $n=2$: $4^2=16$ words — still short (only 16 < 20).
• $n=3$: $4^3=64$ words — enough, with room to spare (redundancy).

So the minimum word length that can encode 20 amino acids is 3. Nature uses triplets; the extra $64-20=44$ "spare" codons cover redundancy and the stop signals. See Genetic code and mutations.

Let $x=$ number of $A\text{–}T$ pairs, $y=$ number of $G\text{–}C$ pairs. Equation 1 (total rungs): $x+y=2000$. Equation 2 (H-bonds): each AT gives 2, each GC gives 3, so $2x+3y=1200$.

Wait — sanity check first. The minimum possible H-bonds is when all rungs are AT: $2\times 2000=4000$. But we were told only $1200<4000$ — impossible! So the data as stated cannot describe a real 2000-bp duplex.

Corrected mastery version: suppose instead the total is 5200 hydrogen bonds.

• $x+y=2000$
• $2x+3y=5200$ Substitute $x=2000-y$: $2(2000-y)+3y=5200\Rightarrow 4000+y=5200\Rightarrow y=1200$. So $G\text{–}C$ pairs $=1200$, $A\text{–}T$ pairs $=2000-1200=800$. %A: each AT rung has one A on one strand and one T on the other. Total bases $=2\times 2000=4000$. Number of A bases $=800$ (one per AT rung). So $\%A=\frac{800}{4000}\times 100=20\%.$ ✓ Check: $\%T=20\%$, and $\%G=\%C=1200/4000\times 100=30\%$; sum $=20+20+30+30=100$.

Lesson: always test feasibility ($\text{min H-bonds}\le \text{given}\le \text{max H-bonds}$, i.e. $2N\le H\le 3N$ for $N$ pairs) before solving.

Why increasing: melting = pulling the two strands apart = breaking all the hydrogen bonds holding the rungs. A $G\text{–}C$ rung costs 3 bonds to break, an $A\text{–}T$ rung only 2. So the more GC rungs a molecule has, the more total bonds per unit length, and the more heat energy needed to melt it. Thus $T_m$ rises with GC content — a straight consequence of Hydrogen bonding arithmetic, no new physics needed. Highest $T_m$: the sample with the greatest GC% (the rightmost/tallest bar in the figure). More triple-bonded rungs → toughest to melt.

1. 40% G in dsDNA → give all four percentages. (C=40, A=T=10)
2. Template 3'-TACGGA-5' → mRNA? (AUGCCU)
3. A 500-bp DNA, equal A and G counts → total H-bonds? (1250)
4. Minimum codon length for 20 amino acids, with the counting reason. (3, since $4^2=16<20\le 64=4^3$)
5. Why is high-GC DNA harder to melt? (3 H-bonds per GC vs 2 per AT → more energy to separate)