please asap 5) Interpreting Data: For the following question, consider the table and graph presented below.AV3 amino acid sequenceB10?7 p24 pg/mlmolecular clone296306317 320 324只日charge*SI phenotype**168.1CTRPNNNTRKSIHIGPGRAFYATGDIIGDIRQAHC3+168.10---T--Q---N-6+Mutagenesis set510168.1/RT--T----4+***168.1/RQ-R-5+104168.1/RN-N-5++168.1/RTQ-T--Q--T--5++168.1/RTN168.1/RQN-R103----5+-R--Q---N------6+1021014 68101214days after transfectionTable SEQ Table \* ARABIC 3 and Graph 2All the lines on the graph represent the effect of one molecular clone. What molecular clone does line Xdescribe? How did come to such a conclusion?Answer: Line X is Molecular clone is the negative control group. We came to this conclusion because thenegative control group is the only one with 3+ charge and a negative SI phenotype.

The Ziehl-Neelson (ZN) stain is the most commonly utilized of the different types of AF staining…

5) Interpreting Data: For the following question, consider the table and graph presented below. p24 pg/ml B10? A V3 amino acid sequence molecular clone 296 306 317 320 324 charge* SI phenotype** 10 168.1 CTRPNNNTRKSIHIGPGRAFYATGDIIGDIRQAHC 3+ 168.10 6+ Mutagenesis set 10° X. 168.1/RT 4+ * *** ...- 168.1/RQ --Q-- 10 5+ 168.1/RN -R- -N------ 5+ 168.1/RTQ 5+ 103 + .--- 168.1/RTN T---N------ 5+ + 168.1/RON -0---N------ 6+ 102 10 2 4 6 10 12 14 days after transfection Table SEQ Table * ARABIC 3 and Graph 2 All the lines on the graph represent the effect of one molecular clone. What molecular clone does line X lescribe? How did come to such a conclusion? Answer: Line X is Molecular clone is the negative control group. We came to this conclusion because the egative control groun is the only one with 3+ charợe and a negative SI phenotype

5) Interpreting Data: For the following question, consider the table and graph presented below. p24 pg/ml B10? A V3 amino acid sequence molecular clone 296 306 317 320 324 charge* SI phenotype** 10 168.1 CTRPNNNTRKSIHIGPGRAFYATGDIIGDIRQAHC 3+ 168.10 6+ Mutagenesis set 10° X. 168.1/RT 4+ * *** ...- 168.1/RQ --Q-- 10 5+ 168.1/RN -R- -N------ 5+ 168.1/RTQ 5+ 103 + .--- 168.1/RTN T---N------ 5+ + 168.1/RON -0---N------ 6+ 102 10 2 4 6 10 12 14 days after transfection Table SEQ Table * ARABIC 3 and Graph 2 All the lines on the graph represent the effect of one molecular clone. What molecular clone does line X lescribe? How did come to such a conclusion? Answer: Line X is Molecular clone is the negative control group. We came to this conclusion because the egative control groun is the only one with 3+ charợe and a negative SI phenotype

