PART 1 Introduction
4.6. METABOLIC REGULATION
oligosaccharide (the dolichol pyrophosphate-oligosaccharide) with 14 sugars is trans- ferred to the amino group of asparagine.
The 14-sugar residue is first “trimmed” by a set of specific glycosidases. In yeast, oligosaccharide processing often stops in the ER, leading to simple glycoforms(or high mannose or oligomannose forms). The initial trimming takes place in the ER, followed by transfer to the Golgi apparatus where final trimming occurs, followed by addition of vari- ous sugars or aminosugars. These units are added through the action of various glycosyl- transferases using nucleotide-sugar cosubstrates as sugar donors. In insect cells high levels of N-acetylglucosaminidase activity results typically in dead-end structures with a mannose cap. Complex glycoformshave sugar residues (N-acetyl glucosamine, galactose, and/or sialic acid) added to all branches of the oligosaccharide structure. Hybrid glyco- formshave at least one branch modified with one of these sugar residues and one or more with mannose as the terminal residue.
b-galactosidase (or lactase), which cleaves lactose to glucose and galactose. The lac y pro- tein is a permease, which acts to increase the rate of uptake of lactose into the cell. Lac- tose is modified in the cell to allolactose, which acts as the inducer. The conversion of lactose to allolactose is through a secondary activity of the enzyme b-galactosidase. Re- pression of transcription in uninduced cells is incomplete, and a low level (basal level) of proteins from the operon is made. Allolactose acts as indicated in Fig. 4.10, but induction by allolactose is not both necessary and sufficient for maximum transcription. Further reg- ulation is exerted through catabolite repression(also called the glucose effect).
When E. colisenses the presence of a carbon–energy source preferred to lactose, it will not use the lactose until the preferred substrate (e.g., glucose) is fully consumed. This control mechanism is exercised through a protein called CAP(cyclic-AMP-activating pro- tein). Cyclic AMP (cAMP) levels increase as the amount of energy available to the cell decreases. Thus, if glucose or a preferred substrate is depleted, the level of cAMP will in- crease. Under these conditions, cAMP will readily bind to CAP to form a complex that binds near the lac promoter. This complex greatly enhances RNA polymerase binding to the lac promoter. Enhancerregions exist in both procaryotes and eucaryotes.
Figure 4.9. Process of enzyme repression.
(a) Transcription of the operon occurs be- cause the repressor is unable to bind to the operator. (b) After a corepressor (small mol- ecule) binds to the repressor, the repressor now binds to the operator and blocks tran- scription. m-RNA and the proteins it codes for are not made. (With permission, from T. D. Brock, K. M. Brock, and D. M. Ward, Basic Microbiology with Applications, 3d ed., Pearson Education, Upper Saddle River, NJ, 1986, p. 143.)
Figure 4.10. Process of enzyme induction.
(a) A repressor protein binds to the operator region and blocks the action of RNA poly- merase. (b) Inducer molecule binds to the re- pressor and inactivates it. Transcription by RNA polymerase occurs and an m-RNA for that operon is formed. (With permission, from T. D. Brock, K. M. Brock, and D. M.
Ward, Basic Microbiology with Applications, 3d ed., Pearson Education, Upper Saddle River, NJ, 1986, p. 142.) In the lac operon, gene 1 is lac z, gene 2 is lac y, and gene 3 is lac a. The repressor is made on a separate gene called lac i.
120 How Cells Work Chap. 4
The reader should now see how the cellular control strategy emerges. If the cell has an energetically favorable carbon–energy source available, it will not expend significant energy to create a pathway for utilization of a less favorable carbon–energy source. If, however, energy levels are low, then it seeks an alternative carbon–energy source. If and only if lactose is present will it activate the pathway necessary to utilize it.
Catabolite repression is a global response that affects more than lactose utilization.
Furthermore, even for lactose, the glucose effect can work at levels other than genetic.
The presence of glucose inhibits the uptake of lactose, even when an active uptake system exists. This is called inducer exclusion.
