DNA and RNA Chemistry Practice

DNA questions mix structural chemistry (sugars, phosphates, hydrogen bonds) with information processing (codons, reading frames, mutations). This set of 14 questions covers both, and includes several calculations of the kind that appear in exams. Try them without notes, then use the answer key to check your working. For background reading, see DNA structure, DNA vs RNA and the genetic code.

Questions

Q1. Draw or describe a DNA nucleotide, naming its three components and stating how they’re bonded together.

Q2. Give three chemical differences between DNA and RNA.

Q3. Explain why adenine pairs with thymine and guanine pairs with cytosine, but adenine doesn’t pair with cytosine.

Q4. DNA from a bacterium contains 31 % guanine. Calculate the percentages of cytosine, adenine and thymine.

Q5. Explain why DNA with a high proportion of G–C pairs has a higher “melting” temperature (the temperature at which the strands separate) than DNA rich in A–T pairs.

Q6. A DNA strand has the sequence 5′-ATGCCTAG-3′. (a) Write the complementary DNA strand, labelling its 5′ and 3′ ends. (b) Write the mRNA that would be transcribed if the given strand is the template.

Q7. Explain what is meant by antiparallel and why it matters during replication.

Q8. Describe the semi-conservative model of DNA replication. What would the Meselson–Stahl experiment show after one generation in ¹⁴N medium, if cells were first grown in ¹⁵N?

Q9. Why is DNA’s sugar–phosphate backbone negatively charged at cell pH? Give one consequence of this charge used in the lab.

Q10. Using this partial codon table (mRNA):

Codon	Amino acid
AUG	Met (start)
UUC	Phe
GGA	Gly
CAU	His
UAG	Stop

translate the mRNA 5′-AUGUUCGGACAUUAG-3′.

Q11. In Q10, the fourth codon’s middle base changes from A to G, giving CGU (Arg). Name this type of mutation and explain how it could affect the protein.

Q12. A gene of 1,500 base pairs codes for a protein. Assuming the whole gene is coding sequence including one stop codon, how many amino acids are in the protein?

Q13. The human genome contains about 3.1 × 10⁹ base pairs. The mean mass of a base pair is about 650 g mol⁻¹ (as the sodium salt). Estimate: (a) the mass in grams of one copy of the human genome (Avogadro constant 6.02 × 10²³ mol⁻¹); (b) the length of this DNA if each base pair adds 0.34 nm.

Q14. A student extracts DNA from strawberries using detergent, salt and cold ethanol. Explain the role of each of these three substances.

Answer key

A1. A nucleotide has a phosphate group, a deoxyribose sugar (a five-carbon pentose) and a nitrogenous base (A, T, G or C). The base is bonded to carbon 1′ of the sugar; the phosphate is bonded to carbon 5′. Nucleotides join through phosphodiester bonds between the 3′ carbon of one sugar and the phosphate on the next — condensation reactions. See nucleic acids.

A2. Any three:

• Sugar: deoxyribose in DNA, ribose in RNA (ribose has an extra –OH on carbon 2′).
• Bases: DNA has thymine; RNA has uracil instead.
• Strands: DNA is usually double-stranded; RNA is usually single-stranded.
• RNA is generally shorter and less stable (the 2′-OH makes it easier to hydrolyse).

A3. Each pair has matching hydrogen-bond donors and acceptors in the right positions, and each pair is one large two-ring base (purine) plus one small one-ring base (pyrimidine), so every pair is the same width and fits the helix. A–T forms 2 hydrogen bonds and G–C forms 3. A and C don’t have complementary donor/acceptor positions, so they can’t form a stable set of hydrogen bonds. See base pairing.

A4. C = G = 31 %. G + C = 62 %, so A + T = 38 %, and A = T = 19 % each.

A5. G–C pairs are held by three hydrogen bonds instead of two, and G–C pairs also stack more strongly with neighbouring bases. More energy is needed to separate G–C-rich strands, so the melting temperature is higher. (Organisms living in hot springs often have G–C-rich DNA in key genes.)

A6. (a) 3′-TACGGATC-5′ (written 5′→3′: 5′-CTAGGCAT-3′). (b) The mRNA is complementary to the template and has U instead of T: 3′-UACGGAUC-5′, i.e. 5′-CUAGGCAU-3′.

A7. The two strands run in opposite directions: one 5′→3′, the other 3′→5′. DNA polymerase can only add nucleotides to a 3′ end, building new strands 5′→3′. So one new strand (the leading strand) is made continuously, while the other (the lagging strand) is made in short pieces (Okazaki fragments) that are later joined by DNA ligase. See DNA replication.

A8. Each new double helix contains one original strand and one new strand. After one generation in ¹⁴N, all DNA would be hybrid (¹⁵N/¹⁴N) and form a single band of intermediate density when centrifuged in a caesium chloride gradient. (After two generations: half hybrid, half light.) This is what Meselson and Stahl observed in 1958.

A9. Each phosphate group in the backbone has an –OH that is acidic (pKa around 1). At pH 7 it’s fully ionised as –O⁻, giving one negative charge per nucleotide. Consequences used in the lab: in gel electrophoresis, DNA moves towards the positive electrode, with smaller fragments moving faster (see electrophoresis); in DNA extraction, salt cations shield the charges so DNA can precipitate.

A10. AUG | UUC | GGA | CAU | UAG → Met–Phe–Gly–His (then stop).

A11. A point (substitution) mutation, specifically a missense mutation. Histidine is replaced by arginine. Both are basic amino acids, but they differ in size and charge behaviour (histidine’s side chain is only partly charged at pH 7; arginine’s is fully positive). If this position is important for folding or the active site, the protein’s shape or function could change; if not, the effect may be small.

A12. 1,500 / 3 = 500 codons. One is the stop codon (which codes for no amino acid), so the protein has 499 amino acids. (The starting methionine is sometimes removed later, but it’s included here.)

A13. (a) Mass = (3.1 × 10⁹ × 650) / 6.02 × 10²³ = 2.015 × 10¹² / 6.02 × 10²³ ≈ 3.3 × 10⁻¹² g (about 3 picograms). (b) Length = 3.1 × 10⁹ × 0.34 × 10⁻⁹ m ≈ 1.05 m. A human cell with two copies holds about 2 m of DNA, packed into a nucleus only a few micrometres across.

A14.

• Detergent breaks down the phospholipid membranes around the cell and nucleus, releasing DNA.
• Salt provides Na⁺ ions that neutralise the negative phosphate charges, helping DNA strands clump together and separate from proteins.
• Cold ethanol: DNA is insoluble in ethanol, so it precipitates as white strands at the boundary; the cold reduces solubility further and slows DNA-cutting enzymes.

See extracting DNA from strawberries.

Common mistakes

• Mixing up 5′ and 3′: always write the ends when giving sequences.
• Forgetting U replaces T in RNA.
• Counting the stop codon as an amino acid.
• Applying Chargaff’s rule (A = T, G = C) to single-stranded RNA — it only holds for double-stranded DNA.

Key takeaways

• Nucleotides = phosphate + sugar + base, linked by phosphodiester bonds.
• A–T (2 H-bonds), G–C (3 H-bonds); strands are antiparallel.
• Replication is semi-conservative; polymerases work 5′→3′.
• Codons are triplets; mutations can be silent, missense, nonsense or frameshift.
• The backbone is negatively charged, which is exploited in electrophoresis and extraction.