Exam 3 - CH 16 Recommended Problems

2. Name six different levels at which gene expression might be controlled.

(1) Alteration or modification of the gene structure at the DNA level (2) Transcriptional regulation (3) Regulation at the level of mRNA processing (4) Regulation of mRNA stability (5) Regulation of translation (6) Regulation by posttranslational modification of the synthesized protein

18. Examine Figure 16.7. What would be the effect of a drug that altered the structure of allolactose so that it was unable to bind to the regulator protein?

Allolactose is produced when lactose is present; allolactose normally binds to the repressor protein and makes it inactive, allowing transcription to occur when lactose is present. If a drug altered the structure of allolactose, it would not bind to the repressor and the repressor would continue to bind to the operator, keeping transcription off. The result would be that transcription was repressed even in the presence of lactose; thus, no -galactosidase or permease would be produced.

8. What is antisense RNA? How does it control gene expression?

Antisense RNA molecules are small RNA molecules that are complementary to other DNA or RNA sequences and that form RNA-protein complexes. In bacterial cells, antisense RNA molecules can bind to a complementary region in the 5' UTR of a mRNA molecule, blocking the attachment of the ribosome to the mRNA and stopping translation or they pair with specific regions of the mRNA and cleave the mRNA stopping translation.

7. What is attenuation? What are the mechanisms by which the attenuator forms when tryptophan levels are high and the antiterminator forms when tryptophan levels are low?

Attenuation is the termination of transcription prior to the structural genes of an operon. It is a result of the formation of a termination hairpin structure or attenuator in the mRNA. Two types of secondary structures can be formed by the mRNA 5′ UTR of the trp operon. If the 5′ UTR forms two hairpin structures from the base pairing of region 1 with region 2 and the pairing of region 3 with region 4, then transcription of the structural genes will not occur. The hairpin structure formed by the pairing of region 3 with region 4 results in a terminator being formed that stops transcription. When region 2 pairs with region 3, the resulting hairpin acts as an antiterminator allowing for transcription to proceed. Region 1 of the 5' UTR also encodes a small protein and has two adjacent tryptophan codons (UGG). Tryptophan levels affect transcription due to the coupling of translation with transcription in bacterial cells. When tryptophan levels are high, the ribosome quickly moves through region 1 and into region 2, thus preventing region 2 from pairing with region 3. Therefore, region 3 is available to form the attenuator hairpin structure with region 4, stopping transcription. When tryptophan levels are low, the ribosome stalls or stutters at the adjacent tryptophan codons in region 1. Region 2 now becomes available to base pair with region 3, forming the antiterminator hairpin. Transcription can now proceed through the structural genes.

*14. A mutation prevents the catabolite activator protein (CAP) from binding to the promoter in the lac operon. What will the effect of this mutation be on the transcription of the operon?

Catabolite activator protein binds the CAP site of the lac operon and stimulates RNA polymerase to bind the lac promoter, thus resulting in increased levels of transcription from the lac operon. If a mutation prevents CAP from binding to the site, then RNA polymerase will bind the lac promoter poorly. This will result in significantly lower levels of transcription of the lac structural genes.

16. Under which of the following conditions would a lac operon produce the greatest amount of β-galactosidase? The least? Explain your reasoning.

Condition 1 will result in the production of the maximum amount of β- galactosidase. For maximum transcription, the presence of lactose and the absence of glucose are required. Lactose (or allolactose) binds to the lac repressor reducing the affinity of the lac repressor to the operator. This decreased affinity results in the promoter being accessible to RNA polymerase. The lack of glucose allows for increased synthesis of cAMP, which can complex with CAP. The formation of CAP-cAMP complexes improves the efficiency of RNA polymerase binding to the promoter, which results in higher levels of transcription from the lac operon. Condition 2 will result in the production of the least amount of β-galactosidase. With no lactose present, the lac repressor is active and binds to the operator, inhibiting transcription. The presence of glucose results in a decrease of cAMP levels. A CAP-cAMP complex does not form, and RNA polymerase will not be stimulated to transcribe the lac operon.

1. Why is gene regulation important for bacterial cells?

Gene regulation allows for biochemical and internal flexibility while maintaining energy efficiency by the bacterial cells.

6. What is catabolite repression? How does it allow a bacterial cell to use glucose in preference to other sugars?

In catabolite repression, the presence of glucose inhibits or represses the transcription of genes involved in the metabolism of other sugars. Because the gene expression necessary for utilizing other sugars is turned off, only enzymes involved in the metabolism of glucose will be synthesized. Operons that exhibit catabolite repression are under the positive control of catabolic activator protein (CAP). For CAP to be active, it must form a complex with cAMP. Glucose affects the level of cAMP. The levels of glucose and cAMP are inversely proportional—as glucose levels increase, the level of cAMP decreases. Thus, CAP is not activated.

