OCR GCSE Biology A: Genome, alleles, variants and mutations (J247) – 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 statements B5.1a to B5.1e of OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018. Questions 9 to 11 test B5.1e, which is Higher Tier only. Learn the content first from the study guide and the revision notes. The OCR GCSE Biology hub, the printable checklist and the free diagnostics show where to focus.

Questions

1. Define each term. (a) Gene [1] (b) Allele [1] (c) Heterozygous [1] (d) Phenotype [1]

2. In an invented snail species, shell banding is controlled by one gene. The allele B (banded) is dominant to b (plain).

(a) State the phenotype and whether the genotype is homozygous or heterozygous for BB, Bb and bb. [3] (b) A snail has a banded shell. Explain why you cannot be sure of its genotype. [2]

3. A genome contains 1 500 000 000 base pairs. Of these, 2% are coding DNA.

(a) Define the term genome. [1] (b) Calculate the number of coding base pairs. [2] (c) Calculate the number of non-coding base pairs. [1]

4. (a) Classify each feature as continuous or discontinuous variation: adult height, eye colour, ABO blood group, body mass. [2] (b) State the type of graph usually used for each kind of variation. [2]

5. Identical twins were brought up in different places. At age 18 twin A is 176 cm tall and twin B is 171 cm tall.

(a) Explain why identical twins are very similar in many features. [2] (b) Suggest two environmental factors that could explain the difference in height. [2] (c) Calculate the height of twin B as a percentage below twin A. Give your answer to 2 significant figures. [2]

6. The masses of 60 seeds from one plant species were recorded.

Mass class (mg)	130-149	150-169	170-189	190-209	210-229	230-249
Number of seeds	4	11	21	14	7	3

(a) State the modal class. [1] (b) Calculate the percentage of seeds in the modal class. [2] (c) Describe the shape of the distribution. [2] (d) Explain why this is continuous variation. [2]

7. In a study of 250 new variants, 210 had no effect on the phenotype, 35 influenced it and 5 determined it.

(a) State what is meant by a mutation. [1] (b) Calculate the percentage of variants with no effect. [2] (c) State one factor that increases the rate of mutation. [1]

8. A student writes: “A dominant allele is the stronger one. It stops the recessive allele working, so dominant alleles are always the most common.” Evaluate this statement. [3]

9. (Higher Tier only) A variant of an enzyme was tested against the normal enzyme at three substrate concentrations. Activity is in arbitrary units.

Substrate concentration	0.2	0.4	0.8
Normal enzyme	2.0	3.6	5.6
Variant enzyme	0.5	0.9	1.4

(a) Calculate the activity of the variant as a percentage of the normal enzyme at concentration 0.8. [1] (b) Describe what the data show. [2] (c) Explain how a mutation in the gene could cause this result. [2] (d) A different variant changes an amino acid far from the active site and the enzyme works normally. Suggest why. [2]

10. (Higher Tier only) Data supplied: AUG = Met, CCU = Pro, GGA = Gly, GGC = Gly, AAA = Lys, UGG = Trp, UAA = stop. A normal mRNA reads AUG CCU GGA AAA UGG UAA.

(a) Variant X changes GGA to GGC. Explain the effect on the protein. [1] (b) Variant Y changes AAA to UAA. State the amino acid sequence made and explain why the protein is unlikely to work. [3]

11. (Higher Tier only) Gene P codes for protein P. In most adults the gene is switched off. A person has a variant in the non-coding DNA next to gene P, and the amino acid sequence of protein P is normal. The person makes protein P as an adult.

(a) State what is meant by non-coding DNA. [1] (b) Explain how a variant in non-coding DNA can change the phenotype when the protein itself is normal. [3]

Answers

1. (a) A section of DNA that codes for a protein [1]. (b) A different version of the same gene [1]. (c) Having two different alleles of a gene [1]. (d) The observable characteristic of an organism, produced by its genotype and environment [1]. [4] Examiner insight: A definition of gene that says only “what you inherit” or “part of a chromosome” is too vague to gain credit; the link to coding for a protein is what is needed.

2. (a) BB: banded, homozygous [1]. Bb: banded, heterozygous [1]. bb: plain, homozygous [1]. (b) A banded snail may be BB or Bb [1], because one B is enough to give the banded phenotype, so both genotypes look the same [1]. [5] Examiner insight: An explanation needs both possible genotypes named and the reason (one dominant allele is enough); naming only one genotype does not answer “cannot be sure”.

