Thymine dimer 5' 3' 3' The UvrA/UvrB complex tracks along the DNA In search of damaged DNA. 3' 3' 5' After damage is detected, UvrA Is released, and Uvr C binds. 5' Uvr C Uvr C makes cuts on both sides of the thymine dimer. Cut Cut 3' UvrC UvrD, which is a helicase (not shown). removes the damaged reglon. UvrB and UvrCare also released. 5° 3' டட 3' 5' DNA polymerase fills in the gap. and DNA ligase seals the region. No thymine dimer 5' 3' 5' FIGURE 19.18 Nudeotide excision repair in E. coli.

Thymine dimer 5' 3' 3' The UvrA/UvrB complex tracks along the DNA In search of damaged DNA. 3' 3' 5' After damage is detected, UvrA Is released, and Uvr C binds. 5' Uvr C Uvr C makes cuts on both sides of the thymine dimer. Cut Cut 3' UvrC UvrD, which is a helicase (not shown). removes the damaged reglon. UvrB and UvrCare also released. 5° 3' டட 3' 5' DNA polymerase fills in the gap. and DNA ligase seals the region. No thymine dimer 5' 3' 5' FIGURE 19.18 Nudeotide excision repair in E. coli.

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 16P: Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins,...

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5' - ATG GGG CCC GTT TTC AAT ATG CAG GTC CAT CCG TAC GTA CAG GCC GGA ATT TGA - 3' There are two introns in this DNA sequence. Remember introns start with GT and end with AG. (a) How many base pairs are in intron 1 and intron 2.
5'-GCGGTACGTTACGGCTTTACTGACCTGCAGGC-3' A. Convert this to a double strand DNA molecule by writing the complementary sequence. Be sure to correctly label the ends of the double stranded molecule B. Pick one of the two strands and write the sequence of an RNA molecule that could be transcribed from it. Again, you must correctly label the ends of the molecule.
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FIGURE 19.16 Direct repair of dam- aged bases in DNA. (a) The repair of thymine dimers by photolyase. (b) The repair of methylguanine by the transfer of the methyl group to Thymine dimer CH3 SH 06-Methylguanine CH2 ONLINE ANIMATION Н. CH3 Нас. N- н- Н Н N. `NH2 Alkyltransferase alkyltransferase. CONCEPT CHECK: Which of these DNA backbone Alkyltransferase catalyzes the removal of the methyl group onto itself. repair systems is particularly valuable to plants? DNA DNA photolyase cleaves the 2 backbone bonds between the thymine dimer. CH3 Guanine На 'N' CHз Нас. -- н CH2 н Н `NH2 The normal structure of the 2 thymines is restored. The normal structure of guanine is restored. (a) Direct repair of a thymine dimer (b) Direct repair of a methylated base O=
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.
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.
Recombination in Immunoglobulin Genes If recombination between a Vkand Jkgene formed a CCA codon at codon 95 (Figure 28.41), which amino acid would appear at this position?
Heteroduplex DNA Formation in Recombination From the information in Figures 28.17 and 28.18, diagram the recombinational event leading to the formation of a heteroduplex DNA region within a bacteriophage chromosome.
Molecules 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?
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?
Using Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw the complementary dinucleotide in an antiparallel fashion. Connect the dinucleotides with the appropriate hydrogen bonds. FIGURE 8.9 The two polynucleotide chains in DNA run in opposite directions. The left strand runs 5 to 3, and the right strand runs 3 to 5. The base sequences in each strand are complementary. An A in one strand pairs with a T in the other strand, and a C in one strand is paired with a G in the opposite strand. FIGURE 8.7 Nucleotides can be joined together to form chains caled polynucleotides. Polynucleotides are polar molecules with a 5 end (at the phosphate group) and a 3 end (at the sugar group). An RNA polynucleotide is shown at the left, and a DNA polynucleotide is shown at the right.
On the 4 diagrams below show the four different ways (which strands are cut and where for each Holliday junction) that the Holliday junction structures can be resolved and denote whether this leads to a recombination event or not." 3 1 IT |? 3 ||| |AL||T|||c| |||| 4 5 6 TIT!!T?L|| ?|||?| |||| TIT|| |AL||T|||C| |||| 4 5 6 3 3' 3 T?L|| ?| ||?| |!!! || |A|||T| |jcj ||| 4 5 6 TI T?L|| ?|||?T ||| jT|| |ALİ|Ti||c| ||| 4 5 6