*19. For E. coli strains with the lac genotypes shown below, use a plus sign (+) to indicate the synthesis of β-galactosidase and permease and a minus sign (−) to indicate no synthesis of the proteins.

In determining if expression of the β-galactosidase and the permease gene will occur, you should consider several factors. The presence of lacZ+ and lacY+ on the same DNA molecule as a functional promoter (lacP+) is required because the promoter is a cis-acting regulatory element. However, the lacI+ gene product or lac repressor is trans-acting and does not have to be located on the same DNA molecule as β-galactosidase and permease genes to inhibit expression. For the repressor to function, it does require that the cis-acting lac operator be on the same DNA molecule as the functional β-galactosidase and permease genes. Finally, the dominant lacIs gene product is also trans-acting and can inhibit transcription at any functional lac operator region.

15. Transformation is a process in which bacteria take up new DNA released by dead cells and integrate it into their own genomes (see p. 265 in Chapter 9). In Streptococcus pneumoniae (which causes many cases of pneumonia, inner-ear infections, and meningitis), the ability to carry out transformation requires from 105 to 124 genes, collectively termed the com regulon. The com regulon is activated in response to a protein called competence-stimulating peptide (CSP), which is produced by the bacteria and exported into the surrounding medium. When enough CSP accumulates, it attaches to a receptor on the bacterial cell membrane, which then activates a regulator protein that stimulates the transcription of genes within the com regulon and sets in motion a series of reactions that ultimately result in transformation. Does the com regulon in Streptococcus pneumoniae exhibit positive or negative control? Explain your answer.

It is regulated through positive control. Increased levels of CSP induce the receptor to stimulate transcriptional activator functions, which allow gene expression to occur similarly to what occurs in positive inducible regulation.

28. Some mutations in the trp 5′ UTR increase termination by the attenuator. Where might these mutations occur, and how might they affect the attenuator?

Mutations that disrupt the formation of the antiterminator will increase termination by the attenuator. Such disruptions could be caused by a deletion in region 2 that prevents region 2 from pairing with region 3. Mutations in region 1 could also affect the antiterminator if the mutations prevented the ribosome from stalling at the adjacent tryptophan codons within region 1. For example, any mutation that blocks translation initiation or stops translation early within region 1 would not allow the ribosome to migrate on the trp operon mRNA. Another type of mutation affecting antiterminator formation in region 1 is one that eliminates or replaces the two adjacent tryptophan codons in the small protein. Elimination of these codons would prevent the ribosome from stalling in region 1, thus increasing the rate of the terminator formation.

4. What is the difference between positive and negative control? What is the difference between inducible and repressible operons?

Positive transcriptional control requires an activator protein to stimulate transcription at the operon. In negative control, a repressor protein inhibits or turns off transcription at the operon. An inducible operon normally is not transcribed. It requires an inducer molecule to stimulate transcription either by inactivating a repressor protein in a negative inducible operon or by stimulating the activator protein in a positive inducible operon. Transcription normally occurs in a repressible operon. In a repressible operon, transcription is turned off either by the repressor becoming active in a negative repressible operon or by the activator becoming inactive in a positive repressible operon.

9. What are riboswitches?

Riboswitches are regulatory sequences in RNA molecules. Most can fold into compact secondary structures consisting of a base stem and several branching hairpins. At riboswitches, regulatory molecules bind and influence gene expression by affecting the formation of secondary structures within the mRNA molecule.

5. Briefly describe the lac operon and how it controls the metabolism of lactose.

The lac operon consists of three structural genes involved in lactose metabolism, the lacZ gene, the lacY gene, and the lacA gene. Each of these three genes has a different role in the metabolism of lactose. The lacZ gene codes for the enzyme β- galactosidase, which breaks the disaccharide lactose into galactose and glucose, and converts lactose into allolactose. The lacY gene, located downstream of the lacZ gene, codes for lactose permease. Permease is necessary for the passage of lactose through the E. coli cell membrane. The lacA gene, located downstream of lacY, encodes the enzyme thiogalactoside transacetylase whose function in lactose metabolism has not yet been determined. All of these genes share a common overlapping promoter and operator region. Upstream from the lactose operon is the lacI gene that encodes the lac operon repressor. The repressor binds at the operator region and inhibits transcription of the lac operon by preventing RNA polymerase from successfully initiating transcription. When lactose is present in the cell, the enzyme β-galactosidase converts some of it into allolactose. Allolactose binds to the lac repressor, altering its shape and reducing the repressor's affinity for the operator. Since this allolactose-bound repressor does not occupy the operator, RNA polymerase can initiate transcription of the lac structural genes from the lac promoter.

22. Which strand of DNA (upper or lower) in Figure 16.8 is the template strand? Explain your reasoning.

The lacI gene encodes the lac repressor protein, which can diffuse within the cell and attach to any operator. It can therefore affect the expression of genes on the same or different molecules of DNA. The lacO gene encodes the operator. The binding of the lac repressor to the operator affects the binding of RNA polymerase to the DNA, and therefore affects only the expression of genes on the same molecule of DNA.

