Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 12P
Interpretation Introduction
Interpretation:
To draw a holliday junction between two duplex DNAs and to show how patch or splice recombinant DNA molecules are obtained.
Concept introduction:
The nuclei acid structure which is branched and has four numbers of double stranded arms which are joined together is a holliday junction.
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Chapter 28 Solutions
Biochemistry
Ch. 28 - Semiconservative or Conservative DNA Replication...Ch. 28 - The Enzymatic Activities of DNA Polymerase I (a)...Ch. 28 - Multiple Replication Forks in E. coli I Assuming...Ch. 28 - Multiple Replication Forks in E. coli II On the...Ch. 28 - Molecules of DNA Polymerase III per Cell vs....Ch. 28 - Number of Okazaki Fragments in E. coli and Human...Ch. 28 - The Roles of Helicases and Gyrases How do DNA...Ch. 28 - Human Genome Replication Rate Assume DNA...Ch. 28 - Heteroduplex DNA Formation in Recombination From...Ch. 28 - Homologous Recombination, Heteroduplex DNA, and...
Ch. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Chemical Mutagenesis of DNA Bases Show the...Ch. 28 - Prob. 14PCh. 28 - Recombination in Immunoglobulin Genes If...Ch. 28 - Helicase Unwinding of the E. coli Chromosome...Ch. 28 - Prob. 17PCh. 28 - Functional Consequences of Y-Family DNA Polymerase...Ch. 28 - Figure 28.11 depicts the eukaryotic cell cycle....Ch. 28 - Figure 28.41 gives some examples of recombination...Ch. 28 - Prob. 21PCh. 28 - Prob. 22P
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- Number of Okazaki Fragments in E. coli and Human DNA Replication Approximately how many Okazaki fragments are synthesized in the course of replicating an E. coli chromosome? How many in replicating an “average� human chromosome?arrow_forwardMolecules of DNA Polymerase III per Cell vs. Growth Rate It is estimated that there are 40 molecules of DNA polymerase III per E. coli cell, is it likely that the growth rate of E. coli is limited by DNA polymerase III availability?arrow_forwardFunctional Consequences of Y-Family DNA Polymerase Structure The eukaryotic translesion DNA polymerases fall into the Y family of DNA polymerases. Structural studies reveal that their fingers and thumb domains are small and stubby (see Figure 28.10). In addition, Y-family polymerase active sites are more open and less constrained where base pairing leads to selection of a dNTP substrate for the polymerase reaction. Discuss the relevance of these structural differences. Would you expect Y-family polymerases to have 3-exonuclease activity? Explain your answer.arrow_forward
- Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.arrow_forwardMultiple Replication Forks in E. coli I Assuming DNA replication proceeds at a rate of 750 base pairs per second, calculate how long it will take to replicate the entire E. coli genome. Under optimal conditions, E. coli cells divide every 20 minutes. What is the minimal number of replication forks per E. coli chromosome in order to sustain such a rate of cell division?arrow_forwardHuman Genome Replication Rate Assume DNA replication proceeds at a rate of 100 base pairs per second in human cells and origins of replication occur every 300 kbp. Assume also that human DNA polymerases are highly processive and only two molecules of DNA polymerase arc needed per replication fork. How long would it take to replicate the entire diploid human genome? How many molecules of DNA polymerase does each cell need to carry out this task?arrow_forward
- B PLEASE second onearrow_forwardRestriction sites of Lambda (A) DNA - In base pairs (bp) The sites at which each of the 3 different enzymes will cut the same strand of lambda DNA are shown in the maps (see figure 3 B-D), each vertical line on the map is where the respective enzymes will cut. A DNA A (bp) 48502 10 000 20 000 30 000 40 000 9162 17 198 B Sal I 7059 14 885 28 338 35 603 42 900 (bp) Hae III 11 826 21 935 29 341 38 016 (bp) 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864 Figure 3: Restrictrion site map showing the following A) inear DNA that is not cut as reference B) DNA CLt with Sal L C) DNA cut with Hae , D) DNA cut with Eco RI 1. Calculate the size of the resulting fragments as they will occur after digestion and write the sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (see figure 3A). Page 3 of 7 9162 17 198 Sal i (bp) 7059 14 885 28 338 35 603 42 900 Hae I (bp) 11 826 21 935 29 341 38 016 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864arrow_forwardnand portable. 6.5 mm micro-edge 16405U processon USE YOUR SMARTPHONE w micro-edge display design See dnclamers on product bo Revi Vide Fea Sameness and Variety (Mitosis and Meiosis) 161 Spe Example Sup SCAN Expressed using apples, DNA replication looks like this. Go look at the model of DNA on the demonstration table. We have been presenting DNA as a straight ladder, but actually it is twisted on itself like a spiral staircase. This shape is called a heliz. Observe The Chromosome Normally DNA exists as loose strands (chromatin) in the nucleus of a cell. This nuclear DNA sends a message (RNA) to the ribosomes where protein and enzymes are synthesized. When stretched out, the length of one DNA molecule in a human cell is almost 4 cm. However, the cell itself is but a tiny fraction of that size. During cell reproduction the DNA must be able to move around. So it shortens its length by tightly coiling up. In doing so, the DNA strands become wider and are visible under a microscope.…arrow_forward
- Explain.arrow_forwardApplication/ Analysis Explain how the anti-parallel structure of DNA predicts its replication mechanism. Identify the major and minor groove of DNA and explain why they are there. Differentiate between semiconservative, conservative, and dispersive replication. Interpret a diagram of a bi-directional replication fork and correctly determine strand polarity and fork direction.arrow_forwardTime (h) 0 1 3 6 12 Mouse thymus cells were treated with 10 micromolar dexamethasone to induce apoptosis. At the indicated times the chromosomal DNA was extracted and analyzed by gel electrophoresis, which separates the DNA fragments based upon size differences. When cells undergo apoptosis their DNA is cut into fragments that are approximately multiples of 200 base pairs. This results in a characteristic “ladder" appearance when the fragments are separated by gel electrophoresis, as shown above. Based upon what you know about eukaryotic chromatin, explain why this pattern of DNA fragments is obtained. 6.arrow_forward
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