Genetics Problem Set. See attached file for full problem description.
1A. The mRNA codon is what codes for the amino acid. The mRNA sequence is complementary to the DNA template (that is A-T, C-G) and identical to the non-template strand (except T is replaced by U). Let's look at Arg since it's the first and last amino acid in the sequence. The mRNA codons for Arg are CGU, CGC, CGG, CGA, AGA, and AGG. The strand that has any of these base sequences at both the beginning and end would be the non-template strand, making the other strand the template.
1B/C. The mRNA strand codes for amino acids in the 5' to 3' direction, so the DNA template strand would need to be in the 3' to 5' direction and the non-template strand in the same direction as the mRNA (5' to 3'). Write out the sequence of the mRNA strand by duplicating the non-template strand, but replace T with U.
1D. 5' end = 7 methyl-guanosine cap (just call it the 5' cap if that's what you were taught), 3' end = poly-A tail.
1E. Codons are represented by three bases, so count the bases and divide by three to get the number of codons.
2. The codons for leucine are UUA, UUG, CUU, CUC, CUA, and CUG. Codons refer to the base sequence on the mRNA so they are 5' to 3'. Anti-codons refer to the complementary base sequence on the tRNA, so they are 3' to 5'. It appears ...