Problem 12-0 for  19th edition (2009) of Problem Book


12-0. Normal bacteria can grow on medium containing lipid L as the only carbon source. The enzymes needed to metabolize lipid L (enzymes 1-5) are coded for by the greasy operon, which is inducible. If there is no lipid L in the medium, normal bacteria do not make enzymes 1-5. Assume there is no feedback inhibition (or activation) of the enzymes of the greasy operon.


A. Which of the following components would you find in the greasy operon? (repressor) (co-repressor) (operator) (promotor)?  Choose all correct answers, and explain the role of each component.


B. What role(s) does lipid L play in the greasy operon? (repressor) (co-repressor) (inducer) (substrate of one of enzymes 1-5)? Choose all correct answers and explain what happens to each one you have chosen when lipid L is added to the medium. 


C. Suppose that lipid L is added to the medium containing the bacteria.

            C-1. Which of the items listed below will be the first to to slow down or decrease?  

            C-2. Which of the items listed below will be the first to speed up or increase?

            C-3. Are there any items on the list that will not change at all?

List of items to consider:

(amount of enzyme 1 per cell) (rate of synthesis of enzyme 1) (enzymatic activity (Vmax) of enzyme 1 per cell) (transcription of repressor gene of greasy operon) (binding of RNA polymerase to promotor of greasy operon) (binding of repressor to operator of greasy operon).

            Explain your answers. List the items that will change in response to the addition of lipid L in the order in which they will occur. Indicate whether each item will increase or decrease. You do NOT need to include any choices that will not change, or any steps not mentioned here.


D. Suppose normal bacteria are growing on lipid L, and the lipid in the medium is used up, but the bacteria continue to grow (on other carbon sources in the medium).  After the lipid L is used up, you wait a while and then measure the amount of enzyme 3 per cell. It is about 0.5 X the maximum level.

            D-1. You would expect the amount of enzyme 4 per cell to be (greater than 0.5 X its maximum level)
(less than 0.5 X its maximum level) (about 0.5 X its maximum level) (beats me).
D-2. You probably waited (significantly less than 1/2 a generation time) (about 1/2 a generation time)
(about 1 generation time) (significantly greater than 1 generation time) (beats me). Explain both answers.



12-0. A. All but co-repressor. The promotor is the DNA sequence where RNA polymerase binds and transcription starts. The operator is the DNA sequence that binds the repressor protein. When repressor protein binds to the operator, it blocks transcription by blocking binding of RNA polymerase to the promotor. See text or operon handout. A co-repressor is needed for repression but not for induction.

            B. Lipid L is an inducer, and is also a substrate of one of the enzymes made from the operon -- the enzyme that catalyzes the first step in the pathway of degradation. When lipid L is added to the medium, it binds to the repressor protein. The complex does not bind to (or falls off) the operator, so the promotor is free to bind RNA polymerase. Transcription of the operon occurs, leading to synthesis of enzymes 1-5. One of these enzymes catalyzes the first step in the breakdown (degradation) of lipid L.

C-1 & C-2. Events are like this: (1) Binding of repressor protein to operator goes down. This is the first and only item mentioned that decreases. (2) Binding of RNA polymerase to promotor goes up. This is the first item listed that increases. (3) Rate of synthesis of enzymes of operon increases, since new mRNA is made and translated. 
(4) Amount of enzyme (number of molecules/cell) and enzyme activity (Vmax) per cell increase in parallel. (Amounts of enzyme and activity per culture also increase in parallel.)

C-3. The rate of transcription of the repressor gene doesn't change -- production of repressor is constitutive.

Note: In this problem, enzyme activity refers to the catalytic capability of the enzyme (if given enough substrate), not to how much catalyzing it is doing at this very minute (which depends on the supply of substrate). The terms "activity" and "amount" are both needed because either the capability of each enzyme molecule or the number of enzyme molecules can change independently. In this problem, enzyme activity and enzyme amount change in parallel. But this is not always the case. For example, if inhibitors or activators are added, the enzyme activity can change while the amount of enzyme stays constant. Also, if enzyme activity changes, it may be because of a change in inhibition/activation or a change in enzyme amount.

D. The amount of enzyme 4 will be about 0.5X its maximum after about one generation time. Both answers assume that the enzymes are all stable, which is generally true in bacteria. The mRNA, but not the protein, is degraded. Details:
D-1. All the enzymes are from the same operon, so they are made from the same polycistronic mRNA. The rates of synthesis of all 5 enzymes should be proportional -- the more mRNA, the more you make of all 5. If every ribosome makes it all the way to the end of the mRNA, then all the enzymes will always be made in equal proportions. Since some ribosomes fall off, and some can start in the middle, the proportions of the enzymes are not always 1:1. However, the rates of synthesis (and final enzyme levels) of the 5 enzymes should always be proportional since all are translated from the same mRNA. Once you stop making new mRNA, and stop making enzyme 3, you stop making all the others too. Whatever the relative amounts, they should all decline together (by dilution -- see part D-2).

D-2. When lipid L is used up, no more enzyme 3 is made, and the amount that is there is divided up among the bacteria as they grow. It should take about 1 generation to cut the enzyme per cell in half. (Same amount of enzyme/culture, but 2X as many bacteria.)