Becker's World of the Cell (9th Edition)
Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 17, Problem 17.1PS

Meselson and Stahl Revisited. For each of the alternative models of DNA replication shown in Figure 17-3, use a sketch to indicate the distribution of DNA bands that Meselson and Stahl would have found in their cesium chloride gradients after one and two rounds of replication.

Chapter 17, Problem 17.1PS, Meselson and Stahl Revisited. For each of the alternative models of DNA replication shown in Figure

Figure 17-3 Alternative Models for DNA Replication. The results expected for two cycles of DNA replication are shown for each model, which were tested in the experiments of Meselson and Stahl.

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Recall the semi discontinuity in DNA replication. Is it biologically possible for DNA to undergo replication in vivo, without the lagging and the leading strands? In PCR, we also start with an antiparallel strand of DNA. Does this also mean that replication is semidiscontinuous in in vitro replication? PLEASE EXPLAIN COMPLETELY AND NOT JUST COPY PASTE ANSWERS FROM GOOGLE.
Calculating Transformation Efficiency Transformation Efficiency is defined as the number of colony forming units (cfu) which would be produced by transforming 1 ug of plasmid DNA into a given volume of competent cells. Transformation Efficiency (TE) = number of cfu/ug DNA You have just transformed 1 ul (100 pg/ul) of control pWasabi (plasmid) DNA into 50 µl of E. coli DH5a by the heat-shock method. You then add 950 µl of SOC medium to outgrow the bacteria. Of this, you plated 50 µl onto a LB plate containing ampicillin. After overnight incubation, you observed that there are 150 colony forming units (cfu) on the plate. Calculate the transformation efficiency (show your step-wise calculations below).
Which statements are true? Explain why or why not.1 The different cells in your body rarely havegenomes with the identical nucleotide sequence.2 In E. coli, where the replication fork travels at 500nucleotide pairs per second, the DNA ahead of the fork—in the absence of topoisomerase—would have to rotate atnearly 3000 revolutions per minute.3 In a replication bubble, the same parental DNAstrand serves as the template strand for leading-strandsynthesis in one replication fork and as the template forlagging-strand synthesis in the other fork.4 When bidirectional replication forks from adja-cent origins meet, a leading strand always runs into a lag-ging strand.5 DNA repair mechanisms all depend on the exis-tence of two copies of the genetic information, one in eachof the two homologous chromosomes
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