You are comparing the process of RNA splicing in an mRNA from normal and mutant cells of the same tissue. You find that the mutant mRNA is longer than the mRNA from the normal cells. Examination of the sequences of the gene encoding this mRNA reveals that there are nucleotide changes at the boundary of one of the exons in the mutant cell gene. Which of the following nucleotides would be expected to be found at the 3′ end of upstream donor exons and the 5′ end of downstream acceptor exons in the normal gene but not in the mutant gene?
A) AG-G
B) AG-AG
C) CA-GA
D) CG-G
E) G-AG

Question

Quizplus · Accepted Answer

The correct answer is A because AG-G is the consensus sequence for the 3' end of upstream donor exons and the 5' end of downstream acceptor exons in the normal gene. In the mutant gene, there are nucleotide changes at the boundary of one of the exons, which would disrupt the consensus sequence and prevent proper splicing.  Analysis of the human genome and the encoded mRNA have revealed highly conserved consensus sequences at the 5′ and 3′ ends of essentially all mRNA exons and introns. The 3′ end of all upstream exons contains AG with the G being the donor nucleotide attached to the 5′ end of the downstream exon. The 5′ end of the acceptor exon is always a G residue.

You Are Comparing the Process of RNA Splicing in an MRNA