The authors describe the underlying mechanisms that control the expression of genes.
The abundance of DNA, chro mosome structure, and the existence of the cell cycle are explained in the text.
The structure of nucleosomes is described by the authors.
The histones are wrapped around the DNA in the nucleosome.
The individual nucleosomes are linked by stretches of DNA.
When genes are not being expressed, the nucleosomes are packed with other proteins to form a highly condensed chromosomes.
The introduction to the topic of the regulation of eukaryotic gene expression is described in the text.
The authors describe how enhancers and combinations of regulatory proteins and histone modifications are made.
The text describes the steroid-hormone estrogen receptor that uses the zinc-finger motif to bind specifically estrogen response elements in DNA.
The mechanisms of coactivators and corepressors are described.
The mechanism that is used by several amino acid biosynthetic operons relies on alternative RNA secondary structures.
The regulation of iron metabolism in animals is shown to show how RNA secondary structures can be bound to specific genes.
Chapter 5, Sections 5.4 and 5.6, and Chapter 28 should be reviewed prior to studying this chapter.
On page 748 of the text, there is a description of the functional groups on DNA.
These examples show the principles by which a specific sequence of DNA could be interacted with.
You should be able to complete the objectives once you have mastered this chapter.
There are other DNA-binding motifs.
The stability of the H3 and H4 histones should be noted and the types of histones assigned to their locations within the nucleosome.
Relate the IRE to aconitase and iron.
The helix-turn-helix motif is a pattern.
The amount of DNA in the haploid genome is listed in the left column.
The 10-methylenetetrahydrofolate is used to form the 5-methyl group of thymine.
Specific interactions between transcription factors and specific DNA sequence allow for specific control of transcription.
The nuclear hormone receptors are made up of two parts.
The tails of histones have a lower affinity for DNA.
Answer (c) is incorrect because it is rare inbacteria.
Although it is an inducer for the synthesis of b-galactosidase, it is not an allosteric activator.
The product of a reaction catalyzed by b-galactosidase and binding to it is called al lolactose.
There are approximately equal numbers of copies of the same genes in each of the three enzymes.
A decrease in the synthesis of cAMP by adenylate cyclase is caused by the preferential metabolism of Glucose.
The concentration of cAMP increases when there is a decrease in the amount of sugar in the water.
The answer is incorrect because the recognition helix inserts into the wider major grooves.
The answer is incorrect because the DNA molecule more than a few kilobases long is sensitive to the shearing forces developed when solutions are stirred.
Answer (c) is correct because the longer a DNA molecule is, the longer it takes to reassume its normal solution after it has been stretched.
The length of the bacterium is expressed in kilobases, not megabases.
Histones make up nearly half the mass of a chromosomes.
A nucleosome core consists of 145 base pairs of DNA wrapped around his tone Octamer.
The exact length of the linker DNA depends on the tissue and the organisms.
The average length is 200 bp.
Answer (b) is incorrect because genes that are transcribed are less compact than those that are not.
The nucleosome core has a disk shape.
The left-handed torroidal supercoil has 145 base pairs of DNA wound about the core of histones.
Two copies of histones H2A, H2B, H3 and H4 are on the inside of the toroidal coil, whereas histone H1 is associated with the core.
A basic tail protrudes from the core structure of each core histone.
200 bp of DNA is present in 888-282-0465 888-282-0465 888-282-0465 888-282-0465 888-282-0465 888-282-0465.
Approximately 200 base pairs of DNA are associated with each nucleosome.
A sphere with a diameter of 100 A would have a packing ratio of 7 if this DNA were coiled into it.
The resulting 360-A coil must be looped and folded to form the chromatin fiber.
Small basic molecules, such as the polyamines, contribute to the ultimate compaction of 104 that is observed in metaphase chromosomes.
The answer is wrong because the methylation takes place at 5.
The answer is incorrect because SAM is the donor of postsynthetic DNA methylation.
A critical feature of combinatorial control is that some components of the transcription complex must interact with DNA in order to locate the proper region on the DNA.
Some transcription factors bind to enhancers far from the site of transcription initiation.
The properties of a more numerous class of hormones that act by triggering cascades within cells after binding outside the cell are incorrect.
bromination has nothing to do with the action of this acetyllysine-binding protein structure.
Many of the sites on DNA are not accessible to the pro teins that must be assembled to form an active transcription complex.
Some of the accessible sites are due to the remodeling of chromatin.
Answer (a) is correct.
