BIO378
Biology: Fundamentals of Bioinformatics (3 credits)

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Lecture Three

DNA

Learning Objectives

     Our objective in this lecture will be to achieve an understanding of the structure and function of the genetic information-containing element, deoxyribonucleic acid (DNA).  After this lecture you will be expected to:

·        Know that genetic materials are made of DNA.

·        Know the DNA structure.

·         Know DNA replication is semiconservative.

·         Know mechanism of DNA replication

Learning Objectives

3.1 DNA as the Genetic Material

3.2 DNA Structure

3.3 DNA Replication

3.4 The Mechanism of DNA Replication

Summary

Practice Test and Answers

Required Readings

Assignment

Note: The material in this lecture has been developed partially based on material and figures that are linked to from the lecture come from material supplied in conjunction with the required texts for the course. The copyright for such material is held by M.J. Farabee, Sinauer Associates, Inc. as well as by W. H. Freeman & Co.s

Lecture Notes:

        Deoxyribonucleic acid (DNA) and associated protein molecules, organized into structures called chromosomes.  It is actually DNA containing most or all of the genetic information of the cell.

3.1 DNA as the Genetic Material

A)   During the first half of the twentieth century, the structure that carried hereditary information from generation to generation was generally assumed to be a protein.

1)      The diversity and specificity of proteins seemed appropriate for genetic material.

2)      DNA seemed too simple to carry complex information.

B)   Circumstantial evidence, however, pointed to DNA as the genetic material.

1)      DNA was found in the nucleus, which was already known to carry genes.

2)      A dye that bound to DNA showed that the amount of DNA in somatic cells was twice that in eggs or sperm, as would be expected from Mendel’s discoveries.

3.1.1 DNA from one type of bacterium genetically transforms another type

1)    In the 1920s, the English physician Frederick Griffith made a landmark discovery about heredity while looking for a vaccine against Streptococcus pneumoniae, one of the bacteria that cause pneumonia in humans.

2)      Griffith worked with two different strains of the bacterium.

·         The S strain produced shiny, smooth colonies when grown in the laboratory.

·         The R strain produced colonies that looked rough.

·         The S strain was virulent (mice injected with the S strain died within a day); a capsule around the S strain bacteria protected them from the host’s immune system.

·         The R strain lacked this capsule and was nonvirulent.

3)      Griffith heated some S strain bacteria to kill them, then injected the bacteria into mice.

4)      The heat-killed bacteria did not kill the mice.

5)      A mixture of heat-killed S strain bacteria and living R strain bacteria, however, did kill the mice.

6)      Griffith found living S strain bacteria in the hearts of the mice killed in this way.

7)      He concluded that some of the living R strain bacteria had been transformed by the presence of the heat-killed S strain bacteria.

8)      Further tests demonstrated that some substance from the dead S strain bacteria could cause a heritable change in the R strain bacteria.

9)      Some scientists concluded that this “transforming principle” carried heritable information and thus was the genetic material.

3.1.2 The transforming principle is DNA

1)   Oswald T. Avery and colleagues spent several years identifying the transforming principle by a process of elimination.

2)   They treated the extract from heat-killed S strain bacteria in various ways to destroy different types of substances but retain others.

3)      Invariably, when DNA was destroyed, the transforming activity was lost, but when DNA was left intact, the transforming activity survived.

4)      This work, published in 1944, was not immediately appreciated.

·         Little was known about bacterial genetics, and it was not yet obvious that bacteria even had genes.

3.1.3 Viral replication experiments confirm that DNA is the genetic material

1)    In 1952, Alfred D. Hershey and Martha Chase performed experiments confirming that DNA is the genetic material.

2)      The T2 bacteriophage, a virus that attacks E. coli, consists almost entirely of a DNA core packed in a protein coat.

3)      When a T2 bacteriophage attacks a bacterium, part but not all of the virus enters the bacterial cell.

4)      The Hershey–Chase experiment determined which part of the virus entered the bacterium.

5)      Some viruses were labeled with radioactive sulfur. Sulfur is present in proteins but not in DNA.

