Chapter 7 and 10.6

Transformation

(1) The bacterial process of gene transfer in which donated DNA fragments originating in a dead donor cell, or plasmid DNA, are taken up across the cell wall and membrane of a recipient cell and recombined into the transformant genome. (2) More generally refers to the process by which exogenous DNA is directly taken up by a cell resulting in a genetic alteration of the cell. (3) The conversion of animal cells to an abnormal unregulated state by an oncogenic virus or by transforming DNA.

Primase

(DnaG) The specialized RNA polymerase that synthesizes the RNA primer during DNA replication

• Why does replication occur continuously on one strand and semidiscontinuously on the other? Maybe better answer

1 DNA pol III synthesizes 1 daughter strand continuously in the same direction as fork progression: leading strand • 1 DNA pol III elongates the other daughter strand discontinuously, in the opposing direction to fork progression, via short segments (Okazaki fragments): lagging strand

• How did the work of Chargaff and Franklin and Wilkins contribute to Watson and Crick's model of DNA structure? *******CHECK*******

A key observation made from Franklin's research was the recognition of two slightly different forms of DNA. These were designated A-form DNA and B-form DNA. B-form DNA was much more common than A-form DNA, and it is now known to predominate in all organisms. A third type of DNA is also known, as we describe at the end of this section. The molecular dimensions of DNA are measured using the unit called an angstrom (Å) or in nanometers (nm). One angstrom is equal to 10−1010−10 meters, or 1 ten-billionth of a meter, and 1 nm equals one-billionth of a meter, or 10−910−9 meters. In B-form DNA, the distance from the axis of symmetry to the outer edge of either sugar-phosphate backbone is 10 Å (1 nm), and the molecular diameter is 20 Å (2 nm) at any point along the length of the helix (Figure 7.7a). The 20-Å molecular symmetry of the double helix was the key observation that told Watson and Crick that DNA structure results from pairing of a purine (A or G) with its complementary pyrimidine (T or C). The purine-pyrimidine base-pair pattern gives each base pair the same dimension. A second key observation derived from Franklin's Photo 51 is that nucleotide base pairs are spaced at intervals of 3.4 Å along DNA duplexes. This tight packing of DNA bases in the duplex leads to base stacking, the slight rotation of adjacent base pairs around the axis of symmetry so that their planes are parallel, imparting a twist to the double helix. Figure 7.7a shows that one complete helical turn spans 34 Å. This span is occupied by approximately 10.5 base pairs. Figure 7.7b is a space-filling model that illustrates base-pair stacking and the twisting of the sugar-phosphate backbones. Figure 7.7c is a ball-and-stick model illustrating how base pairs twist around the axis of symmetry to create the helical spiral. Base-pair stacking creates two grooves in the double helix, gaps between the spiraling sugar-phosphate backbones that partially expose the nucleotides. The alternating grooves, known as the major groove and minor groove, are highlighted in Figures 7.7b and 7.7c. The major groove is approximately 12 Å wide, and the minor groove is approximately 6 Å wide. The major and minor grooves are regions where DNA-binding proteins can most easily make direct contact with nucleotides along one or both strands of the double helix. In this chapter and in later chapters, we discuss many of the important functions DNA-binding proteins perform, such as regulating the initiation of transcription and controlling the onset and progression of DNA replication. Most of these functions depend on the presence of characteristic sequences of DNA nucleotides. DNA-binding proteins gain access to DNA nucleotides in major and minor grooves of the molecule. B-form DNA, overwhelmingly the most common DNA structure in organisms, has a right-handed twisting of the sugar-phosphate backbone. A-form DNA also has a right-handed twist to the helix. It is more compact than B-form DNA, with about 11 base pairs per complete helical twist, although its diameter is a little greater than that of B-form DNA (Table 7.1). A-form DNA is occasionally detected in cells, and it appears to be particularly common in bacteriophage, where its more compact size makes it functional for packaging of bacteriophage DNA. A-form DNA may be less amenable to binding by DNA-binding regulatory proteins, due to alterations of the major and minor grooves in comparison with B-form DNA. A-form DNA major grooves are deeper and narrower than those of B-form DNA, and its minor grooves are wider and shallower than those of B-form DNA. Research subsequent to Franklin's initial discovery of A-form DNA identifies the level of hydration of DNA as the principal factor determining its formation. Dehydration converts B-form DNA to the A form, and it is thought that by assuming the A form, DNA is better protected from damage under desiccation conditions. Such conditions occur in certain bacterial species. The third form of DNA, Z-form DNA, was discovered by Robert Wells and colleagues in 1970, and its structure was determined by Andrew Wang, Alexander Rich, and colleagues in 1979. Z-form DNA is quite different from A-form and B-form DNA in that it has a left-handed twist that gives the sugar-phosphate backbone a zigzag appearance—hence the name Z-form—and other structural differences as well (see Table 7.1). No definitive biological significance has been identified for Z-form DNA; however, it occurs in cells and is particularly common near the start sites for gene transcription. Studies of Z-form DNA have identified certain DNA sequences that are associated with Z-form DNA formation. These too occur most frequently near the starting points of gene transcription. Study of the human genome reveals numerous such sequences where Z-form DNA is detected. Human chromosome 22 appears to have a number of transcription start sequences where Z-form DNA can occur. Two questions about DNA frequently come up in discussions of DNA molecular structure. Why do complementary base pairs consist of one purine and one pyrimidine? Why are the strands of the double helix antiparallel and not parallel? The presence of one purine and one pyrimidine per base pair is a matter of molecular symmetry. Were the molecule to be composed of two paired purines (both double-ringed) the base pair would measure more than 20 Å across. Conversely, if the base pair was two pyrimidines (both single-ringed), the measurement would be much less than 20 Å. This would give the molecule an irregular diameter that might make it more difficult to package in cells and nuclei, or might make the binding of DNA-binding regulatory proteins more difficult. The reason why DNA strands are antiparallel to one another and not parallel is a matter of hydrogen bond formation. For hydrogen bonds to form, the negative charge of an oxygen or nitrogen must occur opposite the positive charge of a hydrogen. This occurs when complementary base pairs align in antiparallel strands. If a purine and a pyrimidine were aligned in parallel strands, positively charged hydrogens would be opposite one another, as would negatively charged nitrogens and oxygens. These repelling forces would prevent hydrogen bond formation.

