11 Regulation of Gene Expression in Bacteria and Their Viruses WORKING WITH THE FIGURES 1. Compare the structure of IPTG shown in Figure 11-7 with the structure of galactose shown in Figure 11-5. Why is IPTG bound by the lac repressor but not broken down by -galactosidase? Answer: The sulfur atom in IPTG prevents hydrolysis by the beta-galactosidase enzyme. 2. Looking at Figure 11-9, why were partial diploids essential for establishing the trans-acting nature of the lac repressor? Could one distinguish cis-acting from transacting genes in haploids? Answer: Partial diploids were essential for distinguishing cis-acting from trans- acting mutants, because by definition one must introduce a second copy of the locus in trans to test this property of the mutants. 3. Why do promoter mutations cluster at positions –10 and –35 as shown in Figure 11-11? Answer: Promoter mutations cluster around the –10 and –35 positions because these are the DNA sequences recognized and bound by the sigma subunit of the polymerase. Alterations to these sequences will affect the ability of the RNA polymerase holoenzyme to recognize the promoter. 4. Looking at Figure 11-16, how large is the overlap between the operator and the lac transcription unit? Answer: The operator sequence overlaps the lac operon transcription unit by 24 base pairs. 314 Chapter Eleven 5. Examining Figure 11-21, what effect do you predict trpA mutations will have on tryptophan levels? Answer: Mutations that inactivate the trpA gene will block the synthesis of tryptophan in the cells, resulting in trp auxotrophs. 6. Examining Figure 11-21, what effect do you predict trpA mutations have on trp mRNA expression? Answer: Because the trpA gene is the last to be transcribed, mutations in the trpA would not be predicted to affect transcription of the operon. BASIC PROBLEMS 7. Explain why I– alleles in the lac system are normally recessive to I+ alleles and why I+ alleles are recessive to IS alleles. Answer: The I gene determines the synthesis of a repressor molecule, which blocks expression of the lac operon and which is inactivated by the inducer. The presence of the repressor I+ will be dominant to the absence of a repressor I–. Is mutants are unresponsive to an inducer. For this reason, the gene product cannot be stopped from interacting with the operator and blocking the lac operon. Therefore, Is is dominant to I+. 8. What do we mean when we say that OC mutations in the lac system are cis- acting? Answer: Oc mutants are changes in the DNA sequence of the operator that impair the binding of the lac repressor. Therefore, the lac operon associated with the Oc operator cannot be turned off. Because an operator controls only the genes on the same DNA strand, it is cis (on the same strand) and dominant (cannot be turned off). Unpacking the Problem 9. The genes shown in the following table are from the lac operon system of E. coli. The symbols a, b, and c represent the repressor (I) gene, the operator (O) region, and the structural gene (Z) for -galactosidase, although not necessarily in that order. Furthermore, the order in which the symbols are written in the genotypes is not necessarily the actual sequence in the lac operon. Activity (+) or inactivity (–) of Z gene Chapter Eleven 315 Genotype Inducer absent Inducer present a– b+ c+ + + a+ b+ c– + + a+ b– c– – – a+ b– c+/a– b+ c– + + a+ b+ c+/a– b– c– – + a+ b+ c–/a– b– c+ – + a– b+ c+/a+ b– c– + + a. State which symbol (a, b, or c) represents each of the lac genes I, O, and Z. b. In the table, a superscript minus sign on a gene symbol merely indicates a mutant, but you know that some mutant behaviors in this system are given special mutant designations. Use the conventional gene symbols for the lac operon to designate each genotype in the table. (Problem 9 is from J. Kuspira and G. W. Walker, Genetics: Questions and Problems. Copyright 1973 by McGraw-Hill.) Answer: a. You are told that a, b, and c represent lacI, lacO, and lacZ, but you do not know which is which. Both a– and c– have constitutive phenotypes (lines 1 and 2) and therefore must represent mutations in either the operator (lacO) or the repressor (lacI). b– (line 3) shows no ß-gal activity and by elimination must represent the lacZ gene. Mutations in the operator will be cis-dominant and will cause constitutive expression of the lacZ gene only if it’s on the same chromosome. Line 6 has c– on the same chromosome as b+ but the phenotype is still inducible (owing to c+ in trans). Line 7 has a– on the same chromosome as b+ and is constitutive even