Attenuation provides a rapid, sensitive fine-tuning mechanism on top of the control exerted by the repressor-operator interaction.
Answer (d) is incorrect when a single polycistronic mRNA is produced when the RNA is not terminated at the attenuator.
The rho-independent transcription-termination structure could no longer be formed because of this deletion.
The fer ritin molecule is expressed in the 5, untranslated region of the IRE.
The estrogen-estrogen receptor complex is involved in trancription by binding to the ERE.
The operator sequence is where some of the known constitutive mutations of the lactose operon occur.
If it's possible to detect whether the enzymes are produced in diploid or haploid amounts, then so be it.
More galactosidase is produced than permease.
A mechanism to account for this is consistent with the facts.
Predict whether active enzymes Z, Y, and A will be produced or not.
If you assume semidominance, a partially diploid cell will have twice the activity of a haploid cell.
Tell your swer briefly.
If you assume semidominance, a partially diploid cell will have twice the activity of a haploid cell.
In systems of genetic regulation involving positive control, a regulatory gene makes a substance that enhances rather than decreases the amount of transcription.
An operon is regulated by a mechanism called attenuation.
Predict the expression of tryptophan genes in the presence or absence of each of the Mutants described below.
The number of differences between the two genes is a measure of the evolutionary divergence.
The minimum number of events that must have happened to cause the differences is a better measure.
There must be changes in both genes.
The globin-produced cells can be treated with micrococcal nuclease under certain conditions.
When the resulting nucleosomes are isolated and their DNA is examined, it is found to contain a sequence for the synthesis of globin.
Each copy of the DNA is translated into a different molecule of the messenger RNA and a different molecule of the mRNA.
There are a number of CG-rich regions upstream from the open reading frame.
Provide an explanation for the observations and how they would be affected by 5-azacytidine.
You amplify the plasmid construct and then isolated it from the bacteria.
The muscle cells in the culture medium lacking 5-azacytosine are formed when the isolated plasmid molecules are used to attack them.
The digests are separated using gel electrophoresis.
You will need to look up the specificities of these restriction endonucleases.
Female is the default state for sex determination in humans.
To become male, genes must be activated that lead to the development of the testes and external male genitalia and suppression of the development of the female sex organs.
The Y chromosome is absent in genotypic (XX) females.
Testosterone is involved in this process.
A group of biologists showed that the three DNA sequences below can be used to amplify the expression of a reporter gene.
The lack of affinity for the promoter of the eukaryotic polymerases makes them dependent on several activator genes for initiation of transcription.
eukaryotic genes are more likely to be under positive regulatory control thanbacterial genes.
The mode of regulation may be related to the size of the genome.
Small amounts of catalysts can be easily detected.
A turnover number of 300,000 s:1 will provide an assays that is 300,000 times more sensitive than that for a structuralProtein, which is a single molecule.
Direct chemical methods can't detect many cellular proteins produced in small amounts.
1,6-allolactose is the inducer of the lactose operon.
The lag shows how long it takes for residual b-galactosidase to convert lactose into 1,6-allolactose.
It acts as an inducer.
No lag is observed.
Only the structural genes on the same chro mosome will be affected, a cis-acting effect.
If the effect is cis, the haploid amount of Z will be produced in the absence of inducer.
The diploid amount of Z will be produced if the effect is trans.
In the absence of an inducer, very low levels of lactose operon enzymes are produced.
chloramphenicol won't prevent induction if zymogen is involved.
The presence of chloramphenicol won't be observed if the synthesis of newProtein is involved.
The expression of the three structural genes in the lactose operon must be at the level of translation and not transcription since a single, polycistronic mRNA molecule is formed.
mRNA transcripts containing genetic information for all three genes are produced.
The messenger might be dropped off at the end of the structural genes by some ribosomes.
The statement that the enzymes are repressible is the clue.
Substance S is a corepressor because it is active in the presence of T.
The percentage of X is 3.3%.
A positive control element is CAP.
The formation of the cAMP-CAP complex occurs when the level of cAMP is high.
Positive control elements are act as coactivators.
The leader sequence should contain codons.
If there is enough X in the cell, there will be enough X-tRNAX for the synthesis of the leader.
There won't be a need to modify the enzymes to produce X.
There will be increased transcription of the trp synthesis genes in both the presence and absence of tryptophan.
The rate of translation will be slowed down by the low levels of Trp-tRNA.