PCR

A laboratory method for controlled replication of a specific target sequence of DNA in successive cycles. Using two short single-stranded primers that bind to sequences on opposite sides of the target sequence, exponential replication of the target sequence occurs.

Consensus sequence

A nucleotide sequence in a DNA segment derived by comparing sequences of similar segments from other genes or organisms. The most commonly occurring nucleotides at each position comprise the sequence.

Base pair

A pair of complementary nitrogenous bases in a DNA molecule

Base-stacking

A phenomenon of DNA base-pair interaction that rotates the base pairs around a central axis of symmetry and imparts twisting to the double helix.

7.4: DNA Replication Precisely Duplicates the Genetic Material • Explain semi-conservative DNA replication at the level of the DNA molecule and at the level of the DNA sequence, including the roles of the involved components and structures: replication bubble, replication fork, origin of replication, ligase, helicase, topoisomerase, single-stranded binding proteins (SSB), DNA pol III, primase, RNA primers, Okazaki fragments, leading strand, lagging strand, DNA pol I, the replisome/sliding clamp, hydrogen bonding, and phosphodiester bonds. DO LATER

Bacteria (eukaryotes) 1. Helicase breaks hydrogen bonds. Topoisomerase relaxes supercoiling 2. Single-stranded binding (SSB) protein prevents reannealing 3. Primase synthesizes RNA primers. 4. DNA polymerase III synthesizes daughter strand. 5. DNA polymerase III elongates the leading strand continuously and the lagging strand discontinuously. 6. DNA polymerase I removes and replaces nucleotides of the RNA primer. 7. DNA ligase joins Okazaki fragments.

• Why is DNA replication considered bidirectional in both prokaryotes and eukaryotes? MAYBE BETTER ANSWER

Because it proceeds in both directions from an origin of replication

Annealing

Binding of primers in PCR

Euchromatin

Chromosome regions containing chromatin that is not densely compacted. Most expressed genes are located within euchromatic regions of chromosomes.

Chromosome scaffold

Composed of numerous nonhistone proteins, the superstructure of eukaryotic chromosomes.

• What is a consensus sequence? o What consensus sequences (names, not specific sequences) are associated with the origin of replication in prokaryote and eukaryotes? o What nucleotides are most common in the origin consensus sequences? Why?

Consensus sequences have similar functions, similar overall length, and similarity of the pattern of base pairs. They feature nucleotides occurring frequently at the same positions in the DNA sequences of many species. Consensus sequences are not, however, identical to one another. Instead, consensus sequences are defined by the nucleotides that occur most often at particular positions in the sequence. The sequence making up a consensus sequence is determined by recognizing the similar sequences in several related species and identifying the most common nucleotide at each position. 13-mer and 9-mer sequences of oriC in bacteria. Saccharomyces cerevisiae (yeast) has the most fully characterized origin-of-replication sequences in eukaryotes. Are called autonomously replicating sequences (ARS) AT rich DNA because A//T, easier to break than C///G!

Nucleic acid

DNA and RNA

10.6: Eukaryotic Chromosomes Are Organized into Chromatin • What makes prokaryotes different from eukaryotes?

DNA in nucleus for eukaryotes

• Briefly describe the differences between the 2 major DNA polymerases in bacteria (DNA pol I and DNA pol III). Include what activities each carries out and the processes they are most typically involved in. à = -->

