AQA AS Level Biology Paper 1, November 2021: Question 1

8 marks · Medium difficulty · Short Answer

Identify bonds in DNA, calculate gene length using a scale diagram, compare tRNA and mRNA structure, and describe pre-mRNA splicing differences.

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Question

Figure 1 shows part of a double-stranded DNA helix with a bracket indicating a length of 1.7 nm spanning 10 base pairs. Four sub-questions follow: 01.1 asks to name the types of bonds between complementary base pairs and adjacent nucleotides; 01.2 asks to calculate the length of a gene containing 4.38 x 10^3 base pairs using Figure 1; 01.3 asks to describe two differences between tRNA and mRNA structure; 01.4 asks to describe and explain a structural difference between pre-mRNA and mRNA in eukaryotic cells.
Question text

01 Figure 1 shows part of a DNA molecule.

Figure 1

01.1 Name the type of bond between:

[2 marks]

complementary base pairs

adjacent nucleotides in a DNA strand

01.2 The length of a gene is described as the number of nucleotide base pairs it contains.

Use information in Figure 1 to calculate the length of a gene containing 4.38 ×10

base pairs.

[2 marks]

Answer nm

01.3 Describe two differences between the structure of a tRNA molecule and the

structure of an mRNA molecule.

[2 marks]

01.4 In a eukaryotic cell, the structure of the mRNA used in translation is different from

*02* the structure of the pre-mRNA produced by transcription.

Describe and explain a difference in the structure of these mRNA molecules.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for questions 01.1 to 01.4. Question 01.1 awards 2 marks for hydrogen bonds and phosphodiester bonds. Question 01.2 awards 2 marks for 1489 or 1489.2 nm (with 1 mark for intermediate calculation steps like 876 or 1861-1862). Question 01.3 awards 2 marks for comparing features such as cloverleaf vs linear shape, presence of hydrogen bonds, amino acid binding site, and anticodon/codon differences. Question 01.4 awards 2 marks for stating fewer nucleotides/shorter length or absence of introns due to splicing.

Question Marking Guidance Mark Comments

01.1 1. Hydrogen (bonds); 2. Accept

2. Phosphodiester (bonds); ester/covalent bond

01.2 Correct answer for 2 marks = 1489/1489.2;;

Incorrect answer but for 1 mark accept:

OR

1861 - 1862

01.3 Must be a comparison

1. tRNA is 'clover leaf shape', mRNA is linear;

1. Reject tRNA is

2. tRNA has hydrogen bonds, mRNA does not; double stranded

3. tRNA has an amino acid binding site, mRNA 2 1. Accept tRNA is

does not; folded for tRNA is

‘clover leaf shaped’

4. tRNA has anticodon, mRNA has codon;

3. Accept ‘CCA end'

– LOGYfor amino acid binding––JUNE2021

site

01.4 1. mRNA fewer nucleotides 1. Accept mRNA is

shorter OR pre-mRNA

OR is longer

Pre-mRNA more nucleotides

OR

mRNA has no introns/has (only) exons

OR

Pre-mRNA has (exons and) introns;

2. (Because of) splicing;

TOTAL 8

How to answer it

Nucleic Acids & Protein Synthesis Study Guide

AQA AS Level Biology

What this question tests

This question assesses your knowledge of the chemical structure of DNA, the ability to interpret biological diagrams to extract dimensional scale factors for calculations, comparative structural biology of different RNA species (tRNA vs mRNA), and post-transcriptional processing in eukaryotic cells (pre-mRNA splicing).

Part (01.1) - Chemical Bonds in DNA

Name the type of bond between complementary base pairs and adjacent nucleotides in a DNA strand.

✅ Correct Answer

  • Complementary base pairs: Hydrogen (bonds)
  • Adjacent nucleotides: Phosphodiester (bonds)
2 marks available (1 mark per correct bond type). Note: 'Ester' or 'covalent' accepted for phosphodiester.

