what happens if a repressor is not tightly linked to operon

Part 4: GENE REGULATION

ANSWERS

Answers to questions from Chapter xv on Positive and negative command of the lac operon

15.1 lac I, lac Z, lac Y, lac A

xv.2 The lac operon is negatively regulated past a repressor, the production of the lac I cistron (additional positive aspects of lac regulation result from activity of military camp-CAP). The lac repressor binds to a specific DNA sequence called the operator ( lac O) and prevents efficient initiation of transcription by RNA polymerase from the promoter ( lac P). An inducer (allolactose or an analog) binds to the repressor and prevents its bounden to the operator, thereby releasing the repression and allowing transcription of the lac operon.

a) Most mutations in the operator, the bounden site for repressor, lead to lower analogousness for the repressor and hence less binding. Thus these mutations let continued transcription (and thus expression) of the lac operon even in the absenteeism of inducer; this is referred to constitutive expression.

b) Mutations in the repressor that prevent its bounden to the operator will atomic number 82 to constitutive expression (no repression in the absence of inducer). Mutations that prevent binding of the inducer without affecting the ability to bind to the operator lead to a non‑inducible phenotype.

c) The lac promoter is not a particularly strong promoter, and mutations take been obtained that either increment or subtract its efficiency of initiating transcription. Base substitutions that brand the promoter sequence more similar to the consensus generate a stronger promoter (promoter "upward" mutations) whereas those that make the promoter less similar to the consensus generate a weaker promoter (promoter "downwardly" mutations). An "upward" mutation would make the lac operon no longer dependent on the positive regulation by the campsite‑CAP circuitous (when the operon is induced). A "down" mutation would not permit expression even in the de‑repressed state (presence of inducer) and hence would prove a not‑inducible phenotype.

xv.3 a) Choice (c) is correct. 0C causes constitutive expression of genes in cis , but may not give total expression without an inducer.

b) Choice (c) is correct. R- causes depression, but B- gives no B activity.

c) Selection (b) is correct. 0C causes constitutive expression in cis .

d) Option (a) is correct. R+ is dominant in trans .

fifteen.four The catabolite regulated operons are examples of positive regulation. Glucose is the preferred carbon source for many bacteria, and the operons for metabolism of other sugars, such as lactose and arabinose, are less active when glucose is available (even in the presence of inducer). The [cAMP] is depression when glucose is available. As glucose is depleted, the [cAMP] increases, and camp forms a complex with the catabolite activator protein (CAP). The camp‑CAP complex binds to a specific site near the promoter for the lac operon and the ara operon, and in both cases army camp‑CAP increases the efficiency of transcription from the promoter under induced conditions.

At least iii different explanations can be offered for the observed loss of activity upon purification.

(i) Using the example of cAMP‑CAP, perhaps the fungal RNA polymerase is agile on its promoter only in the presence of an activator poly peptide. The efficiency of initiation past the polymerase may be low for the purified polymerase, only high in the presence of the activator.

(ii) Some disquisitional subunit of the RNA polymerase is not tightly associated with the balance of the polymerase and is dissociated during purification.

(iii) If the analysis for polymerase activity during the purification scheme is specific initiation of transcription from the promoter, so the decline in activity may reflect a loss of specific initiation only not a decline in nonspecific transcription. In this instance, a specificity factor, such equally the sigma subunit of the E. coli RNA polymerase, may be lost at one step of the purification.

15.5 a) Lyase action Arginase activity

Strain two _(A-)_constitutive_______ ____constitutive_________

Strain 3 _(B-)_defective__________ ____inducible___________

Strain 4 _(C-)_constitutive_______ ____constitutive________

Strain 5 _(D-)_inducible__________ ____defective___________

Strain 6 _(D-/B-)_inducible________ ____inducible___________

(Note that D- and B- complement in trans. )

b) citB encodes argininosuccinate lyase (or lyase)

citD encodes arginase

c) Strain 7 is inducible, so citA+ is dominant to citA- in trans .