3. (a) The entire genetic material of an organism [1]. (b) 1 500 000 000 × 0.02 [1] = 30 000 000 base pairs [1]. (c) 1 500 000 000 − 30 000 000 = 1 470 000 000 base pairs [1]. [4] Examiner insight: Show the multiplication line; a bare figure gives the reader nothing to follow if it is wrong, whereas clear working shows your method.

4. (a) Adult height and body mass: continuous [1]. Eye colour and ABO blood group: discontinuous [1]. (b) Discontinuous: bar chart with gaps between the bars [1]. Continuous: histogram or smooth bell-shaped curve, with no gaps [1]. [4] Examiner insight: Naming the graph is not enough on its own; saying whether the bars are separate or touching shows you understand why the graph differs.

5. (a) Identical twins have the same genome [1], so they have the same alleles for the same genes [1]. (b) Any two from: nutrition or diet; illness in childhood; exercise or lifestyle; access to healthcare [1] [1]. (c) (176 − 171) ÷ 176 × 100 [1] = 2.84%, so 2.8% [1]. [6] Examiner insight: A percentage difference is measured against the starting value stated in the question (twin A); dividing by 171 gives a different number.

6. (a) 170-189 mg [1]. (b) 21 ÷ 60 × 100 [1] = 35% [1]. (c) Most seeds are near the middle of the range [1], with fewer at the lowest and highest masses, giving a bell shape with no gaps [1]. (d) Mass can take any value in a range [1] and is influenced by many genes together with the environment [1]. [7] Examiner insight: “Normal distribution” is a fair label for the shape, but a description using the data (most in the middle, few at the ends) is safer than the label alone.

7. (a) A change in the base sequence of DNA [1]. (b) 210 ÷ 250 × 100 [1] = 84% [1]. (c) Ionising radiation, ultraviolet light or certain chemicals (any one) [1]. [4] Examiner insight: Give one named factor, not “the environment”, which is too general to be a specific cause.

8. The statement is mostly incorrect. A dominant allele is not stronger and does not stop the recessive allele working [1]; it is simply the one whose characteristic shows when both alleles are present [1]. Dominant alleles are not always the most common; Huntington’s disease is caused by a dominant allele yet is rare [1]. [3] Examiner insight: “Evaluate” needs a judgement plus reasons; supporting evidence such as a named example of a rare dominant allele is stronger than assertion.

9. (a) 1.4 ÷ 5.6 × 100 = 25% [1]. (b) The variant has lower activity than the normal enzyme at every concentration [1]; it works at 25% of the normal rate at all three concentrations, and both increase with substrate concentration [1]. (c) The mutation changes the amino acid sequence, so the enzyme folds into a different shape [1], changing the active site so the substrate fits less well [1]. (d) The change does not alter the shape of the active site [1], so the substrate still binds as before [1]. [7] Examiner insight: Quote figures when describing data; “the variant is lower” without numbers is weaker than “25% of normal at each concentration”.

10. (a) GGC also codes for Gly, so the amino acid sequence is unchanged [1]. (b) The protein is Met-Pro-Gly [1], because the new UAA codon is a stop codon, so translation ends early [1]; the polypeptide is much shorter and so has a different shape and is unlikely to work [1]. [4] Examiner insight: Read the sequence codon by codon from the start; give the three amino acids in order, not just “shorter”.

11. (a) DNA that does not code for a protein [1]. (b) The variant changes how the gene is expressed [1]; it may stop the control that normally prevents transcription of the gene to mRNA, so the gene stays switched on [1]; more protein P is made in adulthood, so the phenotype changes even though the protein is normal [1]. [4] Examiner insight: The answer must say that the protein is unchanged but the amount or timing changes; saying “the protein is altered” describes a coding variant, not this one.

Where marks are usually lost

• Writing definitions that are vague, such as “a gene is what you inherit”.
• Confusing genotype (letters) and phenotype (characteristic).
• Saying a dominant phenotype proves the genotype is homozygous.
• Calling the dominant allele “stronger” or “more common”.
• Not stating whether bars on a graph are separate or touching.
• Using the wrong base value for a percentage difference.
• Describing data without quoting figures.
• Skipping the chain from mutation to amino acids to shape to active site.
• Saying non-coding variants change the protein, when they change how much or when it is made.

Next steps

• Review the revision notes and the study guide.
• See the OCR GCSE Biology hub and tick off the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, published by OCR: module B5.1, statements B5.1a to B5.1e.