💡 Key Knowledge

Hydrogen bonds form between nitrogenous bases (A with T, C with G), providing stability to the double helix while allowing easy unzipping during replication and transcription. Phosphodiester bonds form between the sugar of one nucleotide and the phosphate group of the next, building the strong sugar-phosphate backbone.

❌ Common Errors

Students often confuse glycosidic bonds (found between the pentose sugar and the nitrogenous base) with phosphodiester bonds. Ensure you specify phosphodiester for the backbone.

Part (01.2) - Gene Length Calculation

Use information in Figure 1 to calculate the length of a gene containing 4.38 × 10³ base pairs.

📐 Step-by-Step Calculation

  1. Analyze Figure 1: The bracket shows that a span of base pairs measures 1.7 nm . Count the base pair intervals carefully within that bracket (there are 5 base pair steps spanning 4 intervals, or looking closely at standard AQA scaling, 10 base pairs span roughly 3.4 nm, meaning 1 base pair interval = 0.34 nm, or based on the printed figure bracket spanning 5 base pairs = 1.7 nm, giving 1.7 / 5 = 0.34 nm per base pair). Alternatively, standard DNA has 10 base pairs per 3.4 nm turn, yielding 0.34 nm per base pair. Let's use the figure scale: 1.7 nm represents a specific number of base pairs shown. Counting the rungs in the bracket gives 5 base pairs. Thus, 1.7 / 5 = 0.34 nm per base pair.
  2. Calculate total length: Multiply the number of base pairs by the length per base pair.
    4.38 × 10³ × (1.7 / 5) OR using standard biological constant 0.34 nm per base pair:
    4.38 × 10³ × 0.34 = 1489.2 nm
2 marks for correct answer ( 1489 or 1489.2 ). 1 mark awarded for working/error carried forward if using alternative clear base-pair length interpretations from the diagram (e.g., accepting 876 or 1861-1862 based on miscounts).

🧠 Exam Technique

Always inspect scale bars or labeled brackets on diagrams meticulously. If a measurement span covers multiple units, divide the dimension by the number of units to find a single unit value before scaling up to large numbers.

Part (01.3) - Comparing tRNA and mRNA Structure

Describe two differences between the structure of a tRNA molecule and the structure of an mRNA molecule.

✅ Correct Answers (Any two of:)

  • tRNA has a 'cloverleaf' shape (or is folded), whereas mRNA is linear.
  • tRNA has hydrogen bonds between base pairs, whereas mRNA does not (or has fewer).
  • tRNA has an amino acid binding site (CCA end), whereas mRNA does not.
  • tRNA has an anticodon, whereas mRNA has codons.
2 marks available (1 mark per valid, comparative difference).

❌ Common Errors & Examiner Warnings

Crucial Rule: Answers must be comparative! Writing "tRNA has an anticodon" scores 0 marks unless you contrast it with mRNA having codons, or state that mRNA does not have an anticodon. Do not state "tRNA is double stranded" (it is single-stranded folded back on itself).

Part (01.4) - pre-mRNA vs mRNA in Eukaryotes

In a eukaryotic cell, the structure of the mRNA used in translation is different from the structure of the pre-mRNA produced by transcription. Describe and explain a difference in the structure of these mRNA molecules.

✅ Correct Answer

  • Difference (Describe): mRNA has fewer nucleotides / is shorter / has no introns (only exons), whereas pre-mRNA has more nucleotides / is longer / contains introns.
  • Explanation: This difference is caused by splicing, where non-coding introns are removed and coding exons are joined together.
2 marks available: 1 mark for describing the structural difference, 1 mark for explaining it via splicing.

💡 Key Knowledge

Transcription in eukaryotic genes produces pre-mRNA containing both non-coding regions ( introns ) and coding regions ( exons ). Before translation can occur at the ribosome, spliceosomes remove introns and stitch exons together during post-transcriptional modification.

Topics

Biology · 3.1 Biological molecules · 3.4 Genetic information, variation and relationships between organisms

Question and mark scheme from the AQA AS Level Biology examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.