Strain 8 is constitutive, and Strain 9 is constitutive for arginase. Therefore, citC - (the mutant) is ascendant to citC+ (when citC - is in cis to wild-type structural genes.

d) citA encodes a trans -acting regulatory protein, e.g., a repressor.

citC is a cis -interim regulatory site, e.g. an operator.

xv.6

ALA synthetase ALA dehydrase

Strain 2 ____constitutive________ ____constitutive________

Strain 3 ____nonfunctional_______ ____repressible_________

Strain 4 ____repressible_________ ____nonfunctional_______

Strain v ____constitutive________ ____constitutive________

Strain six ____repressible_________ ____repressible_________

b)

porA Strain vii or 9 is constitutive for the A - , B + , C + and D + chromosome , which shows that mutant porA is ascendant to wt, when A - is in cis to B + , C + .

porB Strain 6 is repressible , which shows that wild-type porB is dominant to mutant (or mutant is recessive to wild-blazon).

porC Strain half-dozen (or nine or viii) is repressible (for ALA dehydrase) , which shows that wild-type porC is ascendant to mutant (or mutant is recessive to wild-type) .

porD Strain 8 (or ix) is repressible , which shows that mutant porD is recessive to wild-blazon (or wild-type is dominant to mutant).

For strain ix, note that chromosome A+B-C+D- is repressible when D+ is provided on the other chromosomes.

c) 1 concludes that porA is an operator, porB encodes ALA synthetase, porC encodes ALA dehydrase, and porD encodes an apo-repressor (it forms a repressor when heme binds).

d) This operon is under negative control, since porD - gives full activity of the synthetase and the dehydrase.

Answers to questions from Chapter 16, on Transcriptional regulation by effects on RNA polymerase

16.1

In practice, it is difficult to measure [R], [RDS] and [DS] simultaneously, then a more complicated procedure is used. But this problem illustrates the general idea.

xvi.2. The [TBP] = 1/Ks when the ratio of bound to free probe = ane. From a plot of the data, one obtains [TBP] = 0.47 ten 10 -9 M when spring/gratuitous = 1. Thus Ks = one/ 0.47 x ten-9 M ,

or Ks = 2 x 10nine M-1.

You can also apply the slope of this plot to get Ks = 2 x ten9 M-i

16.3 The lac repressor dissociates from the operator and rebinds to nonspecific sites on the Dna.

xvi.four Past bounden to an operator centered on the sequence at +xi and holding the complex between RNA polymerase and the promoter in the closed conformation.

16.5 . The most accurate value of DG for binding of AP1 to this duplex oligonucleotide is -eleven kcal/mole. The minimum in fault is the maximum in accuracy.

sixteen.half dozen The almost accurate measure of the equilibrium abiding, Ks, is

1.17 � 10eight M-1. Apply the equation DK = -RT ln Ks.

16.7 The value for thousand f in either the presence or absence of the repressor is 2 per sec. The graph shows that in both weather condition, the y-intercept is 0.50 sec, which is 1/thousandf .

xvi.8 The value for G_B is 1.0 � teneight G-i without repressor and ane.25 � xvii M-1 with repressor.

The forward charge per unit constant does not modify, but binding constant is decreased, consistent with a promoter occulsion model for the repressor. The graph shows an increment in gradient with no change in the y-intercept in the presence of the repressor, so this means that the bounden constant decreased. Annotation that d is the simply option with reasonable binding constants that showed this decrease. The binding constants were calculated from the slopes of 0.v � 10-eight sec Chiliad in the absenteeism of repressor, four.0 � ten-8 sec Yard in the presence of repressor, and yardf = 2 sec-ane (previous problem).

Answers to questions from Chapter 17 on Transcriptional regulation in lambda

17.1 Lysogeny results from the integration of a l prophage under atmospheric condition where the expression of all the 50 genes except cI are repressed. The 50 repressor, or CI poly peptide, will demark to the leftward and rightward operators of l to prevent transcription from PL and PR, hence blocking the expression of the genes required for lytic infection.

Leaner that are lysogenic for l are already producing the CI protein, or repressor. Subsequent infection by some other l phage results in the immediate binding of the l repressor to the leftward and rightward operators of the incoming phage, thereby preventing transcription of whatever of its genes, including those required for lytic infection. The newly introduced phage DNA as well cannot integrate (no expression of int and xis ) and it is eventually degraded.

17.two Statements 1, two, and three are correct.

17.three The two phage would complement to class turbid plaques. The fifty phage that are mutated in the cII gene (and are wild-type for the cI gene) volition be constitute in the lysogen.

17.4 Since the l repressor is present in 100 times the molar concentration of a single operator (and thus operator binding tin can use only ane of the 100 repressor dimers), the distribution of the repressor to nonspecific sites will predominate in terms of determining the fraction of repressor molecules that are non bound to Deoxyribonucleic acid. Since each base of operations pair in the E. coli genome is the showtime of a nonspecific binding site for l repressor, there are 4.ii x 106 nonspecific sites per prison cell. This means that [D tot] = [total concentration of nonspecific sites] = 7 x 10-3 One thousand. This is much greater than the concentration of repressor or Cro, and can be treated every bit a constant.

