Genome, DNA, RNA and protein synthesis – IGCSE Biology exam questions with mark scheme

1Grade 1–34 marks

(a) Which of these is the correct description of a genome? [1]

A all of the genes on one chromosome
B the entire DNA of an organism
C a section of DNA that codes for a protein
D the two alleles of a gene

(b) Describe what is meant by a gene. [2]

(a) B(1)
(b) a section / length / part of a molecule of DNA; accept section of a chromosome; ignore 'unit of inheritance' on its own(1)
(b) that codes for a specific / particular protein; accept codes for a sequence of amino acids / a polypeptide; reject 'codes for a characteristic' on its own(1)
(c) nucleus; reject nucleolus / cytoplasm(1)
Answer: (a) B (b) a section of DNA that codes for a specific protein (c) nucleus
Examiner tip: For 'gene' you need both ideas: a section of DNA AND that it codes for a specific protein — 'codes for a characteristic' does not get the second mark.

2Grade 1–3Triple only4 marks

The diagram shows part of a DNA molecule, drawn as if it had been untwisted. The letters W, X, Y and Z represent bases that are not shown.

(a) In a cell, the two strands of DNA are coiled. Name the shape this forms. [1]

(b) Give the letters of the bases W, X, Y and Z. [2]

(a) (double) helix; accept spiral; ignore 'twisted ladder'(1)
(b) all four correct: W = G, X = G, Y = A, Z = C (2 marks)(1)
(b) two or three of W, X, Y, Z correct (1 mark); allow full names guanine / adenine / cytosine(1)
(c) cytosine; ignore spelling if phonetically correct; reject cysteine / cytokine(1)
Answer: (a) double helix (b) W = G, X = G, Y = A, Z = C (c) cytosine
Examiner tip: Pair each unknown with the base directly across the rung from it — A always with T and C always with G.

3Grade 4–5Triple only4 marks

A scientist analysed a sample of double-stranded DNA taken from a bacterium.

She found that 31% of the bases in the sample were adenine.

(a) Calculate the percentage of bases in the sample that were thymine, cytosine and guanine. [3]

(b) Explain why the percentage of adenine is always equal to the percentage of thymine in double-stranded DNA. [1]

(a) thymine = 31%(1)
(a) cytosine + guanine = 100 − 62 = 38%; allow 100 − 31 − 31 shown(1)
(a) cytosine = 19% and guanine = 19%; correct answers with no working = 3 marks(1)
(b) adenine (always) pairs with thymine (between the two strands); accept complementary base pairing of A with T; ignore 'they bond' without stating A–T(1)
Answer: (a) T = 31%, C = 19%, G = 19% (b) each adenine on one strand is paired with a thymine on the other
Examiner tip: Students often subtract only the adenine (100 − 31 = 69) and split that — remember thymine takes another 31% first.

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4Grade 4–5Triple only4 marks

Both DNA and messenger RNA (mRNA) are involved in making proteins.

(a) Give two differences between the structure of a DNA molecule and the structure of an mRNA molecule. [2]

(b) The sequence of bases on part of the DNA strand used to make an mRNA molecule is shown.

T A C G G A T T C

Write down the sequence of bases in the mRNA molecule made from this part of the DNA. [2]

(a) any two from: DNA is double-stranded but (m)RNA is single-stranded / DNA contains thymine (T) but RNA contains uracil (U) / DNA is a (double) helix but mRNA is not / mRNA is (much) shorter; allow DNA contains deoxyribose but RNA contains ribose; each must be comparative or both molecules stated (first difference)(1)
(a) second correct difference from the list above; do not credit 'double-stranded vs single-stranded' and 'helix vs not a helix' as two separate differences(1)
(b) A U G C C U A A G — correct complementary bases opposite every C, G and T on the DNA(1)
(b) U (not T) placed opposite every A on the DNA (positions 2 and 6); a fully correct sequence AUGCCUAAG = 2 marks; AT instead of AU (e.g. ATGCCTAAG) = 1 mark max(1)
Answer: (a) DNA double-stranded / mRNA single-stranded; DNA has T, RNA has U (b) AUGCCUAAG
Examiner tip: Never write T in an RNA sequence — every A on the DNA pairs with U in mRNA.

