15. In Problem 14, cDNAs F and G could not be found in cDNA libraries (from any
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15. In Problem 14, cDNAs F and G could not be found in cDNA libraries (from any tissue prepared sing the method shown in Fig. 10.4. The reason that the corresponding transcripts do not have poly-A tails.
a. why is the lack of poly-A tails ot surprising in light of your answer to part (d) of Problem 14?
b. why does the lack of poly-A tails present a differiculty for the method diagrammed in Fig 10.4?
Genetics From Genes to Genomes (6th Edition) Chapter 10
FIG. 10.4
(a) Obtain mRNA from red blood cell precursors. (b) Create a hybrid cDNA-mRNA molecule using reverse transcript and oligo-dT drimer. (c) Heat the miture to separate mRNA and cDNA strands, and then eliminate the mRNA transcript. The 3' ends of the cDNA strands bind by chance to complementary nuclotides within the same strand, forming a hairpin loop that an prime DNA polymerization. (d) Create a second cDNA strand complementary to the first. The enzyme S1 nuclease is used to cleave the hairpin loop. (e) Insert the newly created double-stranded DNA molecule into a vector for cloning.
blood cell precursors oloase mRNA trom cytoplasm and pur Figure 10.4 Converting RNA transcripts to cDNA. (a) otain mRNA from red blood cell precursors. (b) Create a hybrid CONA mRNA molecule using reverse transcriptase and oligo-al primer (c) Heat the mixture to separate mRNA and cDNA strands, and then eliminate the mRNA transcript The 3 ends of the CONA stands bind by chance to complementary nucleotides within the same strand, forming a hairpin loop that can prime DNA polymerization. (d) Create a second cDNA strand complementary to the first. The entyme 51 nuclease used to cleave the hairpin loop. (e) Insert the newly created double stranded DNA molecule into a vector for cloning IIIIIIIIIIAAAANA TITTLITLAAAARNA IIIIIIIIIIIIAAAARNA S Add oligo-dt prin presence of dat AT primer. Treat with reverse transcriptase in ace of dATP, dCTP, dGTP, and dTTP. Primer in mRNA in ond 5 Growing CDNA AAAAL 3. mRNA Reverse transcriptase ini ino CDNA mRNA iel Denature cDNA-mRNA hybrids ridigast mRNA with RNese. 3' end of cDNA folds back on itself as acts as primer. TITETTTTTTT cDNA Next, because mRNAs constitute only a small fraction of all the RNAs in the cell (1-5% depending on the cell type), it would be desirable to separate the mRNAs from the much more abundant rRNAs and tRNAs. This goal is possible be- cause mRNAs in eukaryotic cells have poly-A tails at their 3' ends. mRNAs will hybridize through their poly-A lails to the oligo-dT (single-stranded fragments of DNA containing about 20 Ts in a row). mRNA will thus bind to magnetic beads linked to oligo-dT, while other kinds of RNA will not This interaction provides the basis for a separation technique (not shown) that will allow you to obtain a purified prepara- tion of mRNA. The preparation will contain all of the mRNAs that are expressed in red blood cell precursors (Fig. 10.4a). The addition of reverse transcriptase to this total mRNA—as well as ample amounts of the four deoxyribo- nucleotide triphosphates and primers to initiate synthesis- generates single-stranded cDNA bound to the mRNA template (Fig. 10.4b). The primers used in this reaction are also oligo-dT so as to initiate polymerization of the first CDNA strand from the 3' ends of all mRNAs. After synthesis is finished, you can denature (separate) the mRNA-CDNA hybrids into single strands by heating the hybrids to high temperature. The addition of an RNase enzyme that digests the original RNA strands leaves intact single strands of CDNA (Fig. 10.4c). Most of these fold back on themselves at their 3' ends to form transient hairpin loops that serve as primers for synthesis of the second DNA strand. Now the addition of DNA polymerase, in the presence of the requi- site deoxyribonucleotide triphosphates, initiates the pro- duction of a second cDNA strand from the just-synthesized single-stranded cDNA template (Fig. 10.4d). The products are double-stranded cDNA molecules. After using restriction enzymes and ligase to insert the double-stranded cDNA into a suitable vector (Fig. 10.4e) and then transforming the vector-insert recombinants into appro- priate host cells, you would have a library of double-stranded cDNA fragments. The cDNA fragment in each individual clone will correspond to an mRNA molecule in the red blood cell precursors that served as your sample. It is important to note that this cDNA library includes only the exons from that part of the genome that these cells were actively tran- scribing for translation into protein. The clones in cDNA libraries do not contain introns because the mature mRNAs from which they were produced do not have introns. You 10 The first cDNA strand acts as a template for synthesis of the second cDNA strand in the presence of the four deoxynucleotides and DNA polymerase. Growing second strand Si nuclease DNA cuts hairpin loop. polymerase 3' cDNA TTT double double 5: helix le) Insert cDNA into vector.Explanation / Answer
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