Question

spliceosomes remove introns exactly. an mrna strand has the sequence 5-uuaggcuaacgg-3. the intron is uuag, and the exon is gcuaacgg. if the spliceosome malfunctions, and forms the intron uuaggc and the exon uaacgg, what is the most likely consequence of this?

○ translation begins with the uua codon

○ translation begins with the gcu codon

○ translation does not occur, as aug is missing

○ translation does not occur, as uaa is the signal for translation termination

○ translation does not occur, as the ribosome does not recognize the new mrna

Explanation:

1. First, recall the role of start and stop codons in translation. The start codon is typically AUG, and stop codons include UAA, UAG, UGA.
2. Analyze the malfunctioning splicing: The new exon is UAACGG. The first codon of the exon would be UAA (since codons are 3 - nucleotide sequences). UAA is a stop codon.
3. Evaluate each option:

• Option 1: The exon starts with UAA, not UUA, so translation doesn't start with UUA. Eliminate.
• Option 2: The exon starts with UAA, not GCU. Eliminate.
• Option 3: AUG is the start codon, but the issue here is a stop codon at the start, not just missing AUG. Eliminate.
• Option 4: The first codon of the exon (UAACGG, first three nucleotides UAA) is a stop codon (UAA signals translation termination). So translation will not occur as UAA is a stop codon at the start.
• Option 5: Ribosomes recognize mRNA based on start codons and other factors, but the main issue here is the stop codon, not recognition failure. Eliminate.

Answer:

D. Translation does not occur, as UAA is the signal for translation termination (assuming the options are labeled A - E with D being this option; if the original options had different labels, adjust accordingly. But based on the content, the correct option is the one stating translation does not occur as UAA is a translation termination signal).