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GENETIC ANALYSIS: INTEGRATED - ACCESS
- This problem investigates issues encountered in sequencing the inserts in cDNA libraries.a. If you sequenced many clones individually, wouldn’tyou spend many of your resources inefficiently sequencing cDNAs for the same type of mRNA molecule over and over again? Explain. Does thisapparently inefficient process provide any useful information beyond the sequences of individual mRNAs?b. Suppose that you identified a clone with a cDNA insert that was 4 kb long. You could determine the entire sequence of the clone by shearing the DNA intosmall random fragments, cloning these fragments intoa vector to make a mini-shotgun library, and then sequencing hundreds of these clones to allow the computer to assemble the full sequence of the 4 kb–longinsert. However, this procedure would be inefficient.An alternative that requires many fewer sequencing reactions is called primer walking. This techniqueinvolves the synthesis of additional oligonucleotideprimers corresponding to cDNA sequences you…arrow_forwardAfter Drosophila DNA has been treated with a restriction enzyme, the fragments are inserted into plasmids and selected as clones in E. coli. With the use of this “shotgun” technique, every DNA sequence of Drosophila in a library can be recovered.a. How would you identify a clone that contains DNA encoding the protein actin, whose amino acid sequence is known?b. How would you identify a clone encoding a specific tRNA?arrow_forwardYou made four mutants for a promoter sequence in DNA and studied them for transcription. The results of the amount of gene expression or transcription (based on beta-Gal activity shown on Y-axis) for these DNAs (X-axis) are shown. The sequence of the wild-type and mutant DNAs, and consensus sequence from many promoters are shown here for your convenience. From this experiment you can conclude that: Nucleotide substitution can identify important bases of the binding sites or promoter in DNA (e.g., -10 and -35 promoter sequences of lac operon). True or false: Spacer (a) -10 region -35 region TTGACA Consensus sequence TATAAT Wild-type Lac promoter GGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGAATT Mutant 1 GGCTTTACACTTTATG-TTCCGGCTCGTATGTTGTGTGGAATT Mutant 2 GGCTTTACACTTTATGCTTCCGGCTCGTATAATGTGTGGAATT Mutant 3 GGCTTTACACTTTATG-TTCCGGCTCGTATAATGTGTGGAATT Mutant 4 GGCTTGACACTTTATG-TTCCGGCTCGTATAATGTGTGGAATT (b) 700 600- 500- 400- 300- 200- 100. 0 ● True O False B-Galactosidase activity Wild-type…arrow_forward
- Although the genetic code is universal, a few organisms such as Paramecium have a slightly modified version in which UGA, a stop codon for most organisms, codes for tryptophan in Paramecium. Suppose that the researcher wanted to make an in vitro translation system using all of the components from Paramecium. Which of the components, if any, would she need to replace in order to have an in vitro system that was universal? Possible Answers: A. She would need to leave out the P site. B. She would need to leave out the termination factor proteins. C. She would need to leave out the tRNA that recognizes UGA. D. She would need to leave out the ubiquitinarrow_forwardKnowing that the genetic code is almost universal, a scientist uses molecular biological methods to insert the human β-globin gene (Shown in Figure 17.11) into bacterial cells, hoping the cells will express it and synthesize functional β-globin protein. Instead, the protein produced is nonfunctional and is found to contain many fewer amino acids than does β-globin made by a eukaryotic cell. Explain why.arrow_forwardThe E. coli genome contains approximately 4639 kb. (a) How many copies of the 6-bp recognition sequence for the trp repressor would be expected to occur in the E. coli chromosome? (b) Explain why it is advantageous for the trp repressor to be a dimer that recognizes two adjacent 6-bp sequences.arrow_forward
- Mutagenesis is a technique in which genetic information of an organism is altered in a stable manner resulting in a mutation. It may occur spontaneously in nature of as a result of exposure to mutagens. It can also be achieved experimentally using optimized laboratory procedures. (i) (ii) What is site directed mutagenesis (SDM)? Explain how SDM can assist in the integration of a His-tag at the end of your gene of interest.arrow_forwardConsider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.arrow_forwardTay Sachs disease is an autosomal recessive disease in which a protein – Hex A - is abnormal. To make the Hex A protein: The promoter and transcription termination sites are 33,000 base pairs apart. The Hex A protein has 600 amino acids 5’ and 3’ UTR’s are each 500 bp long. a)How many base pairs would you expect in the final mRNA? Show your work b)How many bases were spliced out? Show your workarrow_forward
- The IMD2 promoter contains three upstream transcription start sites (TSS) that are utilized under high GTP conditions and a single downstream TSS (-106) that is normally only utilized under low GTP conditions. In a wild type cell, expression of IMD2 mRNA only occurs if transcription initiates from the -106 TSS. In 300 words or less, describe: 1.) The normal function of Ssl2, and 2.) why a mutation in Ssl2, that increases its catalytic rate, would allow expression of the IMD2 ORF under high GTP conditions. (Conditions under which the IMD2 ORF is NOT expressed in the wild type.)arrow_forwardWhich of the following set(s) of primers a-d could you use to amplify the following target DNA sequence, which is part of the last protein-coding exon of the CFTR gene? Explain briefly. (Note: The three dots represent the body of the region to be amplified, whose beginning and end are only being shown.) 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' АССCGTTATTАСАТСGCG 5 d. 5' GCTAAGATCTGAATTTTC 3'; 5' TGGGCAATAATGTAGCGC 3'arrow_forwardSuppose you want to study the transcription in vitro of one particular gene in a DNA molecule that contains several genes and promoters. Without adding specific regulatory proteins, how might you stimulate transcription from the gene of interest relative to the transcription of the other genes on your DNA template? To make all of the complexes identical, you would like to arrest all transcriptional events at the same position on the DNA template before isolating the complex. How might you do this?arrow_forward
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