Genetic code exercises:
Given the following DNA sequences, derive the (a) complementary mRNA, and the (b)
resulting protein:
1. 5’ TTTCATGCCCCGATAUGTACCC 3’
a. to derive the complementary RNA, we simply take note of the pairing
rules (A with T/U, and C with G). Also, the DNA and RNA strands must
be antiparallel (i.e. 5’ and 3’ ends are opposite each other)
5’ TTTCATGCCCCGATAUGCATCC 3’ (this is the given DNA)
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3’ AAAGUACGGGGCUAUACGUAGG 5’ (mRNA)
Take note that the mRNA is antiparallel. Also,
there are uracils (U), instead of thymines (T)
b. to get the amino acid sequence, you read the mRNA from the 5’ end. If
you’ll find it easier, you can first rewrite the mRNA so that it reads from
5’ to 3’
3’ AAAGUACGGGGCUAUACGUAGG 5’ (mRNA)
5’ GGAUGCAUAUCGGGGCAUGAAA 3’ (the same mRNA
sequence, rewritten so that it reads from 5’-3’.
Now it is easier to see the start (AUG) and the
end (UGA) of the sequence).
Taking our bases three at a time, we can now refer to our chart, to get the
corresponding amino acids
AUG – Met
CAU – His
AUC – Ile
GGG – Gly
GCA - Ala
UGA – Stop
Thus, the protein is composed of:
Met – His – Ile- Gly – Ala
For practice (all of the following are DNA sequences):
a. 5’ CTAAAACCCTTTGGGCAT 3’
b. 5’ TTTATGTTGTCATAA 3’
c. 5’ ATCACGCTATCATAAG 3’
Answers:
a.
mRNA: 5’ AUG CCC AAA GGG UUU UAG 3’
protein: Met-Pro-Lys-Gly-Phe
b.
mRNA: 5’ UU AUG ACA ACA UAA A 3’
protein:
Met-Thr-Thr
c.
mRNA: 5’ CUU AUG AUA GCG UGA U 3’
protein:
Met-Ile-Ala