The E. coli cells will produce β-galactosidase when they are subjected to high levels of a DNA-
(a) With the Lac operator on the other side of the lac operon, the lac genes would no longer be subject to repression by the Lac repressor. (b) Inactivation of the binding site for CRP would reduce or eliminate expression of the lac genes under all conditions. (c) Alterations in the promoter sequence could either increase or decrease expression of the lac genes, depending on the particular alteration.
The conformation of AraC is altered by arabinose binding, and the protein binds different sites on the DNA when arabinose is absent versus present. In the absence of arabinose, AraC binding blocks RNA polymerase access to the promoter.
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(a) With high tryptophan concentration (added to the medium), tryptophan synthase levels drop due to attenuation of mRNA translation, even if the mRNA is stable. (b) Tryptophan synthase levels remain high for a considerable time. (c) Tryptophan synthase levels decline rapidly.
(a) An uncleavable LexA protein would permanently repress the SOS genes and block induction of the SOS response. (b) Weakened LexA binding would lead to constitutive expression of the SOS response genes.
∼7,000 copies. (10 copies / E. coli cell) × (3.2 × 109 bp / haploid human cell) / (4.6 × 106 bp / E. coli cell) ≈ 7,000 copies/human cell.
Repressor concentration is ∼8 × 10−9 m. The number of moles of repressor = 10 molecules / 6.02 × 1023 molecules / mol = 1.66 × 10−23 mol. Dividing by the volume of the cell in liters gives the concentration: (1.66 × 10−23 mol) / (2 × 10−15 L) = 8 × 10−9 mol/L. This is about five orders of magnitude greater than the dissociation constant for the repressor; as a result, the operator site will almost always be bound by repressor.
(a), (c), (d), (e) Operon expression decreases. (b) Operon expression remains unchanged, as it is already running at maximum expression.
(a) Decreased attenuation and thus increased transcription. If the ribosome translating sequence 1 did not block sequence 2, sequences 2 and 3 would pair more often. (b) The efficiency of pairing between sequences 2 and 3 could decrease, leading to increased attenuation and thus decreased transcription. (c) No attenuation, and thus transcription would proceed at the maximum level. (d) The attenuation system would respond more to histidine than to tryptophan concentration. (e) Decreased attenuation and thus increased transcription, because sequences 2 and 3 would pair more often. (f) Increased attenuation and thus decreased transcription, because pairing of sequences 3 and 4 would almost always occur.
The TPP-
The repressor protein would probably block translation of the mRNA even when the protein’s binding sites on the ribosome were available. This would tend to slow the assembly of active ribosomes.
The λ prophage produces the λ repressor (cI) and cII proteins, which will bind to any λ phages entering the cell and prevent induction of a lytic pathway.
These mutants indicate that activation is not caused by CRP changing the local DNA structure near a promoter; the activator must do more than simply bind to DNA. Because the mutant CRP cannot activate transcription in response to low glucose levels, cells with this mutation will not grow well on lactose or other secondary sugars for which the metabolizing-
Because genes in an operon are transcribed together from one promoter as a polycistronic mRNA, they can be regulated by one set of activators and/or repressors. In this way, enzymes required for a common pathway can be synthesized together. Translational regulation could be used to express the operon genes at different levels: by alteration of the ribosome-
An advantage is that the signal sensor is contained within the mRNA itself, and thus regulation does not require a separate sensor molecule to be synthesized or maintained. A disadvantage is that coordinated gene expression through integration of different cellular signals is difficult.
Conserving energy is important, but careful modulation of gene expression in response to changing growth conditions is paramount. It is preferable to “waste” energy in synthesizing partial transcripts that will remain unused unless antitermination is triggered, so that the cell is primed to respond quickly to a sudden need for enhanced gene expression.
No. Eukaryotic transcription occurs in the nucleus, and translation in the cytoplasm. The spatial and temporal separation prevents the coupling of transcriptional and translational regulation that can occur in bacterial operons such as the trp operon.
(1) These mRNAs tend to have unusually long sequences upstream from the translation start site that are necessary to form the three-
Cell growth would decline, even in the presence of abundant nutrients, due to the inhibitory effects of ppGpp on RNA polymerase.
(a) Band A is, in part, the undigested attenuated mRNA, and this undigested RNA accounts for one of the three A bands in Figure 3. The two additional A bands in the denaturing gel indicate the presence of a good cleavage site for RNase T1 near the middle of the RNA. Because the two parts of the RNA separate only on the denaturing gel, sequences on either side of the T1 cleavage site must be paired and thus migrate together in the first gel (Figure 2). (b) Bands B and C are segments of the larger attenuated mRNA. (c) Both band B and band C have paired RNA sequences that protect the RNA from RNase T1. Because band C is sometimes cleaved in two (the gel in Figure 3 has one band of undigested RNA and two bands derived from the cleavage), there must be a loop of significant size near the end of the paired sequences in this segment. (d) The band B RNA includes sequences 3 and 4. The loop at the end of this hairpin is small enough to limit T1 cleavage. (e) The band C RNA derives from paired regions in sequences 1 and 2. (f) The pairing of sequences 2 and 3 is not present in this analysis. This hairpin must be less stable than the pairing of sequences 3 and 4 and of sequences 1 and 2.
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