5Grade 6–7Triple only4 marks

The table shows some mRNA codons and the amino acids they code for.

mRNA codon	amino acid
AUG	Met
UAC	Tyr
GCU	Ala
CGA	Arg
UUC	Phe
AAG	Lys
AAA	Lys
UUU	Phe

The base sequence of a section of the DNA strand that is transcribed is:

TAC CGA AAG TTT

(a) Use the information to give the sequence of amino acids coded for by this section of DNA. [2]

(b) Give the anticodon on the tRNA molecule that carries the third amino acid in this sequence. [1]

(a) mRNA AUG GCU UUC AAA worked out / shown; allow any correct mRNA shown even if amino acids wrong(1)
(a) Met – Ala – Phe – Lys (in this order); do not award if amino acids read directly from DNA triplets (Tyr – Arg – Lys – Phe)(1)
(b) AAG; allow ecf from their third mRNA codon; reject UUC (the codon) / TTC(1)
(c) each amino acid is coded for by a triplet / codon of three bases (so 4 × 3 = 12); accept 'three bases code for one amino acid'(1)
Answer: (a) Met–Ala–Phe–Lys (b) AAG (c) each amino acid is coded for by a codon / triplet of three bases, so 4 × 3 = 12
Examiner tip: The codon table is for mRNA — convert the DNA to mRNA first; reading the DNA triplets straight off the table gives Tyr–Arg–Lys–Phe and scores nothing.

6Grade 6–7Triple only4 marks

The section of an mRNA molecule that is translated to make a hormone contains 333 bases.

The last codon in this section is a stop codon. A stop codon does not code for an amino acid.

(a) Calculate the number of amino acids in one molecule of the hormone. [2]

(b) A ribosome adds amino acids to the hormone at a rate of 5 amino acids per second.

Calculate the time taken for one ribosome to make one molecule of the hormone. [2]

(a) 333 ÷ 3 = 111 (codons)(1)
(a) 111 − 1 = 110 (amino acids); 111 with no subtraction = 1 mark only(1)
(b) 110 ÷ 5; allow ecf from (a)(1)
(b) 22 (s); units given in question so unit not required; ecf: 111 ÷ 5 = 22.2 s scores 2 marks if (a) = 111(1)
Answer: (a) 110 amino acids (b) 22 s
Examiner tip: Read the stem carefully: the stop codon is a codon but adds no amino acid, so subtract one after dividing by three.

7Grade 4–5Triple only5 marks

The diagram shows part of the process of translation in the cytoplasm of a cell.

(a) Name the structures labelled P, Q, R and S. [3]

(b) The next codon to be read is ACU. Give the anticodon of the tRNA that will bind to this codon. [1]

(a) P = mRNA / messenger RNA; reject DNA; 'RNA' alone not sufficient(1)
(a) Q = tRNA / transfer RNA(1)
(a) R = amino acid AND S = ribosome (both needed for this mark); reject protein for R(1)
(b) UGA; reject TGA(1)
(c) carries / brings a specific amino acid to the ribosome; accept its anticodon pairs with the complementary codon on the mRNA (so the correct amino acid is added)(1)
Answer: (a) P mRNA, Q tRNA, R amino acid, S ribosome (b) UGA (c) brings a specific amino acid to the ribosome (its anticodon pairs with the codon)
Examiner tip: In an anticodon, A on the mRNA pairs with U, not T — there is no thymine anywhere in RNA.

8Grade 6–74 marks

The table shows the genome size and the number of genes for three organisms.

organism	genome size / million bases	number of genes
human	3 100	20 000
fruit fly	140	14 000
yeast	12	6 000

(a) On average, the part of a human gene that codes for protein is 1 500 bases long.

Calculate the percentage of the human genome that codes for protein. Give your answer to 2 significant figures. [2]

(b) A student says: “Organisms with bigger genomes have proportionally more genes.”

