Eukaryotic chromosomes are linear DNA molecules, yet the DNA of a chromosome retains a high level of underwinding (supercoiling) throughout its length. How does eukaryotic chromosomal DNA maintain its supercoiling? Nonhistone proteins link the free ends of linear chromosomes, and negative supercoils form in response to the helical stress. A type II topoisomerase uses energy from ATP to introduce negative supercoils in the DNA. A negative and positive supercoil form when the DNA wraps around a histone protein. Topoisomerases relax the positive supercoil, leaving a net negative supercoil. Large segments of DNA are organized in loops, and protein binding at the base of the loop restricts free DNA rotation.

Eukaryotic chromosomes are linear DNA molecules, yet the DNA of a chromosome retains a high level of underwinding (supercoiling) throughout its length. How does eukaryotic chromosomal DNA maintain its supercoiling? Nonhistone proteins link the free ends of linear chromosomes, and negative supercoils form in response to the helical stress. A type II topoisomerase uses energy from ATP to introduce negative supercoils in the DNA. A negative and positive supercoil form when the DNA wraps around a histone protein. Topoisomerases relax the positive supercoil, leaving a net negative supercoil. Large segments of DNA are organized in loops, and protein binding at the base of the loop restricts free DNA rotation.

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Most prokaryotes have a circular chromosome, with no ends, so the shortening of DNA does not occur. But what protects the genes of linear eukaryotic chromosomes from being eroded away during successive rounds of DNA replication?
What distinguishes topoisomerase type I from topoisomerase type II (“gyrase”) in E. coli bacteria? Both types of topoisomerases participate in the formation of nucleosomes. Topoisomerase type I cleaves one strand of a DNA duplex to remove DNA supercoils, and topoisomerase type II cleaves both strands of the DNA duplex when introducing negative supercoils into the molecule. Topoisomerase type I works to replicate DNA and topoisomerase type II works to transcribe RNA. Topoisomerase type I works on double-stranded DNA, and topoisomerase type II ("gyrase") works on single-stranded DNA. Topoisomerase type I adds negative DNA supercoils, and topoisomerase type II removes them.
Supercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?
PolyADP-ribose polymerase (PARP) plays a keyrole in the repair of DNA single-strand breaks. In the pres-ence of the PARP inhibitor olaparib, single-strand breaksaccumulate. When a replication fork encounters a sin-gle-strand break, it converts it to a double-strand break,which in normal cells is then repaired by homologousrecombination. In cells defective for homologous recom-bination, however, inhibition of PARP triggers cell death.Patients who have only one functional copy of theBrca1 gene, which is required for homologous recombina-tion, are at much higher risk for cancer of the breast andovary. Cancers that arise in these tissues in these patientscan be treated successfully with olaparib. Explain how it isthat treatment with olaparib kills the cancer cells in thesepatients, but does not harm their normal cells.
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DNA polymerase occasionally incorporates the wrong nucleotide during DNA replication. If left unrepaired, the base-pair mismatch that results will lead to mutation in the next replication. As part of a template strand, the incorporated wrong base will direct the incorporation of a base complementary to itself, so the bases on both strands of the DNA at that position will now be different from what they were before the mismatch event. The MER-minus strain of yeast does not have a functional mismatch excision repair system, but it has normal base excision repair and nucleotide excision repair systems. Which of the following statements is correct about differences in the mutation spectrum between MER-minus and wildtype yeast? More than one answer is correct. Options: More point mutations will arise in MER-minus yeast. Fewer point mutations will arise in MER-minus yeast as compared with wildtype. Of the total point mutations that…
What enzymatic features of DNA polymerase prevent it from replicating one of the DNA strands at the ends of linear chromosomes? Compared with DNA polymerase, how is telomerase different in its ability to synthesize a DNA strand? What does telomerase use as its template for the synthesis of a DNA strand? How does the use of this template result in a telomere sequence that is tandemly repetitive?
As shown, telomerase attaches additional DNA, six nucleotides at a time, to the ends of eukaryotic chromosomes. However, it makes only one DNA strand. Describe how the opposite strand is replicated.
Human Fbh1 helicase is important in the process of DNA replication. When a mutation occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? A - Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triangle because helicase is needed to separate the strands of DNA. B - Topoisomerase would unwind the DNA, but then the process would stop at the blue triangle because helicase, the RNA primer, would not be able to attach to the DNA molecule and initiate replication. C - The process would begin at the blue triangle when topoisomerase unwinds the DNA and an RNA primer attaches to the DNA molecule and initiates replication. DNA polymerase would begin the synthesis…
Eukaryotic licensing factors prevent DNA replication from being initiated at origins more than once in the cell cycle. After replication has begun at an origin, a protein called Geminin inhibits licensing factors that are required for MCM2-7 to bind to an origin and initiate replication. Thus, when Geminin is present, MCM2-7 will not bind to an origin. At the end of mitosis, Geminin is degraded, allowing MCM2-7 to bind once again to DNA and relicense the origin. Marina Melixetian and her colleagues suppressed the expression of Geminin protein in human cells by treating the cells with small interfering RNAs (siRNAs) complementary to Geminin messenger RNA (M. Melixetian et al. 2004. Journal of Cell Biology 165:473–482). (Small interfering RNAs form a complex with proteins and pair with complementary sequences on mRNAs; the complex then cleaves the mRNA, so there is no translation of the mRNA; . Forty-eight hours after treatment with siRNA, the Geminin-depleted cells were enlarged and…
The regulation of replication is essential to genomic stability. Normally, the DNA is replicated just once in every eukaryotic cell cycle (in the S phase). Normal cells produce protein A, which increases in concentration in the S phase. In cells that have a mutated copy of the gene encoding protein A, the protein is not functional, and replication takes place continuously throughout the cell cycle, with the result that cells may have 50 times the normal amount of DNA. Protein B is normally present in G1, but disappears from the cell nucleus during the S phase. In cells with a mutated copy of the gene encoding protein A, the levels of protein B fail to drop in the S phase and, instead, remain high throughout the cell cycle. When the gene for protein B is mutated, noreplication takes place. Propose a mechanism for how protein A and protein B might normally regulate replication so that each cell gets the proper amount of DNA. Explain how mutation of these genes produces the effects just…