Strength 22. You are analyzing a new human neuronal enhancer. You cloned the enhancer in front of a minimal promoter, transfected a neuronal cell line, and quantified GFP expression. You then performed the deletion analysis shown on the right. Which letter designates the minimal region where enhancer activity is located? min. Novel enhancer prom. GFP of expression +++ +/- +++ +/- +++ a. a b. b C. C +++ +/- d. d e. both b and c are needed
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- Enhancer RPA Gene A Use the diagram above, which depicts a chromosomal region less than 500,000 bp, to determine the expression of each gene under the scenarios specified in the table. Regulatory promoters are indicated by "RP". Genes A, B, and C have intermediate expression when no regulatory proteins are bound. Complete the table by choosing high, intermediate, or low to describe the expression level of each gene under each specified scenario (i and ii). No regulatory proteins bound (i) A repressor binds to RPA and an activator binds to RPC (ii) A regulatory protein binds at each of the activator, insulator, and silencer. Expression of A? Intermediate 1. [Select] 4. [Select] Insulator RPB Expression of B? 2. Gene B Intermediate 5. [Select] [Select] Silencer RPC Expression of C? 3. Gene C Intermediate 6. [Select] [Select]E30. An electrophoretic mobility shift assay can be used to study the binding of proteins to a segment of DNA. In the experiment shown here, an EMSA was used to examine the requirements for the bind- ing of RNA polymerase II (from eukaryotic cells) to the promoter of a protein-encoding gene. The assembly of general transcription factors and RNA polymerase II at the core promoter is described in Chapter 12 (Figure 12.14). In this experiment, the segment of DNA containing a promoter sequence was 1100 bp in length. The fragment was mixed with various combinations of proteins and then subjected to an EMSA. Lane 1: No proteins added Lane 2: TFID Lane 3: TFIIB Lane 4: RNA polymerase I| Lane 5: TFID + TFIIB Lane 6: TFID + RNA 1 2 4 5 6 polymerase II| Lane 7: TFIID + TFIIB + RNA polymerase I| 1100 bp Explain which proteins (TFIID, TFIIB, or RNA polymerase II) are able to bind to this DNA fragment by themselves. Which transcrip- tion factors (i.e., TFIID or TFIIB) are needed for the binding of…. What is an enhanceosome? Why could a mutation in anyone of the enhanceosome proteins severely reduce thetranscription rate?
- E27. A cloned gene fragment contains a regulatory element that is recog- nized by a regulatory transcription factor. Previous experiments have shown that the presence of a hormone results in transcriptional acti- vation by this transcription factor. To study this effect, you conduct a electrophoretic mobility shift assay and obtain the following results: Tube: 1 2 3 Transcription factor: Hormone: Explain the action of the hormone.a. How many enhancers were you able to identify with these set of experiments? Explain. b. If you find any enhancer, in what genetic region, number of base pairs upstream from MRPA, are they located? Explain.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?
- Methylation state of cytosines at the QRT gene promoter is tested in one pair of identical twins, Ana and Becca, at 1 year of age, again when they are 25 years old. At 1 year old, the twins' DNA is the same, and this part of the genome is not methylated. At age 25, Ana's DNA remains unmethylated, but Becca's cells show methylation at this site in skin cells. a. What most likely caused this change in Becca? i. A mutation event ii. An environmental stimulus b. When the twins were 1 year old, QRC mRNA was expressed at similar levels in both Ana and Becca. At age 25, which twin likely expresses more of the QRC mRNA? i. Ana, whose QRC promoter DNA is unmethylated ii. Becca, whose QRC promoter DNA is methylated iii. Both are the same C. At age 25, which twin has more tightly packed chromatin in the region of the QRC gene? Explain briefly. i. Ana, whose QRC promoter DNA is unmethylated ii. Becca, whose QRC promoter DNA is methylated iii. Both are the same d. Some of Becca's skin cells, in…draw the p21 promoter. Your drawing should include (1) the start site, (2) the TATA box and (3) the ERE/AP-1 binding sitePlz do explain.Thanks Question:- Many types of breast cancer have chromosomal translocation mutations. What scenario best describes, what occurs during this type of mutation, causing cells to proliferate abnormally? Chromosomal translocations may place the gene downstream near the promoter region, therefore causing over-expression of the gene The gene may be placed in the transcription start site, downstream of the gene, initiating transcription by recruiting polymerase II Translocation mutations will initiate the transcription of mRNA in the cytoplasm of the cell catalyzing protein synthesis. Chromosomal translocations can sometimes place a gene under the control of a powerful enhancer, upstream.
- protein. You create a mouse line with Cas9 under control of a brain-specific enhancer, while the short guide RNA complementary to the first exon of Gene Y is expressed in all tissues. You subsequently sequence Gene Y in both brain and liver tissue. What would expect in each tissue? You can assume that the CRISPRICas9 system will impact both copies of Gene Y in cells, and that the first exon of Gene Y is necessary for Gene Ys function. a. Liver: Functional Gene Y; Brain: Functional Gene Y b. Liver: Nonfunctional Gene Y; Brain: Funtional Gene Y c. Liver: Functional Gene Y; Brain: Nonfunctional Gene Y d. Liver: Nonfunctional Gene Y; Brain: Nonfunctional Gene Y. 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.)S-Cdk is likely to be a [Select] The inhibitor of Ras is likely to be a [Select] The MAP kinase is likely to be a [Select] Q Search 14:31 Protein X is a transcription factor that normally activates the expression of genes needed for S- phase. The activator of protein X is likely to be a [Select] is 4 ******* > 21**113 ***** ****** V f6 99+ 1- Il app.honorlock.com is sharing your screen. **** 1 a hp Stop sha