Use the data for humans and fruit flies to evaluate this statement. [2]

(a) 20 000 × 1 500 = 30 000 000 bases AND ÷ 3 100 000 000 (× 100) shown; allow 30 ÷ 3 100 × 100 in millions(1)
(a) 0.97 (%); allow 0.968 / 0.9677 for 1 mark if not rounded; 1% scores 1 mark only(1)
(b) human genome is about 22 times bigger than fruit fly genome but has only about 1.4 times as many genes; accept 2 960 million more bases but only 6 000 more genes(1)
(b) so statement is not supported / incorrect: the number of genes is not proportional to genome size; accept most of the (human) genome / DNA does not code for proteins / is not genes(1)
Answer: (a) 0.97% (b) not supported: genome 22× bigger but only 1.4× as many genes — much of a large genome is non-coding DNA
Examiner tip: Watch the units in the table — the genome size is in millions of bases, so convert before dividing or keep both numbers in millions.

9Grade 8–9Triple only6 marks

Insulin is a protein hormone made by cells in the pancreas. The gene for insulin is located on a chromosome in the nucleus.

Describe how a pancreas cell uses the insulin gene to make a molecule of insulin. [6]

the DNA (double helix) unwinds / the two strands separate at the insulin gene(1)
transcription: one DNA strand is used as a template to make mRNA(1)
free RNA nucleotides / bases pair with complementary DNA bases (C with G, A with U); U pairs with A because RNA has uracil instead of thymine(1)
mRNA (is single-stranded and) leaves the nucleus (through a nuclear pore) and moves to a ribosome in the cytoplasm(1)
translation: the ribosome moves along / reads the mRNA one codon (three bases) at a time(1)
tRNA molecules each carry a specific amino acid to the ribosome(1)
the tRNA anticodon pairs with the complementary codon on the mRNA(1)
amino acids are joined together (in the order set by the codons / base sequence) to form a chain that makes the insulin protein(1)
the sequence of bases in the gene determines the sequence of amino acids in insulin(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, logically sequenced description of both transcription and translation, correctly located in the cell, using mRNA, ribosome, tRNA, codon and anticodon accurately and linking the base sequence of the gene to the amino acid sequence of insulin.
Level 2	3–4	A description covering both stages with some correct use of key terms; the sequence may be incomplete or one stage may be much less detailed, and there may be minor errors (e.g. tRNA carrying codons).
Level 1	1–2	Isolated correct statements about one stage, e.g. 'mRNA is made from DNA' or 'amino acids are joined at a ribosome', with little or no sequence.
Level 0	0	No rewardable content.

Answer: Transcription in the nucleus (DNA template → complementary mRNA with U), mRNA to ribosome, translation (codons read; tRNA anticodons bring specific amino acids; amino acids joined in codon order).
Examiner tip: Top-level answers name both stages and put them in the right place — transcription in the nucleus, translation at the ribosome — and use codon, anticodon and tRNA precisely.

10Grade 8–9Triple only6 marks

Scientists supplied cells with radioactive uracil for 5 minutes. They then replaced it with non-radioactive uracil.

They measured the percentage of the cell’s radioactivity that was in the nucleus and in the cytoplasm over the next 35 minutes. The graph shows their results.

(a) Explain why the scientists used radioactive uracil rather than radioactive thymine. [2]

(b) Describe how the distribution of radioactivity changed between 5 and 40 minutes. Use data from the graph. [2]

(a) uracil is only found in RNA (not DNA)(1)
(a) thymine is (only) found in DNA, so it would label DNA (in the nucleus) not mRNA / so it could not be used to follow (m)RNA(1)
(b) percentage in nucleus decreases AND percentage in cytoplasm increases(1)
(b) correct use of data: nucleus 95% → 30% / cytoplasm 5% → 70% / lines cross at about 23 minutes; allow ±2% / ±1 minute(1)
(c) (radioactive) mRNA is made in the nucleus by transcription (using DNA as a template)(1)
(c) mRNA leaves the nucleus / moves into the cytoplasm (through nuclear pores) to the ribosomes (for translation)(1)
Answer: (a) U only in RNA; T in DNA so would not track mRNA (b) nucleus falls 95%→30%, cytoplasm rises 5%→70% (c) mRNA made in nucleus by transcription, then moves out to ribosomes in the cytoplasm
Examiner tip: 'Use data from the graph' means quoting at least one pair of values with units — a trend alone caps you at one mark.

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