The role of global regulatory systems is still emerging. One concept is that of a reg- ulon. Many noncontiguous gene products under the control of separate promoters can be coordinately expressed in a regulon. The best studied regulon is the heat shock regulon.
The cell has a specific response to a sudden increase in temperature,†which results in the elevated synthesis of specific proteins. Evidence now exists that this regulon works by employing the induction of an alternative sigma factor, which leads to high levels of tran- scription from promoters that do not readily recognize the normal E. colisigma factor.
Examples of other regulons involve nitrogen and phosphate starvation, as well as a switch from aerobic to anaerobic conditions.
Although many genes are regulated, others are not. Unregulated genes are termed constitutive, which means that their gene products are made at a relatively constant rate ir- respective of changes in growth conditions. Constitutive gene products are those that a cell expects to utilize under almost any condition; the enzymes involved in glycolysis are an example.
Example 4.1.
Diauxic growth is a term to describe the sequential use of two different carbon–energy sources. Industrially, diauxic growth is observed when fermenting a mixture of sugars, such as might result from the hydrolysis of biomass. The classic example of diauxic growth is growth of E. colion a glucose–lactose mixture. Observations on this system led to formula- tion of the operon hypothesis and the basis for a Nobel prize (for J. Monod and F. Jacob).
Consider the plot in Fig. 4.11, where the utilization of glucose and lactose and the growth of a culture are depicted for a batch culture (batch reactor). As we will discuss in more detail in Chapter 6, the amount of biomass in a culture, X, accumulates exponentially. Note that at 2 h after inoculation, cells are growing rapidly, glucose is being consumed, and lactose is not being utilized. At 7 h, cell mass accumulation is zero. All the glucose has been consumed. At 10 h the culture is growing and lactose is being consumed, but the rate of growth (cell mass accumulation) is less than at 2 h.
Explain what is happening with intracellular control to account for the observations at 2, 7, and 10 h. What rate of b-galactosidase formation would you expect to find in the culture at these times in comparison to the basal rate (which is <1%of the maximum rate)?
Solution
At 2 h the lac operon is fully induced, since lactose converted to allolactose com- bines with the lac i repressor protein, inactivating it. With the repressor protein deacti- vated, RNA polymerase is free to bind to the promoter, but does so inefficiently. Glucose
†Or other stresses that result in abnormal protein formation or membrane disruption.
levels are still high, which results in higher levels of ATP and low levels of AMP and cAMP. Consequently, little cAMP–CAP complex is formed, and the interaction of the lac promoter with RNA polymerase is weak in the absence of cAMP–CAP. The rate of β-galactosidase formation would be slightly increased from the basal level—perhaps 5% of the maximal rate.
At 7 h the glucose has been fully consumed. The cell cannot generate energy, and the level of ATP decreases and cAMP increases. The cAMP–CAP complex level is high, which increases the efficiency of binding RNA polymerase to the lac promoter. This increased bind- ing leads to increased transcription and translation. The rate of β-galactosidase formation is maximal and much higher than the basal rate or the 2 h rate. However, the cells have not yet accumulated sufficient intracellular concentrations of β-galactosidase and lac permease to allow efficient use of lactose and rapid growth.
At 10 h the intracellular content of proteins made from the lac operon is sufficiently high to allow maximal growth on lactose. However, this growth rate on lactose is slower than on glucose, since lactose utilization generates energy less efficiently. Consequently, the cAMP level remains higher than when the cell was growing on glucose. The level of produc- tion of β-galactosidase is thus higher than the basal or 2 h level.
Irrespective of whether an enzyme is made from a regulated or constitutive gene, its activity in the cell is regulated. Let us now consider control at the enzymatic level.
Figure 4.11. Diauxic growth curve for E. colion glucose and lactose.
122 How Cells Work Chap. 4
4.6.2. Metabolic Pathway Control
In Chapter 3, we learned how enzyme activity could be modulated by inhibitors or activa- tors. Here we discuss how the activities of a group of enzymes (a pathway) can be con- trolled. The cell will attempt to make the most efficient use of its resources; the fermentation specialist tries to disrupt the cell’s control strategy so as to cause the cell to overproduce the product of commercial interest. An understanding of how cells control their pathways is vital to the development of many bioprocesses.
