Imagine that you have just sequenced an unusually small bacterial gene that enco
ID: 60307 • Letter: I
Question
Imagine that you have just sequenced an unusually small bacterial gene that encodes the following mRNA:
5’-UCGAUGGGCUGAUGUUUGGCACGGCUCAGUCCCUGCCAAG UUCGGGACGAGUACACAGGCCCUCUAUAAGCUAGAACUGGAUUA-3’
A. How many amino acids (including fMet) are present in the most probable (longest) polypeptide encoded by
this mRNA?
B. Which amino acid is located at the C-terminus of this polypeptide?
C. What specific term describes the sequence “AACUGGAUUA” in this mRNA?
D. How many serine (ser) amino acids are present in the polypeptide product of this mRNA?
a:____________________,b:____________________,c:__________________,d:_______________________
Explanation / Answer
5’-UC GAU GGG CUG AUG UUU GGC ACG GCU CAG UCC CUG CCA AGU UCG GGA CGA GUA CAC AGG
27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9
CCC UCU AUA AGC UAG AAC UGG AUU A-3’
8 7 6 5 4 3 2 1
Protein synthesis occurs from 5' to 3' direction. So, it will start from the terminal AUU present at the 3'end. The first codon will be UAA, which is stop codon. So, protein synthesis will start from the second codon i.e. UUA . Total 27 amino acids are present. But AUG (codon for methionine) is present at position no. 11. UGA stop codon is present at position 15. So, total 4 amino acids will be there..
B...At C terminus, means at position 14, GCU is there which codes for alanine.
C....5'UTR
C....One serine is present at position 7