C2. In general, why is it important to regulate genes?
Discuss examples of situations in which it would be advantageous for a bacterial cell to regulate genes.
It is more efficient for a bacteria to express genes when they are need.
C4. Transcriptional regulation often involves a regulatory protein that binds to a segment
of DNA and a small effector molecule that binds to the regulatory protein. Do the following terms apply to a regulatory protein, a segment of DNA, or a small effector molecule?
A. Regulatory protein
B. Effector molecule
C. DNA segment
D. Effector molecule
E. Regulatory protein
F. DNA segment
G. Effector molecule
C6. Some mutations have a cis-effect, whereas others have a transeffect.
Explain the molecular differences between cis- and transmutations.
Which type of mutation (cis or trans) can be complemented in a
merozygote experiment?
A mutation that has a cis-effect is within a
genetic regulatory sequence, such as an
operator site, that affects the binding of a genetic regulatory protein. A cis-effect mutation affects only the adjacent genes that the genetic regulatory sequence controls. A mutation having a trans-effect is usually in a gene that encodes a genetic regulatory protein. A transeffect mutation can be complemented in a merozygote experiment by the introduction of a normal gene that encodes the regulatory protein.
C10. What is diauxic growth? Explain the roles of cAMP and the catabolite activator protein in this process.
Diauxic growth refers to the phenomenon in which a cell first uses up one type of sugar (such as glucose) before it begins to metabolize a second sugar (such as lactose). In this case, it is caused by gene regulation. When a bacterial cell is exposed to both sugars, the uptake of glucose causes the cAMP levels in the cell to fall. When this occurs, the catabolite activator protein (CAP) is removed from the lac operon so that it is not able to be activated by CAP.
C14.
Explain how a mutation would affect the regulation of the ara
operon
if the mutation prevented AraC protein from binding to the following sites:
aria
D. Without araO2, the repression of the ara operon could not occur. The operon would be
constitutively expressed at high levels because AraC protein could still activate
transcription of the ara operon by binding to araI. The presence of arabinose would have no effect. Note: The binding of arabinose to AraC is not needed to form an AraC dimer at araI. The dimer is able to form because the loop has been broken. This point may be figured out if you notice that an AraC dimer is bound to araO1 in the presence and absence of arabinose (see Figure 14.12).
B. Without araO1, the AraC protein would be overexpressed. It would probably require
more arabinose to alleviate repression. In addition, activation might be higher because
there would be more AraC protein available.
C. Without araI, transcription of the ara operon cannot be activated. You might get a very low level of constitutive transcription.
D. Without araO2 the repression of the ara operon could not occur. However, without
araI, transcription of the ara operon cannot be activated. You might get a very low
level of constitutive transcription.
C16. As described in Figure 14.14, four regions within the trpL gene can form stem-loop structures. Let’s suppose that mutations have been previously identified that prevent the ability of a particular region to form a stem-loop structure with a complementary region. For example, a region 1 mutant cannot form a 1–2 stem-loop structure, but it
can still form a 2–3 or 3–4 structure. Likewise, a region 4 mutant can form a 1–2 or 2–3 stem-loop but not a 3–4 stem-loop. Under the following conditions, would attenuation occur?
A. Region 1 is mutant, tryptophan is high, and translation is not occurring.
B.Region 2 is mutant, tryptophan is low, and translation is occurring.
C. Region 3 is mutant, tryptophan is high, and translation is not occurring.
D. Region 4 is mutant, tryptophan is low, and translation is not occurring.
A. Attenuation will not occur because loop 2–3 will form.
B. Attenuation will occur because 2–3 cannot form, so 3–4 will form.
C. Attenuation will not occur because 3–4 cannot form.
D. Attenuation will not occur because 3–4 cannot form.
C26. What key features distinguish the lytic and lysogenic cycles?
In the lytic cycle, the virus directs the bacterial cell to make more virus particles
until eventually the cell lyses and releases them. In the lysogenic cycle, the viral genome is incorporated into the host cell’s genome as a prophage. It remains there in a dormant state until some stimulus causes it to excise itself from the bacterial chromosome and enter the lytic cycle.
E6. A mutant strain has a defective lac operator site that results in the constitutive
expression of the lac operon. Outline an experiment you would carry out to demonstrate that the operator site must
be physically adjacent to the genes that it influences. Based on your knowledge of the lac operon, describe the results you would
expect.
You could mate a strain that has an F′ factor carrying a normal lac operon and a normal lacI gene to this mutant strain. Because the mutation is in the operator site, you would still continue to get expression of β-galactosidase, even in the absence of lactose.
