A short gene gives the mRNA 5′-AUGGCUGAAAAACCUUUGUAA-3′, which has seven codons…
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A short gene gives the mRNA 5′-AUGGCUGAAAAACCUUUGUAA-3′, which has seven codons. The polypeptide has six amino acids: Met, Ala, Glu, Lys, Pro and Leu. Three mutants are made.
- The first base of codon 3 is deleted, and the mRNA is read again from AUG. How many of the amino acids of the original polypeptide are different, counting from the first one? The mRNA now has no stop codon within the bases left.
- All three bases of codon 3 are deleted. How many amino acids does the polypeptide have?
- One C is inserted after the fourth base of the original mRNA. How many amino acids does the polypeptide have?
Need a reminder? The key ideaShow
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. Original: AUG GCU GAA AAA CCU UUG UAA.
Plan. Write each new mRNA, and read the codons from AUG.
Step 1. (a) Deleting the G of GAA gives AUG GCU AAA AAC CUU UGU AA. This reads Met, Ala, Lys, Asn, Leu, Cys. Compare it with Met, Ala, Glu, Lys, Pro, Leu. Positions 3, 4, 5 and 6 are different, so 4. The stop codon is lost.
Step 2. (b) Deleting GAA gives AUG GCU AAA CCU UUG UAA. This reads Met, Ala, Lys, Pro, Leu, which is 5 amino acids. The frame is kept, and only the Glu is lost.
Step 3. (c) The insertion gives AUG GCC UGA AAA ... The third codon is the stop codon UGA. The polypeptide is Met-Ala, which is 2 amino acids.
Answer: ; ; .
Check. A change of one base spoiled the frame in (a) and (c), and the deletion of a whole codon kept it in (b).
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