DNA pol I: replication (5' à 3' direction) • proofreading activity (3' à 5' exonuclease) • 5' à 3' exonuclease activity - important for replication • DNA pol III: replication (5' à 3' direction) • proofreading activity (3' à 5' exonuclease) Bidirectional expansion is driven by DNA synthesis at each replication fork. One replisome containing two DNA pol III enzymes operates at each fork to replicate both daughter strands. (FIGURE 7.18 The Replication Bubble) Each strand of parental DNA acts as a template for the synthesis of a new daughter strand of DNA. In E. coli, daughter DNA strands are synthesized at the replication fork by DNA polymerase III (DNA pol III), the principal DNA-synthesizing enzyme (see Figure 7.14, step 4). DNA pol III begins its work at the 3'-OH3′-OH end of an RNA primer and rapidly synthesizes new DNA by adding one nucleotide at a time in a sequence that is complementary and antiparallel to the template-strand nucleotides. Pol III requires a template nucleotide to add a new nucleotide to a daughter strand. Enzymes with functions identical to DNA pol III are found in archaea and eukaryotes. Experimental evidence indicates that most of the enzymes participating in DNA replication are part of a large protein complex called a replisome. There is one replisome at each replication fork. Replisomes have numerous components, including, in each replisome, two complete molecules of DNA pol III. One of these DNA pol III molecules carries out the 5'-to-3'5′-to-3′ synthesis of one daughter strand continuously, in the same direction in which the replication fork progresses. The second pol III in the replisome carries out synthesis of the other daughter strand. The continuously elongated daughter strand is called the leading strand (Figure 7.18). Notice that Figure 7.18 divides the replication bubble into four quadrants. The upper right and lower left quadrants contain leading strands. When DNA pol III on the lagging strand reaches an RNA primer, thus running out of template, it leaves a single-stranded gap between the last DNA nucleotide of the newly synthesized daughter strand and the first nucleotide of the RNA primer (Figure 7.19). The pol III, having very low affinity for these DNA-RNA single-stranded gaps, is then replaced by DNA polymerase I (DNA pol I), which has high affinity for such gaps (Figure 7.19, 1). The DNA pol I removes nucleotides of the RNA primer one by one and replaces them with DNA nucleotides, beginning with the 5'5′ nucleotide of the RNA primer and progressing in the 3'3′ direction until all the RNA nucleotides in the primer have been replaced by DNA nucleotides complementary to the template strand. The pol I enzyme possesses two activities that accomplish the removal of RNA nucleotides and their replacement by DNA nucleotides. DNA pol I first uses its 5'5′-to-3'3′ exonuclease activity to remove the 5'-most5′-most nucleotide from the RNA primer (see step 6 in Figure 7.14).This creates one open space opposite the template, which is then filled with the correct DNA nucleotide by the 5'5′-to-3'3′ polymerase activity of DNA pol I. As DNA pol I removes each RNA primer nucleotide and replaces it with a DNA nucleotide, the pol I continually pushes the single-stranded gap in the 3'3′ direction. Once the entire RNA primer is replaced, a remaining single-stranded gap sits between two DNA nucleotides. At this point, DNA ligase, having exclusive and very high affinity for DNA-DNA single-stranded gaps, is attracted to the gap and there performs its single task of forming a phosphodiester bond between the two DNA nucleotides that joins two Okazaki fragments (see step 7 in Figure 7.14). Both pol I and DNA ligase are active on leading and lagging strands. The level of activity is greater on lagging strands, however, where every 1000 to 2000 nucleotides, they are needed to join Okazaki fragments during replication of E. coli DNA.

• Why are RNA primase and RNA primers needed for DNA replication?

DNA pols cannot initiate DNA strand synthesis on their own • RNA primers are needed; these are synthesized by primase. DNA polymerase elongates DNA strands by adding nucleotides to the 3¢ end of a preexisting strand

• What discoveries led to our understanding that chromosomes were found in the nucleus and that chromosomes were made of DNA?

DNA was first noticed in 1869 when Friedrich Miescher isolated it from nuclei of white blood cells ---------• He called it "nuclein" • Around the same time, microscopic studies identified fusion of male and female pro-nuclei during reproduction and chromosomes were observed soon after • 1895: Edmund Wilson suggested that DNA might be hereditary material ---------• sperm and eggs contribute the same number of chromosomes during reproduction ---------• connection made between the substance observed by Miescher and the chromatin of chromosomes • 1900: Mendelʼs hereditary principles were rediscovered • 1903: Walter Sutton and Theodor Boveri independently described the parallels between chromosome partitioning into gametes and the inheritance of genes • ~1920: DNA was identified as the principal component of nuclein --------• The basic chemistry of DNA was deciphered: polynucleotide consisting adenine (A), thymine (T), cytosine (C), and guanine (G), held together by covalent bonds • 1923: DNA localized to chromosomes = candidate for hereditary material --------• However, both proteins and RNA are also found in chromosomes --------• Lipids and carbohydrates were also considered to be candidates

DNA polymerase I

Enzyme that relaxes DNA supercoiling by controlled strand nicking and rejoining.

• How many chromosomes do bacteria have? • What is the structure of bacterial DNA?

Most have 1-2, some have 3 or 4. All of them are circular, some have circular and linear.

• In what cells type is telomerase activity highest? Why is that biologically significant/necessary?

Most human somatic cells experience telomere shortening with successive divisions; this does not occur in germ cells • Telomerase: synthesizes telomeres in germ cells and stem cells

4. ___ determine the complementary nature of DNA and confer the information of the genetic material.

Nitrogenous bases

• Why are Okazaki fragments produced and how are these fragments joined to make a continuous DNA daughter strand? o What proteins and what specific enzymatic activities of those proteins are involved in this?

Okazaki fragments are produced by the lagging strand of DNA due to the lagging strand growing in the opposite direction of the replication bubble. These strands are joined together by the enzyme DNA ligase. 1. DNA pol I binds to a single-stranded gap between DNA and an RNA primer 2. Pol I removes an RNA primer nucleotide using its 5'-3' exonuclease capability 3. and fills the gap with a DNA nucleotide using its 5' to 3' polymerase capability. 4. Pol I removes each RNA primer nucleotide... 5. and replaces it with a DNA nucleotide. 6. When primer removal is complete, DNA ligase replaces pol I at DNA-DNA single-stranded gaps and... 7. catalyzes formation of a phosphodiester bond to join Okazaki fragments.

10. The lagging strand produces ___ that must be joined by ___.

Okazaki fragments, DNA ligase

• Explain what is meant by the polarity of a DNA strand. How is this polarity used in formation of a double-stranded molecule of DNA?

One end differs from the other end which is caused by the phosphodiester bonds linking the nucleotides together. The negatively charged O's of the 5' phosphate group contrasted with the positively charged H of the 3' hydroxyl group give the strand polarity. RNA is polar and has 5' and 3' ends, just like DNA,

• Where does DNA replication begin? What structures are produced give the bidirectional nature of replication?

Origin of replication. • Expansion around the origin of replication, forming a replication bubble, once replication gets underway • A replication fork is found at each end of the replication bubble -----• replication is complete when the replication forks meet

7.1: DNA Is the Hereditary Molecule of Life • Why must a heritable material exist? What does it have to do?

