Cover1 - Chromosome Structure2 - Chromosome Compaction3 - Chromosome Variation5 - Nucleic Acid Structure6 - DNA Replication7 - Mutations and DNA Repair8 - Polymerase Chain Reaction (PCR)9 - Transcription10 - RNA Modifications11 - Translation12 - Gene Cloning13 - The lac Operon14 - Gene Regulation in Eukaryotes15 - Epigenetics16 - Genome Editing

5 - Nucleic Acid Structure

In Part 5, you will first learn about the major experiments that established DNA as the genetic material of nearly all organisms (recall that we learned in Part 1 that some viruses use RNA as the genetic material).  In the second portion of Part 5, you will learn about the structure of both DNA and RNA.

A. Experiments that Established DNA as the Genetic Material

Transformation

The physician Frederick Griffith was interested in developing a vaccine against the bacterium Streptococcus pneumoniae, one of the major causes of pneumonia and meningitis in children.

Some strains (varieties) of S. pneumoniae produce a polysaccharide capsule found on the outside of the bacterial cell. These encapsulated strains of S. pneumoniae are more virulent (disease causing) than strains that do not produce a capsule. Further, the encapsulated strains of S. pneumoniae form smooth colonies (called type S) on bacterial culture media, those without capsules form rough colonies (called type R) on culture media.

In 1928, Griffith showed the following in his experiment (see figure 5.1):

In this final experiment, Griffith reasoned that some chemical component released from the heat-killed smooth (type S) bacteria was internalized by the rough (type R) bacteria (transformation).  This chemical could change phenotype, converting the rough (type R) bacteria into smooth (type S) bacteria. The transformed bacteria then passed the type S trait to their progeny. Thus, the chemical responsible for transformation (in other words, the transforming principle) had properties of the genetic material (can change phenotype, is inherited); however, the actual chemical responsible for transformation was unknown.

5.1_The_Griffith_Experiment.jpg
Figure 5.1 The Griffith Experiment --- image used from OpenStax (access for free at https://books.byui.edu/-vuzA)

Key Questions

  • Describe the Griffith experiment.
  • What is meant by the “transforming principle”?
  • When smooth (type S) bacteria were "heat-killed", what major class of molecules was denatured (destroyed)? What major class of molecules was not denatured?

DNA is the Genetic Material of S. pneumoniae

Oswald Avery, Maclyn McCarty, and Colin MacLeod wanted to identify the transforming principle that was responsible for converting rough (type R) into smooth (type S) bacteria in the Griffith experiment. Avery and his colleagues focused on three candidate chemicals: DNA, RNA, and protein (see figure 5.2).

Avery and colleagues concluded from these experiments that DNA (not RNA, not protein) was the chemical responsible for transforming rough bacteria into smooth bacteria in Griffith’s experiment. Therefore, DNA is the genetic material of the bacterium S. pneumoniae.

5.2_Avery_McCarty_MacLeod_Experiment.jpg
Figure 5.2 Avery, McCarty, MacLeod Experiment --- image used from OpenStax (access for free at https://books.byui.edu/-vuzA)

Key Questions

  • Describe the Avery, McCarty, and MacLeod experiment.
  • What was the key finding of the experiment?

DNA is the Genetic Material of Bacteriophage T2

Alfred Hershey and Martha Chase examined a bacteriophage (a type of virus) named T2 that infects the bacterium Escherichia coli (see figure 5.3). T2 bacteriophages contain two molecular components: DNA and protein. The DNA is encased within a bacteriophage head made of protein. T2 also contains other protein structures, including a sheath, tail fibers, and a base plate. Bacteriophage T2 acts like a hypodermic needle, injecting the phage genetic material into an E. coli cell to initiate an infection. The genetic material of the bacteriophage then reprograms the host E. coli cell to shut off many host cell functions and instead produce progeny T2 bacteriophages within minutes. Hershey and Chase were interested in determining which of the two bacteriophage T2 components, DNA or protein, was responsible for producing more bacteriophage particles. In essence, Hershey and Chase were asking whether DNA or protein is the genetic material of bacteriophage T2.

