DNA, genes and chromosomes - AQA A-Level Biology (Paper 1) Exam Questions

Practise 17 exam-style questions on DNA, genes and chromosomes for AQA A-Level Biology (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

17 questions
1–7 marks each
Instant marking
No sign-up needed

17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Gene T codes for an enzyme in a species of plant. In every plant of this species, gene T is found at the same locus on chromosome 4.

Explain what is meant by a locus, and explain how two plants of this species can have different base sequences at this locus.

Model answer

A locus is the fixed position that a gene occupies on a particular DNA molecule (chromosome). Two plants can have different base sequences at this locus because they carry different alleles of gene T: alleles are different versions of the same gene, produced by mutation.

Explanation

Every gene occupies a fixed position, its locus, on a particular DNA molecule, so gene T is always found at the same place on chromosome 4. The base sequence at the locus can still differ between individuals because there can be different alleles of the gene, which arise by mutation. Saying that the plants have 'different genes' is wrong: it is the same gene with a different base sequence.

Mark scheme

1 mark
(A locus is) the fixed position of a gene on a DNA molecule / chromosome
Accept 'the particular place where a gene is found on a chromosome'. Do not accept 'the position of a chromosome (in the nucleus)'.
1 mark
The plants have different alleles / versions of gene T (produced by mutation)
Accept 'a mutation has changed the base sequence of gene T in one plant'. Ignore 'different genes' (it is the same gene at the same locus).

Question 2

3 marks

Scientists isolated chromosomes from the nuclei of plant cells. They treated one sample of the chromosomes with a protease, an enzyme that hydrolyses peptide bonds. They treated a second sample with protease solution that had been boiled, and left a third sample untreated.

Under an electron microscope, the untreated chromosomes and the chromosomes treated with boiled protease were compact structures. In the chromosomes treated with protease, the DNA had spread out into a large tangle of very long threads.

Explain the result for the chromosomes treated with protease.

Suggest why the scientists used a sample treated with boiled protease.

Model answer

The protease hydrolysed the peptide bonds in the histones, the proteins associated with the DNA in a chromosome. The very long DNA molecule is normally wound around histones, which keep it compact, so once they were removed the DNA unwound and spread out. Boiled protease is denatured, so it cannot hydrolyse proteins; the chromosomes treated with it stayed compact, showing that the change was caused by the enzyme's activity and not by another substance in the solution.

Explanation

In eukaryotes, the DNA in the nucleus is very long, linear and associated with histone proteins; together they form a chromosome. Hydrolysing the histones leaves only the long DNA, which spreads out. A boiled-enzyme control shows that the effect needs an active enzyme. A common error is to say that the protease digests DNA: proteases hydrolyse peptide bonds in proteins.

Mark scheme

1 mark
Protease hydrolyses (peptide bonds in) the histones / proteins associated with the DNA
Max 3. Reject 'protease hydrolyses/digests the DNA'. Accept 'breaks down the histones'.
1 mark
Histones (normally) hold the (very long) DNA in a compact form / the DNA is wound around histones, so without them the DNA unwinds and spreads out
Accept 'the DNA is no longer wound around/packed by proteins'. Ignore 'the chromosome falls apart' unless the DNA is mentioned.
1 mark
Boiled protease is denatured / has no activity, so this shows that the change was caused by the action of the enzyme (hydrolysing proteins), not by something else in the solution / the treatment
Accept 'shows it is the active enzyme, not the solution it is in, that causes the change'. Ignore 'to compare' or 'as a control' unqualified.

Question 3

4 marks

In one species of mammal, the DNA molecule in the largest chromosome is 7.48 cm long when it is fully stretched out. In a DNA molecule, adjacent base pairs are 0.34 nm apart.

Calculate the number of base pairs in this DNA molecule. Give your answer in standard form.

This DNA molecule fits inside a nucleus that is about 6 μm in diameter. Describe how the DNA is arranged in the nucleus so that it fits.

Model answer

7.48 cm = 7.48 × 10⁻² m = 7.48 × 10⁷ nm
Number of base pairs = 7.48 × 10⁷ ÷ 0.34 = 2.2 × 10⁸

In the nucleus the DNA is wound tightly around proteins called histones. The DNA and its histones are coiled and folded further into a compact chromosome, so the very long molecule takes up only a small volume.

