Concept explainers
A
a. What length of DNA is bound by the transcriptional proteins? Explain how the gel results support this interpretation.
b. Draw a diagram of this DNA fragment bound by the transcriptional proteins, showing the approximate position of proteins along the fragment. Use the illustration style seen in Research Technique
c. Explain the role of DNase I.
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GENETIC ANALYSIS: AN INTEG. APP. W/MAS
- A recombinant protein corresponding to 52 repeats of the peptide sequence YSPTSPS was covalently linked to agarose beads to make an affinity column. The beads were incubated with a nuclear extract and the protein:beads became phosphorylated. More nuclear extract was run over the column repeatedly and then the beads were washed extensively. The bound proteins were eluted from the column and when they were separated on a SDS-PAGE gel many bands appeared. What cellular process(es) might these proteins be involved in? Options: splicing polyadenylation DNA replication a and b only a, b and carrow_forwardGive Detailed Solution with explanation (no need Handwritten)arrow_forwardThe cellulase gene of Bacillus licheniformis was successfully cloned into the pET21a vector and expressed in Escherichia coli BL21. The pET21a vector consists of ampicillin resistant gene. To screen for the successful transformants, E. coli BL21 was cultivated on LB agar containing ampicillin (100 pg/mL) and 0.5% (w/v) carboxymethylcellulose, and incubated at 37 0C for 6 hours. After that the agar plate was stained with Congo red solution for 15 minutes and washed twice with sodium chloride solution and the observation is as shown in Figure 3. Answer the following: (i.) Briefly explain why ampicillin was added to the LB agar. (ii) Indicate the function of carboxymethylcellulose in the LB agar. (iii) Conclude how the researchers were able to identify the E. coli BL21 that carried the cellulase gene.arrow_forward
- Expressing genes encoding a particular protein’s subunits from a plasmid in E. coli cells (volume of 1 cell = 0.5 um3) results in 1 gram of this protein when the E. coli cells reach a density of 2 x 109 cells/mL in a culture volume of 1.0L. Is this possible? Why or why not?.arrow_forwardplease solve this step by step with clear explainationarrow_forwardIn Northern blot analysis, mRNA samples from tissues are bound to a labeled DNA probe that is complementary to the mRNA, and run on a gel to be visualized. The protein tropomyosin is known to be present in both brain and liver. When brain and liver tissue were assayed for the presence of tropomyosin mature mRNA, bands of two different sizes were seen. Tropomyosin gene diagram (3000 bp total): Shown in attatched image If the band on the Northern blot for mRNA isolated from liver tissue is 2580 bp, whereas from brain tissue the band is 2250 bp, what is most likely? a)The two mRNAs are made from different tropomyosin DNA sequences. b)Exon 2 is alternatively spliced out of the brain mRNA. c)Introns 1 and 2 are spliced out of the brain transcript but not the liver transcript. d)Exons 1 and 3 are spliced out of the brain transcript but not the liver transcript. e)Exon 2 is alternatively spliced out of the liver mRNA.arrow_forward
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- Please express the following in detailarrow_forwardClone number in this case is number 196 which is shown in the images. State whether a BamHI site has been re-created at the forward- and the reverse-end junctions of the human DNA with the plasmid vector band sizes are shown in one of the images. (0.5, 1, 2, 3, 4, 5, 6, 8, 10kb)arrow_forwardThe transformation results below were obtained with 10 ul of intact plasmid DNA at nine concentrations. The following numbers of colonies are obtained when 100 ul of transformed cells are plated on selective medium: Fill in the following table: Concentration # colonies DNA mass of Fraction of Mass Transformation PGREEN (Concentration x volume OR X spread = x 10ul plasmid solution) PGREEN in cell Cell efficiency Y÷ A suspension suspension spread = 100 ul - total vol cell susp. (Colonies - Mass spread) C x Z = A See (510 ul) HINT: this calculation is constant Given= X Given=Y С. Z. 0.00001 ug/ul | 4 0.00005 ug/ul 12 0.0001 ug/ul 0.0005 ug/ul 32 125 0.001 ug/ul 442 0.005 µg/ul 0.01 ug/ul 0.05 ug/ul 0.1 ug/ul 542 507 475 516 0.5 ug/ul 505arrow_forward
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