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter30: Protein Synthesis

Section: Chapter Questions

Problem 1P

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Similar questions

In a genome project, the following genomic DNA sequences were obtained. Assemble the sequences into a contig. Using the assembled sequence, perform a BLASTn search. Does the search produce sequences similar to your assembled sequence? 5’ TCGGGGTCCTGGGATCTCATCACTGCAGCGC 3’ 5’ACTGCAGCGCTTTCCCAGCGGGCGGTGGTAC 3’ 5’GGGCGGTGGTACTCGGGAAGTCAGGAGTGTT 3’ 5’AGGAGTGTTTAAAACCTGGGGACTGGTTTTG 3’ 5’TGGTTTTGGGGGCGCTGAAGGCAGCGCAGGA 3’
Using the formulae for dsDNA to calculate g/mol: You have a 4110 bp cloning vector Want to ligate a 245 bp insert for molecular cloning. efficient cloning requires that a 1:5 molar ratio of vector to insert is optimal for the ligation reaction. What are the relative weights of vector and the insert DNA necessary in g?
Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.
Which of the following set(s) of primers a–d couldyou use to amplify the following target DNA sequence, which is part of the last protein-coding exonof the CFTR gene?5′ GGCTAAGATCTGAATTTTCCGAG ... TTGGGCAATAATGTAGCGCCTT 3′3′ CCGATTCTAGACTTAAAAGGCTC ... AACCCGTTATTACATCGCGGAA 5′a. 5′ GGAAAATTCAGATCTTAG 3′;5′ TGGGCAATAATGTAGCGC 3′b. 5′ GCTAAGATCTGAATTTTC 3′;3′ ACCCGTTATTACATCGCG 5′c. 3′ GATTCTAGACTTAAAGGC 5′;3′ ACCCGTTATTACATCGCG 5′d. 5′ GCTAAGATCTGAATTTTC 3′;5′ TGGGCAATAATGTAGCGC 3′
Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many "reading frames" can be identified for this sequence? How many "open reading frames" can be identified for this sequence? What is the frame of the longest ORF?
Eukaryotic genomes are replete with repetitive sequences that make genome assembly from sequencereads difficult. For example, sequences such asCTCTCTCTCT . . . (tandem repeats of the dinucleotide sequence CT) are found at many chromosomallocations, with variable numbers (n) of the CT repeating unit at each location. Scientists can assemblegenomes despite these difficulties by using the pairedend sequencing strategy diagrammed in Fig. 9.9. Inother words, they can make libraries with genomicinserts of defined size, and then sequence both endsof individual clones. Following are 12 DNA sequence reads from sixcloned fragments analyzed in a genome project. 1Aand 1B represent the two end reads from clone 1, 2Aand 2B the two end reads from clone 2, etc. Clones1–4 were obtained from a library in which the genomic inserts are about 2 kb long, while the inserts inclones 5 and 6 are about 4 kb long. All of these sequences have their 5′ ends at the left and their 3′ endsat the right. To simplify…
The DNA sequence of one strand of a gene from threeindependently isolated mutants is given here (5′ endsare at left). Using this information, what is the sequence of the wild-type gene in this region?mutant 1 ACCGTAATCGACTGGTAAACTTTGCGCGmutant 2 ACCGTAGTCGACCGGTAAACTTTGCGCGmutant 3 ACCGTAGTCGACTGGTTAACTTTGCGCG
Utilizing Hind III and EcoR V Restriction Enzyme with Pet41 and the following gene of interest... a tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg aactgcaatt ggacggtttc 781…
Image 1. Which 2 primers from the choices provided would work to amplify the DNA sequence given below ? 5’ACTGAGTCCATGCGATCATGACTAT 3’ 3’TGACTCAGGTACGCTAGTACTGATA 5’ this is a hypothetical example. In a real experiment Choose 5’ TGAC 3’ 5’ CTAT 3’ 5’ ACTG 3’ 5’ ATAG 3’ Image2. the template strand?The results of a gel-based sequencing experiment are shown below. What is the sequence, written Only include nucleotides (no spaces or numbers )
3a)ClustalX software was used to perform multiple sequence alignment of thefollowing five Nco protein sequences designated as Nco1-Nco5 (the provided figure). A pair of degenerate primers was designed to PCR-amplify a DNA segment with the size of approximately 290 bp. With justification, discuss which amino acidsequence blocks would be suitable to design the forward and reverse degenerate primers.
Answer the following parts: A. When performing classical Sanger or "dideoxy" sequencing, you set up 4 parallel reactions per template to be sequenced from a specific primer, with each of the four reactions containing a different dideoxynucleotide, and then the four reactions were run in a separate, adjacent lanes on a gel. Why couldn't you combine all 4 dideoxynucleotides with the primer and the template and do the whole reaction in one tube, and then run the set of fragments produced by the reaction mixture on a single lane in an acrylamide gel? B. When doing automated sequencing, on the other hand, all 4 dideoxynucleotides are added to the same sequencing reaction, and run together in a single capillary gel. What's the difference - why can an automated sequencing reaction be done with all 4 dideoxynucleotides mixed together and all the resulting fragments run together, but not a conventional sequencing reaction?
Can you help with 1a please HELPFUL INFORMATION: When performing classical Sanger or "dideoxy" sequencing, you set up 4 parallel reactions per template to be sequenced from a specific primer, with each of the four reactions containing a different dideoxynucleotide, and then the four reactions were run in a separate, adjacent lanes on a gel. 1a. Why couldn't you combine all 4 dideoxynucleotides with the primer and the template and do the whole reaction in one tube, and then run the set of fragments produced by the reaction mixture on a single lane in an acrylamide gel?