QUESTION 4 A gene called biol3315 has four exons. The first exon is 99 bp long,

Question

QUESTION 4 A gene called biol3315 has four exons. The first exon is 99 bp long, the second is 108 bp, the third 270 bp, and the fourth 123 bp. In mouse embryos, the gene produces a mature mRNA molecules (excluding the cap and poly-A tail) of 501 nucleotides. In adult mice, the mRNA (again excluding the cap and poly-A tail) is 600 nucleotides. What is the most likely explanation of this observation? A mutation has occurred at the splice site in the adult. The gene produces two alternatively spliced mRNAs: one containing exons 1, 2, and 3, and the other containing exons 2,3, and 4 The gene produces two alternatively spliced mRNAs: one containing exons 2, 3, and 4, and the other containing exons 1, 2, 3, and 4. The gene produces two alternatively spliced mRNAs: one containing introns (NOT exons) 1, 2, and 3, and the other containing introns 1 and 2. You need to know the lengths of the introns to understand the differences in length of mRNA molecules.

Explanation / Answer

The answer would be The Third one.

The gene produces Two alternatively spliced mRNAs. one containing exon 2, 3 and 4 { total no.nucleotides 108+270+123 = 501 } in mouse embryos. And the other containing exons 1,2,3, and 4 { total no.nucleotides 99+108+270+123 = 600 } in adult mice.

Thanks for asking.

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