AQA A-level Biology: DNA, Genes, Chromosomes and Protein Synthesis (7402) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.4.1 (DNA, genes and chromosomes) and 3.4.2 (DNA and protein synthesis) of the AQA specification for AS and A-level Biology (codes 7401 and 7402), first teaching 2015. Both sections are AS content and also part of the A-level course.

Learn the content first from the study guide or the revision notes. See also the subject hub and the checklist to print.

Questions

1. Compare the DNA in a bacterial cell with the DNA in the nucleus of a human liver cell. [3]

2. (a) State what is meant by a gene. [2] (b) State what is meant by the locus of a gene. [1]

3. Explain what is meant by each of these descriptions of the genetic code: universal, non-overlapping and degenerate. [3]

4. Mitochondria contain DNA. (a) Describe two features of mitochondrial DNA. [2] (b) Name one other organelle that contains DNA. [1]

5. A short gene has this template strand: TAC GAA CCT AGG TGC TTA ACT. Use this codon table: AUG = Met (start), CUU = Leu, GGA = Gly, UCC = Ser, ACG = Thr, AAU = Asn, UGA = stop. (a) Write the mRNA sequence. [1] (b) Give the amino acid sequence of the polypeptide. [2] (c) Give the anticodon of the tRNA that brings the fourth amino acid. [1] (d) Explain why the polypeptide has six amino acids although the template has seven triplets. [1]

6. A eukaryotic pre-mRNA is 2,760 nucleotides long. It contains three introns of 318, 402 and 207 nucleotides. The mature mRNA has 84 nucleotides that are not translated. (a) Calculate the length of the mature mRNA. [2] (b) Calculate the number of amino acids in the polypeptide. The translated region includes the stop codon. [2]

7. Describe two similarities and two differences between the structure of mRNA and the structure of tRNA. [4]

8. A bacterium has a genome of 4.6 million base pairs and 4,300 genes with an average coding length of 950 base pairs. A eukaryote has a genome of 120 million base pairs and 14,000 genes with an average coding length of 1,700 base pairs. (a) Calculate the percentage of each genome that codes for polypeptides, to one decimal place. [2] (b) Explain the difference. [2] (c) A skin cell and a pancreatic cell from the same person have the same genome but different proteomes. Explain. [2]

9. Describe how pre-mRNA is made from DNA and converted to mRNA in a eukaryotic cell. [5]

10. A cell-free extract contains ribosomes, tRNA, ATP, enzymes and 14C-labelled leucine. Different additions were made to six tubes and the radioactivity in newly made polypeptide was measured.

Tube	Addition	Radioactivity (counts per minute)
1	mRNA	4,820
2	None	130
3	mRNA, but ATP omitted	410
4	mRNA treated with ribonuclease	150
5	DNA of the same gene	170
6	mRNA, but tRNA removed	220

(a) Calculate the radioactivity in tube 3 as a percentage of tube 1, to the nearest whole number. [1] (b) Explain the results for tubes 2 and 4. [2] (c) Explain the result for tube 5. [2] (d) What do the results for tube 6 show about tRNA? [1]

11. Cultured cells were treated with an inhibitor of RNA polymerase. RNA synthesis and protein synthesis were measured as a percentage of the rates in untreated cells.

Time after adding inhibitor (h)	RNA synthesis (%)	Protein synthesis (%)
0.5	12	85
2	9	41
4	8	14

(a) Calculate the fall in RNA synthesis at 0.5 h, in percentage points. [1] (b) Describe the change in protein synthesis. Use figures. [2] (c) Explain why protein synthesis fell more slowly than RNA synthesis. [2]

12. Describe how a polypeptide is made at a ribosome from mRNA. Include the roles of tRNA and ATP. [6]

Answers

1. Bacterial DNA is short, human nuclear DNA is very long [1]; bacterial DNA is circular, human nuclear DNA is linear [1]; bacterial DNA is not associated with proteins, human DNA is associated with histones [1]. [3] Examiner insight: A comparison answer should state both sides in each point; naming only the bacterial feature, such as “circular”, leaves the comparison incomplete.

2. (a) A base sequence of DNA [1] that codes for the amino acid sequence of a polypeptide (or for a functional RNA) [1]. [2] (b) The fixed position of the gene on a particular DNA molecule [1]. [1] Examiner insight: “A gene codes for a protein” is a looser wording; tying the gene to the sequence of amino acids in a polypeptide is the safer definition.

3. Universal: a given triplet specifies one particular amino acid whichever organism it is in [1]. Non-overlapping: each base is part of only one triplet, so the sequence is read in separate groups of three [1]. Degenerate: most amino acids are coded for by more than one triplet [1]. [3] Examiner insight: Define each term in terms of triplets and amino acids; repeating the word, such as “degenerate means the code is degenerate”, gains nothing.

4. (a) Any two from: short [1]; circular [1]; not associated with protein (any two features score; a third adds nothing). [2] (b) Chloroplast [1]. [1] Examiner insight: “Circular” and “no histones” are separate features; state each one clearly so that it is easy to see two distinct points.