From the equation

Kns,r = = 105 One thousand-1

ane can calculate = =

= 0.14 x 10-ii = 0.0014

This can exist expressed as the fraction of total repressor molecules as follows:

[R] = 0.0014 [RD]

= = = = 0.0014

This ways that only 1.iv in m repressor dimers is not bound to Deoxyribonucleic acid at an operator or at a nonspecific site. Fifty-fifty though the analogousness of l repressor for its operator or for nonspecific Dna is less than seen for the lac repressor, and the amount of l repressor is about 10 times higher, the l repressor is all the same distributed between specific and nonspecific sites. Very little is not jump to DNA.

17.5. The relevant reaction is

R + O

Ks,r = = 1011 Yard-1

Kns,r = = ten5 M-ane

The specificity, which is the ratio of Ks,r to Kns,r can be used to summate the ratio of costless to jump operator.

specificity = = ten

= x = x

since [RD] = [Rtot] - [RO] - [R] = [Rtot] - [Otot], given that the concentration of costless R is negligible (come across 4.35) and [RO] = [Otot] when operator sites are saturated.

[Rtot] = 100 molecules/cell = 17 x ten-viii Thou

[Otot] = one site/cell = 0.17 x 10-8 Thousand

Thus:

= x

= 0.0416

= = 0.04

Thus 4% of the operator sites are free, and 96% are bound past repressor.

17.half dozen. The relevant equations are

C + O

Ks,c = = 1010 K-i

Kns,c = = 105 Thou-1

and the analysis is the same as in 4.19, except that Ks,c = 10ten Thou-i . Using the equation for specificity as in 4.19, one can calculate that

= 0.416

= = 0.29

Thus about 30% of the operator sites would not be jump by Cro, considerably more than the iv% of the sites that would not exist bound by l repressor.

17.7. Dividing eqn 2 by eqn 5, you obtain

= = x

As calculated in 4.eighteen, the concentration of free R and free C is very small, and are equal for equal concentrations of R and C.

[R] = [C] = (0.0014)(17 x 10-viii Chiliad) = 2.four x x-10 Grand

Thus the

= = 10

or there is 10 times more repressor jump to operator than is Cro bound to the operator. The distribution of l repressor and Cro to nonspecific sites was used in this treatment to summate the same low concentration for gratis repressor and Cro (trouble 4.35).

17.8

R + O

Ks,r =

Ks,r = 10eleven M-1 without cooperativity

Ks,r = three x 1012 Yard-1 with cooperativity

[R] = 10

Without cooperativity, [R] = = 9.9 ten 10-10 M

With cooperativity, [R] = = 3.three x x-xi K

Thus information technology takes 30 times more repressor to make full 99% of the operator sites in the absence of cooperativity. The cooperativity at the operator sites ways that less repressor is needed to make full the sites. The sigmoidal shape of the binding curve for cooperative interactons also ways that a smaller decrease in [R] will pb to dissociation of repressor from a greater fraction of operators, giving a more dramatic response to a lowering of [R].

Anwers to questions from Chapter 18: Regulation after initiation of transcription

18.one Statements 1, 2, and 4 are correct.

eighteen.2 In lysogeny, transcription of int is from the pint promoter, close to the gene. There is no nut site in this transcript, hence transcription from pint is not susceptible to Northward-mediated antitermination. Thus the transcript terminates at tint and the secondary structures that target RNases to a transcript are not formed. Thus the int transcript is stable, and Int protein is fabricated during lysogeny.

18.3 See Figures 4.4.8 and 4.4.9 for diagrams of the secondary structures under these weather condition.

The ribosome will progress to the trp codons in the leader and either stall (low trp ) or go on to a termination codon (loftier trp ). Stalling of the ribosome at trp codons (A state of affairs) prevents formation of the termination loop betwixt regions 3 and 4 of the mRNA. Progress to the UGA terminator allows the 3-four loop to class (B state of affairs) and thus terminate transcription.

xviii.4 Statements 1, 4, and 5 are correct.