5.3_A_Typical_Bacteriophage_SEM.jpg

5.3_B_Bacteriophage_Structure.jpg
Figure 5.3 A) Typical Bacteriophage.  The bacteriophage was imaged in a scanning electron microscope.--- licensed under CC BY 4.0 B) Bacteriophage Structure --- Tevenphage by Adenosine licensed under CC BY-SA 2.5

The Hershey and Chase experiment relied on two important ideas:

The Hershey and Chase experiment was done as follows (see figure 5.4):

  1. In one experiment, bacteriophage T2 proteins were labeled with 35S. In another experiment, bacteriophage T2 DNA was labeled with 32P.
  2. The radiolabeled bacteriophages were mixed in two separate reactions with E. coli cells to allow bacteriophage infections to occur.
  3. The reactions were subjected to blending. During blending, the bacteriophage components attached to the surfaces of the E. coli cells were released.
  4. After blending, the bacteria were collected in a centrifuge. The bacteriophage components (phage head, tail, tail fibers) remain in the supernatant (liquid) after centrifugation. The E. coli cells and the bacteriophage genetic material are found in a pellet at the bottom of the centrifuge tube.
  5. The amount of radioactivity in the supernatant and pellet was calculated.
5.4_Hershey_-_Chase_Experiment.jpg
Figure 5.4 Hershey - Chase Experiment --- image used from OpenStax (access for free at https://books.byui.edu/-vuzA)

Blending removed most of the 35S from the bacterial cells. Thus, proteins are not the major type of molecule injected into E. coli to direct the manufacture of progeny T2 bacteriophages. However, most of the 32P remained within the host E. coli cells after blending and centrifugation. Note that after the completion of the bacteriophage life cycle, the progeny bacteriophages contained 32P.  Thus, the 32P labeled DNA is heritable. The Hershey and Chase experiment showed that DNA is injected into E. coli and serves as the genetic material for bacteriophage T2.

Key Questions

  • Describe the Hershey-Chase experiment.
  • What was the key finding of the experiment?

DNA is the Genetic Material of Eukaryotes

Eukaryotic cells are not as easy to work with in the lab as bacteria and bacteriophages. Because of this, it was somewhat easier to determine that DNA is the genetic material of bacteria and bacteriophages than eukaryotic cells. Evidence that DNA is the genetic material of eukaryotes relied on the combination of both indirect evidence and direct evidence.

Many lines of indirect evidence suggested that DNA is the genetic material of eukaryotic cells. Scientists reasoned that the genetic material of eukaryotes should be found within chromosomes because chromosomes are copied and distributed to daughter cells during mitosis and meiosis. Chromosomes contain both proteins and DNA; however, the DNA component is found exclusively in chromosomes (protein is found in the cell cytoplasm as well). In addition, a diploid cell, which contains twice as many chromosomes as a haploid cell, also contains roughly twice as much DNA as a haploid cell. No such correlation was observed when the protein content of haploid and diploid cells was compared. Finally, ultraviolet light (UV light) causes mutations that can affect the phenotype of a cell. The wavelength of UV light that produces the highest frequency of mutations (260 nanometers) corresponds to the wavelength of UV light that is absorbed most strongly by DNA. On the other hand, the wavelength of UV light absorbed most strongly by proteins (280 nanometers) does not produce mutations.

Recombinant DNA technology provided direct evidence that DNA is the genetic material of eukaryotic cells. In this technique, a DNA sequence from a eukaryotic cell can be isolated and then introduced into a bacterial cell. This eukaryotic DNA sequence is then transcribed by the bacterial cell to make a messenger RNA (mRNA); the mRNA is translated by bacterial ribosomes to make a protein. The resulting protein can change the phenotype of the bacterial cell. The introduced eukaryotic DNA sequence is also passed on to the progeny bacterial cells during bacterial cell division. As an example, recombinant DNA technology allows bacterial cells to produce the human insulin protein. The fact that introduced eukaryotic DNA can result in protein production, can alter the phenotype of a bacterial cell, and can be inherited provides strong evidence that DNA is the genetic material of eukaryotes.