First, consider the very simple case of a linear pathway making a product, P1. Most often the first reaction in the pathway is inhibited by accumulation of the product (feed- back inhibitionor end-product inhibition). The enzyme for the entry of substrate into the pathway would be an allosteric enzyme (as described in Chapter 3), where the binding of the end product in a secondary site distorts the enzyme so as to render the primary active site ineffective. Thus, if the cell has a sufficient supply of P1(perhaps through an addition to the growth medium), it will deactivate the pathway so that the substrates normally used to make P1can be utilized elsewhere.
This simple concept can be extended to more complicated pathways with many branch points (see Fig. 4.12). Assuming that P1and P2are both essential metabolites, the cell may use one of several strategies to ensure adequate levels of P1and P2with efficient utilization of substrates.
One strategy is the use of isofunctional enzymes (isozymes). Two separate enzymes are made to carry out the same conversion, while each is sensitive to inhibition by a differ- ent end product. Thus, if P1is added in excess in the growth medium, it inhibits one of the
Figure 4.12. Examples of feedback control of branched pathways. P1and P2are the de- sired end products. M1, M2, . . . , Mjare intermediates, and Ejis the enzyme involved in converting metabolite Mj-1to Mj. Possible paths of inhibition are shown by dashed lines.
isozymes (E2), while the other enzyme (E¢2) is fully active. Sufficient activity remains through isozyme E¢2to ensure adequate synthesis of P2.
An alternative approach is concerted feedback inhibition. Here a single enzyme with two allosteric binding sites (for P1and P2) controls entry into the pathway. A high level of either P1or P2is not sufficient by itself to inhibit enzyme E2, while a high level of both P1and P2will result in full inhibition.
A third possibility is sequential feedback inhibition,by which an intermediate at the branch point can accumulate and act as the inhibitor of metabolic flux into the pathway.
High levels of P1 and P2inhibit enzymes E4 and E5, respectively. If either E4 or E5 is blocked, M3will accumulate, but not as rapidly as when both E4or E5are blocked. Thus, intermediate flux levels are allowed if either P1or P2is high, but the pathway is inacti- vated if both P1and P2are high.
Other effects are possible in more complex pathways. A single allosteric enzyme may have effector sites for several end products of a pathway; each effector causes only partial inhibition. Full inhibition is a cumulative effect, and such control is called cumula- tive feedback inhibitionor cooperative feedback inhibition. In other cases, effectors from related pathways may also act as activators. Typically, this situation occurs when the prod- uct of one pathway was the substrate for another pathway. An example of control of a complex pathway (for aspartate) is described in Section 4.9.
The reader should pause to consider the differences between feedback inhibition and repression. Inhibition occurs at the enzyme level and is rapid; repression occurs at the genetic level and is slower and more difficult to reverse. In bacteria where growth rates are high, unwanted enzymes are diluted out by growth. Would such a strategy work for higher cells in differentiated structures? Clearly not, since growth rates would be nearly zero. In higher cells (animals and plants) the control of enzyme levels is done primarily through the control of protein degradation, rather than at the level of synthesis. Most of our discussion has centered on procaryotes; the extension of these concepts to higher or- ganisms must be done carefully.
Another caution is that the control strategy that one organism adopts for a particular pathway may differ greatly from that adopted by even a closely related organism with an identical pathway. Even if an industrial organism is closely related to a well-studied or- ganism, it is prudent to check whether the same regulatory strategy has been adopted by both organisms. Knowing the cellular regulatory strategy facilitates choosing optimal fer- menter operating strategy, as well as guiding strain improvement programs.
We have touched on some aspects of cellular metabolic regulation. A related form of regulation that we are just now beginning to appreciate has to do with the cell surface.
4.7. HOW THE CELL SENSES ITS EXTRACELLULAR ENVIRONMENT