6)      Other viruses were labeled with radioactive phosphorus. Phosphorus is present in DNA but absent from most proteins.

7)      In separate experiments, viruses with labeled sulfur and labeled phosphorus were combined with bacteria.

·         Blending the resulting bacteria removed the viral material that had not entered the bacteria.

·         Centrifuging revealed that the labeled sulfur (and thus the viral protein) had separated from the bacteria, but the labeled phosphorus (and thus the viral DNA) remained with the bacteria.

8)      Experiments on later generations of bacteria confirmed that the labeled phosphorus remained with subsequent generations while the labeled sulfur was quickly lost.

9)      By this time, it was accepted that bacteria had genes, so the Hershey–Chase experiment convinced most scientists that DNA was the carrier of hereditary information.

3.2 DNA Structure

3.2.1 The chemical composition of DNA was known

1)      By the 1950s it was known that DNA was a polymer of nucleotides.

·         The four nucleotides that make up DNA differ only in their nitrogenous bases.

·         There are two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). (see Fig. 3-1 to 3-3)

2)      In 1950, Erwin Chargaff noted that in DNA from all species tested, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine.

3)      In other words, the total abundance of purines equals the total abundance of pyrimidines, even though the actual proportions of each base vary in different species.

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Fig. 3-1  Structure of a nucleotide, adenosine 5’-monophosphate (AMP).

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Fig. 3-2  Structure of pentose present in nucleic acids.

Fig. 3-3  Structure of the principal bases in nucleic acids.

3.2.2 Four key features define DNA structure

1)      Four features summarize the molecular architecture of DNA.

·         The DNA molecule is a double-stranded helix.

·         The diameter of the DNA molecule is uniform.

·         The twist in DNA is right-handed.

·         The two strands run in different directions, that is, they are antiparallel.

2)      The sugar–phosphate backbones of each strand coil around the outside of the helix.

3)      The nitrogenous bases point toward the center of the helix.

4)      Hydrogen bonds between complementary bases hold the two strands together (see Fig. 3-4)

·         A always pairs with T (two hydrogen bonds).

·         G always pairs with C (three hydrogen bonds).

5)      Each base pair has one purine and one pyrimidine, so the diameter of the double helix remains constant.

6)      The direction of a polynucleotide is defined by the linkages between adjacent nucleotides. (See Fig. 3-4)

·         The phosphate groups link the 3¢ carbon of one deoxyribose molecule to the 5¢ carbon of the next.

·         Thus a single strand of DNA has a 5¢ phosphate group at one end (the 5¢ end) and a free 3¢ hydroxyl group at the other end (the 3¢ end).

·         In a double helix, the 5¢ end of one polypeptide is hydrogen-bonded to the 3¢ end of the other, and vice versa.

 Fig. 3-4  Structure of DNA.

3.2.3 RNA differs from DNA

1)      RNA is single-stranded (Fig. 3-5).

2)      The sugar in RNA is ribose, not deoxyribose.

3)      Wherever thymine is found in DNA, it is replaced by uracil in RNA.

4)      RNA can fold over and base-pair with itself.

Fig. 3-5  Structure of RNA.

3.2.4 Watson and Crick described the double helix

1)      Francis Crick and James D. Watson used the technique of model building to establish the general structure of DNA.

2)      The results of X-ray crystallography from Rosalind Franklin and Maurice Wilkins convinced them that the DNA molecule was helical (Fig. 3-6).

Fig. 3-6 X-ray crystallography diffraction pattern of DNA.

3)      X-ray crystallography also provided the values of certain distances within the helix.

4)      Density measurements and earlier models pointed to a structure with two polynucleotide chains running antiparallel to each other (Fig.3-7).

5)      Although there have been modifications, the principle features of the model they built in 1953 have remained unchanged.

Fig. 3-7  A space-filling model of DNA moecule showing its double-helical structure.

3.2.5 The double helical structure of DNA is essential to its function

1)   The genetic material must perform four important functions:

·         It must be able to store all of an organism’s genetic information.

·         It must be susceptible to mutation.

