Question

The diagram below shows an mRNA molecule that encodes a protein with 202 amino acids. The start and stop codons are highlighted, and a portion of the nucleotide sequence in the early part of the molecule is shown in detail. At position 35, a single base-pair substitution in the DNA has changed that would have been a uracil (U) in the mRNA to an adenine (A).

Based on the genetic code chart below, which of the following would be the result of this single base-pair substitution?

• A. a silent mutation (no change in the amino acid sequence of the protein)
• B. a frameshift mutation causing extensive change in the amino acid sequence of the protein
• C. a missense mutation causing a single amino acid change in the protein
• D. a frameshift mutation causing a single amino acid change in the protein
• E. a nonsense mutation resulting in early termination of translation

 

Answer 1

If we start from the beginning of the RNA and count nucleotides in groups of three (because this is how many nucleotides make a codon), we start with 1-2-3 which is AUG (the start codon), then 4-5-6, 7-8-9, and so on. Of course, we don't see many of these nucleotides on the diagram, but we know that 28-29-30 makes a codon (because, again, codons come in groups of three nucleotides). Then we have 31-32-33 (which is CUA), and then 34-35-36 (which is UUA). The middle U of UUA is substituted with an A. If we look at the codon diagram, UUA makes leucine, but UAA is the stop codon. Thus, because the RNA encoded more amino acids but the mutation switched the codon to read as a stop codon, this is early termination of protein synthesis and therefore a nonsense mutation.

The answer is E: it becomes a nonsense mutation.