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Biology/Genetics Questions. See attached file for full problem description.

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1. Obviously, transcription must occur first in order to make a mRNA. Then, the mRNA leaves the nucleus and goes to the cytoplasm. A small ribosomal subunit binds to the mRNA followed by the binding of the met-tRNA to the mRNA. Then, the large subunit joins in, and only then can translation be intiated.

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2a. The start codon is 5'-AUG. It must be in that orientation. You can see it about 5 bases in from the left side.

2b. Stop codons are 5'-UGA, UAA, and UAG. You can see a UAA about 5 bases in from the right side.

2cd. The coding portion is:

AUG, GGC, AAU, AAA, CCG, GGC, CAG

Looking up these codons, we end up with:

N-Met-Gly-Asn-Lys-Pro-Gly-Gln-C

A protein is always translated from the N-terminus to the C-terminus.

2e. In order to work out the template DNA strand, you have to base pair the mRNA to the original DNA just making sure that you don't have any uracil in it, but rather thymine.

Here is the mRNA:

5' GGCGAUGGGCAAUAAACCGGGCCAGUAAGC

Therefore, here's the DNA:

3'-CCGCTACCCGTTATTT GGCCCGGTCATTCG

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3A. If the operator is mutated so that the repressor can no longer bind, then the genes of the operon will always be "on." Remember, the repressor binds to the operator and turns the operon "off." Therefore, if the operator can't bind the repressor, then the genes of the operon can never be "off." They'd be ...

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