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Viefleur [7K]
4 years ago
5

Given the mRNA codons AGC UUC GAU, what would be the resulting amino acid sequence after translation?

Biology
2 answers:
saul85 [17]4 years ago
7 0

Answer:

Option B) Serine-Phenylalanine-Aspartic acid

Explanation:

mRNA codons AGC code for the amino acid, serine while UUC code for phenylanine and GAU code for aspartic acid.

Thus, the resulting amino acid sequence after translation will be a tripeptide containing Serine-Phenylalanine-Aspartic acid

jeka57 [31]4 years ago
3 0

Answer:

b. Serine-Phenylalanine-Aspartic acid

Explanation:

Gene expression, which is the process of using the information present in a gene to synthesize protein, involves two major stages: transcription and translation. Transcription involves the synthesis of a mRNA molecule using a DNA template. It is the process of copying the information in a DNA into a complementary mRNA molecule.

Once the mRNA has been synthesized, it proceeds for translation in the ribosome (organnelle for protein synthesis). However, in the ribosome, the nuceleotide sequence of the mRNA is read three at a time. This group of three nucleotide sequence is called CODON. The set of all codons make up the GENETIC CODE. Each coding specifies a particular amino acid i.e. reading a codon means addition of a particular amino acid to the polypeptide chain.

This question contains a mRNA with nuceleotide sequence: AGC UUC GAU, which consists of three codons. Hence, three amino acids will be produced.

In the genetic code provided as an attachment to this question, we can obtain which CODON specifies which AMINO ACID.

AGC codes for Serine

UUC codes for Phenylalanine

GAU codes for Aspartic acid

Hence, mRNA sequence, AGC UUC GAU will produce an amino acid sequence Serine-Phenylalanine-Aspartic acid.

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