Question

The following sequence of double stranded DNA encodes a hypothetical protein called GEN in a bacteria. Transcription starts at and includes the C-G base pair in Bold. The underlined T-A base pair indicates the transcription terminator. 3' -AAGCGGATACCTCCCAGTAGATGCTCCGGGGCTAGTGATTTCGAAC-5

A) What are the first 9 bases of the transcribed RNA? Be sure to label the 5' and 3" ends of the RNA 5' CCCCUAUGG3'
B) What are the first 4 amino acids of the subsequent polypeptide? Be sure to label the N- and C- termini. N-MET-GLU-GLY-HIS
C) How many amino acids long is the GEN polypeptide? 10 AMINO ACIDS
D) You identify a strain of bacteria containing a mutant tRNA that is capable of adding a tryptophan residue when it recognizes the stop codon UAG in the mRNA. The GEN polypeptide would be (longer, shorter, the same) in the presence of the mutart (nonsense suppressor) Trp tRNA?

Answer 1

Answer:

A) AUG GAG GGU

B) N-MET-GLU-GLY-HIS

C) 10 amino acids

D) The stop codon here is UAA, instead of UAG.  If a tryptophane residue is added where the stop codon is, the GEN polypeptide would be longer

Explanation:

First we must obtain the the complementary chain or mRNA

A) AUG start codon is complementary to DNA TAC triplet, and GAG is complementary to CTC, and GGU to CCA (running from left to right towards 5'DNA direction (AAGCGGATACCTCCCAGTAGATGCTCCGGGGCTAGTGATTTCGAAC-5)

B) TAC corresponds to Methionine, GAG to Glutamate, GGU to Glycine, and CAU to Histidine

C) UAA is the stop codon, that is complementary where the DNA triplet is ATT.  From AUG to UAA, we can have 10 amino acids transcribed

D) If a trp is included where the stop codon is UAA, then the mRNA would be longer with a consequent traduction of a longer GEN polypeptide