MATCH THE FF. WITH THE CHOICES BELOW TATA sequence CAAT nucleotide within core promoter 3 enhancers silencers heat shock element (HSE) the glucocorticoid response element (GRE) the metal response element (MRE) CHOICES -Response elements (REs) -Core promoter elements -Regulatory elements

MATCH THE FF. WITH THE CHOICES BELOW TATA sequence CAAT nucleotide within core promoter 3 enhancers silencers heat shock element (HSE) the glucocorticoid response element (GRE) the metal response element (MRE) CHOICES -Response elements (REs) -Core promoter elements -Regulatory elements

Name: MATCH THE FF. WITH THE CHOICES BELOWTATA sequenceCAATnucleotide within
Uploaded: 2024-03-20T06:42:41.000Z
Channel: Bartleby
Description: MATCH THE FF. WITH THE CHOICES BELOWTATA sequenceCAATnucleotide within core promoter 3enhancerssilencersheat shock element (HSE) the glucocorticoid response element (GRE)the metal response element (MRE)CHOICES -Response elements (REs)-Core promoter e

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter32: The Reception And Transmission Of Extracellular Information

Section: Chapter Questions

Problem 21P

See similar textbooks

Similar questions

expand this into 1000 words. the context is Identification of Differentially Expressed Genes in Breast Cancer Using RNA-Seq "In investigating the impact of TONSL overexpression on cellular physiology and gene regulation, differential expression analysis emerges as a pivotal analytical approach. By comparing gene expression profiles between primary cells and TONSL-overexpressing cells, researchers can identify genes whose expression levels are significantly altered. This analysis not only sheds light on the direct targets of TONSL but also unveils broader regulatory networks and pathways influenced by TONSL overexpression. Moreover, by pinpointing key genes involved, researchers can discern potential molecular mechanisms underlying the observed phenotypic changes associated with TONSL overexpression, offering valuable insights into TONSL's role in cellular homeostasis and function. Furthermore, functional enrichment analysis stands as a complementary approach to understanding the…
Match the term with its description. cAMP Allows lactose to enter the cell and product of the lacY gene A regulatory sequence between the promoter and the structural genes of the lac operon When glucose levels are high, intracellular levels of this molecule are low Binds to and inhibits the function of the lac repressor when lactose is present
ABOUT Phenylketonuria Explain Potential technical issues and limitations of PCR technology are mentioned Correct information about tissue that can be used to test for a genetic disease and justification of tissue selection Detailed information about the position (exact base pair number) of the new mutation relative to the sequence of the PAH gene. Numbering is based on the start of transcription of the PAH gene. PLEASE ANSWER ALLLL PLEASEE
Answer as Directed. Below is the model of a lac operon. lac I lac Z с promoter operator lac Y lac A DNA +1 1. What are structural genes? Are the lac structural genes transcribed in the absence of lactose? 2. What is the role of the promoter and operator sites in the operon? 3. Is the repressor protein bound to the operator site in the absence of lactose? In its absence? 4. Under what nutritional circumstances (high or low glucose) is CAP bound to cAMP? 5. In the absence of lactose and the presence of glucose in the bacterial growth media, what proteins are bound to the lac control region? Is the operon being transcribed then? 6. In the presence of lactose and the presence of glucose in the bacterial growth media, what proteins are bound to the lac regulatory region? Is the operon being transcribed then? 7. In the presence of lactose and the absence of glucose in the bacterial growth media, what proteins are bound to the lac control region? 8. Why is it adaptive for a bacterium to not…