13. The blob operon produces enzymes that convert compound A into compound B. The operon is controlled by regulator gene S. Normally, the enzymes are synthesized only in the absence of compound B. If gene S is mutated, the enzymes are synthesized in the presence and in the absence of compound B. Does gene S produce a regulator protein that exhibits positive or negative control? Is this operon inducible or repressible?

The operon is controlled by a regulatory gene S. Normally, the enzymes are synthesized only in the absence of compound B. If gene S is mutated, the enzymes are synthesized in the presence and in the absence of compound B. Does gene S produce a regulatory protein that exhibits positive or negative control? Is this operon inducible or repressible?

26. At which level of gene regulation shown in Figure 16.1 does attenuation occur?

Transcription

17. A mutant strain of E. coli produces β-galactosidase in both the presence and the absence of lactose. Where in the operon might the mutation in this strain be located?

Within the operon, the operator region is the most probable location of the mutation. If the mutation prevents the lac repressor protein from binding to the operator, then transcription of the lac structural genes will not be inhibited. Expression will be constitutive. Outside of the operon, a mutation in the lacI gene that inactivates the repressor or keeps it from binding to the operator could also lead to constitutive expression of the structural genes.

*27. Listed in parts a through g are some mutations that were found in the 5′ UTR of the trp operon of E. coli. What will the most likely effect of each of these mutations be on the transcription of the trp structural genes? a. A mutation that prevents the binding of the ribosome to the 5′ end of the mRNA 5′ UTR b. A mutation that changes the Trp codons in region 1 of the mRNA 5′ UTR into codons for alanine c. A mutation that creates a stop codon early in region 1 of the mRNA 5′ UTR d. Deletions in region 2 of the mRNA 5′ UTR e. Deletions in region 3 of the mRNA 5′ UTR f. Deletions in region 4 of the mRNA 5′ UTR g. Deletion of the string of adenine nucleotides that follows region 4 in the 5′ UTR

a.) If the ribosome does not bind to the 5' end of the mRNA, then region 1 of the mRNA 5' UTR will be free to pair with region 2, thus preventing region 2 from pairing with region 3 of mRNA 5' UTR. Region 3 will be free to pair with region 4, forming the attenuator or termination hairpin. Transcription of the trp structural genes will be terminated. Essentially, no gene expression will occur. b.) In the wild-type trp operon, low levels of tryptophan result in the ribosome pausing in region 1 of the mRNA 5' UTR. The pause permits regions 2 and 3 of the mRNA 5' UTR to form the antiterminator hairpin, allowing transcription of the structural genes to continue. If alanine codons have replaced tryptophan codons, then under conditions of low alanine, the stalling of the ribosome will not occur. The attenuator will form, stopping transcription. The ribosome will stall when alanine is low, so transcription of the structural genes will occur only when alanine is low. c.) If region 1 of the mRNA 5' UTR is free to pair with region 2, then regions 3 and 4 of the mRNA 5' UTR can form the attenuator. An early stop codon will result in the ribosome "falling off" region 1, allowing it to form a hairpin structure with region 2. Transcription will not occur because regions 3 and 4 are now free to form the attenuator. d.) If region 2 of the mRNA 5' UTR is deleted, the antiterminator cannot be formed. The attenuator will form and transcription will not occur. e.) The trp operon mRNA 5' UTR will be unable to form the attenuator if region 3 contains a deletion. Attenuation or termination of transcription will not occur, resulting in continued transcription of the trp structural genes. f.) Deletions in region 4 will prevent formation of the attenuator by the 5' UTR mRNA. Transcription will proceed. g.) For the attenuator hairpin to function as a terminator, the presence of a string of uracil nucleotides following region 4 in the mRNA 5' UTR is required. The deletion of the string of adenine nucleotides in the DNA will result in no string of uracil nucleotides following region 4 of the mRNA 5' UTR. No termination will occur, and transcription will proceed.

*11. For each of the following types of transcriptional control, indicate whether the protein produced by the regulator gene will be synthesized initially as an active repressor or as an inactive repressor. a. Negative control in a repressible operon b. Negative control in an inducible operon

a.) Inactive repressor b.) Active repressor

12. A mutation at the operator prevents the regulator protein from binding. What effect will this mutation have in the following types of operons? a. Regulator protein is a repressor of a repressible operon. b. Regulator protein is a repressor of an inducible operon.

a.) The regulator protein-corepressor complex would normally bind to the operator and inhibit transcription. If a mutation prevented the repressor protein from binding at the operator, then the operon would never be turned off and transcription would occur all the time. b.) In an inducible operon, a mutation at the operator site that blocks binding of the repressor would result in constitutive expression and transcription would occur all the time.