·         It must be precisely replicated in the cell division cycle.

·         It must be expressible as the phenotype.

2)      The simple, double-helical structure of DNA, with the two strands linked by complementary base pairs, lends itself well to the first three of these functions.

3)      DNA is also well suited to expression as a phenotype, though this function is not inherent in the structure of the molecule.

3.3 DNA Replication

 3.3.1 Three modes of DNA replication appeared possible

1)    Three years after Watson and Crick published their structure of DNA, the American     biochemist Arthur Kornberg demonstrated that the DNA molecule contains the information needed for its own replication.

2)      Kornberg showed that DNA can replicate in a test tube with only a specific enzyme (DNA polymerase) and a mixture of four precursors: dATP, dCTP, dGTP, and dTTP.

·         These precursors are deoxyribonucleoside triphosphates (dNTP’s).

·         There is one precursor each for adenine, cytosine, guanine, and thymine.

3)      Theoretically, DNA could serve as its own template in one of three different ways:

·         Semiconservative replication would use each parent strand as a template for a new strand. Each new DNA double helix would then have one parent strand and one new strand (Fig. 3-8).

Fig. 3-8 Semiconservative replication model.

·         Conservative replication would build an entirely new double helix based on the template of the old double helix. The new strand would contain none of the original DNA (Fig. 3-9).

Fig. 3-9 Conservative replication model.

·         Dispersive replication would use fragments of the original DNA molecule as templates for assembling two molecules. All the resulting strands would be mixtures of old and new material (Fig. 3-10).

Fig. 3-10 Dispersive replication model.

4)      Watson and Crick’s model suggested but did not prove that replication is semiconservative.

3.3.2 Meselson and Stahl demonstrated that DNA replication is semiconservative

1)   Matthew Meselson and Franklin Stahl demonstrated in 1957 that DNA replication is semiconservative by using a technique they devised called density labeling.

2)      Centrifuging in a cesium chloride (CsCl) solution can separate DNA labeled with “heavy” nitrogen (¹⁵N) from unlabeled DNA.

3)      Meselson and Stahl grew a culture of E. coli for 17 generations in a medium with ¹⁵N instead of ¹⁴N.

4)      As a result, all DNA in the bacteria was “heavy.”

5)      They then transferred bacteria grown on the heavy medium to a normal medium and allowed the bacteria to continue growing.

6)      Under the conditions they used, E. coli replicates its DNA every 20 minutes.

7)      They sampled the DNA at each generation time, starting with the parental, all-heavy generation.

8)      Centrifuging the DNA after the first cell division (20 minutes) yielded a single band of DNA intermediate in density between the heavy and light forms.

·         This ruled out the conservative replication hypothesis, which would have yielded two bands, one heavy and one light.

9)      Centrifuging the DNA after the second cell division (40 minutes) yielded an intermediate band and a light band, the result predicted by the semiconservative replication hypothesis.

·         Dispersive replication would again have yielded a single band with a density less than heavy DNA but greater than light DNA.

10)   Other scientists demonstrated that DNA in eukaryotes also replicates semiconservatively.

3.4 The Mechanism of DNA Replication

A)   There are four requirements for semiconservative replication of DNA:

1)   DNA must act as a template for complementary base pairing.

2)      dATP, dGTP, dCTP, and dTTP must be present.

3)      The enzyme DNA polymerase must be present to bring the substrates to the template and catalyze the reactions.

4)      The reactions involved in replication are endergonic, so a source of chemical energy must be present.

B)      DNA replication takes place in two steps:

1)      The hydrogen bonds between the two strands are broken (the DNA is denatured), making each strand available for base pairing.

2)      The new nucleotides are covalently bonded to each growing strand.

C)     In virtually all DNA replication, nucleotides are added to the 3¢ end of the growing polynucleotide.

D)    The three phosphate groups of the deoxyribonucleoside triphosphate are attached to the 5¢ position of the sugar.

E)     Energy for synthesis of nucleotides to the growing chain comes from breaking the bonds between these three phosphates.