Our contemporary understanding of hereditary transmission and the evolution of species is rooted in this fact. 1. Localized to the nucleus, and a component of chromosomes 2. Present in a stable form in cells 3. Sufficiently complex to contain the genetic information required to direct the structure, function, development, and reproduction of organisms 4. Able to accurately replicate itself so that daughter cells contain the same information as parental cells 5. Mutable, undergoing mutation at a low rate that introduces genetic variation and serves as a foundation for evolutionary change

• Identify the chemical & structural differences between DNA and RNA.

RNA is much shorter than DNA and single-stranded whereas DNA is double stranded. There is a hydroxyl group on RNA's 2' carbon while DNA only has a hydrogen atom on its 2' C. RNA pairs Uracil with Adenine instead of Thymine like in DNA. Additionally, the 5' end of RNA has a triphosphate. Also another difference is the sugar in RNA is ribose and the sugar in DNA is deoxyribose. Another difference between DNA and RNA is the fact that DNA tends to form a double-stranded helix, while RNA is typically single stranded.

Telomere

Repeating DNA sequences, synthesized by telomerase, at the ends of linear chromosomes in eukaryotes; contain dozens to hundreds of copies of specific short DNA sequence repeats that buffer the coding sequence of the chromosome from loss during successive cycles of DNA replication.

• How do prokaryotes organize and condense their genome?

Supercoiling - Additional twisting in the same or opposite direction from the turns in the original helix. Facilitates and maintained by topoisomerase Folding into loops - Compacts together to condense DNA

In the Hershey-Chase experiment, bacteriophages were produced in either 32P-containing or 35S-containing medium. Where were these isotopes eventually detected when the radioactively-labeled bacteriophages were introduced to a fresh bacterial culture?

The 32P was associated with the bacterial cells and 35S was associated with the phage particles.

Template strand

The DNA strand that provides the template for ordering the sequence of nucleotides in an mRNA transcript. The DNA strand serving as a template for synthesis of a complementary nucleic acid strand.

• What is the Hayflick limit?

The Hayflick limit represents the number of cell cycles of a vertebrate cell before the cell succumbs to apoptosis

Proofreading

The capacity of many types of DNA polymerase to utilize a 3'3′ to 5'5′ exonuclease activity to remove and replace mismatched or damaged nucleotides during replication. See also 3'3′ to 5'5′ exonuclease activity.

Chromatin

The complex of nucleic acids and proteins that compose eukaryotic chromosomes.

• What is chromatin and what does it consist of?

The complex of nucleic acids and proteins that compose eukaryotic chromosomes. Chromatin = DNA + proteins (½ histone, ½ non-histone)

• What are histones and what do they do? o How many histones are found in eukaryotes? Where is each found? o Which histones make up the nucleosome core particle?

The five types of histone proteins are small, basic proteins with a positive charge that allows them to bind to DNA. Histone proteins: small basic proteins that tightly bind DNA H1, H2A, H2B, H3, and H4; they are highly conserved among eukaryotes Nucleosome core particles: fundamental units of histone protein organization with 2 molecules each of H2A, H2B, H3, & H4 (octamer) H1 links nucleosomes and allows for tighter packing.

Replisome

The large molecular machine located at the replication fork that coordinates multiple reaction steps during DNA replication.

Major groove

The larger of two grooves formed in the DNA sugar-phosphate backbone by the helical twist of the double helix and exposing certain base pairs.

• What are the major and minor grooves of DNA and what are they used for?

The major and minor grooves are regions where DNA-binding proteins can most easily make direct contact with nucleotides along one or both strands of the double helix. In this chapter and in later chapters, we discuss many of the important functions DNA-binding proteins perform, such as regulating the initiation of transcription and controlling the onset and progression of DNA replication. Most of these functions depend on the presence of characteristic sequences of DNA nucleotides. DNA-binding proteins gain access to DNA nucleotides in major and minor grooves of the molecule.

Hayflick limit

The name given to the observation that normal human cells in cell culture usually experience a limited number of cycles of cell division before they die.

Antiparallel

The opposite arrangement of the sugar-phosphate backbones in a DNA double helix.

Reverse transcription (RT)

The process of DNA synthesis from an RNA template by the enzyme reverse transcriptase.

Telomerase

The ribonucleoprotein complex whose RNA component provides a template used to synthesize repeating DNA segments that form chromosome telomeres.

Minor groove

The smaller of two grooves formed in the sugar-phosphate backbone by the helical twist of the double helix, exposing certain base pairs.

Supercoiling

The superhelical twisting of covalently closed circular DNA. See also positive supercoiling and negative supercoiling.

7.2: The DNA Double Helix Consists of Two Complementary and Antiparallel Strands • What are the three components of a nucleotide? Draw the structure of a nucleotide that shows these three components. (Individual atoms are not required.)

The three components of a nucleotide are a phosphate group (could be one, two, or three), a 5-Carbon sugar, and a base (either A,C,G,T).BE ABLE TO DRAW

• How does the cell prevent supercoiling of DNA when the DNA is unwound at the replication bubble? o What enzyme is important for this process and how does this enzyme generally function to do this?

Unwinding of chromosomes during DNA replication will create torsional stress, potentially leading to supercoiling of DNA • DNA topoisomerases catalyze controlled cleavage and rejoining of DNA to relieve supercoiling 1. Topoisomerase cuts one or both DNA strands. 2. DNA strands rotate to remove supercoils. 3. Topoisomerase rejoins DNA strands

• How does Griffith's experiment show us that a heritable material exists? o What was his experimental question? o What were the key features of his experiment? (What did he do, what were his controls?) o What were the results? What can you conclude from the results? o What is 'transformation'?