Explanation

1 cm = 10⁻² m and 1 nm = 10⁻⁹ m, so 1 cm = 10⁷ nm; dividing the length by the distance per base pair gives the number of base pairs. In the nucleus of a eukaryotic cell each very long, linear DNA molecule is associated with histones, and the DNA-histone complex is coiled into a chromosome, which packs several centimetres of DNA into a nucleus a few micrometres across. Most errors come from converting centimetres to nanometres; 'the DNA is a double helix' does not explain the packing, because every DNA molecule is a double helix.

Mark scheme

1 mark
Correct method, e.g. 7.48 × 10⁷ ÷ 0.34 (length converted to nm, then divided by 0.34); or a correct final answer of 2.2 × 10⁸ with no working
A correct answer of 2.2 × 10⁸ (accept 2.20 × 10⁸) with no working scores 2 marks (points 1 and 2). Accept for 1 mark: evidence of 7.48 × 10⁷ nm (correct conversion) OR 220 000 000 (correct value not in standard form) OR 2.2 × 10⁷ or 2.2 × 10⁹ (method correct but one power-of-ten error in converting cm to nm). Points 1 and 2 are for the calculation; points 3 and 4 are for the arrangement.
1 mark
2.2 × 10⁸ (base pairs)
Accept 2.20 × 10⁸. 220 000 000 scores point 1 only (not in standard form).
1 mark
DNA is wound around / associated with histones (proteins)
Accept 'DNA is coiled around histone proteins' or 'DNA and histones form a chromosome'. Ignore 'DNA is a double helix' on its own. Reject 'the DNA is circular' or 'the DNA is in plasmids'.
1 mark
The DNA and histones are (further) coiled / folded / condensed (into a compact chromosome)
Accept 'supercoiled' or 'tightly packed / condensed to form a chromosome'. Do not accept 'the DNA is coiled into a double helix' (true of all DNA) for this mark.

Question 4

6 marks

Describe the genetic code and how it relates the base sequence of a gene to the primary structure of a polypeptide.

Do not include details of transcription or translation in your answer.

Model answer

The genetic code is a triplet code: a sequence of three DNA bases (a triplet) codes for a specific amino acid, and the order of the triplets in a gene determines the order of amino acids in the polypeptide, which is its primary structure. There are 64 possible triplets but only 20 amino acids, so the code is degenerate: most amino acids are coded for by more than one triplet. Some triplets are stop codes, which do not code for an amino acid and mark the end of the polypeptide. The code is non-overlapping: each base is part of only one triplet, so the triplets are read one after another. The code is universal: the same triplet codes for the same amino acid in all organisms. In eukaryotic genes only the exons code for amino acids; the introns between them are non-coding.

Explanation

Learn a precise meaning for each key word: universal (the same triplet codes for the same amino acid in all organisms), non-overlapping (each base is read in only one triplet) and degenerate (an amino acid can have more than one triplet). Many answers lose marks by naming these features without describing them, or by confusing 'degenerate' with 'non-overlapping'. Link the code to the polypeptide: the sequence of triplets gives the sequence of amino acids.

Mark scheme

1 mark
A sequence of three bases / a triplet codes for a (specific) amino acid
Max 6. For points 3, 4 and 5, the meaning must be described: the term alone does not score. Accept codon for triplet throughout.
1 mark
The sequence of triplets (in the gene) determines the sequence of amino acids / primary structure
1 mark
Universal: the same triplet codes for the same amino acid in all organisms
Do not accept 'DNA is universal'.
1 mark
Non-overlapping: each base is part of only one triplet / triplets are read one after another
1 mark
Degenerate: (most) amino acids are coded for by more than one triplet
Reject 'a triplet codes for more than one amino acid'. Do not accept 'the amino acids are degenerate'.
1 mark
There are 64 (possible) triplets for (only) 20 amino acids
Accept 'more triplets than amino acids'.
1 mark
Some triplets are stop codes / do not code for an amino acid (and mark the end of the polypeptide)
Accept reference to a start codon.
1 mark
(In eukaryotes) only the exons code for amino acids / introns are non-coding

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.