F)      The free hydroxyl group at the 3¢ end of the growing chain reacts with one of the phosphate groups, breaking the bond between the phosphate group attached to the sugar and the two terminal phosphate groups.

G)     The breaking of this bond releases some energy for synthesis.

H)    The one phosphate group still attached to the 5¢ carbon of the new nucleotide bonds to the 3¢ end of the growing chain, becoming part of the sugar–phosphate backbone.

I)       Additional energy is released when the two freed phosphate groups (which constitute a pyrophosphate ion) break apart.

3.4.1 DNA is threaded through a replication complex  [Sound File]

1)   A huge protein complex catalyzes the reactions of DNA replication.

2)      This replication complex recognizes an origin of replication on a chromosome.

3)      DNA replicates in both directions from the origin, forming two replication forks.

4)      In DNA replication, both strands of DNA act as templates.

5)      Until recently, it was believed that the replication complex moved along the strand of DNA.

6)      Recent evidence suggests that the replication complex is stationary, and DNA threads through it.

7)      Replication complexes consist of several proteins with different roles (Fig.3-11).

·         DNA helicase denatures the double helix.

·         Single-strand binding proteins keep the two strands separate.

·         RNA primase makes a primer strand that serves as a starting point for replication.

·         DNA polymerase adds complementary nucleotides to the growing strand, proofreads the DNA, and repairs it.

·         DNA ligase seals up breaks in the sugar–phosphate backbone.

Fig. 3-11 Multiple proteins are collaborated at the replication fork.  DNA helicase (brown mitten); Single-strand binding protein (4 black balls); RNA primase (green bell); DNA polymerase (orange donut).

3.4.2 Small, circular DNA’s replicate from a single origin, while large, linear DNA’s have many origins

1)      The enzyme DNA helicase uses energy from ATP to unwind the two DNA strands and make them available for complementary base pairing.

2)      Special proteins bind to the unwound strands to keep them apart.

3)      Small chromosomes, such as those found in bacteria, have a single origin of replication.

·         Replication in bacteria produces two interlocking circular DNA’s that are separated by an enzyme called DNA topoisomerase.

4)      Large chromosomes can have hundreds of origins of replication.

5)      Replication occurs at many different sites simultaneously.

3.4.3 DNA polymerases need a primer

1)   DNA polymerase is shaped like a hand with “finger” regions that rotate inward.

2)      The finger regions have precise shapes that recognize the shapes of different bases.

3)      DNA polymerases cannot build a strand without having an existing strand of DNA or RNA, called a primer, to start from.

4)      In DNA replication, the primer strand is a short strand of RNA complementary to the DNA template strand.

5)      An enzyme called a primase makes the primer strand.

6)      The primase is part of a protein complex called a primosome.

7)      The RNA primer is later degraded and replaced with DNA, so the final DNA molecule has no RNA.

3.4.4 DNA polymerase III extends the new DNA strands

1)   Bacterial cells contain more than one DNA polymerase.

2)      In prokaryotes, DNA polymerase III is responsible for elongation of new DNA strands.

3)      Recall that new bases are always added to the 3¢ end of a growing DNA strand.

4)      The strands in the template DNA are antiparallel, however.

5)      As a result, as the strands pass through the replication complex, one strand (the leading strand) will be in the correct orientation for addition of new nucleotides, but the other strand (the lagging strand) will be in the reverse orientation. (Fig. 3-12)

Fig. 3-12 Schematic representation of DNA replication at a growing fork.

3.4.5 The lagging strand is synthesized from Okazaki fragments

1)   Because of its backward orientation, the lagging strand must grow in relatively small, discontinuous pieces, called Okazaki fragments after their discoverer, the Japanese biochemist Reiji Okazaki.

2)      Each Okazaki fragment requires an RNA primer strand, which is formed by RNA primase some distance away from the previous Okazaki fragment. (See Figs. 3-11 and 3-12)

3)      DNA polymerase III synthesizes complementary DNA starting from the 3¢ end of the new primer and working toward the previous Okazaki fragment.

4)      When DNA polymerase III reaches the previous Okazaki fragment, it is released.

