(Exam 2) Chapter 8 Study Questions
Some transcription regulators bind to DNA and cause the double helix to bend at a sharp angle. Such "bending proteins" can stimulate the initiation of transcription without contacting either the RNA polymerase, any of the general transcription factors, or any other transcription regulators. Can you devise a plausible explanation for how these proteins might work to modulate transcription? Draw a diagram that illustrates your explanation
Bending proteins help bring together distant DNA regions that normally would contact each other inefficiently. no illustration
Bacterial cells can take up the amino acid tryptophan (Trp) from their surroundings or, if there is an insufficient external supply, they can synthesize tryptophan from other small molecules. The Trp repressor is a transcription regulator that shuts off the transcription of genes that code for the enzymes required for the synthesis of tryptophan (see Figure 8−7). A. What would happen to the regulation of the tryptophan operon in cells that express a mutant form of the tryptophan repressor that (1) cannot bind to DNA, (2) cannot bind tryptophan, or (3) binds to DNA even in the absence of tryptophan? B. What would happen in scenarios (1), (2), and (3) if the cells, in addition, produced normal tryptophan repressor protein from a second, normal gene?
A. Transcription of the tryptophan operon would no longer be regulated by the absence or presence of tryptophan; the enzymes would be permanently turned on in 1 & 2 and permanently shut off in 3. B. In 1 & 2, the normal tryptophan repressor molecules would completely restore the regulation of the tryptophan biosynthesis enzymes. In contrast, expression of the normal protein would have no effect in 3, because the tryptophan operator would remain occupied by the mutant protein, even in the presence of tryptophan.
Which of the following statements are correct? Explain your answers. A. In bacteria, but not in eukaryotes, many mRNAs contain the coding region for more than one gene. B. Most DNA-binding proteins bind to the major groove of the DNA double helix. C. Of the major control points in gene expression (transcription, RNA processing, RNA transport, translation, and control of a protein's activity), transcription initiation is one of the most common.
A. True. Prokaryotic mRNAs are often transcripts of entire operons. Ribosomes can initiate translation at internal AUG start sites of these "polycistronic" mRNAs. B. True. The major groove of double-standard DNA is sufficiently wide to allow a protein surface, such as one face of an alpha helix, access to the base pairs. The sequence of H-bond donors and acceptors in the major groove can then be "read" by the protein to determine the sequence and orientation of the DNA. C. True. It is advantageous to exert control at the earliest possible point in the pathway. This conserves metabolic energy because unnecessary products are not made in the first place.
A virus that grows in bacteria (bacterial viruses are called bacteriophages) can replicate in one of two ways. In the prophage state, the viral DNA is inserted into the bacterial chromosome and is copied along with the bacterial genome each time the cell divides. In the lytic state, the viral DNA is released from the bacterial chromosome and replicates many times in the cell. This viral DNA then produces viral coat proteins that together with the replicated viral DNA form many new virus particles that burst out of the bacterial cell. These two forms of growth are controlled by two transcription regulators, the repressor (product of the cI gene) and Cro, both of which are encoded by the virus. In the prophage state, cI is expressed; in the lytic state, Cro is expressed. In addition to regulating the expression of other genes, cI represses the Cro gene, and Cro represses the cI gene (Figure Q8-4). When bacteria containing a phage in the prophage state are briefly irradiated with UV light, cI protein is degraded. A. What will happen next? B. Will the change in (A) be reversed when the UV light is switched off? C. What advantage might this response to UV light provide to the virus?
A. UV light throws the switch from the prophage to the lytic state: when c1 protein is destroyed, Cro is made and turns off the further production of c1. The virus produces coat proteins, and new virus particles are made B) When the UV light switches off, the virus remains in the lytic state and forms a gene regulatory switch that memorized the previous settings. C) The switch makes sense because a bacteria exposed to UV will be damaged so the virus switches to the lytic state and exits the cell
Explain how DNA-binding proteins can make sequence-specific contacts to a double-stranded DNA molecule without breaking the hydrogen bonds that hold the bases together. Indicate how, through such contacts, a protein can distinguish a T-A from a C-G pair. Indicate the parts of the nucleotide base pairs that could form noncovalent interactions— hydrogen bonds, electrostatic attractions, or hydrophobic interactions (see Panel 2−3, pp. 70-71)—with a DNA-binding protein. The structures of all the base pairs in DNA are given in Figure 5-4.
Contacts can form between the protein and the edges of the base pairs that are exposed in the major groove of the DNA. These sequence-specific contacts can include hydrogen bonds with oxygen, nitrogen and hydrogen atoms, as well as hydrophobic interactions with the methyl group on thymine. Arrangement of hydrogen-bond donors and hydrogen bond acceptors of A-T and G-C pairs would be different from one another which allow recognition of specific DNA sequences via the major groove.
