Shown below is a 240 base pair segment of an E. coli gene. It includes the promo

Question

Shown below is a 240 base pair segment of an E. coli gene. It includes the promoter and the coding region of a gene. Transcription initiation begins at nucleotide position 71 (indicated by bold "a"). What are the first five amino acids of the resulting protein? Use genetic code table from your book. Does translation terminate at the underlined TAA at position 108 (c, bold)? Why or why not? How would your answer to i) change if the C/G base pair at position 95 (d, bold) was deleted? How would your answer to i) change if the A/T base pair were added between 98 & 99 (e, bold)? How would your answer to i) change if the A/T base pair at position 103 (f, bold) were changed to G/C?

Explanation / Answer

i) The first 5 amino acids of the resultant sequence as follows:

DNA sequens:   AATTGTGAGCGGATA

mRNA sequnce: UUA ACA CUC GCC UAU

Amino acids: Leu Thr Leu Ala Tyr

ii) No. Translation does not terminate at TAA (bold c) because in mRNA it is read as AUU and AUU codes for Ile (isoleucine).

iii) When the C/G base pair was deleted the mRNA sequence is written as follows:

DNA sequence: AATTGTGAGCGGATAACAATGTCAC

mRNA: UUAACACUCGCCUAUUGUUACAGU

Amino acid sequnce: Leu Thr Leu Ala Tyr Cys Tyr Ser

vi) DNA sequence: AATTGTGAGCGGATAACAATGTCACACATG(AT instead of AG at 98 and 99 indicated with e)

mRNA sequence:   UUAACACUCGCCUAUUGUUACAGUGUGUAC

Amino acid sequence: Leu Thr Leu Ala Tyr Cys Tyr Ser Val Tyr

vii) DNA sequence: AATTGTGAGCGGATAACAATGTCACACAGGAAG(G/C base pair instead of A/T base pair, indicated with f)

mRNA sequence:    UUAACACUCGCCUAUUGUUACAGUGUGUCCUUAC

Amino acid sequence: Leu Thr Leu Ala Tyr Cys Tyr Ser Cys Pro Tyr.

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