C2005/F2401 '06 -- Recitation Problems #8 -- Hints

1. A. Which end of the chain shown is the 5' end?

B. Is the carboxyl end the first end made? The last end made?

C. Which end of the RNA shown corresponds to the 'first 10 bases of the template strand?'

D. Where does the start codon fit in the ribosome when protein synthesis starts? Where does the stop codon fit when protein synthesis ends?

2. A. Which strand serves as template, the sense strand or its complement?

B. What is shown in the code table, and what's in the mRNA -- the codons or the anticodons?

C-1. What is meant by genotype, phenotype, and wild type?

C-2. Consider the genetic code. Consider both degeneracy -- will the new codon specify the same amino acid? and wobble -- even if the new codon specifies the same amino acid, can the same tRNA be used? If you aren't sure, look up the wobble rules.

C-3. How is this case different than the previous one?

C-4. Consider the properties of the two amino acids found in wild type and mutant 2. What effect would you expect if you substitute one for the other? (If you don't remember the properties of the various amino acids, look them up in your text.)

D. Consider the properties of the amino acids involved. Are the two amino acids more or less different than in the previous case? (See C-4 above.)

E-1. How does each type of mutation change the protein?

E-2. Why is mutation #4 likely to produce a totally nonfunctional protein?

E-3. Draw a picture of the situation -- include the ribosome (A & P sites), mRNA (with mutant codon in A site) and growing chain (& any relevant tRNA's). Now what will happen next?

C2005/F2401 '06 -- Answers to Recitation Questions #9

A-1. Add lactose. You need to induce the lac operon by adding lactose and de-repress the trp operon by removing tryptophan .
A-2. The lac operon. The strength of the promotor determines the rate of synthesis of mRNA. A stronger promotor binds RNA polymerase more strongly so initiation of transcription happens more often, more mRNA is synthesized and the steady state level of mRNA is higher. (In prokaryotes, the mRNA is degraded rapidly, so the rate of synthesis determines the steady state level.) The higher the cellular level of mRNA, the higher the level of protein synthesis and the higher the level of the corresponding protein. Therefore the strength of the promotor (how well it binds RNA polymerase) determines the max. level of the corresponding protein. We know from the information given that cells can make more β-galactosidase/cell than trp synthetase. This indicates that the lac promotor is a "stronger" promotor than the trp promotor -- the lac promotor binds RNA polymerase more strongly so initiation of transcription happens more often and the steady state level of mRNA (and protein production) produced from the lac promotor is higher.

B. Remove lactose. The gene for enzyme X is under the control of the regulatory system used by the lac operon. Therefore the production of mRNA for enzyme X will be inducible, just like the lac operon. Production of mRNA will be turned on by the addition of the inducer, lactose, and conversely, production of mRNA will stop when the lactose is removed. When lactose is present, it binds to the repressor protein, forming an inducer-repressor complex. The complex is inactive -- it does not bind to the operator, and the gene for enzyme X is transcribed. When lactose is removed, the repressor protein (no longer part of the complex) becomes active, binds to the operator, and blocks further transcription of the gene for enzyme X.

C-1. Only lactose. The operator determines which repressor protein will bind, and therefore which co-repressor or inducer must be present to control activity of the repressor protein and turn the operon on or off. In this case, the operator is from the lactose operon, so lactose must be present to inactivate the repressor and allow transcription of the operon, as explained in answer to A. Presence or absence of tryptophan would make no difference.

C-2. Higher with plasmid 1. The promotor determines how effectively RNA polymerase will bind (when the operator is unblocked) and therefore how much mRNA (and protein) will be made per unit time when the operon is fully induced or derepressed.) So plasmid 1, with the lac promotor, should allow a higher level of mRNA and enzyme X per cell than plasmid 2, with the trp promotor.

D. This is a case of complementation; if complementation produces active enzyme X, then the bacteria will grow in the presence of sillimycin. (You need the antibiotic to help you screen large numbers of colonies for ability to make enzyme X. It's very tedious to test individual colonies for presence/absence of the enzyme, but it's easy to see if they grow or not.)

D-1. Neither of these. Adding lac wouldn't make any difference here; adding trp would prevent production of enzyme X. You need both plasmids to be transcribed so that one subunit of enzyme X can be made from one plasmid and the other subunit of enzyme X can be made from the other plasmid. Therefore you want both operons/operators to be "on" -- neither operon should be repressed. The lac operator is constitutive (always on), so no lactose is needed to induce the operon -- transcription is already turned on and cannot be turned off. The trp operon is a repressible operon that would be turned off if you added trp. So you don't want to add trp -- you have to be sure to leave it out, so that the operon will be transcribed.

D-2. In different genes. This is a case of complementation -- one plasmid must have a defective gene for one subunit and the other plasmid must have a defective gene for the other subunit. (Each plasmid has a functional gene for the "other" subunit.) This explains why all bacteria that get both plasmids can make functional enzyme X -- they make one good subunit from each plasmid. Since they all make functional enzyme X, they are all resistant to sillimycin, and all produce colonies. If both plasmids carried mutations affecting the same subunit, then only a few bacteria which got both plasmids would make functional enzyme X. These would be the rare bacteria in which recombination had occurred to produce a recombinant, functional gene. If both plasmids carried identical mutations, there would be virtually no colonies, as there would be no way (except by additional mutations) for bacteria with the two plasmids to make active enzyme X.

D-3. No. If the bacterium is growing on sillimycin, it must make active enzyme X. However, two different molecules of mRNA are needed to make good enzyme X. Plasmid 3 is being transcribed to make mRNA for one good subunit, and plasmid 4 is being transcribed to make mRNA for the other. If you translate only one molecule of mRNA, you will make only one good subunit, and no active enzyme X.
If the bacteria producing enzyme X were recombinants, then the answer would be "yes." In that case, the bacteria should contain one mRNA, transcribed from the recombinant DNA. That mRNA would contain good information for each subunit. Translation of that recombinant mRNA would produce both subunits and therefore working enzyme X.

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