Key Questions

  • What are the three lines of indirect evidence that DNA is the genetic material of eukaryotes?
  • What direct evidence proves that DNA is the genetic material of eukaryotes?

B. The Structure of DNA and RNA

Overview of Nucleic Acid Structure

DNA and RNA molecules have four levels of structural complexity (see figure 5.5):

5.5_Overview_of_DNA_Structure.jpg
Figure 5.5 Overview of DNA Structure.  The basic subunits of DNA and RNA are nucleotides (bottom).  Two nucleic acid strands can interact through hydrogen bonding (upper right).  The general structure of a DNA double helix (upper left).  This image is from OpenStax (access for free at https://books.byui.edu/-vuzA)

Nucleotides

A nucleotide is composed of the following (see figure 5.6):

The nucleotide, consisting of all three parts above, is referred to as a nucleoside triphosphate and is abbreviated as NTP in RNA and dNTP in DNA.  N is the specific name of the nitrogenous base. For example, dTTP is the deoxyribose form of thymine and GTP is the ribose form of guanine.

Key Questions

  • What are the functions of the 1’, 2’, 3’, 4’ and 5’ carbons in deoxyribose and ribose?
  • How are purines and pyrimidines different?
  • Which purines and pyrimidines are found in DNA?
  • Which purines and pyrimidines are found in RNA?
EDI_5.6_Nucleotide_Structure-01_1.jpg
Figure 5.6 Nucleotide Structure --- image used from OpenStax (access for free at https://books.byui.edu/-vuzA)

Nucleic Acid Strands

Nucleotides are covalently linked to form a nucleic acid strand (see figure 5.7). The sugars of two adjacent nucleotides are linked together by phosphodiester linkages to form the backbone of the nucleic acid strand. The phosphate groups in the backbone give the nucleic acid strand a net negative electrical charge.

A strand of DNA or RNA has directionality or polarity. The formation of phosphodiester bonds between nucleotides produces a nucleic acid strand in which the 5’ carbon of the nucleotide at one end of the strand contains a free phosphate group (the 5’ end of the strand). The 3’ carbon of a nucleotide at the other end of the nucleic acid strand (the 3’ end of the strand) is attached to a free hydroxyl group.

Each nucleic acid strand has a particular sequence of nitrogenous bases. The sequence of these bases on a single nucleic acid strand is written from the 5’ end to the 3’ end, for example 5’-TTGCAGG-3’. The sequence of bases allows nucleic acid molecules to carry information (encode proteins).

EDI_5.7_Structure_of_a_DNA_Strand-01_1.jpg
Figure 5.7 Structure of a DNA Strand --- Image created by SL

Key Questions

  • How are adjacent nucleotides linked within a strand?
  • How do you know which end of the strand is the 5’ end and which end is the 3’ end?

Determining the Structure of the Double Helix

James Watson and Francis Crick were awarded the Nobel Prize in 1962 for determining the double helix structure of DNA; however, their work was built upon the contributions of other noteworthy scientists (see figure 5.8).

In the early 1950s, Linus Pauling, who won two Nobel Prizes himself, demonstrated that ball-and-stick models could describe the locations of individual atoms within important biological molecules. Pauling is known for using ball-and-stick models to describe the secondary structures found within proteins. Watson and Crick mimicked Pauling’s approach by building a ball-and-stick model of the DNA double helix.

The structure of the DNA double helix could not have been determined without an important laboratory technique called X-ray diffraction. X-ray diffraction involves subjecting a substance, such as DNA, to X-rays. When the X-rays pass through the DNA, the X-rays diffract to produce a unique pattern. This pattern can be interpreted using complex mathematics to determine the location of every atom within DNA. Rosalind Franklin and Maurice Wilkins used X-ray diffraction to determine that DNA has a helical structure, a diameter of 2 nanometers (2 nm), suggestive of two nucleic acid strands, and 10 base pairs (bp) per turn of the DNA helix. These key observations allowed Watson and Crick to determine the overall double helix structure of DNA.