What mechanisms are used for the cell to confer epigenetic cell memory. Explain in a mechanistic manner how a deficiency in maintenance methyltransferase would affect transcription and cell memory.
Epigenetic cell memory: cells retain epigenetic marks after cell division. Because of the maintenance of these epigenetic marks genes expressed in a particular cell remains the same even after cell division so that cell maintains its identity. Epigenetic memory can be mitotic or meiotic. Mechanisms used by the cell for conferring epigenetic memory are DNA methylation and Histone modifications. After DNA replication hemimethylated DNA is produced in the cell there is maintenance DNA methyltransferase DNMT 1 which can make hemimethylated DNA to fully methylated. Histone modifications may be removed during cell division but after cell division, histone modifications are reestablished but the mechanism for this process is not known. maintenance DNA methyltransferase is required to maintain Methylation marks on the DNA deficiency of Maintenance DNA methyltransferase will affect Epigenetic memory of cell, In a cell, both strands of DNA are methylated, say this cell is an embryonic cell where De novo DNA methyltransferases are active so both strands of DNA are methylated. when cell differentiated de novo DNA methyltransferases are not expressed since these cells are deficient in maintenance DNA methyltransferase in the next round of division cell will contain hemimethylated DNA and in next round of divisions cells having Unmethylated DNA will be produced. Usually, DNA methylation is a repressive mark when these repressive marks are removed, genes which are not supposed to express in the cell usually, are also gets transcribed thus cell loses memory that transcriptome and proteome of cell changes so the identity of the cell also changes
Your task in the laboratory of Professor Quasimodo is to determine how far an enhancer (a binding site for an activator protein) can be moved from the promoter of the straightspine gene and still activate transcription. You systematically vary the number of nucleotide pairs between these two sites and then determine the amount of transcription by measuring the production of Straightspine mRNA. At first glance, your data look confusing (Figure Q8-6). What would you have expected for the results of this experiment? Can you save your reputation and explain these results to Professor Quasimodo?
From our knowledge of enhancers, one would expect their function to be relatively independent of their distance from the promoter—possibly weakening as this distance increases. The surprising feature of the data (which have been adapted from an actual experiment) is the periodicity: the enhancer is maximally active at certain distances from the promoter (50, 60, or 70 nucleotides), but almost inactive at intermediate distances (55 or 65 nucleotides). The periodicity of 10 suggests that the mystery can be explained by considering the structure of double helical DNA, which has 10 base pairs per turn. Thus, placing an enhancer on the side of the DNA opposite to that of the promoter (Figure A8-6) would make It more difficult for the activator that binds to it to interact with the proteins bound at the promoter. At longer distances, there is more DNA to absorb the twist, and the effect is diminished.
Imagine the two situations shown in Figure Q8-12. In cell I, a transient signal induces the synthesis of protein A, which is a transcriptional activator that turns on many genes including its own. In cell II, a transient signal induces the synthesis of protein R, which is a transcriptional repressor that turns off many genes including its own. In which, if either, of these situations will the descendants of the original cell "remember" that the progenitor cell had experienced the transient signal? Explain your reasoning.
In cell I the gene is being activated leading to the synthesis of proteins that inturn turn on many other genes including it. So there will be extensive functioning of the cell and assuming that the gene for cellular memory is activated we can infer that the next generation cells will memorise the kidn of functioning in the progenitor cells. In cell II the mRNA is producing repressor proteins that stops transcription of the genes. So there will be no scope to pass on this information to the next generation cells. So in the case of cell I the descendants will remember that the original cell had experienced transient signal.
Mention the components of eukaryotic transcriptional switches. A scientist is pursuing to increase the transcription of an eukaryotic gene. What component of the transcriptional switch the scientist would select to increase more effectively transcription. Does the scientist need to increase or decrease the expression of the gene encoding such component of the transcriptional switch. If yes, explain why? Give a mechanistic explanation.
In the case of eukaryotic transcription, the components for the transcription switching are chromatin modification, transcription factors, enhancers, regulatory landscapes (Topological Association Domain, Insulators etc.), Ribosomal silencing of the mRNA. Yes, scientist canto increase or decrease the expression of the gene encoding such component of the transcriptional switch by remodelling the chromatin state. To fit into the nucleus the genomic DNA is highly compacted by a protein octamer called Histone. The significant portion of the genomic DNA is silenced through the Histone protein compactness, for that these portions are inaccessible to the RNA polymerase and other transcription factors. The accessibility of the DNA can depend upon the chromatin structure. Which can be altered by the modification in the histone protein by DNA methylation, DNA acetylation, DNA-binding protein or ncRNA etc. Methylation in the Histone or Deacetylation in the histone works together in gene silencing, whereas histone acetylation leads to the activating the gene expression by dissociating the DNA from the histone complex. Methylation in DNA is another method for gene silencing. Methylation in CpG Island in DNA molecule leads to silencing to the active gene. So, through the modification in DNA and histone complex the expression of the gene can be up regulated or down regulated.