When the killed virulent bacteria was mixed with the non virulent bacteria, the mice still died. This shows that something in the virulent bacteria was able to be reproduced by using the non virulent bacteria. This substance (which we now know is DNA) transformed the non virulent bacteria into virulent bacteria, so we know that something must have transferred between the two in order to kill the mice. Can genetic information be transmitted between two strains of bacteria The controls were the virulent bacteria (S) injected, nonvirulent bacteria (R) injected, and the killed virulent were the controls, along with the amount he gave each mice. The variable was the killed virulent bacteria with nonvirulent bacteria. Mice infected with strain SIII developed pneumonia and died • Mice infected with heat-killed strain SIII survived • Mice infected with strain RII survived • Mice infected with heat-killed strain SIII + live strain RII developed pneumonia and died • live type SIII bacteria were recovered from the mice. Dead virulent bacteria somehow caused the nonvirulent bacteria to become virulent. A strand of genetic material that made the bacteria virulent transformed to the other nonvirulent bacteria. Griffith proposed the "transformation factor" as the molecule that transformed the nonvirulent R strain into the virulent S strain • Hypothesized the transforming factor carried hereditary information (but didn't know the molecule • Process of transformation: used by bacteria to transfer hereditary information In Griffith's proposal, the transforming factor was a compound that carried hereditary information. He was unable to identify his transformation factor, but today we know that it is DNA. Today biologists also know that the biological process responsible for the conversion of living RII bacteria by heat-killed SIII is the process of transformation that we describe as a mechanism for gene transfer in bacteria in Section 6.4.

Phosphodiester bond

a chemical bond of the kind joining successive sugar molecules in a polynucleotide.

Purine

a nitrogenous base that has a double-ring structure; one of the two general categories of nitrogenous bases found in DNA and RNA; either adenine or guanine

pyrimidine

a nitrogenous base that has a single-ring structure; one of the two general categories of nitrogenous bases found in DNA and RNA; thymine, cytosine, or uracil

Replication bubble

a region of DNA, in front of the replication fork, where helicase has unwound the double helix

5. Eukaryotic chromosomes are organized into ___ and compacted by ___.

chromatin, histone proteins

2. DNA structure is a ___ consisting of two ___ and ___ strands.

double helix, complementary, antiparallel

9. Due to the antiparallel nature of DNA, one strand of DNA follows ___, while the other does not ___.

fork progression (leading strand), (lagging strand)

1. DNA is the ___ for life.

hereditary molecule

Nucleotide

monomer of nucleic acids made up of a 5-carbon sugar, a phosphate group, and a nitrogenous base

8. Consensus sequences are found at ___.

origins of replication

3. DNA strands are ___ held together by ___

polynucleotides, covalent bonds.

11. DNA polymerase has ___ to prevent accumulation of errors.

proofreading activity

Solenoid

protein molecules around which DNA is tightly coiled in chromatin

Bidirectional replication

replication at both ends of a replication bubble

6. DNA replication occurs in a ___ manner and is initiated at ___

semi-conservative, origins of replication.

DNA polymerase III

synthesizes new DNA only in the 5' to 3' direction

12. The end of linear chromosomes, telomeres, must be lengthened by ___ to prevent DNA loss.

telomerase

• Explain the role of proofreading during DNA replication and what enzyme is important for this process. o What specific enzymatic activities of this proteins are involved in this? MAYBE CHECK

• DNA polymerases undertake DNA proofreading to correct occasional errors • Errors in replication occur 1 in every ~billion bp in E. coli • Proofreading ability of DNA polymerase enzymes is due to a 3¢-to-5¢ exonuclease activity • Replication errors produce a DNA mismatch; mismatched bases cannot form appropriate H bonds • Leads to displacement of 3¢ OH into the 3¢-to-5¢ exonuclease site of the enzyme • Several nucleotides are removed and new nucleotides are incorporated Figure 7.21 DNA polymerase proofreading activity. (a) A replication error by polymerase. (b) Newly synthesized 3 ′ end of daughter strand shifts into exonuclease site, where nucleotides are removed. (c) The polymerase resumes 5'-to-3'5′-to-3′ synthesis.

ORC

• In eukaryotes, a prereplication complex assembles with 14 proteins, six of which are the origin replication complex (ORC)

• Describe the placement and number of origins of replication in prokaryotes and eukaryotes.

• Single origin in bacterial chromosomes • Eukaryotic chromosomes have multiple origins of replication

• Why do the ends of linear DNA become progressively shorter with each round of replication? o What ends become shorter? o What mechanism exists to protect the ends of chromosomes so that the repeated sequences found at the telomeres are not lost? o What is reverse transcription and how does telomerase use this to extend the telomeres of the template strand?

• The leading strand of linear chromosomes can be replicated to the end • Primer on lagging strand prevents these strands from being completely replicated 3' end • When RNA primers at 5' ends are removed, there is nothing for polymerase to attach a base to in order to fill in the gap! • If not fixed, every round of replication will shorten the ends of the chromosomes. Solution: telomerase! • adds repeats to the 3' end (using an RNA template the enzyme carries with it), so that the overhang left by removal of the RNA primer will get "replaced" in the next round of replication by telomerase activity • Example of reverse transcription • Telomerase: synthesizes telomeres in germ cells and stem cells • RNA in telomerase is complementary to the telomere repeat sequence and acts as a template for addition of DNA • template RNA of telomerase allows new DNA replication to lengthen the telomere sequences • Once telomeres are sufficiently elongated, DNA replication fills out the chromosome ends 1. Attachment of telomerase 2. Elongation of DNA 3. Translocation of telomerase 4. Elongation of DNA 5. Telomere completion (by DNA polymerase) • In addition to the repetitive DNA sequence, most eukaryotic telomeres also contain a DNA sequence that forms a knotted fold (T loop) • T loop protects the telomeres from degradation by binding the protein complex shelterin

• Considering the nature of hydrogen bonding, why do the separated template strands not reanneal at the replication bubble during DNA replication?

