Use the following mRNA codon key as needed to answer the next two questions: GCC
ID: 80499 • Letter: U
Question
Use the following mRNA codon key as needed to answer the next two questions:
GCC Alanine
AAU Asparagine
CGU Proline
GGA Glycine
UGG Tryptophan
UGA "stop" (no amino acid)
GAA Glutamic Acid
GAG Glutamic Acid
AGG Arginine
CCC Proline
CAU Histidine
The following DNA sequence (coding strand) occurs near the middle of the coding region of a gene.
DNA
50 55 60 65
5'–A A T G A A T G G G A G C C T G A A G G A G–3'
The corresponding mRNA sequence is shown below. Note that the coding strand of DNA has the same sequence as the mRNA, except that there are U's in the mRNA where there are T's in the DNA. The first triplet of nucleotides AAU (underlined) is in frame for coding, and encodes Asparagine as the codon table above indicates.
mRNA
50 55 60 65
5'–A A U G A A U G G G A G C C U G A A G G A G–3'
1) Which of the following DNA mutations s almost cetain to result in a shorter than normal mRNA?
2) For the same DNA sequence, which of the following DNA mutations is almost certain to result in a shorter than normal protein?
Explanation / Answer
1) For MRNA to be shorter than its parent DNA missense mutations must take place. Missense mutations are point mutations that leads to amino acid change in the polypeptide sequence. If deletion occurs in the DNA then the following mrna product is to be shorter than the normal MRNA.
2) In the given DNA sequence the fourth triplet is GAG. Guanine to thymine transversion occurs via xanthine formation from deamination of Guanine. Xanthine is tautomer analogue of thymine. So sequence becomes TAG in the coding sequence and the transcript becomes UAG and that is a stop codon. So it forms a shorter protein when it is translated than the normal protein