Discuss the following argument: "If the expression of every gene depends on a set of transcription regulators, then the expression of these regulators must also depend on the expression of other regulators, and their expression must depend on the expression of still other regulators, and so on. Cells would therefore need an infinite number of genes, most of which would code for transcription regulators." How does the cell get by without having to achieve the impossible?
Many transcription regulators are continually made in the cell; that is, their expression is constitutive and the activity of the protein is controlled by signals from inside or outside the cell (e.g. the availability of nutrients, as for the tryptophan repressor, or by hormones, as for the glucocorticoid reception). In this way, the transcriptional program is adjusted to the physiological needs of the cell. Moreover, a given transcription regulator usually controls the expression of many different genes. Transcription regulator are often used in various combinations and can affect each others activity, thereby further increasing the possibilities for regulation with a limited set of proteins. Nevertheless, most cells devote a large fraction of their genomes to the control transcription: about 10% of protein coding genes in eukaryotic cells code for transcription regulators
The λ repressor binds as a dimer to critical sites on the λ genome to repress the virus's lytic genes. This is necessary to maintain the prophage (integrated) state. Each molecule of the repressor consists of an N-terminal DNA-binding domain and a C-terminal dimerization domain (Figure Q8-7). Upon viral induction (for example, by irradiation with UV light), the genes for lytic growth are expressed, λ progeny are produced, and the bacterial cell is lysed (see Question 8-4). Induction is initiated by cleavage of the λ repressor at a site between the DNA-binding domain and the dimerization domain, which causes the repressor to dissociate from the DNA. In the absence of bound repressor, RNA polymerase binds and initiates lytic growth. Given that the number (concentration) of DNAbinding domains is unchanged by cleavage of the repressor, why do you suppose its cleavage results in its dissociation from the DNA?
The affinity of the dimeric λ repressor for its binding site depends on the interactions made by each of the two DNA-binding domains. A single DNA-binding domain can make only half the contacts and therefore provide just half the binding energy compared with the dimer. Although cleavage of the repressor does not change the concentration of binding domains, the affinity that each repressor monomer has for DNA is sufficiently weak that the repressors do not remain bound. As a result, the genes for lytic growth are turned on.
Mention the components of bacterial transcriptional switches. The Lac operon is regulated by both activator CAP and Lac repressor. A scientist introduces a mutation in the gene encoding the activator CAP, making this allosteric protein insensitive to glucose levels. In this case, the mutant CAP is always active. What outcomes do you expect if you culture the mutated bacteria in a culture medium with glucose and lactose. Give your explanation based on the mechanism of the Lac operon.
The bacterial transcription unit consists of - a regulator site, a promoter site, an operator site and the structural components of the unit which has gene encoding for protiens. The catabolite regulating protien (CAP) regulates the amount of glucose in the cell of the bacteria. If low amount is present it usually becomes activated by detecting the levels of cAMP in the cell which is a hunger signal molecule. The preferred molecule for energy production is glucodg and lactose is only used when glucose level is low. When glucose level is low, cyclic AMP is high and hence it binds to CAP and then this whole complex binds to the CAP site and the transcription occurs. Hence lactose is now used since the glucose level was low. Hence the CAP usualy is a check for the determination of which molecule to be used up by the bacteria. So when the mutation occur in the CAP , there is no longer differentiation of amount of glucose and lactose present and lactose now becomes the preferred substrate inspite of when the glucose level is adequate.
Gene expression shapes the formation of specialized cells by determining cell fate. Gene expression is controlled at various points. Mention the types of controls for gene expression and explain what type of control can more effectively affect the differentiation of stem cells into specialized cells.
The fate of a cell describes what it will become in the course of normal development differentiation result from differential gene expression. DNA except in the case of the immune system that 18 genetic information is not lost as cells become determined and begin to differentiate. any step of gene expression may be modulated from the DNA-RNA transcription step to post translational modification of a protein. gene expression is regulated the most extensively utilized point is transcription initiation chromatin domains. RNA or protein also contributes to the expression level of the gene an unstable product results in a low expression level when unspecialized stem cells give rise specialized cells the process is called differentiation. the interaction of signal during differentiation causes the cell's DNA to acquire epigenetic marks the restrict DNA expression in the cell and can be passed on through cell division
The Arg genes that encode the enzymes for arginine biosynthesis are located at several positions around the genome of E. coli, and they are regulated coordinately by a transcription regulator encoded by the ArgR gene. The activity of the ArgR protein is modulated by arginine. Upon binding arginine, ArgR alters its conformation, dramatically changing its affinity for the DNA sequences in the promoters of the genes for the arginine biosynthetic enzymes. Given that ArgR is a repressor protein, would you expect that ArgR would bind more tightly or less tightly to the DNA sequences when arginine is abundant? If ArgR functioned instead as an activator protein, would you expect the binding of arginine to increase or to decrease its affinity for its regulatory DNA sequences? Explain your answers.