5. (a) AUG CUU GGA UCC ACG AAU UGA [1]. [1] (b) Reads from the start codon AUG [1]; Met - Leu - Gly - Ser - Thr - Asn [1]. [2] (c) The fourth codon is UCC, so the anticodon is AGG [1]. [1] (d) The seventh codon, UGA, is a stop codon and does not code for an amino acid [1]. [1] Examiner insight: Show the mRNA before the amino acids; if the mRNA contains T, or is not complementary, every later step built on it will be wrong.

6. (a) Total introns = 318 + 402 + 207 = 927 [1]; mRNA = 2,760 - 927 = 1,833 nucleotides [1]. [2] (b) Translated region = 1,833 - 84 = 1,749; 1,749 / 3 = 583 codons [1]; minus the stop codon gives 582 amino acids [1]. [2] Examiner insight: Setting the working out line by line makes it easy to see where an error began; forgetting the stop codon is the commonest slip in part (b).

7. Similarities: both are single polynucleotide strands [1]; both contain ribose and uracil [1]. Differences: mRNA is linear whereas tRNA is folded into a clover-leaf shape held by hydrogen bonds between bases [1]; mRNA has codons whereas tRNA has one anticodon and an amino acid binding site [1]. [4] Examiner insight: If asked for similarities and differences, label them; a list of unlabelled facts about each molecule does not show the comparison.

8. (a) Bacterium: 4,300 x 950 = 4,085,000; 4,085,000 / 4,600,000 x 100 = 88.8% [1]. Eukaryote: 14,000 x 1,700 = 23,800,000; 23,800,000 / 120,000,000 x 100 = 19.8% [1]. [2] (b) The eukaryotic genome contains introns within genes [1] and non-coding multiple repeats between genes [1]. [2] (c) Genome is the complete set of genes and is the same in both cells [1]; different genes are expressed in each cell type, so they make different proteins (different proteomes) [1]. [2] Examiner insight: For a calculation, give the unrounded step before the answer so that a slip in rounding can be seen; part (c) needs both halves, same genes and different proteins made.

9. RNA polymerase attaches to the DNA and the strands separate; one strand acts as the template [1]. Free RNA nucleotides pair with the exposed bases by complementary base pairing [1]. RNA polymerase joins adjacent nucleotides by condensation reactions [1]. This forms pre-mRNA [1]. Introns are removed and exons joined (splicing) to form mRNA [1]. [5] Examiner insight: A sequence of numbered stages helps; the most-missed point is that splicing acts on pre-mRNA, not on the DNA.

10. (a) 410 / 4,820 x 100 = 8.5, so 9% [1]. [1] (b) Without mRNA there is no code for the ribosome to follow, so little polypeptide is made [1]; ribonuclease breaks down the mRNA, so the result is similar to having none [1]. [2] (c) Ribosomes translate mRNA, not DNA [1]; the DNA must first be transcribed to mRNA, which the extract cannot do here [1]. [2] (d) tRNA is needed for translation (to bring amino acids to the ribosome) [1]. [1] Examiner insight: Residual counts in tubes 2 to 6 are small but not zero; say “very little” rather than “none”, and use a figure from the table when you compare.

11. (a) 100 - 12 = 88 percentage points [1]. [1] (b) Protein synthesis falls steadily [1] from 85% at 0.5 h to 14% at 4 h, a fall of 71 percentage points [1]. [2] (c) mRNA already made continues to be translated [1]; protein synthesis falls as the existing mRNA is broken down and no new mRNA is made [1]. [2] Examiner insight: Quote figures from the table for “describe” questions, and use the idea of existing mRNA to turn a description into an explanation.

12. mRNA binds to a ribosome [1]. A tRNA with an anticodon complementary to the first codon binds, carrying its specific amino acid [1]. A second tRNA binds to the next codon [1]. A peptide bond forms between the amino acids by a condensation reaction [1]. ATP provides the energy [1]. The ribosome moves along the mRNA codon by codon until a stop codon, when the polypeptide is released [1]. [6] Examiner insight: For a six-mark description, a clear ordered account that names codon, anticodon, ribosome and ATP covers the needed points; vague phrases such as “tRNA helps” usually earn little.

Where marks are usually lost

• Putting thymine into an mRNA sequence, or giving a DNA complement when mRNA was asked for.
• Giving the codon when the anticodon was asked, or the other way round.
• Including the stop codon when counting amino acids.
• Describing only one type of DNA when asked to compare.
• Saying “no protein” for the control tubes when the data show small non-zero values.
• Describing a trend without figures from the table.
• Saying introns are cut out of DNA.
• Forgetting to say that a tRNA carries one specific amino acid.

Next steps

• Return to the revision notes and study guide.
• Visit the subject hub and mark progress on the checklist to print.
• Use the 10-minute diagnostics to choose your next topic.
• Book a free trial class.

Official syllabus

Source document: the AQA specification for AS and A-level Biology (codes 7401 and 7402), first teaching 2015, version 1.5, published by AQA. Sections 3.4.1 and 3.4.2 only.