QUESTION 95 (8 marks)
In a hypothetical beetle, functional allele G is dominant and produces a green-pigment enzyme. Allele g changes the mRNA codon CAA (glutamine) to UAA (stop), producing a non-functional enzyme. Both parents are green and have genotype Gg. A 600 bp DNA test fragment is cut into 400 bp and 200 bp fragments for G; g lacks the restriction site and remains 600 bp. Digestion is complete.
a) Classify the mutation and explain its likely effect on the polypeptide. [2 marks]
b) Predict the offspring genotype probabilities. [2 marks]
c) A randomly selected green offspring is tested. Determine the probability that its digest shows all three bands: 600, 400 and 200 bp. [2 marks]
d) Explain how the three bands indicate a heterozygote rather than three alleles in one diploid individual. [2 marks]
Practice marking scheme
Answer
a) Nonsense mutation; premature stop/truncated chain. b) GG 1/4, Gg 1/2, gg 1/4. c) 2/3. d) The G copy makes two fragments and the g copy makes one.
Working
CAA becomes a stop codon, potentially terminating translation early. The cross gives . Among green offspring, and remain in ratio 1:2, so the conditional heterozygote probability is . A Gg individual has two alleles; digestion of G produces two smaller fragments, and digestion of g leaves one larger fragment. Band number therefore need not equal allele number.
Marking criteria
- Identifies nonsense mutation. [1 mark]
- Explains premature termination/truncation. [1 mark]
- Gives the correct three genotype outcomes. [1 mark]
- Gives probabilities 1/4, 1/2 and 1/4. [1 mark]
- Conditions on the green offspring group. [1 mark]
- Calculates 2/3. [1 mark]
- Explains two fragments from the G allele. [1 mark]
- Explains one uncut fragment from g and distinguishes fragments from alleles. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.