Answer as Directed. Below is the model of a lac operon. lac I lac Z с promoter operator +1 lac Y lac A DNA 1. In the absence of lactose and the presence of glucose in the bacterial growth media, what proteins are bound to the lac control region? Is the operon being transcribed then? 2. In the presence of lactose and the presence of glucose in the bacterial growth media, what proteins are bound to the lac regulatory region? Is the operon being transcribed then? 3. In the presence of lactose and the absence of glucose in the bacterial growth media, what proteins are bound to the lac control region? 4. Why is it adaptive for a bacterium to not express the genes that encode for that lactose utilization proteins when lactose is not available or when glucose is present? 5. Why is it adaptive for the structural genes for using lactose to be under the control of a single promoter, i.e., synthesize a polycistronic message rather than three monocistronic message?
Match the following. Write correct answer in the form of 1a, 5b, etc. in the space provided. 1. Gene Replacement A. No active gene is present 2. Gene Knockout B. Only mutant gene is active 3. Gene Addition C. Both normal and mutant genes are active
Hi, can you please explain the clinical significance of G protein mutations.
. Mutations in an autosomal gene in humans cause aform of hemophilia called von Willebrand disease(vWD). This gene specifies a blood plasma proteincleverly called von Willebrand factor (vWF). vWFstabilizes factor VIII, a blood plasma protein specified by the wild-type hemophilia A gene. Factor VIIIis needed to form blood clots. Thus, factor VIII is rapidly destroyed in the absence of vWF.Which of the following might successfully be employed in the treatment of bleeding episodes in hemophiliac patients? Would the treatments workimmediately or only after some delay needed forprotein synthesis? Would the treatments have only ashort-term or a prolonged effect? Assume that allmutations are null (that is, the mutations result in thecomplete absence of the protein encoded by the gene)and that the plasma is cell-free.a. transfusion of plasma from normal blood into avWD patientb. transfusion of plasma from a vWD patient into adifferent vWD patientc. transfusion of plasma from a hemophilia A…
Retinoblastoma: The Hits Just Keep Coming case study Part III – The Second Hit Questions A second hit might occur through epigenetic alterations. In the promoter of RB1, there is a CpG island. Knowing this, how might you predict that a cell could epigenetically inactivate RB1 transcription? A second hit might also occur through loss of heterozygosity (LOH). An example of how LOH may occur by reciprocal crossing over during mitosis is diagrammed in Figure 1 (next page). Discuss and interpret this model with your group. Write a brief explanation of (a) what LOH means and (b) how LOH by mitotic reciprocal crossing over can give rise to a cell lineage with functional loss of the wild-type copy of a tumor suppressor gene. . One of the ways that we know what the RB protein does in cells is that its inactivation is a common priority of tumor-initiating viruses. What advantage would a virus gain by inactivating RB function in its host cell? One example of a DNA virus (a virus that…
Q1. Explain why GDP cannot dissociate from the alpha subunit of the Trimeric G-protein even though the receptor is activated by ligand binding and there is enough GTP for the exchange process.
Q38. The ΔF508 mutation of the CFTR gene is the cause of most cases of cystic fibrosis (CF). One focus of research into a cure for CF is "gene therapy" — inserting normal copies of the CFTR gene into the cells of CF patients. Which of the following areas of CF research is most likely to improve symptoms for CF patients? A. Inserting extra copies of genes that code for the proteosome. B. Studying the chaperone proteins of the endoplasmic reticulum (ER). C. Improving the lung function of people who are carriers of the CF allele (Cc). D. Stabilizing the plasma membrane were the ΔF508 form of the protein accumulates.
. An interesting mutation in lacI results in repressorswith 110-fold increased binding to both operator andnonoperator DNA. These repressors display a “reverse”induction curve, allowing β-galactosidase synthesis inthe absence of an inducer (IPTG) but partly repressingβ-galactosidase expression in the presence of IPTG. Howcan you explain this? (Note that, when IPTG binds a repressor, it does not completely destroy operator affinity,but rather it reduces affinity 110-fold. Additionally, ascells divide and new operators are generated by thesynthesis of daughter strands, the repressor must findthe new operators by searching along the DNA, rapidlybinding to nonoperator sequences and dissociating fromthem.)