Erwin Chargaff isolated DNA from many different organisms (for example, bacteria, yeast, chickens, and humans) and then studied the nitrogenous base composition of this isolated DNA. Chargaff found that the total percentage of adenine (A) within any DNA molecule was nearly identical to the total percentage of thymine (T). Likewise, the percentage of cytosine (C) was nearly identical to the total percentage of guanine (G). These relationships are known as Chargaff’s rule.

Watson and Crick used the observations/approaches of Pauling, Franklin, Wilkins, and Chargaff to build a ball-and-stick DNA model with the following features:

Watson and Crick also proposed a mechanism by which the DNA double helix could be copied prior to cell division (semi-conservative replication). We will examine semi-conservative replication in Part 6.

Key Questions

  • What was the contribution of Watson and Crick to the discovery of DNA structure?
  • What was the contribution of Pauling?
  • What was the contribution of Franklin and Wilkins?
  • What was the contribution of Chargaff?

Structure of the DNA Double Helix

Watson and Crick showed that the phosphates and sugars (the backbone) within the DNA molecule are found on the outside of the DNA molecule, and the nitrogenous bases are found on the inside of the DNA molecule. The backbone of the DNA molecule is directly exposed to water within the cell (see figure 5.9).

DNA is very stable because of hydrogen bonding. Hydrogen bonds are formed between pairs of nitrogenous bases, called base pairs (bp). Adenine always forms two hydrogen bonds with thymine and guanine always forms three hydrogen bonds with cytosine. This is called the AT/GC rule or Chargaff’s rule. Thus, the bases in one strand of DNA are complementary to the bases in the other strand. Because of these hydrogen bonding interactions, DNA sequences with a higher proportion of GC base pairs are more stable than DNA molecules that are rich in AT base pairs.

There are 10 base pairs per complete turn of the DNA double helix. Every turn is 3.4 nm in length, meaning that each base pair within the DNA double helix is separated by 0.34 nm. The DNA double helix is 2 nm wide.

5.9_DNA_Double_Helix_Structure.jpg
Figure 5.9 DNA Double Helix Structure.  The backbones of each DNA strand are represented by purple ribbons.  3D Science DNA Structure by 3DScience.com used under license CC BY 2.5

The two nucleic acid strands of DNA are antiparallel. One strand starts with the free 5’ phosphate group at the top of the molecule and ends with a free 3’ hydroxyl group at the bottom of the molecule. The other DNA strand runs in the opposite direction; the free 5’ phosphate group is at the bottom end of the molecule, the free 3’ hydroxyl group is at the top of the molecule.

You can predict the sequence of one strand of DNA if you know the sequence of the other strand. If one DNA strand is 5’-GCCATG-3’, then the opposite strand is 3’-CGGTAC-5’. The two strands are said to be complementary (see figure 5.10).

5.10_Complementary_Base_Pairs.jpg
Figure 5.10 Complementary Base Pairs --- Image used from OpenStax (access for free at https://books.byui.edu/-vuzA)

Key Questions

  • What is the distance between adjacent nucleotide pairs?
  • How many nucleotide pairs are found per helical turn in DNA?
  • How does strand polarity relate to the antiparallel structure of DNA?

Other Features of DNA

The backbone of the DNA double helix is right-handed, meaning that the backbone turns in a clockwise fashion as you look down the axis of the DNA molecule.

Within the central part of the double helix, the bases form hydrogen bonds (A with T; G with C). The base pairs themselves are flat (planar) and appear to stack on top of each other, much like the stairs of a spiral staircase.

On the outside of the DNA double helix, there are two grooves where the nitrogenous bases are directly exposed to the environment. These are called the major and minor grooves, (see figure 5.9 above). Proteins that bind to base pairs within the DNA interact mainly with the major groove and to a lesser extent with the minor groove. These proteins bind to the DNA to control DNA replication and transcription.

Key Questions

  • How do DNA-binding proteins recognize the nitrogenous bases in DNA?

Alternative Forms of DNA

The structure of the DNA double helix detailed above is the standard form of DNA (B DNA). B DNA is the predominant form of DNA that is found in aqueous environments, including living cells. There are at least two alternative forms of DNA, called A DNA and Z DNA (see figure 5.11).

