Imagine you have just sequenced a small bacterial mRNA: 1. Imagine you have just

Question

Imagine you have just sequenced a small bacterial mRNA:
1. Imagine you have just sequenced a small bacterial mRNA: (5 Points) 5 - UCGCUGGGCUGAUGUUUGGCACGGCUCAGUUCAUGCCAAGUUCGCGUA GGCCGCUCGAAGAGCGAGUACACAGGCCCUCUUGCUAGUAGCGCUGGAUG-3' A) How many amino acids (including fMet) are present in the most probable (longest) polpeptide encoded by this mRNA? B) which amino acid is located at the C-terminus of this polypeptide? C) What does "UAGCGCUGGAUG" represent in this mRNA? D) How many phenylalanine (phe) molecules are present in the polypeptide product of this mRNA? E) How many proline (pro) molecules are present in the polypeptide product of this mRNA? 2. Which of the following typically occurs during RNA processing in cukaryotes? List all correct answers. A) addition of the 5' methyl-G-cap: B) association of hnRNA with the ribosome; C) addition of 3' Poly(A)tail; D) binding of anticodons to codons E) translation of introns; F) Joining of adjacent introns (2 Points)

Explanation / Answer

UCG CUG GGC UGA UGU UUG GCA CGG CUC AGU UCA UGC CAA GUU CGC GUA GGC CGC UCG AAG AGC GAG UAC ACA GGC CCU CUU GCU AGU AGC GCU GGA UG 3

fmethionine will not be in this mRNA because it do not have the code for n-formyl methionine that is 5’AUG3’.

The protein synthesis will stops after synthesizing 3 amino acids because after synthesizing 3 amino acids it would encounter with the stop codon that is UGA in this mRNA sequence.

There will be 3 amino acids.

At the c terminal end there will be Glycine (GGC)

UAG CGC UGG AUG

In this mRNA there is stop codon, codon for arginine, tryptophan and ignition codon are present.

There will not be phenylalanine molecule in the product of this mRNA as it does not contain the code UUC or UUU which code for phenylalanine

There will be no proline in this mRNA product because the synthesis of protein will stop after the synthesis of 3 amino acid as the stop codon encounters, therefore there will not be the proline molecule.

2. A) Addition of 5’ methyl G cap

C) Addition of 3’ Poly A tail

D) Binding of anticodon to codon

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