5)      DNA polymerase I then replaces the RNA primer of the previous Okazaki fragment with DNA.

6)      Finally, DNA ligase catalyzes formation of the phosphodiester linkage that joins the two Okazaki fragments (Fig. 3-13).

7)      Okazaki fragments are 100 to 200 nucleotides long in eukaryotes, and 1,000 to 2,000 nucleotides long in prokaryotes.

8)      In E. coli, the replication complex makes new DNA at a rate in excess of 1,000 base pairs per second.

9)      Errors in replication are fewer than one base in a million.

Fig. 3-13 The lagging strand is synthesized from Okazaki fragments.  DNA polymerase I and DNA ligase join DNA polymerase III to complete the complex task of the synthesizing lagging strand.

3.5 Organization of Genes in DNA

A)    The DNA in prokaryote is a double-stranded circular molecule and is supercoiled, complexed to several histone-like proteins. (Fig. 3-14)

Fig. 3-14 These long loop of DNA fiber from a E. coli cell is a continuous circular chromosome.

B)     Eukaryotic cells contain much more DNA than prokaryotes:

1)      In the nucleus, the DNA is packed into chromosomes that consist mainly of DNA and proteins called histones.

2)      Each chromosome contain a single linear double-stranded DNA. (Fig.3-15)

Fig. 3-15 DNA packs into a chromosome.

C)     Organization of genes in DNA is different in prokaryotes and eukaryotes

1)       In prokaryotic DNA, related genes are clustered into a functional region, an operon.  

2)     In eukaryotic DNA each gene is transcribed from its own start site.  The coding regions (exons) are often separated by noncoding regions (introns). (Fig. 3-16)

Fig. 3-16 Comparison of gene organization, transcription, and translation in prokaryotes and eukaryotes. (a) The tryptophan (trp) operon is a continuous segment of the E. coli chromosome, containing five genes that encode the enzymes necessary for the stepwise synthesis of tryptophan. (b) The five genes encoding the enzymes required for tryptophan synthesis in yeast (saccharomyces cerevisiae) are carried on four different chromosomes. On chromosome IV the coding regions (exons) trp1and trp4 are separated by noncoding regions (introns).

a)       DNA from one type of bacterium genetically transforms another type.

b)       The transforming principle is DNA.

c)       Viral replication experiments confirm that DNA is the genetic material.

a)       DNA differs to RNA in its sugar (deoxyribose vs. ribose) and its base (thymine vs. uracil). DNA has double strand but RNA only has single strand.

b)   Each strand of  DNA has the linkage (3’5’ phosphodiester bonds) of alternating 2’-deoxyribose and phosphates as its backbone.

c)    In DNA there are four types of bases in the polydeoxyribonucleotides: adenine (A), guanine (G), thymine (T) and cytosine (C).

d)   The two strands of DNA molecule are held together by hydrogen bonds between pairs of bases. Adenine always pairs with thymine (A-T) and guanine always pairs with cytosine (G-C).

e)   A double helical DNA structure composed of two strands running antiparallel direction.

a) Three modes of DNA replication appeared possible:

i)                    Semiconservative
ii)                   Conservative
iii)                  Dispersive

b)       Meselson and Stahl demonstrated that DNA replication is semiconservative.

a)    DNA is threaded through a replication complex.

b)    Small, circular DNA’s replicate from a single origin, while large, linear DNA’s have many origins.
c)    DNA polymerases need a primer.
d)    DNA polymerase III extends the new DNA strands.
e)    The lagging strand is synthesized from Okazaki fragments

a)       The DNA in prokaryote is a double-stranded circular molecule and is supercoiled, complexed to several histone-like proteins.

b)    In the eukaryotic nucleus, the DNA is packed into chromosomes that consist mainly of DNA and proteins called histones.  Each chromosome contain a single linear double-stranded DNA.

c)   Organization of genes in DNA differs in prokaryotes and eukaryotes.  In prokaryotic DNA, related genes are clustered into a functional region, an operon.  In eukaryotic DNA each gene is transcribed from its own start site.  The coding regions (exons) are often separated by noncoding regions (introns).