The function of these Arg genes is to synthesize arginine. When arginine id abundant, expression of the biosynthetic genes should be turned off. If ArgR acts as a gene repressor, then binding of arginine should increase its affinity for its regulatory sites, allowed it to bind and shut off gene expression. If ArgR acted as a gene activator instead, then the binding of arginine would be predicted to reduce its affinity for its regulatory DNA, preventing its binding and thereby shutting off expression of the Arg genes.
Explain why the control of the expression of a master regulator is the most effective mechanism to regulate cell fate of specialized cells.
The master regulator regulate the functional properties and fates of a cell.The knowledge of master regulator and it's binding sites in a specific cell is very important to understand controlling the expression on a cell.master regulator allows better knowledge of a specific genome in a particular cell type,which is prone to undergo the effects of mutation.so for this reason the control of the expression of a master regulator is the most effective mechanism to regulate cell fate date of specialised cells.
Differentiated cells of an organism contain the same genes. (Among the few exceptions to this rule are the cells of the mammalian immune system, in which the formation of Figure Q8-6 amount of mRNA produced 50 60 70 80 90 100 110 number of nucleotides between enhancer and promoter ECB5 EQ8.06/Q8.06 Figure Q8-7 + C N C N C N C N C N C N repressor monomers repressor dimer DNA binding site cleavage site Figure Q8-9 ECB5 eQ8.09/Q8.09 enhancer biotin attached to one end of each DNA molecule β-globin gene enhancer β-globin gene + avidin transcription promoter 295 Figure Q8-12 ECB5 eQ8.12/Q8.12 transient signal transient signal OFF gene repressor OFF gene activator (A) CELL I (B) CELL II A R turns on transcription of repressor mRNA R repressor protein turns off its own transcription R R R turns on transcription of activator mRNA A activator protein turns on its own transcription A A A specialized cells is based on limited rearrangements of the genome.) Describe an experiment that substantiates the first sentence of this question, and explain why it does.
The most definitive result is one showing that a single differentiated cell taken from a specialized tissue can re-create a whole organism. This proves thatthe cell must contain all the information required to produce a whole organism, including all of its specialized celltypes. Experiments of this type are summarized in Figure 8-2.
When enhancers were initially found to influence transcription from many thousands of nucleotide pairs away from the promoters they control, two principal models were invoked to explain this action at a distance. In the "DNA looping" model, direct interactions between proteins bound at enhancers and promoters were proposed to stimulate transcription initiation. In the "scanning" or "entry-site" model, RNA polymerase (or another component of the transcription machinery) was proposed to bind at the enhancer and then scan along the DNA until it reached the promoter. These two models were tested using an enhancer on one piece of DNA and a β-globin gene and promoter on a separate piece of DNA (Figure Q8-9). The β-globin gene was not expressed when these two separate pieces of DNA were introduced together. However, when the two segments of DNA were joined via a linker (made of a protein that binds to a small molecule called biotin), the β-globin gene was expressed. Does this experiment distinguish between the DNA looping model and the scanning model? Explain your answer.
The results of this experiment favor DNA looping, which should not be affected by the protein bridge. The scanning or entry-site model, however, is predicted to be affected by the nature of the linkage between the enhancer and the promoter. If the proteins enter at the enhancer and scan to the promoter, they would have to transverse the protein linkage. If such proteins are geared to scan on DNA, they would likely have difficulty scanning across such a barrier.
Figure 8−17 shows a simple scheme by which three transcription regulators are used during development to create eight different cell types. How many cell types could you create, using the same rules, with four different transcription regulators? As described in the text, MyoD is a transcription regulator that by itself is sufficient to induce muscle-specific gene expression in fibroblasts. How does this observation fit the scheme in Figure 8−17?
a) The number of cell types that can be produced using the same rules with 4 different transcription regulators is 16. To calculate use formula 2n where 'n' is the number of regulators. b) MyoD is a protein which belongs to a family of proteins nown as Myogenic regulatory factors which play a major role in the muscle differentiation. MyoD enhances the transcription of p21 and myogenin and thereby removes cells from the cell cycle. As per the current scheme the MyoD fits in by converting fibroblasts into skeletal muscle cells.
