2. Below is the coding strand of a gene A: 5 -GACACCATGACATGGTACGACTTTACGAACCCGC
ID: 197265 • Letter: 2
Question
2. Below is the coding strand of a gene A: 5 -GACACCATGACATGGTACGACTTTACGAACCCGCAGGAATAGCTAACAACCGACA-3" Write the sequence of the complementary strand below the coding strand, indicating the 5' and 3 ends. a. b. In eukaryotes, GT and AG mark the beginning and end of introns, respectively. Assuming that the entire sequence is initially transcribed, write the sequence of the mature mRNA that would result from this gene. Write the amino acid sequence that would be translated from this gene in a eukaryote. You can assume that translation starts at the first AUG. c. d. If you cloned this gene into a prokaryote, write the mRNA that would be produced as a result e. If you cloned this gene into a prokaryote, write the amino acid sequence that would be produced as a result. You can assume that translation starts at the first AUG. (1pt)
Explanation / Answer
a. According to base pairing, A pairs with T and G pairs with C. So the complementary strand will be as below.
5’–GACACCATGACATGGTACGACTTTACGAACCCGCAGGAATAGCTAACAACCGACA–3’ => Coding strand
3’–CTGTGGTACTGTACCATGCTGAAATGCTTGGGCGTCCTTATCGATTGTTGGCTGT–5’ =>Complementary strand
b. mRNA will be transcribed will be complement to the complementary strand/template strand. In RNA U is present instead of T. So the Pre-mRNA after transcription will be as follows.
5’ – GACACCAUGACAUGGUACGACUUUACGAACCCGCAGGAAUAGCUAACAACCGACA –3’
In eukaryotes, splicing takes place to remove introns. The beginning and the ending of the introns are indicated by underlining and the entire intron is indicated by bold bases in the above Pre-mRNA strand. So the mature mRNA resulting after splicing is as below.
5’ – GACACCAUGACAUGGAAUAGCUAACAACCGACA –3’
c. The amino acid sequence resulting from the above mRNA (Start codon indicated by bold case, Stop sodon indicated by italics) is as follows.
5’ – GACACC AUG ACA UGG AAU AGC UAA CAACCGACA –3
Resulting amino acid sequence in eukaryotes : Met-Thr-Trp-Asn-Ser
d. In prokaryots, splicing does not occur. So the resulting mRNA will be,
5’ – GACACCAUGACAUGGUACGACUUUACGAACCCGCAGGAAUAGCUAACAACCGACA –3’
e. The amino acid sequence resulting from the above mRNA (Start codon indicated by bold case, Stop sodon indicated by italics) is as follows
5’ – GACACCAUG ACA UGG UAC GAC UUU ACG AAC CCG CAG GAA UAGCUAACAACCGACA –3’
Resulting amino acid sequence in prokaryotes: Met-Thr-Trp-Tyr-Asp-Phe-Thr-Asn-Pro-Gln-Glu
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