• Unwound DNA strands are kept from reannealing by single-stranded binding protein (SSB)

Exonuclease

3'3′ to 5'5′ exonuclease activity - DNA- and RNA-digesting activity that progresses in the 3'3′ to 5'5′ direction to remove nucleotides. See also DNA proofreading. 5'5′ to 3'3′ exonuclease activity - DNA- or RNA-digesting activity that progresses in the 5'5′ to 3'3′ direction to remove nucleotides.

Replication fork

A Y-shaped region on a replicating DNA molecule where new strands are growing.

Heterochromatin

A chromosome region containing densely compacted chromatin and few, if any, expressed genes.

DNA Ligase

An enzyme active in DNA replication that joins together segments of a DNA strand by catalyzing formation of a phosphodiester bond.

Single-stranded binding protein (SSB)

In DNA replication, a protein that adheres to each template strand following unwinding by helicase to prevent strand reannealing before the arrival of the replication fork.

Leading strand

In DNA replication, the continuously synthesized strand. Compare with lagging strand.

Lagging strand

In DNA replication, the discontinuously synthesized strand whose Okazaki fragments are ligated to complete new strand synthesis. Compare with leading strand.

Helicase

In DNA replication, the enzyme responsible for breaking hydrogen bonds between complementary nucleotides of a DNA duplex. Unwinding of the strands occurs ahead of the advancing replication fork

RNA primer

In DNA replication, the short, single-stranded RNA segment synthesized by primase. The 3'3′ end of the RNA primer is used by DNA polymerase to begin synthesis of DNA.

Sliding clamp

In bacterial DNA replication, the multisubunit protein complex that joins with DNA polymerase to hold polymerase on the template and helps drive polymerase along the template.

• How do the experiments from Avery, MacLeod, and McCarty lead to the conclusion that DNA is the heritable material? o What was their experimental question? o What were the key features of the experimental set-up? (What did they do, what was their control?) o What were the results? What can you conclude from the results?

In their experiment, DNA was the only common factor in the solutions that produced virulent bacteria. When it was destroyed in another solution, the virulent bacteria (SIII) was not produced because there was no genetic info allowing it to do so. They wanted to determine the molecule responsible for turning nonvirulent bacteria into virulent bacteria Extract from heat killed SIII bacteria put in... - Control, no components destroyed + Type RII --> No changes --> Live type SIII bacteria recovered - Lipids and polysaccharides destroyed + Type RII --> No Lipids and polysaccharides --> Live type SIII bacteria recovered - Protease added, protein destroyed + Type RII --> No protein --> Live type SIII bacteria recovered - RNase added, RNA destroyed + Type RII --> No RNA --> Live type SIII bacteria recovered - DNase added, DNA destroyed + Type RII --> No DNA --> No Live type SIII bacteria recovered The DNase solution was the only one to produce nonvirulent bacteria only. The others produced virulent and nonvirulent. This shows that the only solution in which DNA was destroyed by the DNase enzyme stayed nonvirulent. With DNA present in the other solutions, virulent bacteria was formed. This lead Avery and his colleagues to assume that DNA carried the genetic material from the virulent bacteria along and resulted in virulent bacteria. When DNA was not present, this did not occur. Thus, DNA is the genetic material carrier in the cell.

7. Replication bubbles consist of ___, each of containing the proteins needed to carry out DNA replication.

2 replication forks

7.3: DNA Replication Is Semiconservative and Bidirectional • Interpret and explain data from the Meselson and Stahl experiment studying the mode of DNA replication. o Describe the three possible models of DNA replication that were tested in this experiment. o How are different isotopes used in this experiment? o Predict what the newly synthesized DNA molecules and the density gradient centrifugation data from the experiment would look like for the different models of replication. o What was observed after each round of DNA replication? What mode of replication was supported by the data at each round?

1. Semi-conservative: the daughter DNA strands are each made up of half the parent DNA strand and the other half is the new copy2. Conservative: the parent DNA strand stays the same and the new DNA is formed from that3. Dispersive: there are pieces of old and new DNA in both strands The two isotopes were nitrogen atoms. The old heavy strands were labeled 15N and the newer lighter strands were labeled 14N. The experiment started out with both strands labeled as heavy nitrogen and then the molecules form a band with the density of heavy DNA. After one round with the absence of heavy nitrogen the parental strand contained heavy nitrogen, while the daughter strand contained light nitrogen. Then the daughter DNA molecule forms a band with an intermediate density. After two rounds of replication, half the duplex DNA molecules have one strand with heavy nitrogen and one with light nitrogen. The other half have two strands with light nitrogen. This causes the daughter DNA molecules to form two bands, where one has an intermediate density and one with a light density. Conservative: after the first round of replication one DNA molecule will contain all N15 while the other will have all N14. This will create two distinct bands. By the second round of replication there will still be one molecule of entirely N15 but now there will be 3 with N14. This will still create 2 bands, but the one at N14 will be slightly darker. Semi-Conservative: after the first round, two DNA molecules each containing N15 and N14 will be produced. This results in one band for N14:N15. In the next round of replication, two molecules will contain a mix of N15 and N14, but the other two will contain all N14. This creates two bands, one for N15:N14 and one for N14. Dispersive: for this model of replication each molecule will contain a mix of N15 and N14, but as generations go by there will be less and less N15 in each molecule. This will a result in one band, that is slightly less dense each time. After one round of replication, the DNA molecules all had the densities expected from 14N/15N molecules • After two rounds of replication, ½ the molecules had densities expected of 14N/15N molecules and ½ had densities expected for 14N/14N molecules • These results are consistent only with the semiconservative model of replication ORRRRRRR After each round of replication, the density of the DNA was observed. Semiconservative replication because after each round the DNA duplex consisted something originally from the parental strand (heavy nitrogen) and one newly synthesized strand (light nitrogen)This is true. But we didn't know this after just 1 round of replication. What did we learn after each round? After the first round of replication, conservative was ruled out. After the second round of replication, dispersive was ruled out. This left semi-conservative as the only option.