xviii.5 The trp operon is subject to regulation both past repression and by attenuation. Attenuation depends on the tight coupling between transcription and translation in bacteria. When the [Trp] is high, translation of the trp leader is completed and the ribosome blocks sequence ii. This allows the transcribed sequences 3 and four to class the stem‑loop attenuator structure. Germination of the iii:4 loop, which resembles a rho‑independent transcription terminator, results in termination of transcription the trp operon earlier the structural genes (EDCBA) are transcribed, and the enzymes for Trp biosynthesis are not produced. When the [Trp] is low, translation of the trp leader is stalled at two Trp codons. In this position, the ribosome does not cover sequence 2. Sequence ii can at present base of operations‑pair with sequence 3 in an culling secondary construction. Germination of the two:3 stem‑loop precludes formation of the iii:4 attenuator loop, and transcription proceeds on through the trp EDCBA genes. Thus when the bacteria has a depression [Trp], the biosynthetic genes are expressed and more Trp is synthesized.

a) Increasing the distance betwixt sequence 1 (encoding the trp leader peptide) and sequence two will decrease attenuation under conditions of high [Trp]. In this situation, the ribosome, subsequently completing translation of the leader, will not cover sequence 2. Hence the 2:3 stem‑loop can form, preventing germination of the three:iv stem loop and thereby losing the normal attenuation with a greater altitude between sequences ane and 2.

b) A large increase in the distance between sequences two and iii could disfavor their germination of a stalk‑loop and hence formation of the 3:iv attenuator structure. Thus when the [Trp] was depression, even though the ribosome has stalled, the 2:iii loop would non form, assuasive the 3:iv attenuator structure to course with the outcome of a decrease in trp operon expression (due to attenuation) even in low [Trp].

c) Since sequence four is required to form the 3:4 attenuator stem‑loop, in its absenteeism no attenuation would be observed.

Answers to questions from Chapter nineteen. Regulation of eukaryotic genes

19.ane The discrete Dna binding domains of transcriptional regulatory proteins form specific complexes with divers sequences of DNA. Their analogousness for these divers sequences is about 105 to xvi greater than their affinity for other sequences.

Using the case of the lac repressor, the bounden site (operator) is 22 base pairs (bp) long. X molecules of the lac repressor are sufficient to go on this operator in a leap state even in the context of iv.half-dozen �106 bp of nonspecific Dna (the rest of the Due east. coli genome). This amounts to finding 1 specific site in a bounding main of four.half-dozen �106 bp, since you can consider each nucleotide pair to exist the beginning of a nonspecific binding site. The man repressor has an even harder job - information technology must detect its specific site within the iii�109 nonspecific sites (the haploid genome size). Thus the ratio of nonspecific to specific sites is 652 times greater in the human cell (3�109 divided by iv.vi�106). Extrapolating from the lac repressor information, we estimate that 6520 molecules of repressor will be needed per cell (10 molecules per prison cell 10 652).

xix.ii If the radius of the nuclear sphere is v microns ( thousandgrand), then the volume of that sphere is 5 10 10-13 Fifty. (Think that the volume of a sphere is (4/3)pr3). The concentration of specific sites is most vi.6 x 10-12 One thousand and the concentration of nonspecific sites is about 6.6 x 10-3 Grand.

19.iii . The fraction of OBF1 not jump to Dna is 0.0015. This is calculated from equation 3, using the [Dns] = 6.vi ten 10 -iii K and Kns = xv M-1.

19.4. From the equation for specificity, or the ratio of equations 2 and 3, one can calculated that [PDs]/[Ds] = 9 when [Ptot] = 60 nM. For a nuclear book of 5 ten 10 -13 L, this requires 18,060 (or near 18,000) molecules of OBF1.

nineteen.5. Deletion of the region betwixt -250 and -200 causes an increase in expression of the reporter in adipocytes, just no consequence on expression in melanocytes. Thus it has a negative effect in adipocytes.

19.6. Deletion of the region between -200 and -150 has no upshot on expression in melanocytes or adipocytes.

19.7. Deletion of the region betwixt -150 and -100 causes an decrease in expression of the reporter in adipocytes and melanocytes. Thus information technology has a positive result in both types of cells.

19.8. The formation of complex A is specific to the sequence between -150 and -100, as shown past the loss of signal in the self-competition but lack of competition by the E. coli DNA. The protein forming complex A is not Sp1, since the bounden site for Sp1 did not compete (lanes 12-fourteen). A candidate for the protein forming circuitous A is AP1 or its relatives; the binding site for AP1 competes too as the self DNA (lanes 9-11).

19.9. Since binding site for AP1 competes every bit well as the self DNA (lanes 9-xi), then you would expect to observe a sequence similar to this binding site in the -150 to -100 fragment, and that it would be bound specifically by a poly peptide. The bounden site for AP1 is TGASTCA.

nineteen.10. All of the Dna fragments compete equally well for both complexes (B and C), thus these complexes are not specific for any item DNA sequence.

