Drill 26 ·
AP Biology: Unit 6, Mutations (Drill 26) is a practice drill. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Analyze the effects of different point mutation types on protein structure and function, using experimental data to evaluate claims about mutation mechanisms.
| Wild-type: Codon GAG | Amino Acid: Glu (negatively charged) | Enzyme Activity: 100% |
|---|---|---|
| Mutation A: Codon GAG->UAG | Amino Acid: Glu->Stop | Enzyme Activity: 0% |
| Mutation B: Codon GAG->GAA | Amino Acid: Glu->Glu | Enzyme Activity: 98% |
| Mutation C: Codon GAG->AAG | Amino Acid: Glu->Lys (positively charged) | Enzyme Activity: 4% |
Question 1. What type of mutation is represented by Mutation B (GAG->GAA)?
Explanation: Mutation B changes GAG to GAA, both of which code for glutamic acid -- this is a synonymous substitution producing no change in amino acid sequence. This is the definition of a silent mutation. The 98% activity reflects experimental variation, not a functional change. A is incorrect because a missense mutation changes the amino acid; Mutation B does not. B is incorrect because a nonsense mutation introduces a stop codon -- that describes Mutation A. D is incorrect because a frameshift requires insertion or deletion of nucleotides, not a substitution.
Question 2. Based on the table, which conclusion is most directly supported by comparing the effects of the stop codon mutation and the charge-reversal mutation at position 47?
Explanation: The stop codon mutation (0% activity) is more disruptive than the charge-reversal mutation (4% activity) -- this conclusion is directly supported by the table without requiring assumptions beyond the data. B is not supported because the table does not isolate amino acid size as a variable. C is incorrect because the charge-reversal mutation retains 4% activity, so it does not eliminate enzyme function entirely. D is contradicted by the data -- the two mutations produce clearly different levels of disruption (0% vs. 4%).
Question 3. A researcher claims that the 4% residual activity in the charge-reversal mutation is due to the positive charge of lysine partially compensating for the loss of negative charge at position 47. Which additional experiment would most directly test this claim?
Explanation: If positive charge partially compensates for lost negative charge, then substituting a nonpolar, uncharged amino acid at position 47 should produce activity lower than 4% -- because alanine provides neither positive nor negative charge, eliminating even partial ionic interaction. This directly tests whether charge drives the residual activity. A confirms the mutation exists but does not test whether charge is the mechanism. C tests pH effects on the wild-type but does not isolate the role of position 47 charge in the mutant. D provides structural information but does not directly measure the functional role of charge.
Question 4. The stop codon mutation at position 47 produces a truncated polypeptide of 46 amino acids. A student predicts this truncated protein will have no enzymatic activity. Which reasoning best supports this prediction?
Explanation: The active site requires a negatively charged residue at position 47, but the truncated polypeptide terminates at position 46 -- one amino acid before the critical residue is incorporated. Without position 47 and the remaining 74 amino acids, the three-dimensional structure required for substrate binding cannot form. A is incorrect because release factors facilitate polypeptide release, not denaturation. C is incorrect because all three stop codons function equivalently in standard translation termination. D is incorrect because ribosomes do not degrade polypeptides -- that is the role of proteasomes, and degradation is not immediate or automatic.
Question 5. Which mutation type, if it occurred earlier in the coding sequence rather than at position 47, would most likely affect the largest number of downstream amino acids?
Explanation: A frameshift mutation caused by a single nucleotide insertion shifts the reading frame for every codon downstream of the insertion point, typically altering every subsequent amino acid before often encountering a premature stop codon. A silent mutation changes a codon but not the amino acid, leaving all downstream sequence unaffected. A missense mutation changes one amino acid but leaves all downstream codons intact. Frameshifts typically alter many more downstream codons, whereas a nonsense mutation stops translation at a single defined position.