• How does a DNA double helix form? o What two specific bonds are responsible for establishing this structure? How do these bonds form and where? o What bases are considered complementary with each other? o What is the difference between purines and pyrimidines? What bases are in each category? o How do the pyrimidines and purines result in a helical structure?

1. The bases of one strand are complementary to the bases in the corresponding strand (A pairs with T and G pairs with C) 2. The two strands are antiparallel with respect to their 5- and 3- ends phosphodiester bonds linking the nucleotides together. Hydrogen bond between the bases. Pyrimidines (thymine & cytosine): single ring • Purines (adenine & guanine): double ring Complementary base pairing combines 1 purine with 1 pyrimidine • Formation of hydrogen (H) bonds between bases on antiparallel strands • 2 H-bonds form between A and T; 3 Hbonds form between G and

• Explain the levels of chromosome condensation. ???

1st level (nucleosome) - DNA (146 bp) wrapping around the nucleosome particle = Nucleosome (11 nm) 2nd level (Beads on a string) - H1 serves as a linker protein to hold the nucleosomes tighter. Electron micrographs of chromatin fibers in a highly decondensed state show a regular series of circular structures strung together by connecting filaments (see Figure 10.24b). This form of chromatin is identified as the "beads on a string" morphology of chromatin. The "beads" are nucleosomes that are a little more than 11 nm in diameter, and the "string" is called linker DNA. Linker DNA is the DNA between regions of core DNA. 3rd level (Solenoid) - The 10-nm fiber is an unnatural state for chromatin. To achieve it, chromatin must be chemically treated and held in conditions that are not found in cells. Under in vitro conditions, chromatin forms the 30-nm fiber, although it is not certain this structure forms in vivo (see Figure 10.24c). Electron micrographs and molecular modeling help us visualize how the 30-nm fiber is assembled. It is produced by coalescence of the 10-nm fiber into a cylindrical filament of coiled nucleosomes that is hollow in the middle. Due to its coiled structure and open middle, the 30-nm fiber is often also called the solenoid structure (like the coil of wire in the starter of a car). Each turn of the solenoid structure contains six to eight nucleosomes. The diameter of the solenoid is approximately 34 nm. Research examining in vivo chromatin structures will soon be able to determine if these occur in cells or only in vitro. The histone protein H1 plays a key role in stabilizing the solenoid structure. The long N-terminal and C-terminal ends of the H1 protein attach to adjacent nucleosome core particles. H1 protein pulls the nucleosomes into an orderly solenoid array and lines the inside of the structure. Experimental analysis shows that chromatin from which H1 has been removed can form 10-nm fibers but not 30-nm fibers. Chromatin exists in a 30-nm-fiber state or a more condensed state during interphase. 4th level (Coiled Chromosome) - Beyond the 30-nm stage, chromatin compaction and the presence of nonhistone proteins are integral to the structure of chromosomes and the process of chromosome condensation that initiates with the onset of prophase in the M phase of the cell cycle. Nonhistone proteins perform multiple roles in influencing chromosome structure and in facilitating M phase chromosome condensation. Interphase chromosome structure results from the formation of looped domains of chromatin similar to supercoiled bacterial DNA (see Figure 10.24d). 5th level (condensed chromatin) - The loops are variable in size, containing from tens to hundreds of kilobase pairs and consisting of 30-nm-fiber DNA looped on a category of nonhistone proteins that are the foundation of chromosome shape. The diameter of looped chromatin is approximately 300 nm, so looped chromatin is called the 300-nm fiber. With continued condensation, the chromatin loops form the sister chromatids. In metaphase, chromosome condensation reaches its zenith, resulting in chromosomes that are easily visualized by microscopy (see Figure 10.24e).

• How do the experiments from Hershey and Chase support the results of Avery and colleagues, showing that DNA is the heritable material? o What is a radioactive isotope? o What are the relative abundances of sulfur and phosphorus in DNA and protein? Why is this relevant to the experimental design? o What is a phage? Why were phage well-suited for this experiment? o What results did Hershey and Chase obtain? What can you conclude from the results?