though the other chromosome is a+. Therefore a is lacO, c is lacI, and b is lacZ. b. Another way of labeling mutants of the operator is to denote that they lead to a constitutive phenotype; lacO– (or a–) can also be written as lacOC. There are also mutations of the repressor that fail to bind inducer (allolactose) as opposed to fail to bind DNA. These two classes have quite different phenotypes and are distinguished by lacIS (fails to bind allolactose and leads to a dominant uninducible phenotype in the presence of a wild-type operator) and lacI– (fails to bind DNA and is recessive). It is possible that line 3, line 4, and line 7 have lacIS mutations (because dominance cannot be ascertained in a cell that is also lacOC), but the other c– alleles must be lacI–. 10. The map of the lac operon is POZY 316 Chapter Eleven The promoter (P) region is the start site of transcription through the binding of the RNA polymerase molecule before actual mRNA production. Mutationally altered promoters (P–) apparently cannot bind the RNA polymerase molecule. Certain predictions can be made about the effect of P– mutations. Use your predictions and your knowledge of the lactose system to complete the following table. Insert a “+” where an enzyme is produced and a “–” where no enzyme is produced. The first one has been done as an example. -Galactosidase Permease Genotype No lactose Lactose No lactose Lactose I+ P+ O+ Z+ Y+/I+ P+ O+ Z+ Y+ – + – + a. I– P+ OC Z+ Y–/ I+ P+ O+ Z– Y+ b. I+ P– OC Z– Y+/I– P+ OC Z+ Y– c. IS P+ O+ Z+ Y–/I+ P+ O+ Z– Y+ d. IS P+ O+ Z+ Y+/I– P+ O+ Z+ Y+ e. I– P+ OC Z+ Y–/I– P+ O+ Z– Y+ f. I– P– O+ Z+ Y+/I– P+ OC Z+ Y– g. I+ P+ O+ Z– Y+/I– P+ O+ Z+ Y– Answer: ß-Galactosidase Permease Part No lactose Lactose No lactose Lactose a + + – + b + + – – c – – – – d – – – – e + + + + f + + – – g – + – + a. The OC mutation leads to the constitutive synthesis of ß-galactosidase because it is cis to a lacZ+ gene, but the permease is inducible because the lacY+ gene is cis to a wild-type operator. b. The lacP– mutation prevents transcription so only the genes cis to lacP+ will be transcribed. These genes are also cis to OC so the lacZ+ gene is transcribed constitutively. c. The lacIs is a trans-dominant mutation and prevents transcription from either operon. d. Same as part c. e. There is no functional repressor made (and one operator is mutant as well). Chapter Eleven 317 f. Same as part b. g. Both operators are wild type and the one functional copy of lacI will direct the synthesis of enough repressor to control both operons. 11. Explain the fundamental differences between negative control and positive control. Answer: A gene is turned off or inactivated by the “modulator” (usually called a repressor) in negative control, and the repressor must be removed for transcription to occur. A gene is turned on by the “modulator” (usually called an activator) in positive control, and the activator must be added or converted to an active form for transcription to occur. 12. Mutants that are lacY– retain the capacity to synthesize -galactosidase. However, even though the lacI gene is still intact, -galactosidase can no longer be induced by adding lactose to the medium. Explain. Answer: The lacY gene produces a permease that transports lactose into the cell. A cell containing a lacY– mutation cannot transport lactose into the cell, so ß-galactosidase will not be induced. (In wild-type cells, even when the repressor is present, there is still a small amount of transcription. This allows a small amount of baseline permease (and β-gal) expression. It is this low level of permease that allows trace amounts of lactose to enter the cell and initiate induction of the lac operon.) 13. What are the analogies between the mechanisms controlling the lac operon and those controlling phage genetic switches? Answer: Both the lac operon and phase genetic switches can be interpreted as a simple switch between two states. For , there is competition between Cro and cI proteins binding at an operator to control the choice between the lysogenic and lytic life cycles. For the lac operon, a repressor binds at the operator to prevent transcription of the structural genes in the absence of lactose. Both also are capable of interpreting their environments and directing an “appropriate” response. The lac operon responds to the presence or absence of lactose and glucose. For , if resources are abundant, or the bacterial cell sustains DNA damage, the lytic cycle prevails, but if resources are not abundant, will enter the lysogenic cycle.