The stalling or slowing of the ribosome in segment 1 is important for regulation, not just the inability to synthesise the leader peptide.
The ribosome will never start translation, so transcription will always end regardless of the levels of tryptophan.
The results will be the same in Mutant C as in Mutant B.
The leader would not have an effect on the level of attenuation.
Trp synthesis would be regulated by the levels of leucine if the Trp codons were removed.
The Trp synthesis genes would not be transcribed even in the absence of tryptophan, and this strain would always require tryptophan to grow.
There would be expression of the Trp synthesis genes even in the presence of leucine.
The Mutant F would never express the genes for Trp synthesis.
In highly ionic solutions, the interaction between prota mines and DNA is diminished because of the salt in solution.
The positively charged guanidinium groups can associate with the negatively charged phophodiester bonds in a DNA helix.
The negatively and positively charged ion that result from the addition of a salt to a solution compete with the DNA-ligand interactions and weaken them.
HATs remove their positive charges and weaken the interaction between the histone and the DNA by acetylating the primary amino groups on the side chains of lysine.
The reduction of charge-charge interactions is the same regardless of the action of HATs or the addition of salts.
A change from A to G in the first position of the codon would cause a substitution of Val for Ile.
The change from an A-T to a G-C base pair is related to this.
A single change from G to A in the second position of the codon would account for the difference.
This shows a change from a G-C to an A-T base pair.
These are conservative changes.
Both Ile and Val have positively charged side chains.
We wouldn't expect any structural or functional difference in the H4 of calf and pea seedlings.
The two organisms are not the same.
In cells that are active in synthesizing globins, the globin genes can be transcribed into messenger RNA.
This region is sensitive to DNase.
Neither embryonic globins nor ovalbumin are produced to a significant extent in an adult cell specialized for the production of globins.
The regions of DNA that carry the information for these genes are not sensitive to DNase.
The globin genes are not destroyed in those oviduct cells that make ovalbumin.
The results are the same.
Many of the globin genes are required to cover the nu cleosomes.
The linker regions between the nucleosomes become susceptible to cleavage by micrococcal nuclease when these genes become transcriptionally active.
The number of molecules that are converted into product per second is given.
Each of the zinc fingers contains an 30-residue-long sequence of two cysteines and two histidines that are coordinated to a zinc ion.
Zinc is involved in the structure of the nuclear hormone receptors.
It is involved in a number of other reactions, including the conversion of acetaldehyde to ethanol, the formation of bicarbonate ion, and the cleavage of peptides by chymotrypsin.
It is likely that zinc deficiency could lead to significant developmental abnormality because of its essential roles in gene expression and cellular metabolism.
Three-fourths of the CG sequence is methylated in many cells.
In your experiments, cells grown in the presence of 5-azacytidine incorporate the azacytosine into DNA, but the azacytosine is not present in the normally present C. Cells grown without the analog are less likely to express the genes you are studying and are more likely to have the methylated CG-rich sequence in the promoter region.
The activity of the promoter is what determines the expression of the gene.
I, which does not cleave the CmCGG sequence, is at the 5 position of the interior C.
Specific genes are allowed to associate with certain genes from cell extracts or fractionated samples.
The control samples of labeled fragment are also subjected to gel electrophoresis.
The gels are visualized by autoradiography on film or in a machine.
Retarded samples are candidates for sequence-specific complexes.
Large amounts of unlabeled, random sequences of DNA can be added to the sample to minimize retardation caused by binding of histones to the labeled fragment.
The gel doesn't see this DNA because it isn't labeled.
The gel-shift assays don't usually interfere with the addition of extra DNA because the sequence-specific proteins have a higher affinity for a specific sequence than the nonspecific ones.
An excess amount of unlabeled DNA can be included in an experiment to assess the specificity of the interaction.
If the interaction is sequence-specific, the unlabeled DNA should bind theProtein and by specific competition abolish the shift of the labeled sequence on the gel.
Gel-shift analysis is used a lot.
Although it is not a true equilibrium technique and can't provide true equilibrium binding constants, it reliably indicates the relative affinities of certain genes.
You can also detect coactivators or corepressors, if you know how they work.
You wouldn't be surprised because the distant sequences might be enhancers that bind transcription factors that themselves associate with the core transcription machinery by looping the DNA to achieve proximity to the transcription start site.
The testosterone-nuclear hormone receptor complex could not be formed, and the genes that promote and suppress feminization would not function properly.
A male who developed into a female would be a genetic result.
