Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Textbook Question
Chapter 28, Problem 18P
Functional 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
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Restriction 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
41 864
Please don't provide handwriting solution
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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- 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_forwardNumber 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_forwardHomologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesn’t the mismatch repair system of E. coli eliminate these mismatches?arrow_forward
- Multiple 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_forwardThe Enzymatic Activities of DNA Polymerase I (a) What are the respective roles of the 5 -exonudease and 3 -exonuclease activities of DNA polymerase I? (b) What might be a feature of an E. coli strain that lacked DNA polymerase I 3 -exonuclease activity?arrow_forwardMultiple 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_forward
- Human 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_forwardB PLEASE second onearrow_forwardPlease do allarrow_forward
- B Relates to the ribosome codon question please do ii ribosome coding questionsarrow_forward4a in context to taking genomic DNA from eukaryotic cells and randomly shearing it into pieces of a constant size, why do some of the genomic DNA fragments re-nature so much more quickly than other fragmentsarrow_forward34) What amino acid sequence is coded for by the following DNA coding strand? (Recall: the DNA template strand runs 5' to 3' and the mRNA strand runs antiparallel to the DNA template strand. Recall that DNA is translated with a start codon.) 5'-TTATGCGACCAGACCAGTTT-3' Coding strandarrow_forward
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