mRNA and amino acid sequence
To find mRNA and amino acid sequence, Consider the DNA sequence below:
3’ – T A C C A A G G A G C A T T A G A T A C T – 5’
5’ – A T G G T T C C T C G T A A T C T A T G A – 3’
A) Write the mRNA sequence that would be produced if the TOP DNA strand were used as the template strand (label the 5’ and 3’ ends):
B) What is the amino acid sequence of the protein that would be made using the mRNA above in part a ?
C) Write the mRNA sequence that would be produced if the BOTTOM DNA strand were used as the template strand
D) What is the amino acid sequence of the protein that would be made using the mRNA above in part c ?
ANSWERS:
A) mRNA sequence:
Out of two DNA strands, one strand acts as the template strand that runs in a 3′ to 5′ direction and is used for synthesis of a complementary RNA transcript. Other strand acts as the coding strand runs in 5′ to 3′ direction and is same as RNA transcript but it has uracil (U) bases instead of thymine (T) bases.
Answer:
5′ – U C A U A G A U U A C G A G G A A C C A U – 3′
B) Amino acid sequence:
Answer:
Ser-STOP
After encoding Serine, UAG is a stop codon and does not code for anything. Stop codon terminates the translation.
C) mRNA sequence:
Transcription is the process of production of RNA molecule from DNA sequence of a gene. mRNA is coded from 5′->3′ as explained in option (a). It is same sequence as sense strand except in DNA Thymine is present and in RNA Uracil is present instead of thymine.
5′- A U G G U U C C U C G U A A U C U A U G A – 3′
D) Amino acid sequence :
Met-Val-Pro-Arg-Asn-Leu-STOP
Because UGA is a stop codon
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