Review Questions:

1.        What is the major difference between DNA and RNA?

Answer: 

1) RNA is single-stranded.

2) The sugar in RNA is ribose, not deoxyribose.

3) Wherever thymine is found in DNA, it is replaced by uracil in RNA.

4) RNA can fold over and base-pair with itself.

2.        What is the structure of DNA?

        Answer: 

DNA cosists of two polynucleotide chains forming a double helix.  The chains are held together by hydrogen bonding between their nitrogenous bases. Base pairing is complementary: A-T and G-C are only possible pairs.  The double strands run opposite direction, that is, antiparallel.

3.        What does happen at replication fork?

        Answer: (1)  The unwinding of the double helix is mediated by two related enzymes called helicases, one of which attaches to each of the parent DNA strands. Energy to separate the strands comes from the hydrolysis of ATP.

    (2)  The separated strands would tend to fold back on themselves because of occasional short regions of complementarity, but this is prevented by the attachment of single-stranded DNA-binding proteins to each of the separated DNA strands.

    (3)  These binding proteins hold the single strands in a configuration that binds readily to DNA polymerase III.

    (4)  A pair of DNA polymerase III molecules at the replication fork catalyzes the elongation of the leading and lagging strands.

    (5)  The discontinuous production of the lagging strand results in a repeated need for a new primer to start the synthesis of the next Okazaki fragment.

    (6)  The primer is a short single strand of RNA, rather than of DNA; it is formed, complementary to the template DNA strand, by an enzyme called a primase, which is one of several polypeptides bound together in an aggregate called a primosome.

    (7)  DNA polymerase III extends the primer. DNA polymerase I later acts to replace the RNA primer segments with DNA segments.

    (8)  Finally, each newly completed Okazaki fragment is linked to the completed portion of  the lagging strand in a reaction catalyzed by DNA ligase, another enzyme.

Practice Test:

1. Which statement about complementary base pairing is NOT true? a) In DNA, T pairs with A b) In DNA, G pairs with C c) Purines pair with purines, and pyrimidine pair with pyrimidine d) It plays a role in DNA replication
2. Which model of DNA replication is proved to be correct? a) conservative b) semiconservative c) dispersive d) none of above
3. Which of the following does NOT occur during DNA replication? a) Unwinding of the parent double helix b) Formation of short pieces that are united by DNA ligase c) Use a primer d) Polymerization in the direction from 3’ to 5’
4. Griffith’s studies of Streptococcus pnemonae: a) demonstrate the phenomenon of bacterial transformation b) prove that DNA is the genetic material of bacteria c) prove that protein is not the genetic material d) prove that bacteria reproduce sexually
5. Which of these materials is NOT a major component of the eukaryotic chromosome? a) DNA b) RNA c) histone d) centromere

        Answers to practice test questions: 1. c) 2. b) 3. d) 4. a) 5. b)

Required Readings:

TEXTBOOK:

1.        M.J. Farabee, On-line Biology Book; Estrella Mountain Community College, Avondale, AZ; 2002

        http://gened.emc.maricopa.edu/bio/bio181/BIOBK/BioBookTOC.html ( OBB for reading assign.)

 2.    H. Lodish, et al. Molecular Cell Biology 4^th ed.; New York: W.H. Freeman & Co.; 2000

         http://www.ncbi.nlm.nih.gov/books/bv.fcgi?call=bv.View..ShowTOC&rid=mcb.TOC ( MCB )

3.     B. Purves, et al. Life: The Science of Biology 6^th ed.; Sinauer Associates, Inc., Sunderland, MA; 2001

( LSB )

Please read Chapters 3 and 16 of OBB, Chapters 4, 9 and 12 of MCB and Chapter 11 of LSB.

Assignment

1.  Draw a chemical structure of double-stranded DNA to show the sugar-phosphate backbone and base-paired bases with hydrogen bonds between A:T and G:C base pairs.

2.  Sketch a diagram of DNA replication steps in prokaryotes on the basis of your understanding of the DNA replication mechanism.