The A form of DNA can be produced in the laboratory under high salt conditions. The A DNA structure is more compact than B DNA, has 9 base pairs (bp) per turn of the helix, and is 2.3 nm wide. Like B DNA, A DNA is a right-handed double helix; however, while the base pairs are perpendicular to the axis of the double helix in B DNA, the base pairs in A DNA are tilted relative to the axis of the molecule.

The Z form of DNA is a left-handed double helix, has 12 base pairs per turn of the helix, and is only 1.8 nm wide. Base pairing still occurs in Z DNA; however, the base pairs are tilted when compared to the base pairs that form in the more conventional B DNA. Z DNA is favored over B DNA when cytosine bases are modified by the addition of methyl (-CH3) chemical groups. The functional significance of Z DNA is unknown; however, Z DNA may form within living cells.  Z DNA formation near a particular gene may influence whether the gene can be activated by transcription to produce an RNA molecule.

DNA can occasionally form a triple helix. A DNA triple helix can be produced when a short single-stranded DNA molecule composed of C and T bases, synthesized artificially in the laboratory, is mixed with the DNA double helix. The synthetic single-stranded DNA binds to the major groove of the double-stranded DNA in a sequence specific way, meaning the C and T bases in the single-stranded DNA interact with specific base pairs in the double-stranded DNA. When a synthetic single-stranded DNA binds to DNA, transcription of nearby genes may be inhibited. Thus, the formation of triplex DNA may provide scientists a means to inhibit overactive genes, for example, in cancerous cells.

5.11_Left_A-DNA_Middle_B-DNA_Right_Z-DNA.jpg
Figure 5.11 Left) A-DNA, Middle) B-DNA, Right) Z-DNA  --- DnaConformations by Mauroesguerroto licensed under CC BY-SA 4.0

Key Questions

  • How do A-DNA, B-DNA, and Z-DNA differ from one another?
  • What effect does the triple helix have on gene transcription?

RNA Structure

Nucleic acid strands composed of RNA nucleotides are very similar to DNA strands. RNA strands have a backbone composed of ribose sugars and phosphate groups, adjacent nucleotides within an RNA strand are linked by phosphodiester bonds, and the RNA strand has polarity (5’ and 3’ ends). However, RNA differs from DNA in that RNA molecules are usually single-stranded and RNA strands are usually much shorter than DNA strands.

Since RNA molecules are single-stranded, there is the possibility that nitrogenous bases in one part of an RNA molecule can form base pairs via hydrogen bonding with nitrogenous bases in another part of the same RNA molecule. This base pairing forms short regions of double-stranded RNA. One important RNA structure formed in this way is a stem-loop (hairpin loop). Transfer RNA (tRNA) molecules, which play a critical role in the translation process, are noteworthy because each tRNA contains three stem-loop structures (see figure 5.12).

Edi_5.12_tRNA_Structure-01_1.jpg
Figure 5.12 tRNA Structure --- Image used from OpenStax (access for free at https://books.byui.edu/-vuzA)

Key Questions

  • What are the structural differences between DNA and RNA?

Review Questions

Fill in the Blank:

  1. In the Griffith experiment, the ______________ strain of bacteria kills the mouse.
  2. The bacterium ___________________ was used in the Hershey-Chase experiment, while the bacterium ______________________ was examined in the Frederick Griffith experiment.
  3. Avery, McCarty, and MacLeod found that the enzyme _______________ destroyed the transforming principle in bacteria, whereas the enzymes _____________ and _______________ did not.
  4. The _______________________________ experiment showed that DNA is the genetic material of bacteriophage T2.
  5. In eukaryotes, a __________ cell has twice the amount of DNA as a __________ cell, but both cells have similar amounts of protein.
  6. Each base pair within the DNA double helix is separated by ________ nanometers (nm).
  7. The two purines found in RNA are ________________ and ___________________.
  8. The pentose sugar in dCTP is ______________________, while the pentose sugar in ATP is ______________________.
  9. Proteins that interact directly with DNA primarily bind to the _____________ groove.
  10. The _____ form of DNA exists in aqueous environments while the _____ form of DNA exists when cytosines are methylated.
  11. The ______________________ experiment is sometimes called the “blender experiment”.

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