A radioactive isotope is an atom that has more than normal levels of nuclear energy. This excess energy makes the atom unstable. Phosphorous makes up the backbone of a DNA molecule and Sulfur is present in the various possible R groups of amino acids. There is no sulfur in DNA and no phosphorous in proteins. By marking the radioactive isotopes of phosphorous and sulfur in DNA and proteins respectively, we are able to track them to see which is the structure that enters and infects a cell. Bacteriophage (phage): viruses that infect bacteria • Phages have a protein shell with a tail segment that attaches to the host cell and a head that contains DNA • Phages must infect hosts to reproduce • Infection begins when the phage injects DNA into the bacterial cell and leaves its protein shell (capsid) on the surface • The phage DNA replicates in the bacterium and produces proteins that are assembled into progeny phage—these are released by lysis of the host cell Bacteriophages are very simple structures made up of basically DNA and proteins. Phages are what attack cells and inject their genetic material making it a virulent cell. Since they only have DNA and proteins, we can track one or the other in the experiment to see what part of the phage is the transformative agent that makes a cell virulent (i.e. what is the genetic material). If protein was the heritable material, then sulfur would be present within the intracellular space in the results. If DNA were the heritable material, phosphorus would be present. Hershey and Chase were able to provide evidence that DNA is the genetic material and not other enzymes or protein. They were able to conclude this because of the method they used (the radioactive isotopes incorporated in each phage) to track each phage.It can be concluded that DNA is genetic material and protein is not.

Okazaki fragment

A short segment of newly synthesized DNA that is part of a lagging strand and is ligated to other Okazaki fragments to complete lagging strand synthesis.

Chromosome

A structure composed of DNA and associated proteins that in total contain the genome of an organism.

• What different models of DNA exist and what are the major differences between them?

A-DNA • Compressed B form • 11 bp per turn • Right-handed • Common in bacteriophage B-DNA • "Standard form" • 10 bp per turn • Right-handed • Most common Z-DNA • "Zigzag form" • 12 bp per turn • Left-handed • found near transcription start sites

• Why is chromosome condensation needed? Why is it important for DNA to have different levels of compaction?

Amazingly, although the DNA is very tightly folded, it is compacted in a way that allows it to easily become available to the many enzymes in the cell that replicate it, repair it, and use its genes to produce proteins. Chromatin in interphase: 30-nm fiber • forms when the 10-nm fiber coils into a solenoid structure, with 6-8 nucleosomes per turn and histone H1 stabilizing the solenoid • Chromatin in metaphase: maximally condensed ----• loops of 30-nm fibers attached to a nonhistone protein chromosome scaffold ----• loops on the scaffold form the 300-nm fiber

• Why are histones evolutionarily conserved in eukaryotes?

Among eukaryotes, there is very strong evolutionary conservation of the amino acid sequences of histone proteins. This consistency among eukaryotes suggests that there is significant evolutionary pressure to retain the structure and function of each histone protein. A comparison of the amino acid sequences of H4 in cows and pea plants, for example, demonstrates this high degree of evolutionarily retained identity. Cows and pea plants last shared a common ancestor more than 500 million years ago, when the animal and land plant lineages diverged. Over those hundreds of millions of years of evolutionary change, there are just two amino acid differences among the 102 amino acids in the protein. The comparison tells us that since the time when plants and animals last shared a common ancestor, extraordinarily strong evolutionary pressure has maintained H4 DNA and its amino acid sequence identity in organisms. This example of evolutionary conservation speaks to the importance of histones in eukaryotic chromosome organization.

• Compare DNA replication on the leading strand versus the lagging strand

Experimental evidence indicates that most of the enzymes participating in DNA replication are part of a large protein complex called a replisome. There is one replisome at each replication fork. Replisomes have numerous components, including, in each replisome, two complete molecules of DNA pol III. One of these DNA pol III molecules carries out the 5'-to-3'5′-to-3′ synthesis of one daughter strand continuously, in the same direction in which the replication fork progresses. The second pol III in the replisome carries out synthesis of the other daughter strand. The continuously elongated daughter strand is called the leading strand (Figure 7.18). Notice that Figure 7.18 divides the replication bubble into four quadrants. The upper right and lower left quadrants contain leading strands. The daughter strands in the upper left and lower right quadrants shown in Figure 7.18 have a 5'-to-3'5′-to-3′ direction of elongation that runs opposite to the direction of movement of the replication fork. These daughter strands are elongated discontinuously, in short segments, each of which is initiated by an RNA primer. The discontinuously synthesized daughter strand is called the lagging strand. Thus in Figure 7.18, the lower right and upper left quadrants of the replication bubble contain lagging strands (see also step 5 of Figure 7.14).

• What proteins make up the replisome? What is the role of the sliding clamp? How does this structure affect the processivity of replication?

Replisome: large protein complex found at each replication fork; 2 pol III & other proteins • 1 DNA pol III synthesizes 1 daughter strand continuously in the same direction as fork progression: leading strand • 1 DNA pol III elongates the other daughter strand discontinuously, in the opposing direction to fork progression, via short segments (Okazaki fragments): lagging strand • closes around the double-stranded DNA during replication • has a "donut hole" of about 35Å, into which the DNA fits • anchors the DNA pol III core enzyme to the template • is key to the high level of pol III activity The "processivity" of DNA polymerases on their own—that is, the ability of DNA polymerases to drive their own movement along template strands during replication—is comparatively low. This means that by themselves they are unable to provide the momentum required to both synthesize new DNA and progress along the template strand. To enhance the processivity of these polymerases, they are associated with an auxiliary protein complex known as a sliding clamp. The sliding clamp, with its diameter of approximately 75Å, has a "doughnut hole" of about 35 Å that encircles double-stranded DNA (Figure 7.20a). Each sliding clamp locks onto a DNA template strand and affiliates with DNA pol III core enzyme, firmly anchoring the enzyme to the template to carry out the bulk of replication (Figure 7.20b). The clamp is the key to the enzyme's high level of activity. When no more template nucleotides are available, the DNA pol III is dropped by the sliding clamp and replaced by DNA pol I, which as we have seen removes RNA primers and replaces them with DNA.

Origin of replication

Site where the replication of a DNA molecule begins, consisting of a specific sequence of nucleotides.

Hydrogen bond

weak attraction between a hydrogen atom and another atom