The C termini is near the ligand-binding domain of nuclear hormone receptors.
The inability to form the testosterone-receptor complex results in a male genotype expressing a phenotype similar to one arising from a missing Y chromosomes, which is, being a female.
testicular feminization is a disorder that arises from the situation described, and many other factors beyond those mentioned here are involved.
The most likely scenario is that each of the two monomers can bind to one of the sequences above.
A Heterodimer is formed by two different monomers binding to a third sequence.
You would expect a Heterodimer to bind to an asymmetrical sequence, which has dyad symmetry, and a Homodimer to bind to an inverted sequence, which has dyad symmetry.
Sequences X and Z have dyad symmetry, whereas sequence Y is asymmetrical and contains half of X and half of Z.
One person associates with the symmetric sequence Z while another person associates with the asymmetric sequence X.
Half of the sequence X and Z are contained in the asymmetric sequence Y.
The potential for regulation of expression is increased by this problem because of the important principle that Heterodimer formation can allow recognition of DNA sequences that do not have dyad symmetry.
A Homodimer DNA bindingProtein is an example of the CREB.
Negative regulation requires synthesis of a specific repressor that blocks tran scription.
40,000 repressor proteins would need to be synthesised in order to carry out negative regulation of genes in a human genome.
Because most eukaryotic genes are not in operons and are normally inactive with regard to transcription,selective activation through synthesis of a small group of activator proteins is used to promote transcription of a particular array of genes needed by the cell at a certain time.
Positive control may be related to the fact that a larger genome may mean that a relatively short DNA sequence for a regulatoryProtein could be present in multiple and possibly wrong locations, bringing about inappropriate or unneeded gene activation.
It is possible to avoid this by requiring that several positive regulatory proteins form a complex that can specifically activation a gene and promote its transcription.
The cell volume of a liter is 10:15 liter.
Avogadro's number in 10:15 liter corresponds to a mole cule concentration of 1.7 M.
The single molecule will be bound to the operator's DNA if the concentration of the repressor is higher than the dissociation constant.
In the table below, a charge of :1 is assigned to each Asp and each Glu, and a charge of ;1 is assigned to each Lys and each histones.
Estimated net charges of ;17 for H2A, ;18 for H2B, ;20 for H3 and ;18 for H4 are summed.
The histone net charge would equal 2.
A 150-base sequence of DNA with one negative charge per linkage has a net charge of about 2.
A single-stranded probe could be used to hybridize the isolated mixture of DNA fragments.
The known probe DNA could be attached to a filter to prepare for the analysis.
The strands of the double helix could be separated using the polymerase chain reaction, labeled with 32P, if necessary.
Under "stringent" hybridization conditions, the fragments would be incubated with the filter-attached probe.
The extent of binding can be determined by counting the specific radioactivity that the filter acquired during the reaction.
The high content of 5-methylcytosine is found in the transcriptionally inactive regions of DNA.
It is possible to prevent methylation by porating 5-azacytidine into the DNA.
Some normally inactive genes will be activated because of the lack of methylation.
The 5-methylcytosine signal might play a role in the inactivation of genes.
The domain might be able to block transcription by binding to regulatory regions of double-stranded DNA.
The major grooves are where the protein domain would bind.
The 5-methyl group will protrude into the major grooves of double-stranded DNA on the outside of a GC base pair.
The anti-inducer could be a competitor.
The anti-inducer would bind to the repressor at a similar or overlap site to that of the inducer, but wouldn't cause the conformational change necessary to release the repressor from the operator DNA.
The anti-inducer would have to be removed from its site by higher concentrations of inducer.
Striking symmetry is a recurring theme for interactions between genes.
There is a chance that the DNA sequence could be used as a binding site.
Inverted repeat sequence can serve as a hot spot for genetic rearrangements because they can form hairpin secondary structures that block DNA polymerases or are processed by structure-specific endonucleases.
The carbonyl carbon of the thioester of acetyl-CoA can be attacked by the lysine group.
The leaving group CoASH could be eliminated by the tetrahedral intermediate.
CREB may be prevented from binding to its true target sites by the injected DNA fragments.
CREB would not be able to perform its role in stimulating the synthesis of new proteins for this reason.
A pathway for the stimulation of long-term memory by Serotonin is proposed.
The coactivator CBP is bound by phosphorylated CREB.
There are genes that are methylated.
The mouse's DNA is cut into pieces of average size about 50,000 base pairs.