Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 29, Problem 12P
Interpretation Introduction
(a)
Interpretation:
Auto radiographic pattern of unidirectional replication is to be sketched.
Concept introduction:
Replication is the process of DNA to copy itself with the help of different enzymes. Replication starts from a single point known as origin of replication. If from the starting point it proceeds in one direction, it will be known as unidirectional. If proceeds in both directions known as bidirectional replication.
Interpretation Introduction
(b)
Interpretation:
Auto radiographic pattern of bidirectional replication is to be sketched.
Concept introduction:
Replication is the process of DNA to copy itself with the help of different enzymes. Replication starts from a single point known as origin of replication. If from the starting point it proceeds in one direction it will be known as unidirectional and if proceeds in both directions known as bidirectional replication.
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Fill in the blank. As helicase unwinds closed circular DNA, it compensates for positive supercoiling. The stress of this force must be overcome or the replication machinery will be unable to proceed. DNA gyrase, a bacterial __________, is the leading candidate for this role in coli.
Pstl.
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pBR322
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Based on the recombinant plasmid growth pattern (bottom row of blue table), which of
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you can tell.
Draw a replication bubble. Be sure to label the directionality of all strands of DNA. For one of the two replication forks, draw and label all of the proteins required the text describes as being important for DNA synthesis, and label the leading and lagging strands.
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- Close contact. Examination of the structure of DNA polymerases bound to nucleotide analogs reveals that conserved residues come within van der Waals contact of C-2'C-2' of the bound nucleotide. What is the potential significance of this interaction?arrow_forwardComprehensine. mfometion regordng boctenal replication Should be provided..arrow_forwardSemiconservative or Conservative DNA Replication If 15N-Iabeled E. coli DNA has a density of 1.724 g/mL, 14N-labeled DNA has a density of 1.710 g/mL, and E. coli cells grown for many generations on 14NH4+as a nitrogen source are transferred to media containing 15NH4+as the sole N-source, (a) What will be the density of the DNA after one generation, assuming replication is semiconservative? (b) Suppose replication took place by a conservative mechanism in which the parental strands remained together and the two progeny strands were paired. Design an experiment that could distinguish between semiconservative and conservative modes of replication.arrow_forward
- Multiple Replication Forks in E. coli II On the basis of Figure 28.2, draw a simple diagram illustrating replication of the circular E. coli chromosome (a) at an early stage, (b) when one-third completed, (c) when two-thirds completed, and (d) when almost finished, assuming the initiation of replication at oriC has occurred only once. Then, draw a diagram showing the E. coli chromosome in problem 3 where the E. coli cell is dividing every 20 minutes.arrow_forwardTrue or False. Do single-stranded DNA strands stabilize other strands?arrow_forwardPen and Paper Exercise. A new virus was causing a localized epidemic in South Africa, affecting the population living along the Limpopo River Basin. Scientists working on the elucidation of more details about the virus have finally characterized it as a DNA virus (containing DNA as a genetic material and not RNA). Upon sequencing, a short DNA segment, speculated to be the structural gene of the viral genome responsible for viral replication, is shown with the following sequence: DNA sequence: 5'-CTACACTITATCGTTAATCTGCCAGAAGCACCCCTCCAGGTGTCTACTGTATGGTGTCGCCAT -3' A. "Transcribe" and form an RNA transcript based on the DNA sequence. B. If the RNA transcript formed is a messenger RNA, illustrate the amino acid sequence of the resulting polypeptide using the genetic code. *Please follow the proper directionality of both the RNA strand and the polypeptide. Indicate clearly and label correctly the ends of the molecules mentioned. C. What is/are the characteristic feature/s of the resulting…arrow_forward
- Need answer ASAP. The double-stranded molecule of DNA: AG/TC is a very good representation of the much larger chromosome. Diagram and label this molecule, then describe how would this molecule look if it were a DNA/RNA hybrid using the AG for the DNA strand.arrow_forwardPick a plasmid . What was its approximate transformation? Express it in # colonies per microgram of DNA transformed. Assume the original DNA was about .001 ug/ul . Count how many colonies you got on one plate (or estimate that number) and figure out how much of the total solution you plated on that plate. Multiply by all the plates, if you plated all of it. OR, if you only plated some of it, figure out how many colonies you would have gotten had you plated all of it. Divide by the number of ug used.arrow_forwardQuestion. What would the forward primer sequence look like if it were intended to bind the area of the DNA template?arrow_forward
- please help me with this question. As this is a non-directional cloning, recombinant plasmids can contain an insert ligated into the vector in two different orientations. Provide two diagrams to illustrate the two potential recombinant plasmids, with the inserts ligated in opposite orientations. Include all RE sites and distances between sites on the diagram.arrow_forwardgene. If the JM109 strain is transformed by the PBKSK plasmid, the strain will produce the B-galactosidase (from the lac gene) and will hydrolyze X-gal to produce the blue compound. Therefore, colonies that were transformed and contain the pBSKS wil you appear blue. IPTG & X-Gal & NO colonies Amp E. coli JM109 E. coli JM109 50 mM calcium chloride-15% glycerol lac lac lac IPTG & I Recovery X-Gal solution at -702C PBSKS White colonies E. coli JM109 E. coli JM109 ampR amp I amp lac lac Heat Shock Non-transformed 42°C E. coli JM109 E. coli JM109 amps amps lac lac IPTG & X-Gal lac I Recovery lac PBSKS BLUE colonies PBSKS ampRI (amp Transformed IPTG & X-Gal & BLUE colonies Amp Hypotheses: Circle the correct answer 1. If PBSKS is transformed into JM109 cells, colonies will be (able/not able) to grow in the presence of ampicillin. a. Why? _ 2. If PBSKS is transformed into JM109 cells, colonies in media with IPTG (will/will not) induce the production the B- galactosidase enzyme. a. Why?_ 3. If…arrow_forwardWhich 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 chromosomesarrow_forward
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