19.11. Since this region contains a site needed for negative regulation in adipocytes, mutation and then that the site is no longer functional should allow expression in adipose tissue, i.due east. ectopic expression due to loss of office of a tissue-specific, cis -acting negative regulator.

19.12. Transcriptional activators take at least two domains which frequently function separately: the Deoxyribonucleic acid‑binding domain and the activation domain. The DNA‑bounden domain is required for the sequence‑specific binding of the protein to DNA; two familiar structural motifs found in different proteins are a helix‑turn‑helix and a Zn‑finger. A different portion of the poly peptide is responsible for activation; this domain may make direct contact with the RNA polymerase or it may facilitate the action of co‑activators or other proteins that stimulate transcription. Three unlike activation domains have been identified: acidic, proline‑rich and glutamine‑rich, and their mechanisms of activity are the bailiwick of much current research.

Our biochemist has done role of a domain‑swap experiment ‑ she has the activation domain of GAL4 fused to the DNA binding domain of the 50 repressor. This new hybrid will no longer recognize the GAL 4 binding site in DNA (called UASG), since that Deoxyribonucleic acid binding domain is no longer present. All the same, replacement of UASG with the l operator (bounden site for the 50 repressor) in the GAL operon should allow the l repressor‑GAL 4 hybrid poly peptide to function as a transcriptional activator of GAL genes in yeast.

4.40 Cys2-His2 zinc fingers.

19.14. A leucine zipper.

19.15 (ASC) The action of transcription factors tin be controlled by their differential synthesis, resulting in the presence or absence of specific factors in specific cell types. The activity transcription factors already present within cells can be controlled by the presence or absence of inhibitor (as with NF Thou B) and by the activation of pre-existing factors. An example of the final mechanism is the protein-protein interaction that forms an agile complex of MyoD with E12. Some other mechanism for activating pre-existing factors is covalent modification, such as phosphorylation. In at least some cells, AP1 is activated by phosphorylation.

19.xvi (ASC)

a) The data in table 4.49 show that insertions in four different regions of the receptor coding sequence cause decreased induction. This finding suggests that the glucocorticoid receptor has four separate functional domains. These iv domains correspond to the following insertions: domain ane = insertion D, Due east, and F; domain two = insertion I; domain 3 = insertion K, L, Thousand, and N; and domain 4 = insertion Q, R, and S.

b) Merely insertions at Q, R, and Southward produce receptor proteins with decreased steroid-binding ability. Therefore, the region of the protein respective to these insertions is the steroid-binding domain.

c) The data bespeak that insertions in three domains block induction without affecting steroid binding. An EMSA-type assay could be used to determine whether changes in any of these three functional domains impact DNA binding of dexamethasone-complexed receptor. Alternatively, the Dna-bounden domain could be inferred by comparing the sequence of glucocorticoid receptor with that of known DNA-binding domains in other steroid hormone receptors.

Answers to questions from Chapter 20. Regulation past changes in chromatin structure

20.1. vi kb from the right terminate of the restriction fragment, which is about ane kb 5' to the get-go site for transcription.

20.2. The DNase hypersensitive site could marker the position of an enhancer located 1 kb 5' to the first site of transcription. The hypersenstive site is as well far 5' to the starting time site for transcription to be a candidate for a promoter. Alternatively, it could be a locus control region, a matrix binding site, a silencer, a replication origin, or other potential regulatory sites.

twenty.iii. Gcn5p aids in the activation of target genes by interacting with other transcriptional activators which bind to specific Dna sequences.

20.4. histone acetyltransferase

20.v. chromatin remodeling/ activation

4.55. Several functions accept been assigned to the locus control region, including to abilities to:

Generate an open, active domain of chromatin.

Insulate from negative effects of adjacent sequences (negative position effects).

Enhance transcription of genes within the domain in a developmentally regulated fashion.

20.7. a) The cytoplasm of uninduced cells and nucleus of induced cells.

b) The PRE-binding activeness is in the nucleus of induced cells.

c) It must be phosphorylated and interact with something in the nucleus in order to bind the PRE site.

d) Binding requires phosphorylated NFL2 plus UBF3.

due east) The protein kinase is in the induced cell cytoplasm.

f) Later exposure of lung cells to pulmonin, NFL2 in the cytoplasm is phosphorylated by a kinase that is activated by the pulmonin treatment. Phospho‑NFL2 translocates to the nucleus where information technology binds UBF3 and after the heterodimer binds to the PRE.

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Source: http://www.bx.psu.edu/~ross/workmg/PartFourAnswers.htm

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