Biology Question: Please use the image attached below to answer the questions un
ID: 149644 • Letter: B
Question
Biology Question: Please use the image attached below to answer the questions under it (A through F). Thank you.
A. What would be the nucleotide sequence of the mRNA molecule produced by transcription of this gene? Please indicate directionality.
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B. Using the three-letter abbreviations for the amino acids, what would be the amino acid sequence of the polypeptide after translation? Please indicate directionality.
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C. Imagine the sequence listed above underwent a mutation in which the AT pair indicated by bold blue was deleted. What effects would this mutation have on expression of the gene? Please briefly explain.
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D. Now imagine a second mutation in which the GC pair indicated by bold red was changed to AT. What effects would this mutation have on expression of the gene? Please briefly explain.
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E. Now imagine a third mutation in which the CG pair indicated by bold green was changed to AT. What effects would this mutation have on expression of the gene? Please briefly explain.
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F. Now consider the three mutations described in C-E. Based on what you know about the relative frequencies of different kinds of mutations, rank the three mutations in decreasing order of likelihood (probability of occurrence), and please briefly explain.
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Thank you!
2. The sequence below shows the first 35 nucleotides of the first exon of a eukaryotic protein-coding gene. The GC pair in bold black indicates the start of transcription, and the ellipsis (...) indicates the gene continues indefinitely: 5' GGTTCACAATGGGATCGGCAGCTCGTAGTGACTAT. .. 3' 3 CCAAGTGTTACCCTAGCCGTCGAGCATCACTGATA 5'Explanation / Answer
5’ GGTTCACAATGGGATCGGCAGCTCGTAGTGACTAT 3’
3’ CCAAGTGTTACCCTAGCCGTCGAGCATCACTGATA 5’
TRANSCRIPTION
5’ GGU UCA CAA UGG GAU CGG CAG CUC GUA GUG ACU AU 3’
POLYPEPTIDE CHAIN
N - Gly – Ser – Gln – Trp – Asp – Arg – Gln – Leu – Val – Val – Ile – C
If AT pair was deleted
5’ GGTTCCAATGGGATCGGCAGCTCGTAGTGACTAT 3’
3’ CCAAGGTTACCCTAGCCGTCGAGCATCACTGATA 5’
5’ GGU UCC AAU GGG AUC GGC AGC UCG UAG UGA CUA U 3’
We have,
N- Gly – Ser – Asn – Gly – Ile – Gly – Ser – Ser- Stop codon – C’
An incomplete protein will form.
When GC changes to AT
5’ GGTTCCAATAGGATCGGCAGCTCGTAGTGACTAT 3’
3’ CCAAGGTTATCCTAGCCGTCGAGCATCACTGATA 5’
5’ GGU UCC AAU AGG AUC GGC AGC UCG UAG UGA CUA U 3’
N – Gly – Ser – Asn – Arg – Ile – Gly – Ser – Ser – Stop codon – C
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